PHYSICAL
CHEMISTRY
A
www:€ boak7°77 . com
Physical
Chemistry
THIRD EDITION
Thomas Engel
University of Washington
Philip Reid
University of Washington
Chapter 26, "Computational Chemistry,"
was contributed by
Warren Hehre
CEO, Wavefunction, Inc.
PEARSON
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Library of Congress Cataloging-in-Publication Data
Engel, Thomas
Physical chemistry / Thomas Engel, Philip Reid, Warren Hehre. — 3rd ed. p. cm.
Includes index.
ISBN 978-0-321-81200-1 (casebound)
1. Chemistry, Physical and theoretical — Textbooks. I. Reid, Philip (Philip J.) II. Engel, Thomas.
III. Hehre, Warren. IV. Title.
QD453.3.E54 2012 541-dc23
2011046907
123456789 10— CRK— 15 14 13 12 11
www.pearsonhighered.com
ISBN-10: 0-321-81200-X; ISBN-13: 978-0-321-81200-1
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To Walter and Juliane, my first teachers, and to Gloria, Alex,
and Gabrielle.
Thomas Engel
To my family. Philip Reid
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Brief Contents
Fundamental Concepts of Thermodynamics 1
Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics 17
The Importance of State Functions: Internal Energy and Enthalpy 45
4 Thermochemistry 67
Entropy and the Second and Third Laws of Thermodynamics 85
6 Chemical Equilibrium 125
The Properties of Real Gases 165
8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases 181
Ideal and Real Solutions 209
Electrolyte Solutions 243
11 Electrochemical Cells, Batteries, and Fuel Cells 259
12 From Classical to Quantum Mechanics 293 The Schrodinger Equation 309
14 The Quantum Mechanical Postulates 331
15 Using Quantum Mechanics on Simple Systems 343
16 The Particle in the Box and the Real World 361
Commuting and Noncommuting Operators and the Surprising Consequences of Entanglement 383
18 A Quantum Mechanical Model for the Vibration and Rotation of Molecules 405
19 The Vibrational and Rotational Spectroscopy of Diatomic Molecules 43 1
20 The Hydrogen Atom 465
Many -Electron Atoms 483
Quantum States for Many-Electron Atoms and Atomic Spectroscopy 507
23 The Chemical Bond in Diatomic Molecules 537
24 Molecular Structure and Energy Levels for Polyatomic Molecules 567
25 Electronic Spectroscopy 601
26 Computational Chemistry 631
27 Molecular Symmetry 687
28 Nuclear Magnetic Resonance Spectroscopy 715
Probability 747
30 The Boltzmann Distribution 771
31 Ensemble and Molecular Partition Functions 793
32 Statistical Thermodynamics 825 Kinetic Theory of Gases 857
>4 Transport Phenomena 877
Elementary Chemical Kinetics 909 36 Complex Reaction Mechanisms 955
APPENDIX A Math Supplement 1007
APPENDIX B Data Tables 1029
APPENDIX C Point Group Character Tables 1047
APPENDIX D Answers to Selected End-of-Chapter Problems 1055
CREDITS 1071 INDEX 1073
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Contents
PREFACE xiii
1 Fundamental Concepts of Thermodynamics 1
1 . 1 What Is Thermodynamics and Why Is It Useful? 1
1.2 The Macroscopic Variables Volume, Pressure, and Temperature 2
1.3 Basic Definitions Needed to Describe Thermodynamic Systems 6
1 .4 Equations of State and the Ideal Gas Law 7
1.5 A Brief Introduction to Real Gases 10
2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics n
2. 1 The Internal Energy and the First Law of Thermodynamics 17
2.2 Work 18
2.3 Heat 21
2.4 Doing Work on the System and Changing the System Energy from a Molecular Level Perspective 23
2.5 Heat Capacity 25
2.6 State Functions and Path Functions 28
2.7 Equilibrium, Change, and Reversibility 30
2.8 Comparing Work for Reversible and Irreversible Processes 31
2.9 Determining A U and Introducing Enthalpy, a New State Function 34
2.10 Calculating q, w, A U, and A H for Processes Involving Ideal Gases 35
2.11 The Reversible Adiabatic Expansion and Compression of an Ideal Gas 39
3 The Importance of State Functions: Internal Energy and Enthalpy 45
3.1 The Mathematical Properties of State Functions 45
3.2 The Dependence of U on V and T 50
3.3 Does the Internal Energy Depend More Strongly on Vor 77 52
3.4 The Variation of Enthalpy with Temperature at Constant Pressure 55
3.5 How Are CP and Cv Related? 57
3.6 The Variation of Enthalpy with Pressure at Constant Temperature 58
3.7 The Joule-Thomson Experiment 60
3.8 Liquefying Gases Using an Isenthalpic Expansion 63
4 Thermochemistry 67
4.1 Energy Stored in Chemical Bonds Is Released or Taken Up in Chemical Reactions 67
4.2 Internal Energy and Enthalpy Changes Associated with Chemical Reactions 68
4.3 Hess’s Law Is Based on Enthalpy Being a State Function 71
4.4 The Temperature Dependence of Reaction Enthalpies 73
4.5 The Experimental Determination of AU and A H for Chemical Reactions 75
4.6 (Supplemental) Differential Scanning Calorimetry 77
5 Entropy and the Second and Third Laws of Thermodynamics 85
5.1 The Universe Has a Natural Direction of Change 85
5.2 Heat Engines and the Second Law of Thermodynamics 86
5.3 Introducing Entropy 90
5.4 Calculating Changes in Entropy 91
5.5 Using Entropy to Calculate the Natural Direction of a Process in an Isolated System 96
5.6 The Clausius Inequality 97
5.7 The Change of Entropy in the Surroundings and
total ~ A 5 + AS surroundings 98
5.8 Absolute Entropies and the Third Law of Thermodynamics 101
5.9 Standard States in Entropy Calculations 104
5.10 Entropy Changes in Chemical Reactions 105
5.11 (Supplemental) Energy Efficiency: Heat Pumps, Refrigerators, and Real Engines 106
5.12 (Supplemental) Using the Fact that S Is a State Function to Determine the Dependence of S on V and T 115
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5.13 (Supplemental) The Dependence of S on TandP 117
5.14 (Supplemental) The Thermodynamic Temperature Scale 118
6 Chemical Equilibrium 125
6.1 The Gibbs Energy and the Helmholtz Energy 125
6.2 The Differential Forms of U, H, A, and G 130
6.3 The Dependence of the Gibbs and Helmholtz Energies on P, V , and T 132
6.4 The Gibbs Energy of a Reaction Mixture 134
6.5 The Gibbs Energy of a Gas in a Mixture 135
6.6 Calculating the Gibbs Energy of Mixing for Ideal Gases 136
6.7 Calculating A GR for a Chemical Reaction 138
6.8 Introducing the Equilibrium Constant for a Mixture of Ideal Gases 139
6.9 Calculating the Equilibrium Partial Pressures in a Mixture of Ideal Gases 141
6.10 The Variation of KP with Temperature 142
6.11 Equilibria Involving Ideal Gases and Solid or Liquid Phases 145
6.12 Expressing the Equilibrium Constant in Terms of Mole Fraction or Molarity 146
6. 13 The Dependence of the Extent of Reaction on T and P 147
6.14 (Supplemental) A Case Study: The Synthesis of Ammonia 148
6.15 (Supplemental) Expressing U and H and Heat Capacities Solely in Terms of Measurable Quantities 153
6.16 (Supplemental) Measuring AG for the Unfolding of Single RNA Molecules 157
6.17 (Supplemental) The Role of Mixing in Determining Equilibrium in a Chemical Reaction 158
7 The Properties of Real Gases 165
7.1 Real Gases and Ideal Gases 165
7.2 Equations of State for Real Gases and Their Range of Applicability 166
7.3 The Compression Factor 170
7.4 The Law of Corresponding States 173
7.5 Fugacity and the Equilibrium Constant for Real Gases 175
8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases isi
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases? 181
8.2 The Pressure-Temperature Phase Diagram 184
8.3 The Phase Rule 190
8.4 The Pressure- Volume and Pressure-Volume- Temperature Phase Diagrams 191
8.5 Providing a Theoretical Basis for the P-T Phase Diagram 193
8.6 Using the Clausius-Clapeyron Equation to Calculate Vapor Pressure as a Function of T 194
8.7 The Vapor Pressure of a Pure Substance Depends on the Applied Pressure 196
8.8 Surface Tension 197
8.9 (Supplemental) Chemistry in Supercritical Fluids 201
8.10 (Supplemental) Liquid Crystal Displays 202
9 Ideal and Real Solutions 209
9.1 Defining the Ideal Solution 209
9.2 The Chemical Potential of a Component in the Gas and Solution Phases 211
9.3 Applying the Ideal Solution Model to Binary Solutions 212
9.4 The Temperature-Composition Diagram and Fractional Distillation 216
9.5 The Gibbs-Duhem Equation 218
9 . 6 Colligative Properties 219
9.7 The Freezing Point Depression and Boiling Point Elevation 220
9.8 The Osmotic Pressure 222
9.9 Real Solutions Exhibit Deviations from Raoult’s Law 224
9.10 The Ideal Dilute Solution 227
9.11 Activities Are Defined with Respect to Standard States 229
9.12 Henry’s Law and the Solubility of Gases in a Solvent 232
9.13 Chemical Equilibrium in Solutions 233
9.14 Solutions Formed from Partially Miscible Liquids 237
9.15 The Solid-Solution Equilibrium 238
10 Electrolyte Solutions 243
10.1 The Enthalpy, Entropy, and Gibbs Energy of Ion Formation in Solutions 243
10.2 Understanding the Thermodynamics of Ion Formation and Solvation 246
10.3 Activities and Activity Coefficients for Electrolyte Solutions 248
10.4 Calculating y± Using the Debye-Huckel Theory 250
10.5 Chemical Equilibrium in Electrolyte Solutions 254
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CONTENTS
VII
Electrochemical Cells, Batteries, and Fuel Cells 259
11.1 The Effect of an Electrical Potential on the Chemical Potential of Charged Species 259
11.2 Conventions and Standard States in Electrochemistry 261
11.3 Measurement of the Reversible Cell Potential 264
11.4 Chemical Reactions in Electrochemical Cells and the Nernst Equation 264
11.5 Combining Standard Electrode Potentials to Determine the Cell Potential 266
11.6 Obtaining Reaction Gibbs Energies and Reaction Entropies from Cell Potentials 267
11.7 The Relationship between the Cell EMF and the Equilibrium Constant 268
11.8 Determination of E° and Activity Coefficients Using an Electrochemical Cell 270
11.9 Cell Nomenclature and Types of Electrochemical Cells 270
11.10 The Electrochemical Series 272
11.11 Thermodynamics of Batteries and Fuel Cells 272
11.12 The Electrochemistry of Commonly Used Batteries 273
11.13 Fuel Cells 277
11.14 (Supplemental) Electrochemistry at the Atomic Scale 280
11.15 (Supplemental) Using Electrochemistry for Nanoscale Machining 286
11.16 (Supplemental) Absolute Half-Cell Potentials 287
From Classical to Quantum Mechanics 293
12.1 Why Study Quantum Mechanics? 293
12.2 Quantum Mechanics Arose out of the Interplay of Experiments and Theory 294
12.3 Blackbody Radiation 295
12.4 The Photoelectric Effect 296
12.5 Particles Exhibit Wave-Like Behavior 298
12.6 Diffraction by a Double Slit 300
12.7 Atomic Spectra and the Bohr Model of the Hydrogen Atom 303
The Schrodinger Equation 309
13.1 What Determines If a System Needs to Be Described Using Quantum Mechanics? 309
13.2 Classical Waves and the Nondispersive Wave Equation 313
13.3 Waves Are Conveniently Represented as Complex Functions 317
13.4 Quantum Mechanical Waves and the Schrodinger Equation 318
13.5 Solving the Schrodinger Equation: Operators, Observables, Eigenfunctions, and Eigenvalues 320
13.6 The Eigenfunctions of a Quantum Mechanical Operator Are Orthogonal 322
13.7 The Eigenfunctions of a Quantum Mechanical Operator Form a Complete Set 324
13.8 Summing Up the New Concepts 326
14 The Quantum Mechanical Postulates 331
14.1 The Physical Meaning Associated with the Wave Function Is Probability 332
14.2 Every Observable Has a Corresponding Operator 333
14.3 The Result of an Individual Measurement 334
14.4 The Expectation Value 334
14.5 The Evolution in Time of a Quantum Mechanical System 338
14.6 Do Superposition Wave Functions Really Exist? 338
15 Using Quantum Mechanics on Simple Systems 343
15.1 The Free Particle 343
15.2 The Particle in a One-Dimensional Box 345
15.3 Two- and Three-Dimensional Boxes 349
15.4 Using the Postulates to Understand the Particle in the Box and Vice Versa 350
16 The Particle in the Box and the Real World 361
16.1 The Particle in the Finite Depth Box 361
16.2 Differences in Overlap between Core and Valence Electrons 362
16.3 Pi Electrons in Conjugated Molecules Can Be Treated as Moving Freely in a Box 363
16.4 Why Does Sodium Conduct Electricity and Why Is Diamond an Insulator? 364
16.5 Traveling Waves and Potential Energy Barriers 365
16.6 Tunneling through a Barrier 367
16.7 The Scanning Tunneling Microscope and the Atomic Force Microscope 369
16.8 Tunneling in Chemical Reactions 374
16.9 (Supplemental) Quantum Wells and Quantum Dots 375
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VIII CONTENTS
17 Commuting and Noncommuting Operators and the Surprising Consequences of Entanglement 383
17.1 Commutation Relations 383
17.2 The Stern-Gerlach Experiment 385
17.3 The Heisenberg Uncertainty Principle 388
17.4 (Supplemental) The Heisenberg Uncertainty Principle Expressed in Terms of Standard Deviations 392
17.5 (Supplemental) A Thought Experiment Using a Particle in a Three-Dimensional Box 394
17.6 (Supplemental) Entangled States, Teleportation, and Quantum Computers 396
18 a Quantum Mechanical Model for the Vibration and Rotation of Molecules 405
18.1 The Classical Harmonic Oscillator 405
18.2 Angular Motion and the Classical Rigid Rotor 409
18.3 The Quantum Mechanical Harmonic Oscillator 411
18.4 Quantum Mechanical Rotation in Two Dimensions 416
18.5 Quantum Mechanical Rotation in Three Dimensions 419
18.6 The Quantization of Angular Momentum 421
18.7 The Spherical Harmonic Functions 423
18.8 Spatial Quantization 425
19 The Vibrational and Rotational Spectroscopy of Diatomic Molecules 431
19.1 An Introduction to Spectroscopy 431
19.2 Absorption, Spontaneous Emission, and Stimulated Emission 433
19.3 An Introduction to Vibrational Spectroscopy 435
19.4 The Origin of Selection Rules 438
19.5 Infrared Absorption Spectroscopy 440
19.6 Rotational Spectroscopy 443
19.7 (Supplemental) Fourier Transform Infrared Spectroscopy 449
19.8 (Supplemental) Raman Spectroscopy 451
19.9 (Supplemental) How Does the Transition Rate between States Depend on Frequency? 453
20 The Hydrogen Atom 465
20.1 Formulating the Schrodinger Equation 465
20.2 Solving the Schrodinger Equation for the Hydrogen Atom 466
20.3 Eigenvalues and Eigenfunctions for the Total Energy 467
20.4 The Hydrogen Atom Orbitals 473
20.5 The Radial Probability Distribution Function 475
20.6 The Validity of the Shell Model of an Atom 479
21 Many-Electron Atoms 483
21.1 Helium: The Smallest Many-Electron Atom 483
21.2 Introducing Electron Spin 485
21.3 Wave Functions Must Reflect the Indistinguishability of Electrons 486
21 .4 Using the Variational Method to Solve the Schrodinger Equation 490
21.5 The Hartree-Fock Self-Consistent Field Method 491
21.6 Understanding Trends in the Periodic Table from Hartree-Fock Calculations 499
22 Quantum States for Many-Electron Atoms and Atomic Spectroscopy 507
22.1 Good Quantum Numbers, Terms, Levels, and States 507
22.2 The Energy of a Configuration Depends on Both Orbital and Spin Angular Momentum 509
22.3 Spin-Orbit Coupling Breaks Up a Term into Levels 516
22.4 The Essentials of Atomic Spectroscopy 517
22.5 Analytical Techniques Based on Atomic Spectroscopy 519
22.6 The Doppler Effect 522
22.7 The Helium-Neon Laser 523
22.8 Laser Isotope Separation 526
22.9 Auger Electron and X-Ray Photoelectron Spectroscopies 527
22. 10 Selective Chemistry of Excited States:
0(3P) and (X'D) 530
22. 1 1 (Supplemental) Configurations with Paired and Unpaired Electron Spins Differ in Energy 531
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CONTENTS
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The Chemical Bond in Diatomic Molecules 537
23.1 Generating Molecular Orbitals from Atomic Orbitals 537
23.2 The Simplest One-Electron Molecule:
Hj 541
23.3 The Energy Corresponding to the Ho Molecular Wave Functions i/jg and i/ju 543
23.4 A Closer Look at the H2 Molecular Wave Functions i/jg and ifju 546
23.5 Homonuclear Diatomic Molecules 548
23.6 The Electronic Structure of Many-Electron Molecules 552
23.7 Bond Order, Bond Energy, and Bond Length 555
23.8 Heteronuclear Diatomic Molecules 557
23.9 The Molecular Electrostatic Potential 560
Molecular Structure and Energy Levels for Polyatomic Molecules 567
24.1 Lewis Structures and the VSEPR Model 567
24.2 Describing Localized Bonds Using Hybridization for Methane, Ethene, and Ethyne 570
24.3 Constructing Hybrid Orbitals for Nonequivalent Ligands 573
24.4 Using Hybridization to Describe Chemical Bonding 576
24.5 Predicting Molecular Structure Using Qualitative Molecular Orbital Theory 578
24.6 How Different Are Localized and Delocalized Bonding Models? 581
24.7 Molecular Structure and Energy Levels from Computational Chemistry 584
24.8 Qualitative Molecular Orbital Theory for Conjugated and Aromatic Molecules: The Hiickel Mode 586
24.9 From Molecules to Solids 592
24.10 Making Semiconductors Conductive at Room Temperature 593
Electronic Spectroscopy 601
25.1 The Energy of Electronic Transitions 601
25.2 Molecular Term Symbols 602
25.3 Transitions between Electronic States of Diatomic Molecules 605
25.4 The Vibrational Fine Structure of Electronic Transitions in Diatomic Molecules 606
25.5 UV- Visible Light Absorption in Polyatomic Molecules 608
25.6 Transitions among the Ground and Excited States 610
25.7 Singlet-Singlet Transitions: Absorption and Fluorescence 611
25.8 Intersystem Crossing and Phosphorescence 613
25.9 Fluorescence Spectroscopy and Analytical Chemistry 614
25 . 1 0 Ultraviolet Photoelectron Spectroscopy 615
25 . 1 1 Single Molecule Spectroscopy 617
25 . 1 2 Fluorescent Resonance Energy Transfer (FRET) 619
25.13 Linear and Circular Dichroism 623
25.14 Assigning + and — to 2 Terms of Diatomic Molecules 625
26 Computational Chemistry 631
26.1 The Promise of Computational Chemistry 631
26.2 Potential Energy Surfaces 632
26.3 Hartree-Fock Molecular Orbital Theory: A Direct Descendant of the Schrodinger Equation 636
26.4 Properties of Limiting Hartree-Fock Models 638
26.5 Theoretical Models and Theoretical Model Chemistry 643
26.6 Moving Beyond Hartree-Fock Theory 644
26.7 Gaussian Basis Sets 649
26.8 Selection of a Theoretical Model 652
26.9 Graphical Models 666
26.10 Conclusion 674
27 Molecular Symmetry 687
27.1 Symmetry Elements, Symmetry Operations, and Point Groups 687
27.2 Assigning Molecules to Point Groups 689
27.3 The H20 Molecule and the C2v Point Group 691
27.4 Representations of Symmetry Operators, Bases for Representations, and the Character Table 696
27.5 The Dimension of a Representation 698
27.6 Using the C2v Representations to Construct Molecular Orbitals for H20 702
27.7 The Symmetries of the Normal Modes of Vibration of Molecules 704
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CONTENTS
27.8 Selection Rules and Infrared versus Raman Activity 708
27.9 (Supplemental) Using the Projection Operator Method to Generate MOs That Are Bases for Irreducible Representations 709
28 Nuclear Magnetic Resonance Spectroscopy 715
28. 1 Intrinsic Nuclear Angular Momentum and Magnetic Moment 715
28.2 The Energy of Nuclei of Nonzero Nuclear Spin in a Magnetic Field 717
28.3 The Chemical Shift for an Isolated Atom 719
28.4 The Chemical Shift for an Atom Embedded in a Molecule 720
28.5 Electronegativity of Neighboring Groups and Chemical Shifts 721
28.6 Magnetic Fields of Neighboring Groups and Chemical Shifts 722
28.7 Multiplet Splitting of NMR Peaks Arises through Spin-Spin Coupling 723
28.8 Multiplet Splitting When More Than Two Spins Interact 728
28.9 Peak Widths in NMR Spectroscopy 730
28.10 Solid-State NMR 732
28.11 NMR Imaging 732
28.12 (Supplemental)The NMR Experiment in the Laboratory and Rotating Frames 734
28.13 (Supplemental) Fourier Transform NMR Spectroscopy 736
28.14 (Supplemental) Two-Dimensional NMR 740
29 Probability 747
29.1 Why Probability? 747
29.2 Basic Probability Theory 748
29.3 Stirling’s Approximation 750
29.4 Probability Distribution Functions 757
29.5 Probability Distributions Involving Discrete and Continuous Variables 759
29.6 Characterizing Distribution Functions 762
30 The Boltzmann Distribution n\
30.1 Microstates and Configurations 771
30.2 Derivation of the Boltzmann Distribution 777
30.3 Dominance of the Boltzmann Distribution 782
30.4 Physical Meaning of the Boltzmann Distribution Law 784
30.5 The Definition of (3 785
31 Ensemble and Molecular Partition Functions 793
31.1 The Canonical Ensemble 793
31.2 Relating Q to q for an Ideal Gas 795
31.3 Molecular Energy Levels 797
31.4 Translational Partition Function 797
31.5 Rotational Partition Function: Diatomics 800
31.6 Rotational Partition Function: Polyatomics 807
31.7 Vibrational Partition Function 809
31.8 The Equipartition Theorem 814
31.9 Electronic Partition Function 815
31.10 Review 819
32 Statistical Thermodynamics 825
32.1 Energy 825
32.2 Energy and Molecular Energetic Degrees of Freedom 829
32.3 Heat Capacity 833
32.4 Entropy 837
32.5 Residual Entropy 842
32.6 Other Thermodynamic Functions 843
32.7 Chemical Equilibrium 847
33 Kinetic Theory of Gases 857
33.1 Kinetic Theory of Gas Motion and Pressure 857
33.2 Velocity Distribution in One Dimension 858
33.3 The Maxwell Distribution of Molecular Speeds 862
33.4 Comparative Values for Speed Distributions: ^avei Vmp-> and T^rms $64
33.5 Gas Effusion 866
33.6 Molecular Collisions 868
33.7 The Mean Free Path 872
34 Transport Phenomena 877
34.1 What Is Transport? 877
34.2 Mass Transport: Diffusion 879
34.3 The Time Evolution of a Concentration Gradient 882
34.4 (Supplemental) Statistical View of Diffusion 884
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34.5 Thermal Conduction 886
34.6 Viscosity of Gases 890
34.7 Measuring Viscosity 892
34.8 Diffusion in Liquids and Viscosity of Liquids 894
34.9 (Supplemental) Sedimentation and Centrifugation 896
34.10 Ionic Conduction 899
35 Elementary Chemical Kinetics 909
35.1 Introduction to Kinetics 909
35.2 Reaction Rates 910
35.3 Rate Laws 912
35.4 Reaction Mechanisms 917
35.5 Integrated Rate Law Expressions 918
35.6 Numerical Approaches 923
35.7 Sequential First-Order Reactions 924
35.8 Parallel Reactions 929
35.9 Temperature Dependence of Rate Constants 931
35.10 Reversible Reactions and Equilibrium 933
35.11 (Supplemental) Perturbation-Relaxation Methods 936
35.12 (Supplemental) The Autoionization of Water:
A Temperature- Jump Example 938
35.13 Potential Energy Surfaces 940
35.14 Activated Complex Theory 942
35.15 Diffusion Controlled Reactions 946
36 Complex Reaction Mechanisms 955
36.1 Reaction Mechanisms and Rate Laws 955
36.2 The Preequilibrium Approximation 957
36.3 The Lindemann Mechanism 959
36.4 Catalysis 961
36.5 Radical-Chain Reactions 972
36.6 Radical-Chain Polymerization 975
36.7 Explosions 976
36.8 Feedback, Nonlinearity, and Oscillating Reactions 978
36.9 Photochemistry 981
36.10 Electron Transfer 993
APPENDIX A Math Supplement 1007
APPENDIX B Data Tables 1029
APPENDIX Point Group Character Tables 1047
APPENDIX D Answers to Selected End-of-Chapter Problems 1055
CREDITS 1071 INDEX 1073
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About the Authors
Thomas Engel has taught chemistry at the University of Washington for more than 20 years, where he is currently professor emeritus of chemistry. Professor Engel received his bachelor’s and master’s degrees in chemistry from the Johns Hopkins University, and his Ph.D. in chemistry from the University of Chicago. He then spent 1 1 years as a researcher in Germany and Switzerland, in which time he received the Dr. rer. nat. habil. degree from the Ludwig Maximilians University in Munich. In 1980, he left the IBM research laboratory in Zurich to become a faculty member at the University of Washington.
Professor Engel’s research interests are in the area of surface chemistry, and he has published more than 80 articles and book chapters in this field. He has received the Sur¬ face Chemistry or Colloids Award from the American Chemical Society and a Senior Humboldt Research Award from the Alexander von Humboldt Foundation.
Philip Reid has taught chemistry at the University of Washington since 1995. Professor Reid received his bachelor’s degree from the University of Puget Sound in 1986, and his Ph.D. from the University of California, Berkeley in 1992. He performed postdoctoral research at the University of Minnesota, Twin Cities before moving to Washington.
Professor Reid’s research interests are in the areas of atmospheric chemistry, ultra¬ fast condensed-phase reaction dynamics, and organic electronics. He has published more than 100 articles in these fields. Professor Reid is the recipient of a CAREER Award from the National Science Foundation, is a Cottrell Scholar of the Research Corporation, and is a Sloan Fellow. He received the University of Washington Distinguished Teaching Award in 2005.
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Preface
The third edition of this book builds on user and reviewer comments on the previous editions. Our goal remains to provide students with an accessible overview of the whole field of physical chemistry while focusing on basic principles that unite the subdisciplines of the field. We continue to present new research developments in the field to emphasize the vibrancy of physical chemistry. Many chapters have been extensively revised as described below. We include additional end-of-chapter concept problems and most of the numerical problems have been revised. The target audience remains undergraduate students majoring in chemistry, biochemistry, and chemical engineering, as well as many students majoring in the atmospheric sciences and the biological sciences. The following objectives, illustrated with brief examples, outline our approach to teaching physical chemistry.
• Focus on teaching core concepts. The central principles of physical chemistry are explored by focusing on core ideas, and then extending these ideas to a variety of problems. The goal is to build a solid foundation of student understanding rather than cover a wide variety of topics in modest detail.
• Illustrate the relevance of physical chemistry to the world around us. Many students struggle to connect physical chemistry concepts to the world around them. To address this issue, example problems and specific topics are tied together to help the student develop this connection. Fuel cells, refrigerators, heat pumps, and real engines are discussed in connection with the second law of thermodynamics. The particle in the box model is used to explain why metals conduct electricity and why valence electrons rather than core electrons are important in chemical bond forma¬ tion. Examples are used to show the applications of chemical spectroscopies. Every attempt is made to connect fundamental ideas to applications that are familiar to the
U.S. 2002 Carbon Dioxide Emissions from Energy Consumption - 5,682* Million Metric Tons of C02**
from U.S. territories, less 90.2 MtC02 from international and military bunker fuels.
**Previous versions of this chart showed emissions in metric tons of carbon, not of C02.
***Municipal solid waste and geothermal energy.
Note: Numbers may not equal sum of components because of independent rounding.
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XIII
student. Art is used to convey complex information in an accessible manner as in the images here of U.S. carbon dioxide emissions.
• Present exciting new science in the field of physical chemistry. Physical chem¬ istry lies at the forefront of many emerging areas of modern chemical research. Recent applications of quantum behavior include band-gap engineering, quantum dots, quantum wells, teleportation, and quantum computing. Single-molecule spec¬ troscopy has led to a deeper understanding of chemical kinetics, and heterogeneous catalysis has benefited greatly from mechanistic studies carried out using the techniques of modem surface science. Atomic scale electrochemistry has become possible through scanning tunneling microscopy. The role of physical chemistry in these and other emerging areas is highlighted throughout the text. The following figure shows direct imaging of the arrangement of the atoms in pentacene as well as imaging of a delocalized molecular orbital using scanning tunneling and atomic force miscroscopies.
• Web -based simulations illustrate the concepts being explored and avoid math overload. Mathematics is central to physical chemistry; however, the mathemat¬ ics can distract the student from “seeing” the underlying concepts. To circumvent this problem, web-based simulations have been incorporated as end-of-chapter problems throughout the book so that the student can focus on the science and avoid a math overload. These web-based simulations can also be used by instructors dur¬ ing lecture. An important feature of the simulations is that each problem has been designed as an assignable exercise with a printable answer sheet that the student can submit to the instructor. The Study Area in MasteringChemistry® also includes a graphing routine with a curve-fitting capability, which allows students to print and submit graphical data. The 50 web-based simulations listed in the end-of-chapter
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PREFACE
XV
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problems are available in the Study Area of MasteringChemi stry ® for Physical Chemistry. MasteringChemistry® also includes a broad selection of end-of-chapter problems with answer- specific feedback.
• Show that learning problem-solving skills is an essential part of physical chemistry. Many example problems are worked through in each chapter. They introduce the student to a useful method to solve physical chemistry problems.
[ EXA
EXAMPLE PROBLEM 2.5
A system cun laming 2.50 mol of an ideal gas Tot which = 20,79 J mol-1 K- 1 is taken through the cycle in the following diagram in the direction indicated by the arrows. The curved path corresponds to FV — nRTt where T = T \ =■ TV
a. Calculate q, if/, and AH for each segment and for the cycle assuming that the heat capacity is independent of temperature.
h. Calculate q , yv, AU, and A H for each segment and for the cycle in which the direction of each process is reversed.
The End-of-Chapter Problems cover a range of difficulties suitable for students at all levels.
P8.6 A F-T phase diagram for potassium is shown next.
Source: Phase Diagrams of the Elements by David A. Young. © 1991 Regents of the University of California. Published by the University of California Press.
a. Which phase has the higher density, the fee or the bcc phase? Explain your answer.
b. Indicate the range of P and T in the phase diagram for which fee and liquid potassium are in equilibrium. Does fee potassium float on or sink in liquid potassium? Explain your answer.
c. Redraw this diagram for a different pressure range and indicate where you expect to find the vapor phase. Explain how you chose the slope of your liquid vapor coexistence line.
• Conceptual questions at the end of each chapter ensure that students learn to express their ideas in the language of science.
Conceptual Problems
Q2L1 Why does the effective nuclear charge for the Is orbital increase by 0.99 in going from oxygen to fluorine but only increases by 0.65 for the 2 p orbital?
Q21.2 There arc more electrons in the n — 4 shell than for the n = 3 shell in krypton. However, the peak in the radial distribution in Figure 2 1.6 is smaller for the n = 4 shell than for the n = 3 shell. Explain this fact.
Q21J How is the effective nuclear charge related to the size of the basis set in a Harlree-Fock calculation?
Q2 1 .4 The angu ! ar f u net ions. O (tf } d> ( tf> ) . for the one-electron Hartree-Fock orbitals are the same as for the hydrogen atom, and the radial functions and radial probability functions are similar to those for the hydrogen atom. The contour coloring is explained in the caption to figure 20,7. The following figure shows (a) a contour plot in the xy plane with the y axis being the vertical axis, (h) the radial function, and (c) the radial proba¬ bility distribution for a one-electron orhitah Identify the orbital ( Is, 4dxz, and so on).
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XVI
PREFACE
Integrate computational chemistry into the standard curriculum. The teaching of quantum mechanics has not taken advantage of the widespread availability of Ab Initio Software. Many chapters include computational problems for which detailed instruc¬ tions for the student are available in the Study Area in MasteringChemistry®. It is our experience that students welcome this material, (see L. Johnson and T. Engel, Journal of Chemical Education 2011, 88 [569-573]) which transforms the teaching of chemical bonding and molecular structure from being qualitative to quantitative. For example, an electrostatic potential map of acetonitrile built in Spartan Student is shown here.
Key equations. Physical chemistry is a chemistry subdiscipline that is mathemat¬ ics intensive in nature. Key equations that summarize fundamental relationships between variables are colored in red for emphasis.
Green boxes. Fundamental principles such as the laws of thermodynamics and the quantum mechanical postulates are displayed in green boxes.
Updated graph design. Color is used in graphs to clearly display different rela¬ tionships in a single figure as shown in the heat capacity for oxygen as a function of temperature and important transitions in the electron spectroscopy of molecules.
Si
This text contains more material than can be covered in an academic year, and this is entirely intentional. Effective use of the text does not require a class to proceed sequen¬ tially through the chapters, or to include all sections. Some topics are discussed in sup¬ plemental sections that can be omitted if they are not viewed as essential to the course. Also, many sections are self contained so that they can be readily omitted if they do not serve the needs of the instructor. This text is constructed to be flexible to your needs, not the other way around. We welcome the comments of both students and instructors how the material was used and how the presentation can be improved.
Thomas Engel University of Washington
Philip Reid
University of Washington
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New to This Edition
The third edition of Physical Chemistry includes changes at several levels. The most far- reaching change is the introduction of MasteringChemistry® for Physical Chemistry. Over 460 tutorials will augment the example problems in the book and enhance active learning and problem solving. Selected end of chapter problems are now assignable within MasteringChemistry® and numerical, equation and symbolic answer types are automati¬ cally graded.
The art program has been updated and expanded, and several levels of accuracy checking have been incorporated to increase accuracy throughout the text. Many new conceptual problems have been added to the book and most of the numerical problems have been revised. Significant content updates include moving part of the kinetic gas theory to Chapter 1 to allow a molecular level discussion of P and T. The heat capac¬ ity discussion previously in sections 2.5 and 3.2 have been consolidated in Chapter 2, and a new section on doing work and changing the system energy from a molecular level perspective has been added. The discussion of differential scanning calorimetry in Chapter 4 has been expanded and a molecular level discussion of entropy has been added to Chapter 5. The discussion of batteries and fuel cells in Chapter 11 has been revised and updated. Problems have been added to the end of Chapter 14 and a new section entitled on superposition wave functions has been added. A new section on traveling waves and potential energy barriers has been added to Chapter 16. The dis¬ cussion of the classical harmonic oscillator and rigid rotor has been better integrated by placing these sections before the corresponding quantum models in Chapter 18. Chapter 23 has been revised to better introduce molecular orbital theory. A new sec¬ tion on computational results and a set of new problems working with molecular orbitals has been added to Chapter 24. The number and breadth of the numerical prob¬ lems has been increased substantially in Chapter 25. The content on transition state theory in Chapter 32 has been updated. A discussion of oscillating reactions has been added to Chapter 36 and the material on electron transfer has been expanded.
Acknowledgments
Many individuals have helped us to bring the text into its current form. Students have provided us with feedback directly and through the questions they have asked, which has helped us to understand how they learn. Many of our colleagues including Peter Armentrout, Doug Doren, Gary Drobny, Graeme Henkelman, Lewis Johnson, Tom Pratum, Bill Reinhardt, Peter Rosky, George Schatz, Michael Schick, Gabrielle Varani, and especially Wes Borden and Bruce Robinson have been invaluable in advising us. Paul Siders generously provided problems for Chapter 24. We are also fortunate to have access to some end-of-chapter problems that were originally presented in Physical Chemistry, 3rd edition, by Joseph H. Noggle and in Physical Chemistry, 3rd edition, by Gilbert W. Castellan. The reviewers, who are listed separately, have made many suggestions for improvement, for which we are very grateful. All those involved in the production process have helped to make this book a reality through their efforts. Special thanks are due to Jim Smith, who helped initiate this project, to our editors Jeanne Zalesky and Jessica Neumann, and to the staff at Pearson, who have guided the production process.
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Fundamental Concepts of Thermodynamics
I hermodynamics provides a description of matter on a macroscopic scale using bulk properties such as pressure, density, volume, and temper¬ ature. This chapter introduces the basic concepts employed in thermody¬ namics including system, surroundings, intensive and extensive variables, adiabatic and diathermal walls, equilibrium, temperature, and thermome¬ try. The macroscopic variables pressure and temperature are also dis¬ cussed in terms of a molecular level model. The usefulness of equations of state, which relate the state variables of pressure, volume, and tempera¬ ture, is also discussed for real and ideal gases.
l.l
1.2
1.3
1.4
1.5
IWhat Is Thermodynamics and Why Is It Useful?
Thermodynamics is the branch of science that describes the behavior of matter and the transformation between different forms of energy on a macroscopic scale, or the human scale and larger. Thermodynamics describes a system of interest in terms of its bulk prop¬ erties. Only a few such variables are needed to describe the system, and the variables are generally directly accessible through measurements. A thermodynamic description of matter does not make reference to its structure and behavior at the microscopic level. For example, 1 mol of gaseous water at a sufficiently low density is completely described by two of the three macroscopic variables of pressure, volume, and temperature. By con¬ trast, the microscopic scale refers to dimensions on the order of the size of molecules. At the microscopic level, water would be described as a dipolar triatomic molecule, H20, with a bond angle of 104.5° that forms a network of hydrogen bonds.
In this book, we first discuss thermodynamics and then statistical thermodynamics. Statistical thermodynamics (Chapters 31 and 32) uses atomic and molecular properties to calculate the macroscopic properties of matter. For example, statistical thermody¬ namics can show that liquid water is the stable form of aggregation at a pressure of 1 bar and a temperature of 90°C, whereas gaseous water is the stable form at 1 bar and 110°C. Using statistical thermodynamics, the macroscopic properties of matter are cal¬ culated from underlying molecular properties.
What Is Thermodynamics and Why Is It Useful?
The Macroscopic Variables Volume, Pressure, and Temperature
Basic Definitions Needed to Describe Thermodynamic Systems
Equations of State and the Ideal Gas Law
A Brief Introduction to Real Gases
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1
2
CHAPTER 1 Fundamental Concepts of Thermodynamics
z
X
FIGURE 1.1
Cartesian components of velocity. The particle velocity v can be decomposed into three velocity components: v*, \y, and vz.
Given that the microscopic nature of matter is becoming increasingly well under¬ stood using theories such as quantum mechanics, why is a macroscopic science like thermodynamics relevant today? The usefulness of thermodynamics can be illustrated by describing four applications of thermodynamics which you will have mastered after working through this book:
• You have built a plant to synthesize NH3 gas from N2 and H2. You find that the yield is insufficient to make the process profitable and decide to try to improve the NH3 output by changing the temperature and/or the pressure. However, you do not know whether to increase or decrease the values of these variables. As will be shown in Chapter 6, the ammonia yield will be higher at equilibrium if the temper¬ ature is decreased and the pressure is increased.
• You wish to use methanol to power a car. One engineer provides a design for an internal combustion engine that will burn methanol efficiently according to the reaction CH3OH(/) + 3/202(g) — > C02(g) + 2H20(/). A second engineer designs an electrochemical fuel cell that carries out the same reaction. He claims that the vehicle will travel much farther if powered by the fuel cell than by the inter¬ nal combustion engine. As will be shown in Chapter 5, this assertion is correct, and an estimate of the relative efficiencies of the two propulsion systems can be made.
• You are asked to design a new battery that will be used to power a hybrid car. Because the voltage required by the driving motors is much higher than can be gen¬ erated in a single electrochemical cell, many cells must be connected in series. Because the space for the battery is limited, as few cells as possible should be used. You are given a list of possible cell reactions and told to determine the number of cells needed to generate the required voltage. As you will learn in Chapter 11, this problem can be solved using tabulated values of thermodynamic functions.
• Your attempts to synthesize a new and potentially very marketable compound have consistently led to yields that make it unprofitable to begin production. A supervi¬ sor suggests a major effort to make the compound by first synthesizing a catalyst that promotes the reaction. How can you decide if this effort is worth the required investment? As will be shown in Chapter 6, the maximum yield expected under equilibrium conditions should be calculated first. If this yield is insufficient, a cata¬ lyst is useless.
IThe Macroscopic Variables Volume,
•A Pressure, and Temperature
We begin our discussion of thermodynamics by considering a bottle of a gas such as He or CH4. At a macroscopic level, the sample of known chemical composition is com¬ pletely described by the measurable quantities volume, pressure, and temperature for which we use the symbols V, P, and T. The volume V is just that of the bottle. What physical association do we have with P and 77
Pressure is the force exerted by the gas per unit area of the container. It is most eas¬ ily understood by considering a microscopic model of the gas known as the kinetic the¬ ory of gases. The gas is described by two assumptions: the atoms or molecules of an ideal gas do not interact with one another, and the atoms or molecules can be treated as point masses. The pressure exerted by a gas on the container confining the gas arises from collisions of randomly moving gas molecules with the container walls. Because the number of molecules in a small volume of the gas is on the order of Avogadro’s number NA, the number of collisions between molecules is also large. To describe pres¬ sure, a molecule is envisioned as traveling through space with a velocity vector v that can be decomposed into three Cartesian components: vx, \y, and \z as illustrated in Figure 1.1.
The square of the magnitude of the velocity v2 in terms of the three velocity components is
v2 = V • V = v2 + v2 + v2 (1.1)
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1 .2 THE MACROSCOPIC VARIABLES VOLUME, PRESSURE, AND TEMPERATURE
3
The particle kinetic energy is 1/2 mv2 such that
1 2 1 2 1
= — mv; H — mv; H — mv 2 2 J 2
2
z
- sTrx + + £t>z
(1.2)
where the subscript Tr indicates that the energy corresponds to translational motion of the particle. Furthermore, this equation states that the total translational energy is the sum of translational energy along each Cartesian dimension.
Pressure arises from the collisions of gas particles with the walls of the container; therefore, to describe pressure we must consider what occurs when a gas particle col¬ lides with the wall. First, we assume that the collisions with the wall are elastic collisions, meaning that translational energy of the particle is conserved. Although the collision is elastic, this does not mean that nothing happens. As a result of the collision, linear momentum is imparted to the wall, which results in pressure. The definition of pressure is force per unit area and, by Newton’s second law, force is equal to the product of mass and acceleration. Using these two definitions, the pressure arising from the col¬ lision of a single molecule with the wall is expressed as
p = F_ = rm = mf f/vA = J_ ( dmvj\ = J_ ( dpA A A A\dt ) A\ dt ) A\dt )
In Equation (1.3), F is the force of the collision, A is the area of the wall with which the particle has collided, m is the mass of the particle, v, is the velocity component along the i direction (/ = x, y, or z), and pt is the particle linear momentum in the i direction. Equation (1.3) illustrates that pressure is related to the change in linear momentum with respect to time that occurs during a collision. Due to conservation of momentum, any change in particle linear momentum must result in an equal and opposite change in momentum of the container wall. A single collision is depicted in Figure 1.2. This fig¬ ure illustrates that the particle linear momentum change in the x direction is —2 m\x (note there is no change in momentum in the y or z direction). Given this, a correspon¬ ding momentum change of 2 m\x must occur for the wall.
The pressure measured at the container wall corresponds to the sum of collisions involving a large number of particles that occur per unit time. Therefore, the total momentum change that gives rise to the pressure is equal to the product of the momen¬ tum change from a single particle collision and the total number of particles that collide with the wall:
A Z9 , = - X (number of molecules) (1.4)
Total molecule v '
How many molecules strike the side of the container in a given period of time? To answer this question, the time over which collisions are counted must be considered. Consider a volume element defined by the area of the wall A times length Ax as illus¬ trated in Figure 1.3. The collisional volume element depicted in Figure 1.3 is given by
V = AAx (1.5)
FIGURE 1.2
Collision between a gas particle and a wall. Before the collision, the particle has a momentum of m\x in the x direction, and after the collision the momentum is —mxx. Therefore, the change in particle momen¬ tum resulting from the collision is —2 m\x. By conservation of momentum, the change in momentum of the wall must be 2 m\x. The incoming and outgoing trajectories are offset to show the individual momentum components.
The length of the box Ax is related to the time period over which collisions will be counted At and the component of particle velocity parallel to the side of the box (taken to be the x direction):
Ax = \xAt (1.6)
In this expression, v* is for a single particle; however, an average of this quantity will be used when describing the collisions from a collection of particles. Finally, the number of particles that will collide with the container wall Ncou in the time interval At is equal to the number density N. This quantity is equal to the number of particles in the container N divided by the container volume V and multiplied by the collisional vol¬ ume element depicted in Figure 1.3:
Ncoll = N X (AvxAt)(^J = ^ (AVj[Af)^ (1.7)
FIGURE 1.3
Volume element used to determine the num¬ ber of collisions with the wall per unit time.
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4
CHAPTER 1 Fundamental Concepts of Thermodynamics
We have used the equality N = n NA where NA is Avogadro’s number and n is the number of moles of gas in the second part of Equation (1.7). Because particles travel in either the -hr or -x direction with equal probability, only those molecules traveling in the -hr direction will strike the area of interest. Therefore, the total number of collisions is divided by two to take the direction of particle motion into account. Employing Equation (1.7), the total change in linear momentum of the container wall imparted by particle collisions is given by
APTolal = (.2mvx)(NcoU)
= (2 mvx)
f nN A A\xAt \V 2~
nNA
V
AAt m(vx)
(1.8)
In Equation (1.8), angle brackets appear around \x2 to indicate that this quantity represents an average value since the particles will demonstrate a distribution of veloc¬ ities. This distribution is considered in detail later in Chapter 30. With the total change in linear momentum provided in Equation (1.8), the force and corresponding pressure exerted by the gas on the container wall [Equation (1.3)] are as follows:
Total
At
nNA
V
Am(\x)
P
F
A
nNA
V
™(v£>
(1.9)
Equation (1.9) can be converted into a more familiar expression once 1/2 m(\x) is rec¬ ognized as the translational energy in the x direction. In Chapter 31, it will be shown that the average translational energy for an individual particle in one dimension is
m{v2x) = IcbT 2 ~ 2
(1.10)
where T is the gas temperature.
Substituting this result into Equation (1.9) results in the following expression for pressure:
P =
nRT
(l.H)
We have used the equality N A kB = R where kB is the Boltzmann constant and R is the ideal gas constant in the last part of Equation (1.11). Equation (1.11) is the ideal gas law. Although this relationship is familiar, we have derived it by employing a clas¬ sical description of a single molecular collision with the container wall and then scaling this result up to macroscopic proportions. We see that the origin of the pressure exerted by a gas on its container is the momentum exchange of the randomly moving gas mole¬ cules with the container walls.
What physical association can we make with the temperature 77 At the microscopic level, temperature is related to the mean kinetic energy of molecules as shown by Equation (1.10). We defer the discussion of temperature at the microscopic level until Chapter 30 and focus on a macroscopic level discussion here. Although each of us has a sense of a “temperature scale” based on the qualitative descriptors hot and cold , a more quantitative and transferable measure of temperature that is not grounded in indi¬ vidual experience is needed. The quantitative measurement of temperature is accom¬ plished using a thermometer. For any useful thermometer, the measured temperature, T, must be a single- valued, continuous, and mono tonic function of some thermometric system property such as the volume of mercury confined to a narrow capillary, the elec¬ tromotive force generated at the junction of two dissimilar metals, or the electrical resistance of a platinum wire.
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1 .2 THE MACROSCOPIC VARIABLES VOLUME, PRESSURE, AND TEMPERATURE
5
The simplest case that one can imagine is when T is linearly related to the value of the thermometric property x:
T(x) = a + bx (1.12)
Equation (1.12) defines a temperature scale in terms of a specific thermometric prop¬ erty, once the constants a and b are determined. The constant a determines the zero of the temperature scale because T(0) = a and the constant b determines the size of a unit of temperature, called a degree.
One of the first practical thermometers was the mercury-in-glass thermometer. This thermometer utilizes the thermometric property that the volume of mercury increases monotonically over the temperature range -38.8°C to 356. 7°C in which Hg is a liquid. In 1745, Carolus Linnaeus gave this thermometer a standardized scale by arbitrarily assigning the values 0 and 100 to the freezing and boiling points of water, respectively. Because there are 100 degrees between the two calibration points, this scale is known as the centigrade scale.
The centigrade scale has been superseded by the Celsius scale. The Celsius scale (denoted in units of °C) is similar to the centigrade scale. However, rather than being determined by two fixed points, the Celsius scale is determined by one fixed reference point at which ice, liquid water, and gaseous water are in equilibrium. This point is called the triple point (see Section 8.2) and is assigned the value 0.01 °C. On the Celsius scale, the boiling point of water at a pressure of 1 atmosphere is 99.975°C. The size of the degree is chosen to be the same as on the centigrade scale.
Although the Celsius scale is used widely throughout the world today, the numeri¬ cal values for this temperature scale are completely arbitrary, because a liquid other than water could have been chosen as a reference. It would be preferable to have a tem¬ perature scale derived directly from physical principles. There is such a scale, called the thermodynamic temperature scale or absolute temperature scale. For such a scale, the temperature is independent of the substance used in the thermometer, and the constant a in Equation (1.12) is zero. The gas thermometer is a practical thermometer with which the absolute temperature can be measured. A gas thermometer contains a dilute gas under conditions in which the ideal gas law of Equation (1.11) describes the relationship among P, T, and the molar density pm = n/V with sufficient accuracy:
P = pmRT (1.13)
Equation (1.13) can be rewritten as
showing that for a gas thermometer, the thermometric property is the temperature dependence of P for a dilute gas at constant V. The gas thermometer provides the inter¬ national standard for thermometry at very low temperatures. At intermediate tempera¬ tures, the electrical resistance of platinum wire is the standard, and at higher temperatures the radiated energy emitted from glowing silver is the standard. The absolute temperature is shown in Figure 1 .4 on a logarithmic scale together with associ¬ ated physical phenomena.
Equation (1.14) implies that as T — » 0, P —> 0. Measurements carried out by Guillaume Amontons in the 17th century demonstrated that the pressure exerted by a fixed amount of gas at constant V varies linearly with temperature as shown in Figure 1.5. At the time of these experiments, temperatures below -30°C were not attainable in the laboratory. However, the P versus Tc data can be extrapolated to the limiting Tc value at which P — » 0. It is found that these straight lines obtained for dif¬ ferent values of V intersect at a common point on the Tc axis that lies near -273°C.
The data in Figure 1.5 show that at constant V , the thermometric property P varies with temperature as
P = a + bTc (1.15)
where Tc is the temperature on the Celsius scale, and a and b are experimentally obtained proportionality constants. Figure 1.5 shows that all lines intersect at a single
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1012K
Ambient temperature 10-4s after Big Bang
1010 K —
108K -
-Core of red giant star
106K
104K
102K
1 K
10 2 K
10“4K
Core of sun Solar corona
Surface of sun
Mercury boils
H20 is liquid
Oxygen boils
Helium boils
Average temperature of universe
3He superfluid
FIGURE 1.4
The absolute temperature is shown on a logarithmic scale together with the temper¬ ature of a number of physical phenomena.
Temperature/Celsius FIGURE 1.5
The pressure exerted by 5.00 X 10~3 mol of a dilute gas is shown as a function of the temperature measured on the Celsius scale for different fixed volumes. The dashed por¬ tion indicates that the data are extrapolated to lower temperatures than could be achieved experimentally by early investigators.
6
CHAPTER 1 Fundamental Concepts of Thermodynamics
Nucleus
Cell wall Chloroplast Mitochondrion
Plant cell
Nucleus
Mitochondrion Plasma membrane
Animal cell
FIGURE 1.6
Animal and plant cells are open systems. The contents of the animal cell include the cytosol fluid and the numerous organelles (e.g., nucleus, mitochondria, etc.) that are separated from the surround¬ ings by a lipid-rich plasma membrane.
The plasma membrane acts as a boundary layer that can transmit energy and is selectively permeable to ions and various metabolites. A plant cell is surrounded by a cell wall that similarly encases the cytosol and organelles, including chloro- plasts, that are the sites of photosynthesis.
point, even for different gases. This suggests a unique reference point for temperature, rather than the two reference points used in constructing the centigrade scale. The value zero is given to the temperature at which P — » 0, so that a = 0. However, this choice is not sufficient to define the temperature scale, because the size of the degree is undefined. By convention, the size of the degree on the absolute temperature scale is set equal to the size of the degree on the Celsius scale. With these two choices, the absolute and Celsius temperature scales are related by Equation (1.16). The scale measured by the ideal gas thermometer is the absolute temperature scale used in thermodynamics. The unit of tem¬ perature on this scale is called the kelvin, abbreviated K (without a degree sign):
T/ K = Tc/° C + 273.15
(1.16)
Basic Definitions Needed to Describe Thermodynamic Systems
1.3
Having discussed the macroscopic variables pressure, volume, and temperature, we introduce some important concepts used in thermodynamics. A thermodynamic system consists of all the materials involved in the process under study. This material could be the contents of an open beaker containing reagents, the electrolyte solution within an electrochemical cell, or the contents of a cylinder and movable piston assembly in an engine. In thermodynamics, the rest of the universe is referred to as the surroundings. If a system can exchange matter with the surroundings, it is called an open system; if not, it is a closed system. Living cells are open systems (see Figure 1.6). Both open and closed systems can exchange energy with the surroundings. Systems that can exchange neither matter nor energy with the surroundings are called isolated systems.
The interface between the system and its surroundings is called the boundary. Boundaries determine if energy and mass can be transferred between the system and the surroundings and lead to the distinction between open, closed, and isolated systems. Consider Earth’s oceans as a system, with the rest of the universe being the surround¬ ings. The system-surroundings boundary consists of the solid-liquid interface between the continents and the ocean floor and the water-air interface at the ocean surface. For an open beaker in which the system is the contents, the boundary surface is just inside the inner wall of the beaker, and it passes across the open top of the beaker. In this case, energy can be exchanged freely between the system and surroundings through the side and bottom walls, and both matter and energy can be exchanged between the system and surroundings through the open top boundary. The portion of the boundary formed by the beaker in the previous example is called a wall. Walls can be rigid or movable and permeable or nonpermeable. An example of a movable wall is the surface of a bal¬ loon. An example of a selectively permeable wall is the fabric used in raingear, which is permeable to water vapor, but not liquid water.
The exchange of energy and matter across the boundary between system and sur¬ roundings is central to the important concept of equilibrium. The system and sur¬ roundings can be in equilibrium with respect to one or more of several different system variables such as pressure (P), temperature (7), and concentration. Thermodynamic equilibrium refers to a condition in which equilibrium exists with respect to P, 7, and concentration. What conditions are necessary for a system to come to equilibrium with its surroundings? Equilibrium is established with respect to a given variable only if that variable does not change with time, and if it has the same value in ah parts of the sys¬ tem and surroundings. For example, the interior of a soap bubble1 (the system) and the surroundings (the room) are in equilibrium with respect to P because the movable wall (the bubble) can reach a position where P on both sides of the wall is the same, and because P has the same value throughout the system and surroundings. Equilibrium with respect to concentration exists only if transport of ah species across the boundary in both directions is possible. If the boundary is a movable wall that is not permeable to
!For this example, the surface tension of the bubble is assumed to be so small that it can be set equal to zero. This is in keeping with the thermodynamic tradition of weightless pistons and frictionless pulleys.
Nucleus
Cell wall Chloroplast Mitochondrion
Plant cell
Nucleus
Mitochondrion Plasma membrane
Animal cell
FIGURE 1.6
Animal and plant cells are open systems. The contents of the animal cell include the cytosol fluid and the numerous organelles (e.g., nucleus, mitochondria, etc.) that are separated from the surround¬ ings by a lipid-rich plasma membrane.
The plasma membrane acts as a boundary layer that can transmit energy and is selectively permeable to ions and various metabolites. A plant cell is surrounded by a cell wall that similarly encases the cytosol and organelles, including chloro- plasts, that are the sites of photosynthesis.
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1 .4 EQUATIONS OF STATE AND THE IDEAL GAS LAW
7
all species, equilibrium can exist with respect to P, but not with respect to concentra¬ tion. Because N2 and 02 cannot diffuse through the (idealized) bubble, the system and surroundings are in equilibrium with respect to P, but not to concentration. Equilibrium with respect to temperature is a special case that is discussed next.
Two systems that have the same temperature are in thermal equilibrium. We use the concepts of temperature and thermal equilibrium to characterize the walls between a system and its surroundings. Consider the two systems with rigid walls shown in Figure 1.7a. Each system has the same molar density and is equipped with a pressure gauge. If we bring the two systems into direct contact, two limiting behav¬ iors are observed. If neither pressure gauge changes, as in Figure 1.7b, we refer to the walls as being adiabatic. Because Pi ^ P2, the systems are not in thermal equilib¬ rium and, therefore, have different temperatures. An example of a system surrounded by adiabatic walls is coffee in a Styrofoam cup with a Styrofoam lid.2 Experience shows that it is not possible to bring two systems enclosed by adiabatic walls into thermal equilibrium by bringing them into contact, because adiabatic walls insulate against the transfer of “heat.” If we push a Styrofoam cup containing hot coffee against one containing ice water, they will not reach the same temperature. Rely on experience at this point regarding the meaning of heat; a thermodynamic definition will be given in Chapter 2.
The second limiting case is shown in Figure 1.7c. In bringing the systems into inti¬ mate contact, both pressures change and reach the same value after some time. We conclude that the systems have the same temperature, T\ = T2, and say that they are in thermal equilibrium. We refer to the walls as being diathermal. Two systems in contact separated by diathermal walls reach thermal equilibrium because diathermal walls conduct heat. Hot coffee stored in a copper cup is an example of a system sur¬ rounded by diathermal walls. Because the walls are diathermal, the coffee will quickly reach room temperature.
The zeroth law of thermodynamics generalizes the experiment illustrated in Figure 1.7 and asserts the existence of an objective temperature that can be used to define the condition of thermal equilibrium. The formal statement of this law is as follows:
f >
Two systems that are separately in thermal equilibrium with a third system are
also in thermal equilibrium with one another. v J
The unfortunate name assigned to the “zeroth” law is due to the fact that it was formu¬ lated after the first law of thermodynamics, but logically precedes it. The zeroth law tells us that we can determine if two systems are in thermal equilibrium without bringing them into contact. Imagine the third system to be a thermometer, which is defined more precisely in the next section. The third system can be used to compare the temperatures of the other two systems; if they have the same temperature, they will be in thermal equilibrium if placed in contact.
(a) Two separated systems with rigid walls and the same molar density have different temperatures, (b) Two systems are brought together so that their adiabatic walls are in intimate contact. The pressure in each system will not change unless heat transfer is possible, (c) As in part (b), two systems are brought together so that their diathermal walls are in intimate contact. The pressures become equal.
Equations of State and the Ideal Gas Law
Macroscopic models in which the system is described by a set of variables are based on experience. It is particularly useful to formulate an equation of state, which relates the state variables. A dilute gas can be modeled as consisting of point masses that do not interact with one another; we call this an ideal gas. The equation of state for an ideal gas was first determined from experiments by the English chemist Robert Boyle. If the pressure of He is measured as a function of the volume for different values of tempera¬ ture, the set of nonintersecting hyperbolas as shown in Figure 1.8 is obtained. The curves in this figure can be quantitatively fit by the functional form
PV = aT (1.17)
2In this discussion, Styrofoam is assumed to be a perfect insulator.
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8
CHAPTER 1 Fundamental Concepts of Thermodynamics
Volume/10 2 m3
where T is the absolute temperature as defined by Equation (1.16), allowing a to be determined. The constant a is found to be directly proportional to the mass of gas used. It is useful to separate this dependence by writing a = nR, where n is the number of moles of the gas, and R is a constant that is independent of the size of the system. The result is the ideal gas equation of state
PV = NkBT = nRT (1.18)
as derived in Equation (1.11). The equation of state given in Equation (1.18) is familiar as the ideal gas law. Because the four variables P, V, T, and n are related through the equation of state, any three of these variables is sufficient to completely describe the ideal gas.
Of these four variables, P and T are independent of the amount of gas, whereas V and n are proportional to the amount of gas. A variable that is independent of the size of the system (for example, P and T) is referred to as an intensive variable, and one that is proportional to the size of the system (for example, V) is referred to as an extensive variable. Equation (1.18) can be written in terms of intensive variables exclusively:
P = pmRT (1.13)
FIGURE 1.8
Illustration of the relationship between pressure and volume of 0.010 mol of He for fixed values of temperature, which dif¬ fer by 100 K.
For a fixed number of moles, the ideal gas equation of state has only two independent intensive variables: any two of P, T, and pm.
For an ideal gas mixture
PV = 2 niRT
i
(1.19)
because the gas molecules do not interact with one another. Equation (1.19) can be rewritten in the form
^ rii RT ^
p = 2 -V- = = Pi + p2 + P3 + ... (i.20)
i * i
In Equation (1.20), Pt is the partial pressure of each gas. This equation states that each ideal gas exerts a pressure that is independent of the other gases in the mixture. We also have
niRT niRT
Pi V V Hi
-i = - -jSF = — = ~ = xi (1.21)
P ^ rifR! nRT n
i "T”
which relates the partial pressure of a component in the mixture PL with its mole fraction, X{ — n-J n, and the total pressure P.
In the SI system of units, pressure is measured in Pascal (Pa) units, where 1 Pa = 1 N/m2. The volume is measured in cubic meters, and the temperature is meas¬ ured in kelvin. However, other units of pressure are frequently used, and these units are related to the Pascal as indicated in Table 1.1. In this table, numbers that are not exact have been given to five significant figures. The other commonly used unit of volume is the liter (L), where 1 m3 = 103 L and 1 L = 1 dm3 = 10-3 m3.
TABLE 1.1 Units of Pressure and Conversion Factors
Unit of Pressure
Symbol
Numerical Value
Pascal
Pa
1 Nm“2 = 1 kg m 1 s-2
Atmosphere
atm
1 atm = 101,325 Pa (exactly)
Bar
bar
1 bar = 105 Pa
Torr or millimeters of Hg
Torr
1 Torr = 101,325/760 = 133.32 Pa
Pounds per square inch
psi
1 psi = 6,894.8 Pa
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1 .4 EQUATIONS OF STATE AND THE IDEAL GAS LAW
9
EXAMPLE PROBLEM 1.1
Starting out on a trip into the mountains, you inflate the tires on your automobile to a recommended pressure of 3.21 X 105 Pa on a day when the temperature is -5.00°C. You drive to the beach, where the temperature is 28.0°C. (a) What is the final pressure in the tires, assuming constant volume? (b) Derive a formula for the final pressure, assuming more realistically that the volume of the tires increases with increasing pres¬ sure as V f = Vi(\ + y[pf — P;]) where y is an experimentally determined constant.
Solution
a. Because the number of moles is constant,
PiVi PfVf D _ PiViTf
_ ; r r —
Ti
VfTi ’
PiViTf Vi (273.15 + 28.0) s
Pf = - - = 3.21 X 105Pa X — X - - - = 3.61 X 105Pa
VfTi
Vi (273.15 - 5.00)
b.
w PM: + ipf ~ p'
Ti
PiTf = PfTi( 1 + y[Pf - />■])
PjT.y + PfTi{ 1 - Pi y) ~ PiTf = 0
-Tt{ 1 - P(y) ± Vr?( 1 - Pj-y)2 + VTiTfyPi
pf =
2Tty
We leave it to the end-of-chapter problems to show that this expression for Pj has the correct limit as y — > 0.
the j
In the SI system, the constant R that appears in the ideal gas law has the value 8.314 J K-1 mol-1, where the joule (J) is the unit of energy in the SI system. To sim¬ plify calculations for other units of pressure and volume, values of the constant R with different combinations of units are given in Table 1.2.
EXAMPLE PROBLEM 1.2
Consider the composite system, which is held at 298 K, shown in the following figure. Assuming ideal gas behavior, calculate the total pressure and the partial pressure of each component if the barriers separating the compartments are removed. Assume that the volume of the barriers is negligible.
TABLE 1.2 The Ideal Gas
Constant, /?, in Various Units
R
= 8.314 JK'1 mol-1
R
= 8.314 Pam3 K-1 mol-1
R
= 8.314 X 10-2 L bar KT1
mol-1
R
= 8.206 X 10"2 L atm K“
1 mol-1
R
= 62.36 LTorrKT1 mol-1
° ° ° ^^RO(Qr>Poc(XQOQQQQ
o o
°0°0'
TWo 8^0o°°QGor
oocp^yoWi
b°ort§^!'
He
2.00 L 1.50 bar
Ne
3.00 L 2.50 bar
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10 CHAPTER 1 Fundamental Concepts of Thermodynamics
273.6 -I
2 4 6 8 10
10-4 P/Pa
272.8 -
FIGURE 1.9
The temperature measured in a gas ther¬ mometer is independent of the gas used only in the limit that P — > 0.
Solution
The number of moles of He, Ne, and Xe is given by
PV
1.50 bar
X
2.00L
RT "
" 8.314
X
10-2 Lbar
K
_1 mo r1
X
298
K
PV
2.50 bar
X
3.00 L
RT "
" 8.314
X
10-2 L bar
K
_1 mol-1
X
298
K
PV
1.00 bar
X
1.00 L
RT ~
" 8.314
X
10-2 Lbar
K
_1 mol-1
X
298
K
n = nHe + nNe + nXe = 0.464
0.121 mol
0.303 mol
0.0403 mol
The mole fractions are
xHe ~
xNe -
xXe
nHe 0.121
n
0.464
nNe
0.303
n
” 0.464
nXe
0.0403
0.464
0.261
0.653
0.0860
The total pressure is given by
_ {nHe + nNe + nXe)RT V
0.464 mol X 8.314 X 10-2 L bar K_1 mol- 1 X 298 K 6.00 L
= 1.92 bar
The partial pressures are given by
P Re = xHeP = 0.261 X 1.92 bar = 0.501 bar PNe — xNeP = 0.653 X 1.92 bar = 1.25 bar PXe - xxeP ~ 0.0860 X 1.92 bar = 0.165 bar
A Brief Introduction to Real Gases
The ideal gas law provides a first look at the usefulness of describing a system in terms of macroscopic parameters. However, we should also emphasize the downside of not taking the microscopic nature of the system into account. For example, the ideal gas law only holds for gases at low densities. In practice, deviations from the ideal gas law that occur for real gases must be taken into account in such applications as a gas thermometer. If data were obtained from a gas thermometer using He, Ar, and N2 for a temperature very near the temperature at which the gas condenses to form a liquid, they would exhibit the behavior shown in Figure 1.9. We see that the temperature only becomes independent of P and of the gas used in the thermometer if the data are extrapolated to zero pressure. It is in this limit that the gas thermometer provides a measure of the thermodynamic temperature. In practice, gas-independent T values are only obtained below P ~ 0.01 bar.
For most applications, calculations based on the ideal gas law are valid to much higher pressures. Real gases will be discussed in detail in Chapter 7. However, because we need to take nonideal gas behavior into account in Chapters 2 through 6, we intro¬ duce an equation of state that is valid to higher densities in this section.
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1 .5 A BRIEF INTRODUCTION TO REAL GASES 1 1
The ideal gas assumptions that the atoms or molecules of a gas do not interact with one another and can be treated as point masses have a limited range of validity, which can be discussed using the potential energy function typical for a real gas, as shown in Figure 1.10. This figure shows the potential energy of interaction of two gas molecules as a function of the distance between them. The intermolecular potential can be divided into regions in which the potential energy is essentially zero (r > rtransition), negative (attractive interaction) (rtransition > r > rv=0), and positive (repulsive interaction) (r < rv=(:i'). The distance rtmnsition is not uniquely defined and depends on the energy of the molecule. It is on the order of the molecular size.
As the density is increased from very low values, molecules approach one another to within a few molecular diameters and experience a long-range attractive van der Waals force due to time-fluctuating dipole moments in each molecule. This strength of the attractive interaction is proportional to the polarizability of the elec¬ tron charge in a molecule and is, therefore, substance dependent. In the attractive region, P is lower than that calculated using the ideal gas law. This is the case because the attractive interaction brings the atoms or molecules closer than they would be if they did not interact. At sufficiently high densities, the atoms or mole¬ cules experience a short-range repulsive interaction due to the overlap of the elec¬ tron charge distributions. Because of this interaction, P is higher than that calculated using the ideal gas law. We see that for a real gas, P can be either greater or less than the ideal gas value. Note that the potential becomes repulsive for a value of r greater than zero. As a consequence, the volume of a gas even well above its boiling tem¬ perature approaches a finite limiting value as P increases. By contrast, the ideal gas law predicts that V — > 0 as P —> oo .
Given the potential energy function depicted in Figure 1.10, under what conditions is the ideal gas equation of state valid? A real gas behaves ideally only at low densities for which r > rtransition, and the value of rtmnsition is substance dependent. The van der Waals equation of state takes both the finite size of molecules and the attractive poten¬ tial into account. It has the form
Real gas Ideal gas
FIGURE 1.10
The potential energy of interaction of two molecules or atoms is shown as a function of their separation r. The red curve shows the potential energy function for an ideal gas. The dashed blue line indicates an approximate r value below which a more nearly exact equation of state than the ideal gas law should be used. V(r) = 0 at r = rv= o and as r — » oo.
nRT n2a V - nb V2
(1.22)
This equation of state has two parameters that are substance dependent and must be experimentally determined. The parameters b and a take the finite size of the molecules and the strength of the attractive interaction into account, respectively. (Values of a and b for selected gases are listed in Table 7.4.) The van der Waals equation of state is more accurate in calculating the relationship between P, V, and T for gases than the ideal gas law because a and b have been optimized using experimental results. However, there are other more accurate equations of state that are valid over a wider range than the van der Waals equation, as will be discussed in Chapter 7.
EXAMPLE PROBLEM 1.3
Van der Waals parameters are generally tabulated with either of two sets of units:
a. Pa m6 mol-2 or bar dm6 mol-2
b. m3 mol-1 or dm3 mol-1
Determine the conversion factor to convert one system of units to the other. Note that 1 dm3 = 1(T3 m3 = 1 L.
Solution
f. _9 bar 106 dm6 , _9
Pam6 mol 2 X — - — X - - — = 10 bar dm6 mol 2
105 Pa
m
o 103 dm3 o q
m3 mol 1 X - t — = 103 dm3 mol 1
m
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12
CHAPTER 1 Fundamental Concepts of Thermodynamics
In Example Problem 1 .4, a comparison is made of the molar volume for N2 calcu¬ lated at low and high pressures, using the ideal gas and van der Waals equations of state.
EXAMPLE PROBLEM 1.4
a. Calculate the pressure exerted by N2 at 300. K for molar volumes of 250. L mol-1 and 0.100 L mol-1 using the ideal gas and the van der Waals equations of state. The values of parameters a and b for N2 are 1.370 bar dm6 mol-2 and
0.0387 dm3 mol-1, respectively.
b. Compare the results of your calculations at the two pressures. If P calculated using the van der Waals equation of state is greater than those calculated with the ideal gas law, we can conclude that the repulsive interaction of the N2 molecules outweighs the attractive interaction for the calculated value of the density. A similar statement can be made regarding the attractive interaction. Is the attrac¬ tive or repulsive interaction greater for N2 at 300. K and Vm = 0.100 L?
Solution
a. The pressures calculated from the ideal gas equation of state are nRT
P =
lmol X 8.314 X 10-2 L bar mol-1K-1 X 300. K
V
250. L
= 9.98 X 10-2 bar
P
nRT
lmol X 8.314 X 10-2 L bar mol-1K-1 X 300. K 0.100 L
= 249 bar
The pressures calculated from the van der Waals equation of state are
nRT n2a
P = -
V - nb V2
_ lmol X 8.314 X 10~2Lbarmor1K~1 X 300. K 250. L — lmol X 0.0387 dm3 mol-1 (lmol)2 X 1.370 bar dm6 mol-2 (250. L)2
= 9.98 X 10"2 bar
lmol X 8.314 X 10~2 L bar moP'K"1 X 300 K 0.100 L - lmol X 0.0387 dm3 mol"1 (lmol)2 X 1.370 bar dm6 mol"2 (0.100 L)2
= 270. bar
b.
Note that the result is identical with that for the ideal gas law for Vm = 250. L, and that the result calculated for Vm = 0.100 L deviates from the ideal gas law result. Because Preal > pldea\ we conclude that the repulsive interaction is more
important than the attractive interaction for this specific value of molar volume and temperature.
ime
Vocabulary
absolute temperature scale adiabatic
Boltzmann constant boundary
Celsius scale centigrade scale closed system diathermal
elastic collision equation of state equilibrium extensive variable
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NUMERICAL PROBLEMS 13
Freed
gas thermometer ideal gas
ideal gas constant ideal gas law intensive variable isolated system kelvin
macroscopic scale
macroscopic variables mole fraction open system partial pressure surroundings system
system variables temperature
temperature scale thermal equilibrium thermodynamic equilibrium thermodynamic temperature scale thermometer
van der Waals equation of state wall
zeroth law of thermodynamics
Conceptual Problems
Ql.l Real walls are never totally adiabatic. Use your experience to order the following walls in increasing order with respect to their being diathermal: 1-cm-thick concrete,
1 -cm- thick vacuum, 1-cm-thick copper, 1-cm-thick cork.
Q1.2 The parameter a in the van der Waals equation is greater for H20 than for He. What does this say about the difference in the form of the potential function in Figure 1 . 10 for the two gases?
Q1.3 Give an example based on molecule-molecule interac¬ tions excluding chemical reactions, illustrating how the total pressure upon mixing two real gases could be different from the sum of the partial pressures.
Q1.4 Can temperature be measured directly? Explain your answer.
Q1.5 Explain how the ideal gas law can be deduced for the measurements shown in Figures 1.5 and 1.8.
Q1.6 The location of the boundary between the system and the surroundings is a choice that must be made by the thermo- dynamicist. Consider a beaker of boiling water in an airtight room. Is the system open or closed if you place the boundary just outside the liquid water? Is the system open or closed if you place the boundary just inside the walls of the room?
Q1.7 Give an example of two systems that are in equilib¬ rium with respect to only one of two state variables.
Q1.8 At sufficiently high temperatures, the van der Waals equation has the form P ~ RT /( Vm — b). Note that the
attractive part of the potential has no influence in this expres¬ sion. Justify this behavior using the potential energy diagram of Figure 1.10.
Q1.9 Give an example of two systems separated by a wall that are in thermal but not chemical equilibrium.
Q1.10 Which of the following systems are open? (a) a dog, (b) an incandescent light bulb, (c) a tomato plant, (d) a can of tomatoes. Explain your answers.
Ql.ll Which of the following systems are isolated? (a) a bottle of wine, (b) a tightly sealed, perfectly insulated ther¬ mos bottle, (c) a tube of toothpaste, (d) our solar system. Explain your answers.
Q1.12 Why do the z and y components of the velocity not change in the collision depicted in Figure 1.2?
Q1.13 If the wall depicted in Figure 1.2 were a movable piston, under what conditions would it move as a result of the molecular collisions?
Q1.14 The mass of a He atom is less than that of an Ar atom. Does that mean that because of its larger mass, Argon exerts a higher pressure on the container walls than He at the same molar density, volume, and temperature? Explain your answer.
Q1.15 Explain why attractive interactions between mole¬ cules in gas make the pressure less than that predicted by the ideal gas equation of state.
Numerical Problems
Problem numbers in red indicate that the solution to the prob¬ lem is given in the Student’s Solutions Manual.
Pl.l Approximately how many oxygen molecules arrive each second at the mitochondrion of an active person with a mass of 84 kg? The following data are available: Oxygen con¬ sumption is about 40. mL of 02 per minute per kilogram of body weight, measured at T = 300. K and P = 1.00 atm. In an adult there are about 1.6 X 1010 cells per kg body mass. Each cell contains about 800. mitochondria.
P1.2 A compressed cylinder of gas contains 2.74 X 103 g of N2 gas at a pressure of 3.75 X 107 Pa and a temperature of 18.7°C. What volume of gas has been released into the
atmosphere if the final pressure in the cylinder is 1.80 X 105 Pa? Assume ideal behavior and that the gas temperature is unchanged.
PI. 3 Calculate the pressure exerted by Ar for a molar vol¬ ume of 1.31 L mol-1 at 426 K using the van der Waals equa¬ tion of state. The van der Waals parameters a and b for Ar are 1.355 bar dm6 mol-2 and 0.0320 dm3 mol-1, respectively. Is the attractive or repulsive portion of the potential dominant under these conditions?
P1.4 A sample of propane (C3H8) is placed in a closed ves¬ sel together with an amount of 02 that is 2.15 times the amount needed to completely oxidize the propane to C02 and
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14 CHAPTER 1 Fundamental Concepts of Thermodynamics
H20 at constant temperature. Calculate the mole fraction of each component in the resulting mixture after oxidation, assuming that the H20 is present as a gas.
PI. 5 A gas sample is known to be a mixture of ethane and butane. A bulb having a 230.0 cm3 capacity is filled with the gas to a pressure of 97.5 X 103 Pa at 23.1°C. If the mass of the gas in the bulb is 0.3554 g, what is the mole percent of butane in the mixture?
PI. 6 One liter of fully oxygenated blood can carry 0.18 liters of 02 measured at T = 298 K and P = 1.00 atm. Calculate the number of moles of 02 carried per liter of blood. Hemoglobin, the oxygen transport protein in blood has four oxygen binding sites. How many hemoglobin mole¬ cules are required to transport the 02 in 1.0 L of fully oxy¬ genated blood?
P1.7 Yeast and other organisms can convert glucose (C6H1206) to ethanol (CH3CH2OH) by a process called alchoholic fermentation. The net reaction is
C6H1206 (Y) — > 2C2H5OH (/) + 2C02 (g)
Calculate the mass of glucose required to produce 2.25 L of C02 measured at P = 1.00 atm and T = 295 K.
P1.8 A vessel contains 1.15 g liq H20 in equilibrium with water vapor at 30. °C. At this temperature, the vapor pressure of H20 is 31.82 torr. What volume increase is necessary for all the water to evaporate?
P1.9 Consider a 3 1 .0 L sample of moist air at 60. °C and one atm in which the partial pressure of water vapor is 0.131 atm. Assume that dry air has the composition 78.0 mole percent N2, 21.0 mole percent 02, and 1.00 mole percent Ar.
a. What are the mole percentages of each of the gases in the sample?
b. The percent relative humidity is defined as %RH = Ph2o/Ph2o where Ph2o is the partial pressure of water in the sample and Ph2o = 0.197 atm is the equilibrium vapor pressure of water at 60. °C. The gas is compressed at 60. °C until the relative humidity is 100.%. What volume does the mixture contain now?
c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 81.0 atm?
PI. 10 A typical diver inhales 0.450 liters of air per breath and carries a 25 L breathing tank containing air at a pressure of 300. bar. As she dives deeper, the pressure increases by 1 bar for every 10.08 m. How many breaths can the diver take from this tank at a depth of 35 m? Assume that the tempera¬ ture remains constant.
PI. 11 Use the ideal gas and van der Waals equations to cal¬ culate the pressure when 2.25 mol H2 are confined to a vol¬ ume of 1.65 L at 298 K. Is the gas in the repulsive or attractive region of the molecule-molecule potential?
PI. 12 A rigid vessel of volume 0.400 m3 containing H2 at 21.25°C and a pressure of 715 X 103 Pa is connected to a sec¬ ond rigid vessel of volume 0.750 m3 containing Ar at 30.15°C at a pressure of 203 X 103 Pa. A valve separating the two
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vessels is opened and both are cooled to a temperature of 12.2°C. What is the final pressure in the vessels?
PI. 13 A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO according to the reaction CuO(s) + H2(g) — » Cu(T) + H20(/), and oxygen reoxidizes the copper formed according to Cxx(s) + 1/2 02(g) — » CuO(T). At 25 °C and 750. Torr, 172.0 cm3 of the mixture yields 77.5 cm3 of dry oxygen measured at 25 °C and 750. Torr after passage over CuO and the drying agent. What is the original composition of the mixture?
PI. 14 An athlete at high performance inhales ~3.75 L of air at 1.00 atm and 298 K. The inhaled and exhaled air contain 0.50 and 6.2% by volume of water, respectively. For a respira¬ tion rate of 32 breaths per minute, how many moles of water per minute are expelled from the body through the lungs?
PI. 15 Devise a temperature scale, abbreviated G, for which the magnitude of the ideal gas constant is 5.52 J G-1 mol-1.
P1.16 Aerobic cells metabolize glucose in the respiratory system. This reaction proceeds according to the overall reaction
602 (g) + C6H1206 (s) —> 6C02(g) + 6H20(/)
Calculate the volume of oxygen required at STP to metabo¬ lize 0.025 kg of glucose (C6H1206). STP refers to standard temperature and pressure, that is, T = 273 K and P = 1.00 atm. Assume oxygen behaves ideally at STP.
PI. 17 An athlete at high performance inhales ~3.75 L of air at 1.0 atm and 298 K at a respiration rate of 32 breaths per minute. If the exhaled and inhaled air contain 15.3 and 20.9% by volume of oxygen respectively, how many moles of oxy¬ gen per minute are absorbed by the athlete’s body?
P1.18 A mixture of 2.10 X 10-3 g of 02, 3.88 X 10-3 mol ofN2, and 5.25 X 1020 molecules of CO are placed into a vessel of volume 5.25 L at 12.5°C.
a. Calculate the total pressure in the vessel.
b. Calculate the mole fractions and partial pressures of each gas.
PI. 19 Calculate the pressure exerted by benzene for a molar volume of 2.00 L at 595 K using the Redlich-Kwong equation of state:
_ RT _ a _ 1
Vm-b VrVm{Vm + b)
nRT n2a 1 ~ V - nb VT V(V + nb )
The Redlich-Kwong parameters a and b for benzene are 452.0 bar dm6 mol-2 K1/2 and 0.08271 dm3 mol-1, respec¬ tively. Is the attractive or repulsive portion of the potential dominant under these conditions?
P1.20 In the absence of turbulent mixing, the partial pres¬ sure of each constituent of air would fall off with height above sea level in Earth’s atmosphere as Pt = p^e~Mi§z^RT where Pt is the partial pressure at the height z, Pf is the partial pressure of component i at sea level, g is the acceleration of
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NUMERICAL PROBLEMS 15
gravity, R is the gas constant, T is the absolute tempera¬ ture, and Mi is the molecular mass of the gas. As a result of turbulent mixing, the composition of Earth’s atmosphere is constant below an altitude of 100 km, but the total pressure decreases with altitude as P = p°e~Mave8Z^RT where Mave is the mean molecular weight of air. At sea level, xNl = 0.78084 and xHe = 0.00000524 and T = 300. K.
a. Calculate the total pressure at 8.5 km, assuming a mean molecular mass of 28.9 g mol-1 and that T = 300. K throughout this altitude range.
b. Calculate the value that xNJxHe would have at 8.5 km in the absence of turbulent mixing. Compare your answer with the correct value.
P1.21 An initial step in the biosynthesis of glucose C^H^O^ is the carboxylation of pyruvic acid CH3COCOOH to form oxaloacetic acid HOOCCOCH2COOH
CH3COCOOH 0) + C02 (g) —> HOOCCOCH2COOH (s)
If you knew nothing else about the intervening reactions involved in glucose biosynthesis other than no further car- boxylations occur, what volume of C02 is required to produce 1.10 g of glucose? Assume P = 1 atm and T = 298 K.
PI. 22 Consider the oxidation of the amino acid glycine NH2CH2COOH to produce water, carbon dioxide, and urea NH2CONH2:
NH2CH2COOH(s) + 302(g) — >
NH2CONH2 (s) + 3C02 (g) + 3H2O(0
Calculate the volume of carbon dioxide evolved at P = 1.00 atm and T = 305 K from the oxidation of 0.022 g of glycine.
P1.23 Assume that air has a mean molar mass of 28.9 g mol-1 and that the atmosphere has a uniform temperature of 25.0°C. Calculate the barometric pressure in Pa in Santa Fe, for which z — 7000. ft. Use the information contained in Problem PI. 20.
P1.24 When Julius Caesar expired, his last exhalation had a volume of 450. cm3 and contained 1.00 mole percent argon. Assume that T = 300. K and P = 1.00 atm at the location of his demise. Assume further that T has the same value through¬ out Earth’s atmosphere. If all of his exhaled Ar atoms are now uniformly distributed throughout the atmosphere, how many inhalations of 450. cm3 must we make to inhale one of the Ar atoms exhaled in Caesar’s last breath? Assume the radius of Earth to be 6.37 X 106 m. [ Hint: Calculate the number of Ar atoms in the atmosphere in the simplified geometry of a plane of area equal to that of Earth’s surface. See Problem PI. 20 for the dependence of the barometric pressure and the composi¬ tion of air on the height above Earth’s surface.
PI. 25 Calculate the number of molecules per m3 in an ideal gas at the standard temperature and pressure conditions of 0.00°C and 1.00 atm.
P1.26 Consider a gas mixture in a 1 .50 dm3 flask at 22.0°C. For each of the following mixtures, calculate the partial pressure
of each gas, the total pressure, and the composition of the mix¬ ture in mole percent:
a. 3.06 g H2 and 2.98 g 02
b. 2.30 g N2 and 1.61 g 02
c. 2.02 g CH4 and 1.70 g NH3
P1.27 A mixture of H2 and NH3 has a volume of 139.0 cm3 at 0.00°C and 1 atm. The mixture is cooled to the temperature of liquid nitrogen at which ammonia freezes out and the remaining gas is removed from the vessel. Upon warming the vessel to 0.00°C and 1 atm, the volume is 77.4 cm3. Calculate the mole fraction of NH3 in the original mixture.
P1.28 A sealed flask with a capacity of 1.22 dm3 contains 4.50 g of carbon dioxide. The flask is so weak that it will burst if the pressure exceeds 9.500 X 105 Pa. At what temperature will the pressure of the gas exceed the bursting pressure?
P1.29 A balloon filled with 1 1 .50 L of Ar at 1 8.7°C and 1 atm rises to a height in the atmosphere where the pressure is 207 Torr and the temperature is -32.4°C. What is the final volume of the balloon? Assume that the pressure inside and outside the balloon have the same value.
P1.30 Carbon monoxide competes with oxygen for bind¬ ing sites on the transport protein hemoglobin. CO can be poisonous if inhaled in large quantities. A safe level of CO in air is 50. parts per million (ppm). When the CO level increases to 800. ppm, dizziness, nausea, and unconscious¬ ness occur, followed by death. Assuming the partial pressure of oxygen in air at sea level is 0.20 atm, what proportion of CO to 02 is fatal?
PI. 31 The total pressure of a mixture of oxygen and hydro¬ gen is 1.65 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.190 atm when measured at the same values of T and V as the original mixture. What was the composition of the original mixture in mole percent?
P1.32 Suppose that you measured the product PV of 1 mol of a dilute gas and found that PV = 24.35 L atm at 0.00°C and 33.54 L atm at 100.°C. Assume that the ideal gas law is valid, with T = t(°C) + a , and that the values of R and a are not known. Determine R and a from the measurements provided.
P1.33 Liquid N2 has a density of 875.4 kg m-3 at its normal boiling point. What volume does a balloon occupy at 298 K and a pressure of 1.00 atm if 3.10 X 10-3 L of liquid N2 is injected into it? Assume that there is no pressure difference between the inside and outside of the balloon.
PI. 34 Calculate the volume of all gases evolved by the complete oxidation of 0.375 g of the amino acid alanine CH3CH(NH2)COOH if the products are liquid water, nitro¬ gen gas, and carbon dioxide gas; the total pressure is 1.00 atm; and T = 298 K.
P1.35 As a result of photosynthesis, an acre of forest (1 acre = 4047 square meters) can take up 1000. kg of C02. Assuming air is 0.0314% C02 by volume, what volume of air is required to provide 350. kg of C02? Assume T = 310 K and P = 1.00 atm.
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16
CHAPTER 1 Fundamental Concepts of Thermodynamics
P1.36 A glass bulb of volume 0.198 L contains 0.457 g of gas at 759.0 Torr and 134.0°C. What is the molar mass of the gas?
P1.37 Use L’Hopital’s rule, lim[/(x)/g(x)]x^0 =
~ df(x)/dx 1
lim - - — to show that the expression derived for
_dg(x)/dx Jx^o
Pj in part (b) of Example Problem 1 . 1 has the correct limit as y —> 0.
P1.38 A 455 cm3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.245 g at 25.0°C and the pressure is 760. Torr, calculate the mole fraction of Xe in the mixture.
PI. 39 Many processes such as the fabrication of integrated circuits are carried out in a vacuum chamber to avoid reac¬ tion of the material with oxygen in the atmosphere. It is dif¬ ficult to routinely lower the pressure in a vacuum chamber below 1.0 X 10-10 Torr. Calculate the molar density at this pressure at 300. K. What fraction of the gas phase molecules initially present for 1 .0 atm in the chamber are present at 1.0 X 10-10 Torr?
PI. 40 Rewrite the van der Waals equation using the molar volume rather than V and n.
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Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
I n this chapter, the internal energy U is introduced. The first law of ther¬ modynamics relates AU to the heat (g) and work (w) that flows across the boundary between the system and the surroundings. Other important concepts introduced include heat capacity, the difference between state and path functions, and reversible versus irreversible processes. The enthalpy H is introduced as a form of energy that can be directly meas¬ ured by the heat flow in a constant pressure process. We show how AU, AH, q, and w can be calculated for processes involving ideal gases.
2 The Internal Energy and the First Law of ■ [ Thermodynamics
This section focuses on the change in energy of the system and surroundings during a thermodynamic process such as an expansion or compression of a gas. In thermo¬ dynamics, we are interested in the internal energy of the system, as opposed to the energy associated with the system relative to a particular frame of reference. For exam¬ ple, a container of gas in an airplane has a kinetic energy relative to an observer on the ground. However, the internal energy of the gas is defined relative to a coordinate sys¬ tem fixed on the container. Viewed at a molecular level, the internal energy can take on a number of forms such as
• the translational energy of the molecules.
• the potential energy of the constituents of the system; for example, a crystal consist¬ ing of polarizable molecules will experience a change in its potential energy as an electric field is applied to the system.
• the internal energy stored in the form of molecular vibrations and rotations.
• the internal energy stored in the form of chemical bonds that can be released through a chemical reaction.
• the potential energy of interaction between molecules.
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2.1
The Internal Energy and the First Law of Thermodynamics
2.2
Work
2.3
Heat
2.4
Doing Work on the System and Changing the System Energy from a Molecular
Level Perspective
2.5
Heat Capacity
2.6
State Functions and Path
Functions
2.7
Equilibrium, Change, and Reversibility
2.8
Comparing Work for Reversible and Irreversible
Processes
2.9
Determining AL/and Introducing Enthalpy, a New State Function
2.10
Calculating q, w, AU, and
AH for Processes Involving Ideal Gases
2.11
The Reversible Adiabatic Expansion and Compression of an Ideal Gas
17
18 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
The total of all these forms of energy for the system of interest is given the symbol U and is called the internal energy.1
The first law of thermodynamics is based on our experience that energy can be neither created nor destroyed, if the energies of both the system and the surroundings are taken into account. This law can be formulated in a number of equivalent forms. Our initial formulation of this law is as follows:
( "
The internal energy U of an isolated system is constant.
\ _ _ _ /
This form of the first law looks uninteresting because it suggests that nothing happens in an isolated system when viewed from outside the system. How can the first law tell us anything about thermodynamic processes such as chemical reactions? Consider sep¬ arating an isolated system into two subsystems, the system and the surroundings. When changes in U occur in a system in contact with its surroundings, A Utotai is given by
Af? total
A U
system
+ A U
surroundings
= o
Therefore, the first law becomes
(2.1)
A U
system
= -A U
surroundings
(2.2)
For any decrease of Usystem, U surroundings must increase by exactly the same amount. For example, if a gas (the system) is cooled, the temperature of the surroundings must increase.
How can the energy of a system be changed? There are many ways to alter U, several of which are discussed in this chapter. Experience has shown that all changes in a closed system in which no chemical reactions or phase changes occur can be classified only as heat, work, or a combination of both. Therefore, the internal energy of such a system can only be changed by the flow of heat or work across the boundary between the system and surroundings. For example, U for a gas can be increased by putting it in an oven or by compressing it. In both cases, the temperature of the system increases. This important recognition leads to a second and more useful formulation of the first law:
A U = q + w (2.3)
where q and w designate heat and work, respectively. We use A U without a subscript to indicate the change in internal energy of the system. What do we mean by heat and work? In the following two sections, we define these important concepts and discuss how they differ.
The symbol A is used to indicate a change that occurs as a result of an arbitrary process. The simplest processes are those in which one of P, V, or T remains constant. A constant temperature process is referred to as isothermal, and the corresponding terms for constants P and V are isobaric and isochoric, respectively.
Piston
PM
Final state
FIGURE 2.1
A system is shown in which compression work is being done on a gas. The walls are adiabatic.
Work
In this and the next section, we discuss the two ways in which the internal energy of a system can be changed. Work in thermodynamics is defined as any quantity of energy that “flows” across the boundary between the system and surroundings as a result of a force acting through a distance. Examples are moving an ion in a solution from one region of electrical potential to another, inflating a balloon, or climbing stairs. In each of these examples, there is a force along the direction of motion. Consider another example, a gas in a piston and cylinder assembly as shown in Figure 2.1. In this exam¬ ple, the system is defined as the gas alone. Everything else shown in the figure is in the surroundings. As the gas is compressed, the height of the mass in the surroundings is lowered and the initial and final volumes are defined by the mechanical stops indicated in the figure.
!We could include other terms such as the binding energy of the atomic nuclei but choose to include only the forms of energy that are likely to change in simple chemical reactions.
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2.2 WORK 19
Work has several important characteristics:
Work is transitory in that it only appears during a change in state of the system and surroundings. Only energy, and not work, is associated with the initial and final states of the systems.
The net effect of work is to change U of the system and surroundings in accordance with the first law. If the only change in the system results from a force acting through a distance (as for example the movement of the mass in Figure 2.1), work has flowed between the system and the surroundings. Work can usually be represented by a mass in the surroundings that has been raised or lowered in Earth’s gravitational field.
The quantity of work can be calculated using the definition
rxf
w =
F • dx
(2.4)
Note that because of the scalar product in the integral, the work will be zero unless the force has a component along the displacement direction.
• The sign of the work follows from evaluating the preceding integral. If w > 0, AU > 0 for an adiabatic process. It is common usage to say that if w is positive, work is done on the system by the surroundings. If w is negative, work is done by the system on the surroundings. The quantity of work can also be calculated from the change in potential energy of the mass in the surroundings, A Epotentiai = mg Ah = —w, where g is the gravitational acceleration and Ah is the change in the height of the mass m.
Using the definition of pressure as the force per unit area (A), the work done in moving the mass in Figure 2.1 is given by
rxf
rxf
w =
F • dx = —
external
Adx = -
external
dV
(2.5)
The minus sign appears because F and dx are vectors that point in opposite directions. Note that the pressure that appears in this expression is the external pressure P external which need not equal the system pressure P.
An example of another important kind of work, electrical work, is shown in Figure 2.2, in which the content of the cylinder is the system. Electrical current flows through a con¬ ductive aqueous solution and water undergoes electrolysis to produce H2 and 02 gas. The current is produced by a generator, like that used to power a light on a bicycle through the mechanical work of pedaling. As current flows, the mass that drives the generator is low¬ ered. In this case, the surroundings do the electrical work on the system. As a result, some of the liquid water is transformed to H2 and 02. From electrostatics, the work done in transporting a charge Q through an electrical potential difference cj) is
W electrical Qt
For a constant current I that flows for a time t, Q = It. Therefore,
W electrical ~
(2.6)
(2.7)
The system also does work on the surroundings through the increase in the volume of the gas phase at the constant external pressure Pi9 as shown by the raised mass on the piston. The total work done is
fVf
W — Wp—y + W electrical ~ ^ ~ I P external ~ ~ P externaliY f ~ l^z)( 2.8)
JVi
Other forms of work include the work of expanding a surface, such as a soap bub¬ ble, against the surface tension. Table 2.1 shows the expressions for work for four dif¬ ferent cases. Each of these different types of work poses a requirement on the walls separating the system and surroundings. To be able to carry out the first three types of work, the walls must be movable, whereas for electrical work, they must be conductive. Several examples of work calculations are given in Example Problem 2.1.
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FIGURE 2.2
Current produced by a generator is used to electrolyze water and thereby do work on the system as shown by the lowered mass linked to the generator.
20 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
TABLE 2.1 Types of Work
Types of Work
Variables
Equation for Work
Conventional Units
Volume expansion
Pressure (P), volume (V)
w =
rvf
~ 1 P external dV
JVi
Pam3 = J
Stretching
Force (F), length (/)
w =
!>«
Nm = J
Surface expansion
Surface tension (y), area (a)
w =
raf
/ y • d(r
(Nm'1)(m2) = J
Electrical
Electrical potential (</>), electrical charge ( Q )
w =
[Q 4>dQ'
Jo
VC = J
| EXAMPLE PROBLEM 2.1
a. Calculate the work involved in expanding 20.0 L of an ideal gas to a final volume of 85.0 L against a constant external pressure of 2.50 bar.
b. An air bubble in liquid water expands from a radius of 1.00 cm to a radius of 3.25 cm. The surface tension of water is 71.99 N m-1. How much work is done in increasing the area of the bubble? Assume that the system is the contents of the bubble.
c. A current of 3.20 A is passed through a heating coil for 30.0 s. The electrical potential across the resistor is 14.5 V. Calculate the work done on the coil.
d. If the force to stretch a fiber a distance v is given by F = —kx with k = 100. N cm-1, how much work is done to stretch the fiber 0.15 cm?
Solution
fVf
— / P external dV — P externaliY f ~ Vi)
JVi
105 Pa 10-3 m3
= -2.50 bar X - X (85.0 L - 20.0 L) X - = -16.3 kJ
bar L
b. A factor of 2 is included in the following calculation because a bubble has an inner and an outer surface. We consider the bubble and its contents to be the sys¬ tem. The vectors y and a point in opposite directions, giving rise to the negative sign in the second integral.
r°7
rdf
w
y • dxr = — / y da = 2y Air{rj — rf )
jdi Jdi
= -477 X 71.99 Nm_1(3.252 cm2 - 1.002cm2) X
10 4m2
cm
= -0.865 J
fQ
c. w= 4>dQf = <f>Q = I<f>t = 14.5 V X 3.20 A X 30.0 s = 1.39 kJ
Jo
d. We must distinguish between F, the restoring force on the fiber, and F', the force exerted by the person stretching the fiber. They are related by F = — F'. If we calculate the work done on the fiber, F' • d\ = F'dl because the vectors F' and dl point in the same direction
XL
w = J F' • d\ = J kxdx =
1
?7-
*
to
_ 1
xf
100. Nm-1 X X2
2
x0
2
0.15
- 1.1 J
Xq
If we calculate the work done by the fiber, the sign of w is reversed because F and dl point in opposite directions.
tnd j
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2.3 HEAT 21
Heat
Heat2 is defined in thermodynamics as the quantity of energy that flows across the boundary between the system and surroundings because of a temperature difference between the system and the surroundings. Heat always flows spontaneously from regions of high temperature to regions of low temperature. Just as for work, several important characteristics of heat are of importance:
• Heat is transitory, in that it only appears during a change in state of the system and surroundings. Heat is not associated with the initial and final states of the system and the surroundings.
• The net effect of heat is to change the internal energy of the system and surround¬ ings in accordance with the first law. If the only change in the surroundings is a change in temperature of a reservoir, heat has flowed between the system and the surroundings. The quantity of heat that has flowed is directly proportional to the change in temperature of the reservoir.
• The sign convention for heat is as follows. If the temperature of the system is raised, q is positive; if it is lowered, q is negative. It is common usage to say that if q is pos¬ itive, heat is withdrawn from the surroundings and deposited in the system. If q is negative, heat is withdrawn from the system and deposited in the surroundings.
Defining the surroundings as the rest of the universe is impractical because it is not realistic to search through the whole universe to see if a mass has been raised or low¬ ered and if the temperature of a reservoir has changed. Experience shows that in gen¬ eral only those parts of the universe close to the system interact with the system. Experiments can be constructed to ensure that this is the case, as shown in Figure 2.3. Imagine that we are interested in an exothermic chemical reaction that is carried out in a rigid sealed container with diathermal walls. We define the system as consisting solely of the reactant and product mixture. The vessel containing the system is immersed in an inner water bath separated from an outer water bath by a container with rigid diathermal walls. During the reaction, heat flows out of the system (q < 0), and the temperature of the inner water bath increases to Tf. Using an electrical heater, the temperature of the outer water bath is increased so that at all times, Touter = Tinner. Because of this condition, no heat flows across the boundary between the two water baths, and because the container enclosing the inner water bath is rigid, no work flows across this boundary. Therefore, A U = q + w = 0 + 0 = 0 for the composite system made up of the inner water bath and everything within it. Therefore, this composite sys¬ tem is an isolated system that does not interact with the rest of the universe. To deter¬ mine q and w for the reactant and product mixture, we need to examine only the composite system and can disregard the rest of the universe.
To emphasize the distinction between q and w and the relationship between q , w, and A U, we discuss the two processes shown in Figure 2.4. They are each carried out in an isolated system, divided into two subsystems, I and II. In both cases, system I con¬ sists solely of the liquid in the beaker, and everything else including the rigid adiabatic walls is in system II. We refer to system I as the system and system II as the surround¬ ings in the following discussion. We assume that the temperature of the liquid is well below its boiling point so that its vapor pressure is negligibly small. This ensures that no liquid is vaporized in the process, and the system is closed. We also assume that the change in temperature of the system is very small. System II can be viewed as the sur¬ roundings for system I and vice versa.
Rest of universe
Thermometers
Heating coil
FIGURE 2.3
An isolated composite system is created in which the surroundings to the system of interest are limited in extent. The walls surrounding the inner water bath are rigid.
2Heat is perhaps the most misused term in thermodynamics as discussed by Robert Romer [“Heat is not a Noun.” American Journal of Physics, 69 (2001), 107-109]. In common usage, it is incorrectly referred to as a substance as in the phrase “Close the door; you’re letting the heat out!” An equally inappropriate term is heat capacity (discussed in Section 2.5), because it implies that materials have the capacity to hold heat, rather than the capacity to store energy. We use the terms heat flow or heat transfer to emphasize the transitory nature of heat. However, you should not think of heat as a fluid or a substance.
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22 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
FIGURE 2.4
Two subsystems, I and II, are enclosed in a rigid adiabatic enclosure. System I con¬ sists solely of the liquid in the beaker for each case. System II consists of every¬ thing else in the enclosure, and is the sur¬ roundings for system I. (a) The liquid is heated using a flame, (b) The liquid is heated using a resistive coil through which an electrical current flows.
Propane Bunsen burner
(a)
In Figure 2.4a, a Bunsen burner fueled by a propane canister is used to heat the liq¬ uid (system). The boundary between system and surroundings is the surface that encloses the liquid, and heat can flow all across this boundary. It is observed that the temperature of the liquid increases in the process. The temperature of the surroundings also increases because the system and surroundings are in thermal equilibrium. From Section 1.2, we know that the internal energy of a monatomic gas increases linearly with T. This result can be generalized to state that U is a monotonically increasing function of T for a uniform single-phase system of constant composition. Therefore, because A T > 0, A U > 0.
We next consider the changes in the surroundings. From the first law, A U surroundings = “At/ < 0. No forces oppose changes in the system. We conclude that w = 0. Therefore, if A U > 0 ,q > 0 and q surroundings < 0.
We now consider Figure 2.4b. In this case, the boundary between system and sur¬ roundings lies just inside the inner wall of the beaker, across the open top of the beaker, and just outside of the surface of the heating coil. Note that the heating coil is entirely in the surroundings. Heat can flow across the boundary surface. Upon letting the mass in the surroundings fall, electricity flows through the heating coil. It is our experience that the temperature of the liquid (system) will increase. We again conclude that AU > 0. What values do q and w have for the process?
To answer this question, consider the changes in the surroundings. From the first law, A U surr0undings = “At/ < 0. We see that a mass has been lowered in the sur¬ roundings. Can we conclude that w > 0? No, because work is being done only on the heating coil, which is in the surroundings. The current flow never crosses the boundary between system and surroundings. Therefore, w = 0 because no work is done on the system. However, A U > 0, so if w = 0 we conclude that q > 0. The increase in U is due to heat flow from the surroundings to the system caused by the difference between the temperature of the heating filament and the liquid and not by the electrical work done on the filament. Note that because A U + A U sur roundings = 0, the heat flow from the surroundings to the system can be calculated from the electrical work done entirely within the surroundings, q = ~wsurroundings = I<f>t.
These examples show that the distinction between heat and work must be made carefully with a clear knowledge of the position and nature of the boundary between the system and the surroundings.
f EXA
EXAMPLE PROBLEM 2.2
A heating coil is immersed in a 100. g sample of H20 liquid which boils at 99.61 °C in an open insulated beaker on a laboratory bench at 1 bar pressure. In this process, 10.0% of the liquid is converted to the gaseous form at a pressure of 1 bar. A current of 2.00 A flows through the heater from a 12.0 V battery for 1.00 X 103 s to
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2.4 DOING WORK ON THE SYSTEM AND CHANGING THE SYSTEM ENERGY FROM A MOLECULAR LEVEL PERSPECTIVE
23
effect the transformation. The densities of liquid and gaseous water under these condi¬ tions are 997 and 0.590 kg m-3, respectively.
a. It is often useful to replace a real process by a model that exhibits the important features of the process. Design a model system and surroundings, like those shown in Figures 2.1 and 2.2, that would allow you to measure the heat and work associated with this transformation. For the model system, define the system and surroundings as well as the boundary between them.
b. How can you define the system for the open insulated beaker on the laboratory bench such that the work is properly described?
c. Calculate q and w for the process.
Solution
a. The model system is shown in the following figure. The cylinder walls and the piston form adiabatic walls. The external pressure is held constant by a suit¬ able weight.
b. Define the system as the liquid in the beaker and the volume containing only molecules of H20 in the gas phase. This volume will consist of disconnected volume elements dispersed in the air above the laboratory bench.
c. In the system shown, the heat input to the liquid water can be equated with the work done on the heating coil. Therefore,
q = I(f>t = 2.00 A X 12.0 V X 1.00 X 103 s = 24.0kJ
As the liquid is vaporized, the volume of the system increases at a constant exter¬ nal pressure. Therefore, the work done by the system on the surroundings is
Vi) = — 105 Pa
W = —Pp
al(Y f
X
= — 1.70kJ
Note that the electrical work done on the heating coil is much larger than the P-V \ work done in the expansion.
e P-V ^
P external TOO at ID
Heating coil
/ 10.0 X 10“3kg 90.0 X 10“3kg
100.0 X 10“3kg\
V 0.590 kgm-3 997 kgm-3
997 kgm-3 )
Doing Work on the System and Changing the System Energy from a Molecular Level Perspective
Our discussion so far has involved changes in energy for macroscopic systems, but what happens at the molecular level if energy is added to the system? In shifting to a molecular perspective, we move from a classical to a quantum mechanical description of matter. For this discussion, we need two results that will be discussed elsewhere in this textbook. First, as will be discussed in Chapter 15, in general the energy levels of quantum mechanical systems are discrete and molecules can only possess amounts of energy that correspond to these values. By contrast, in classical mechanics the energy is a continuous variable. Second, we use a result from statistical mechanics that the relative probability of a molecule being in a state corresponding to the allowed energy values Ei and s2 at temperature T is given by e~^S2~Sl^kBT\ This result will be discussed in Chapter 30.
To keep the mathematics simple, consider a very basic model system: a gas con¬ sisting of a single He atom confined in a one-dimensional container with a length of 10. nm. Quantum mechanics tells us that the translational energy of the He atom confined in this box can only have the discrete values shown in Figure 2.5a. We lower the temperature of this one-dimensional He gas to 0.20 K. The calculated probability of the atom being in a given energy level is shown in Figure 2.5a. If we now do work on the system by compressing the box to half its original length at con¬ stant temperature, the energy levels will change to those shown in Figure 2.5b. Note
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24 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
FIGURE 2.5
Energy levels are shown for the box of length (a) 10. nm, and (b) 5.0 nm. The circles indicate the probability that the He atom has an energy correspon¬ ding to each of the energy levels at 0.2 K. Each circle indicates a probability of 0.010. For example, the prob¬ ability that the energy of the He atom corre¬ sponds to the lowest energy level in Figure 2.5a is 0.22. Note the different scales for energy in each graph.
8 x 1 0'24
6 x 1 0'24
111 4 x 1 0'2
2x1 O'24
0
(a)
that all the energy levels are shifted to higher values as the container is made smaller (an effect that will be fully explored in Chapter 15). If we keep the temperature con¬ stant at 0.20 K during this compression, the distribution of the atoms among the energy levels changes to that shown in Figure 2.5b. This redistribution occurs because the total translational energy of the He atom remains constant in the com¬ pression if the temperature is kept constant, assuming ideal gas behavior.
What happens if we keep the container at the smaller length and raise the system energy by increasing T to 0.40 K? In this case, the energy levels are unchanged because they depend on the container length, but not on the temperature of the gas. However, the energy of the system increases and as shown in Figure 2.6, this occurs by a redistri¬ bution of the probability of finding the He atom in the energy levels. The increase in system energy comes from an increase in the probability of the He atom populating higher-energy levels and a corresponding decrease in the probability of populating lower-energy levels.
Just as for a calculation of the energy of a gas using classical mechanics, the quan¬ tum mechanical system energy increases through an increase in the translational energy
FIGURE 2.6
Energy levels are shown for the 5.0 nm box. The circles indi¬ cate the probability that a He atom has an energy corresponding to each of the energy levels at (a) 0.20 K and (b) 0.40 K.
4x1 0‘23 -i
3 x 1 0'23 -
2x1 O'23 “
1 x IQ'23
M9.
(a)
0
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of the atom. However, the discrete energy level structure influences how the system can take up energy. This will become clearer in the next section when we consider how molecules can take up energy through rotation and vibration.
2.5 HEAT CAPACITY
25
Heat Capacity
The process shown in Figure 2.4b provides a way to quantify heat flow in terms of the easily measured electrical work done on the heating coil, w = I(pt. The response of a single-phase system of constant composition to heat input is an increase in T as long as the system does not undergo a phase change such as the vaporization of a liquid.
The thermal response of the system to heat flow is described by a very important thermodynamic property called the heat capacity, which is a measure of energy needed to change the temperature of a substance by a given amount. The name heat capacity is unfortunate because it implies that a substance has the capacity to take up heat. A much better name would be energy capacity.
Heat capacity is a material-dependent property, as will be discussed later. Mathematically, heat capacity is defined by the relation
C =
lim
Ar^o
dq
~dT
(2.9)
where C is in the SI unit of J K-1. It is an extensive quantity that doubles as the mass of the system is doubled. Often, the molar heat capacity Cm is used in calculations. It is an intensive quantity with the units of J K-1 mol-1. Experimentally, the heat capacity of fluids is measured by immersing a heating coil in the gas or liquid and equating the electrical work done on the coil with the heat flow into the sample. For solids, the heat¬ ing coil is wrapped around the solid. The significance of the notation d q for an incre¬ mental amount of heat is explained in the next section.
The value of the heat capacity depends on the experimental conditions under which it is determined. The most common conditions are constant V or P, for which the heat capacity is denoted Cv and CP , respectively. Values of CPm at 298.15 K for pure sub¬ stances are tabulated in Tables 2.2 and 2.3 (see Appendix B, Data Tables), and formulas for calculating CPm at other temperatures for gases and solids are listed in Tables 2.4 and 2.5, respectively.
We next discuss heat capacities using a molecular level model, beginning with gases. Figure 2.7 illustrates the energy level structure for a molecular gas. Molecules can take up energy by moving faster, by rotating in three-dimensional space, by peri¬ odic oscillations (known as vibrations) of the atoms around their equilibrium structure, and by electronic excitations. These energetic degrees of freedom are referred to as
energy
levels
Rotational
energy
levels
Translational
energy
levels
Electronic
energy
levels
FIGURE 2.7
Energy levels are shown schematically for each degree of freedom. The gray area between electronic energy levels on the left indicates what appear to be a continu¬ ous range of allowed energies. However, as the energy scale is magnified stepwise, discrete energy levels for vibration, rota¬ tion, and translation can be resolved.
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26 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
TABLE 2.2
Energy Level
Spacings for Different Degrees
of Freedom
Degree of
Energy Level
Freedom
Spacing
Electronic
5 X 10“19J
Vibration
2 X 10_20J
Rotation
2 X 10"23 J
Translation
2 X 10“41 J
translation, rotation, vibration, and electronic excitation. Each of the degrees of free¬ dom has its own set of energy levels and the probability of an individual molecule occupying a higher energy level increases as it gains energy. Except for translation, the energy levels for atoms and molecules are independent of the container size.
The amount of energy needed to move up the ladder of energy levels is very differ¬ ent for the different degrees of freedom: AEelectronic » A £ vibration ^AErotation ^^translation- Values for these A E are molecule dependent, but order of magnitude num¬ bers are shown in Table 2.2.
Energy is gained or lost by a molecule through collisions with other molecules. An order of magnitude estimate of the energy that can be gained or lost by a molecule in a collision is kBT , where k = R/N A is the Boltzmann constant, and T is the absolute temperature. A given degree of freedom in a molecule can only take up energy through molecular collisions if the spacing between adjacent energy levels and the temperature satisfies the relationship A E ~ kBT , which has the value 4.1 X 10-21 J at 300 K. At 300 K, A E ~ kBT is always satisfied for translation and rotation, but not for vibration and electronic excitation. We formulate the following general rule relating the heat capacity Cym and the degrees of freedom in a molecule, which will be discussed in more detail in Chapter 32.
s N
The heat capacity Cym for a gas at temperature T not much lower than 300 K is
R/2 for each translation and rotational degree of freedom, where R is the ideal gas constant. Each vibrational degree of freedom for which the relation A E/kT < 0.1 is obeyed contributes R to Cym • If A E/kBT > 10, the degree of freedom does not contribute to Cym. For 10 > A E/kBT > 0.1, the degree of freedom contributes partially to Cym.
Temperature/K
FIGURE 2.8
Molar heat capacities Cym are shown for a number of gases. Atoms have only translational degrees of freedom and, therefore, have comparatively low values for Cv m that are independent of tempera¬ ture. Molecules with vibrational degrees of freedom have higher values of Cv m at temperatures sufficiently high to activate the vibrations.
Figure 2.8 shows the variation of CVm for a monatomic gas and several molecu¬ lar gases. Atoms only have three translational degrees of freedom. Linear molecules have an additional 2 rotational degrees of freedom and 3^-5 vibrational degrees of freedom where n is the number of atoms in the molecule. Nonlinear molecules have 3 translational degrees of freedom, 3 rotational degrees of freedom, and 3n-6 vibra¬ tional degrees of freedom. A He atom has only 3 translational degrees of freedom, and all electronic transitions are of high energy compared to kT. Therefore, Cy m = 3R/2 over the entire temperature range as shown in the figure. CO is a lin¬ ear diatomic molecule that has two rotational degrees of freedom for which A E/kBT < 0.1 at 200. K. Therefore, Cy m = 5R/2 at 200. K. The single vibrational degree of freedom begins to contribute to CV m above 200. K, but does not contribute fully for T < 1000. K because 10 > \E/kBT below 1000. K. C02 has 4 vibrational degrees of freedom, some of which contribute to Cy m near 200. K. However, Cy m does not reach its maximum value of 13R/2 below 1000. K. Similarly, Cy m for C2H4, which has 12 vibrational degrees of freedom, does not reach its maximum value of 15 R below 1000. K, because 10 > A E/kBT for some vibrational degrees of freedom. Electronic energy levels are too far apart for any of the molecular gases to give a contribution to CV m.
To this point, we have only discussed Cy m for gases. It is easier to measure CPm than CVm for liquids and solids because liquids and solids generally expand with increasing temperature and exert enormous pressure on a container at constant vol¬ ume (see Example Problem 3.2.) An example of how CPm depends on T for solids and liquids is illustrated in Figure 2.9 for Cl2. To make the functional form of CPjn(T) understandable, we briefly discuss the relative magnitudes of CPm in the solid, liquid, and gaseous phases using a molecular level model. A solid can be thought of as a set of interconnected harmonic oscillators, and heat uptake leads to the excitations of the collective vibrations of the solid. At very low temperatures these vibrations cannot be activated, because the spacing of the vibrational energy levels is large compared to kBT. As a consequence, energy cannot be taken up by the solid. Hence, CPm approaches zero as T approaches zero. For the solid, CPm rises rapidly with T because the thermal energy available as T increases is sufficient to activate the vibrations of
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2.5 HEAT CAPACITY 27
the solid. The heat capacity increases discontinuously as the solid melts to form a liq¬ uid. This is the case because the liquid retains all the local vibrational modes of the solid, and more low-energy modes become available upon melting. Therefore, the heat capacity of the liquid is greater than that of the solid. As the liquid vaporizes, the local vibrational modes present in the liquid are converted to translations that cannot take up as much energy as vibrations. Therefore, CPm decreases discontinu¬ ously at the vaporization temperature. The heat capacity in the gaseous state increases slowly with temperature as the vibrational modes of the individual mole¬ cules are activated as discussed previously. These changes in CPm can be calculated for a specific substance using a microscopic model and statistical thermodynamics, as will be discussed in detail in Chapter 32.
Once the heat capacity of a variety of different substances has been determined, we have a convenient way to quantify heat flow. For example, at constant pressure, the heat flow between the system and surroundings can be written as
T sysj T.surrj
qP = Jcpstem{T)dT = - J Cfrroundings{T)dT (2.10)
T T
1 sts,i 1 surr,i
By measuring the temperature change of a thermal reservoir in the surroundings at con¬ stant pressure, qP can be determined. In Equation (2.10), the heat flow at constant pres¬ sure has been expressed both from the perspective of the system and from the perspective of the surroundings. A similar equation can be written for a constant vol¬ ume process. Water is a convenient choice of material for a heat bath in experiments because CP is nearly constant at the value 4.18 J g-1 K-1 or 75.3 J mol-1 K-1 over the range from 0°C to 100.°C.
FIGURE 2.9
The variation of CP>m with temperature is shown for Cl2.
Constant pressure heating
EXAMPLE PROBLEM 2.3
The volume of a system consisting of an ideal gas decreases at constant pressure. As a result, the temperature of a 1.50 kg water bath in the sur¬ roundings increases by 14.2°C. Calculate qP for the system.
Solution
T
1 surrj
_ / ^surroundings = —^surroundings
= -1.50 kg X 4.18 J g_1 K-1 X 14.2 K = -89
1 kJ _ J
Initial state
Final state
Constant volume heating
How are CP and Cy related for a gas? Consider the processes shown in Figure 2.10 in which a fixed amount of heat flows from the surroundings into a gas. In the constant pressure process, the gas expands as its tem¬ perature increases. Therefore, the system does work on the surround¬ ings. As a consequence, not all the heat flow into the system can be used to increase A U. No such work occurs for the corresponding constant volume process, and all the heat flow into the system can be used to increase A U. Therefore, dTP < dTv for the same heat flow d q. For this reason, CP > Cy for gases.
The same argument applies to liquids and solids as long as V increases with T. Nearly all substances follow this behavior, although notable excep¬ tions occur, such as liquid water between 0°C and 4°C, for which the vol¬ ume increases as T decreases. However, because AVm upon heating is much smaller than for a gas, the difference between CPm and Cym for a liq¬ uid or solid is much smaller than for a gas.
FIGURE 2.10
Not all the heat flow into the system can be used to increase A U in a constant pressure process, because the system does work on the surroundings as it expands. However, no work is done for constant volume heating.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
The preceding remarks about the difference between CP and Cv have been qualita¬ tive in nature. However, the following quantitative relationship, which will be proved in Chapter 3, holds for an ideal gas:
Cp — Cy - nR or CPm Cym - R
(2.11)
2.6
State Functions and Path Functions
An alternate statement of the first law is that AU is independent of the path connecting the initial and final states, and depends only on the initial and final states. We make this statement plausible for the kinetic energy, and the argument can be extended to the other forms of energy listed in Section 2.1. Consider a single molecule in the system. Imagine that the molecule of mass m initially has the speed v^ We now change its speed incrementally following the sequence Vi —> v2 —> v3 — > v4. The change in the kinetic energy along this sequence is given by
(2.12)
Even though v2 and v3 can take on any arbitrary values, they still do not influence the result. We conclude that the change in the kinetic energy depends only on the initial and final speed and that it is independent of the path between these values. Our conclusion remains the same if we increase the number of speed increments in the interval between Vx and v2 to an arbitrarily large number. Because this conclusion holds for all molecules in the system, it also holds for AU.
This example supports the assertion that A U depends only on the final and initial states and not on the path connecting these states. Any function that satisfies this condi¬ tion is called a state function, because it depends only on the state of the system and not the path taken to reach the state. This property can be expressed in a mathematical form. Any state function, for example U, must satisfy the equation
/
(2.13)
where i and/denote the initial and final states. This equation states that in order for A U to depend only on the initial and final states characterized here by i and/, the value of the integral must be independent of the path. If this is the case, U can be expressed as an infinitesimal quantity, dU , that when integrated, depends only on the initial and final states. The quantity dU is called an exact differential. We defer a discussion of exact differentials to Chapter 3.
It is useful to define a cyclic integral, denoted by the symbol j>, as applying to a cyclic path such that the initial and final states are identical. For U or any other state function,
(2.14)
because the initial and final states are the same in a cyclic process.
We next show that q and w are not state functions. The state of a single-phase sys¬ tem at fixed composition is characterized by any two of the three variables P, T, and V. The same is true of U. Therefore, for a system of fixed mass, U can be written in any of the three forms U(V,T ), U(P,T), or U(P,V). Imagine that a gas characterized by V\ and T i is confined in a piston and cylinder system that is isolated from the surroundings. There is a thermal reservoir in the surroundings at a temperature T 3 < T\. We do com¬ pression work on the system starting from an initial state in which the volume is V\ to
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2.6 STATE FUNCTIONS AND PATH FUNCTIONS 29
an intermediate state in which the volume is V2 using a constant external pressure P external where V2 < V The work is given by
external dV ~ ^ external
Vi Vt
Because work is done on the system in the compression (see Figure 2.1 1), w is positive and U increases. Because the system consists of a uniform single phase, U is a monoto¬ nic function of T, and T also increases. The change in volume AV has been chosen such that the temperature of the system T2 in the intermediate state after the compression sat¬ isfies the inequality Tx <T2<T3.
We next lock the piston in place and let an amount of heat q flow between the system and surroundings at constant V by bringing the system into contact with the reservoir at temperature T3. The final state values of T and V after these two steps are T3 and V2.
This two-step process is repeated for different values of the external pressure by changing the mass resting on the piston. In each case the system is in the same final state characterized by the variables V2 and T3. The sequence of steps that takes the system from the initial state V\,T\ to the final state V2,T3 is referred to as a path. By changing the mass, a set of different paths is generated, all of which originate from the state V\ ,7\, and end in the state V2,T3. According to the first law, A U for this two-step process is
7
dV = -P„
al(Vf ~ Vt) = ~P,
external
A V (2.15)
AU = U(T3,V2) ~ U(T i,Vi) = q + w (2.16)
Because A U is a state function, its value for the two-step process just described is the same for each of the different values of the mass.
Are q and w also state functions? For this two step process,
w = -PexternaAV (2.17)
and P external is different for each value of the mass or for each path, whereas AV is constant. Therefore, w is also different for each path; we can choose one path from V\,T\ to V2,T3 and a different path from V2,T3 back to V\ ,T\. Because the work is dif¬ ferent along these paths, the cyclic integral of work is not equal to zero. Therefore, w is not a state function.
Using the first law to calculate q for each of the paths, we obtain the result
q= au -w = AU + PexternaiAV (2.18)
Because AU is the same for each path, and w is different for each path, we conclude that q is also different for each path. Just as for work, the cyclic integral of heat is not equal to zero. Therefore, neither q nor w are state functions, and they are called path functions.
Because both q and w are path functions, there are no exact differentials for work and heat unless the path is specified. Incremental amounts of these quantities are denoted by d q and d w , rather than dq and dw, to emphasize the fact that incremental amounts of work and heat are not exact differentials. Because d q and d w are not exact differentials, there are no such quantities as A q, qp qt and Aw, wp wt. One cannot refer to the work or heat possessed by a system or to the change in work or heat associated with a process. After a transfer of heat and/or work between the system and surroundings is completed, the system and surroundings possess internal energy, but not heat or work.
The preceding discussion emphasizes that it is important to use the terms work and heat in a way that reflects the fact that they are not state functions. Examples of systems of interest to us are batteries, fuel cells, refrigerators, and internal combustion engines. In each case, the utility of these systems is that work and/or heat flows between the system and sur¬ roundings. For example, in a refrigerator, electrical energy is used to extract heat from the inside of the device and to release it in the surroundings. One can speak of the refrigerator as having the capacity to extract heat, but it would be wrong to speak of it as having heat. In the internal combustion engine, chemical energy contained in the bonds of the fuel mole¬ cules and in 02 is released in forming C02 and H20. This change in internal energy can be used to rotate the wheels of the vehicle, thereby inducing a flow of work between the vehi¬ cle and the surroundings. One can refer to the capability of the engine to do work, but it would be incorrect to refer to the vehicle or the engine as containing or having work.
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Final state
FIGURE 2.11
A system consisting of an ideal gas is contained in a piston and cylinder assembly. An external pressure is generated by the mass resting on the piston.The gas in the initial state V\,T\ is compressed to an intermediate state, whereby the temperature increases to the value T2. It is then brought into contact with a thermal reservoir at T3, leading to a further rise in temperature. The final state is V2,T3. The mechanical stops allow the system volume to be only V\ or V2.
30 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
03
CL
CD
o
FIGURE 2.12
All combinations of pressure, volume, and temperature consistent with 1 mol of an ideal gas lie on the colored surface. All combinations of pressure and volume consistent with T = 800 K and all combi¬ nations of pressure and temperature con¬ sistent with a volume of 4.0 L are shown as black curves that lie in the P-V-T sur¬ face. The third curve corresponds to a path between an initial state i and a final state /that is neither a constant tempera¬ ture nor a constant volume path.
Pulley
1 kg
FIGURE 2.13
Two masses of exactly 1 kg each are con¬ nected by a wire of zero mass running over a frictionless pulley. The system is in mechanical equilibrium and the masses are stationary.
Equilibrium, Change, and Reversibility
Thermodynamics can only be applied to systems in internal equilibrium, and a require¬ ment for equilibrium is that the overall rate of change of all processes such as diffusion or chemical reaction be zero. How do we reconcile these statements with our calcula¬ tions of q , w, and A U associated with processes in which there is a macroscopic change in the system? To answer this question, it is important to distinguish between the sys¬ tem and surroundings each being in internal equilibrium, and the system and surround¬ ings being in equilibrium with one another.
We first discuss the issue of internal equilibrium. Consider a system made up of an ideal gas, which satisfies the equation of state, P = nRT /V. All combinations of P, V , and T consistent with this equation of state form a surface in P-V-T space as shown in Figure 2.12. All points on the surface correspond to states of internal equilibrium, mean¬ ing that the system is uniform on all length scales and is characterized by single values of T, P, and concentration. Points that are not on the surface do not correspond to any equi¬ librium state of the system because the equation of state is not satisfied. Nonequilibrium situations cannot be represented on such a plot, because T , P, and concentration do not have unique values for a system that is not in equilibrium. An example of a system that is not in internal equilibrium is a gas that is expanding so rapidly that different regions of the gas have different values for the density, pressure, and temperature.
Next, consider a process in which the system changes from an initial state charac¬ terized by Pt, Vt, and Tt to a final state characterized by Pf, Vf, and Tf as shown in Figure 2.12. If the rate of change of the macroscopic variables is negligibly small, the system passes through a succession of states of internal equilibrium as it goes from the initial to the final state. Such a process is called a quasi-static process, in which inter¬ nal equilibrium is maintained in the system. If the rate of change is large, the rates of diffusion and intermolecular collisions may not be high enough to maintain the system in a state of internal equilibrium. Thermodynamic calculations for such a process are valid only if it is meaningful to assign a single value of the macroscopic variables P, V, T, and concentration to the system undergoing change. The same considerations hold for the surroundings. We only consider quasi-static processes in this text.
We now visualize a process in which the system undergoes a major change in terms of a directed path consisting of a sequence of quasi-static processes, and distin¬ guish between two very important classes of quasi-static processes, namely reversible and irreversible processes. It is useful to consider the mechanical system shown in Figure 2.13 when discussing reversible and irreversible processes. Because the two masses have the same value, the net force acting on each end of the wire is zero, and the masses will not move. If an additional mass is placed on either of the two masses, the system is no longer in mechanical equilibrium, and the masses will move. In the limit in which the incremental mass approaches zero, the velocity at which the initial masses move approaches zero. In this case, one refers to the process as being reversible, meaning that the direction of the process can be reversed by placing the infinitesimal mass on the other side of the pulley.
Reversibility in a chemical system can be illustrated by a system consisting of liq¬ uid water in equilibrium with gaseous water that is surrounded by a thermal reservoir. The system and surroundings are both at temperature T. An infinitesimally small increase in T results in a small increase of the amount of water in the gaseous phase, and a small decrease in the liquid phase. An equally small decrease in the temperature has the opposite effect. Therefore, fluctuations in T give rise to corresponding fluctua¬ tions in the composition of the system. If an infinitesimal opposing change in the vari¬ able that drives the process (temperature in this case) causes a reversal in the direction of the process, the process is reversible.
If an infinitesimal change in the driving variable does not change the direction of the process, one says that the process is irreversible. For example, if a large stepwise temperature increase is induced in the system using a heat pulse, the amount of water in the gas phase increases abruptly. In this case, the composition of the system cannot be
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2.8 COMPARING WORK FOR REVERSIBLE AND IRREVERSIBLE PROCESSES
31
returned to its initial value by an infinitesimal temperature decrease. This relationship is characteristic of an irreversible process. Although any process that takes place at a rapid rate in the real world is irreversible, real processes can approach reversibility in the appropriate limit. For example, a slow increase in the electrical potential in an elec¬ trochemical cell can convert reactants to products in a nearly reversible process.
Comparing Work for Reversible and Irreversible Processes
We concluded in Section 2.6 that w is not a state function and that the work associated with a process is path dependent. This statement can be put on a quantitative footing by comparing the work associated with the reversible and irreversible expansion and the compression of an ideal gas. This process is discussed next and illustrated in Figure 2.14.
Consider the following irreversible process, meaning that the internal and external pressures are not equal. A quantity of an ideal gas is confined in a cylinder with a weight¬ less movable piston. The walls of the system are diathermal, allowing heat to flow between the system and surroundings. Therefore, the process is isothermal at the temper¬ ature of the surroundings, T. The system is initially defined by the variables T, P\, and V\. The position of the piston is determined by Pexternal = Pb which can be changed by adding or removing weights from the piston. Because the weights are moved horizontally, no work is done in adding or removing them. The gas is first expanded at constant tem¬ perature by decreasing Pextemai abruptly to the value P2 (weights are removed), where P2 < P\. A sufficient amount of heat flows into the system through the diathermal walls to keep the temperature at the constant value T. The system is now in the state defined by
T, P2, and V2, where V2 > V\. The system is then returned to its original state in an
isothermal process by increasing Pextemai abruptly to its original value Px (weights are added). Heat flows out of the system into the surroundings in this step. The system has been restored to its original state and, because this is a cyclic process, AU = 0. Are qtotai and wtotai also zero for the cyclic process? The total work associated with this cyclic process is given by the sum of the work for each individual step:
W total ~ P external, ~ W expansion W compression
i
= -P2(V2 - vx) - Pl(vl - v2)
= ~(P2 - Px) X (v2 - Vi) > 0 because P2 < Px andV2 > Vx (2.19)
The relationship between P and V for the process under consideration is shown graphi¬ cally in Figure 2.14, in what is called an indicator diagram. An indicator diagram is useful because the work done in the expansion and contraction steps can be evaluated from the appropriate area in the figure, which is equivalent to evaluating the integral w = — f Pexternai dV. Note that the work done in the expansion is negative because av > 0, and that done in the compression is positive because XV < 0. Because Pi < Pi, the magnitude of the work done in the compression process is more than that done in the expansion process and wtotai > 0. What can one say about qtotaP The first law states that because A U = qtotai + w total = Qtotai < 0-
The same cyclical process is carried out in a reversible cycle. A necessary condition for reversibility is that P = P external at every step of the cycle. This means that P changes during the expansion and compression steps. The work associated with the expansion is
w expansion = ~ J P external dV = ~ f P dV = ~nRT Jy = ~nRT In y (2.20)
This work is shown schematically as the red area in the indicator diagram of Figure 2.15.
If this process is reversed and the compression work is calculated, the following result is obtained:
W compression
V 1
-nRT In — V2
(2.21)
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y v2
v
FIGURE 2.14
The work for each step and the total work can be obtained from an indicator diagram. For the compression step, w is given by the total area in red and yellow; for the expan¬ sion step, w is given by the red area. The arrows indicate the direction of change in V in the two steps. The sign of w is opposite for these two processes. The total work in the cycle is the yellow area.
32
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
The magnitudes of the work in the forward and reverse processes are equal. The total work done in this cyclical process is given by
V 2
W W expansion W compression ~ ~nRTXlCl TlRTXw
y i y 2
v2 v2
= -nRT In — + nRT In — = 0 (2.22)
Vi Vi
Therefore, the work done in a reversible isothermal cycle is zero. Because A U = q + w is a state function, q = — w = 0 for this reversible isothermal process. Looking at the heights of the weights in the surroundings at the end of the process, we find that they are the same as at the beginning of the process. To compare the work for reversible and irreversible processes, the state variables need to be given specific values as is done in Example Problem 2.4.
10 15
VIL
20 25
EXAMPLE PROBLEM 2.4
FIGURE 2.15
Indicator diagram for a reversible process. Unlike Figure 2.14, the areas under the P-V curves are the same in the forward and reverse directions.
In this example, 2.00 mol of an ideal gas undergoes isothermal expansion along three different paths: (1) reversible expansion from Pt = 25.0 bar and Vt = 4.50 L to Pf = 4.50 bar, (2) a single-step irreversible expansion against a constant external pressure of 4.50 bar, and (3) a two-step irreversible expansion consisting initially of an expansion against a constant external pressure of 11.0 bar until P = Pexternah followed by an expansion against a constant external pressure of 4.50 bar until
R ~ R external'
Calculate the work for each of these processes. For which of the irreversible processes is the magnitude of the work greater?
Solution
The processes are depicted in the following indicator diagram:
We first calculate the constant temperature at which the process is carried out, the final volume, and the intermediate volume in the two-step expansion:
T
PjVj
nR
Vf
Vint
nRT
nRT
R\ nt
25.0 bar X 4.50 L
- 7 - t - T - = 677 K
8.314 X 10-2 L bar mol-1 K-1 X 2.00 mol
8.314 X 10~2Lbarmor1 K-1 X 2.00 mol X 677 K 4.50 bar
8.314 X 10~2Lbarmor1 KT1 X 2.00 mol X 677 K 11.0 bar
25.0 L
10.2 L
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2.8 COMPARING WORK FOR REVERSIBLE AND IRREVERSIBLE PROCESSES
33
The work of the reversible process is given by
v f
w = —nRT 1 1n —
Vi
= -2.00 mol X 8.314 J mol-1 K_1 X 677 K X In
25.0 L
= -19.3 X 1(E J
4.50 L
We next calculate the work of the single-step and two-step irreversible processes:
v single
= -p.
external
= -9.23 X 1(EJ
1q5 Pa
AV = -4.50 bar X — - X (25.0 L - 4.50 L) X
10“3 m3
bar
105 Pa
^ two-step P externally I 1.0 bclf X
X (10.2 L - 4.50 L) X
10“3 m3
105 Pa
-4.50 bar X - X (25.0 L - 10.2 L) X
bar
10“3 m3
= -12.9 X 103J
The magnitude of the work is greater for the two-step process than for the single-step process, but less than that for the reversible process.
Example Problem 2.4 shows that the magnitude of w for the irreversible expansion is less than that for the reversible expansion, but also suggests that the magnitude of w for a multistep expansion process increases with the number of steps. This is indeed the case, as shown in Figure 2.16. Imagine that the number of steps n increases indefinitely.
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FIGURE 2.16
The work done in an expansion (red plus yellow areas) is compared with the work done in a multistep series of irreversible expansion processes at constant pressure (yellow area) in the top panel. The bottom panel shows analogous results for the compression, where the area under the black curve is the reversible compression work. Note that the total work done in the irreversible expansion and compression processes approaches that of the reversible process as the number of steps becomes large.
34
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
As n increases, the pressure difference P external ~ P f°r each individual step decreases. In the limit that n —> oo, the pressure difference Pexternal ~ P ^ 0, and the total area of the rectangles in the indicator diagram approaches the area under the reversible curve. In this limit, the irreversible process becomes reversible and the value of the work equals that of the reversible process.
By contrast, the magnitude of the irreversible compression work exceeds that of the reversible process for finite values of n and becomes equal to that of the reversible process as n — > oo. The difference between the expansion and compression processes results from the requirement that Pexternal < P at the beginning of each expansion step, whereas P external > P at the beginning of each compression step.
On the basis of these calculations for the reversible and irreversible cycles, we introduce another criterion to distinguish between reversible and irreversible processes. Suppose that a system undergoes a change through one or more individual steps, and that the system is restored to its initial state by following the same steps in reverse order. The system is restored to its initial state because the process is cyclical. If the surroundings are also returned to their original state (all masses at the same height and all reservoirs at their original temperatures), the process is reversible. If the surround¬ ings are not restored to their original state, the process is irreversible.
We are often interested in extracting work from a system. For example, it is the expan¬ sion of the fuel-air mixture in an automobile engine upon ignition that provides the torque that eventually drives the wheels. Is the capacity to do work similar for reversible and irre¬ versible processes? This question is answered using the indicator diagrams of Figures 2.14 and 2.15 for the specific case of isothermal expansion work, noting that the work can be calculated from the area under the P-V curve. We compare the work for expansion from V\ to V2 in a single stage at constant pressure to that for the reversible case. For the single- stage expansion, the constant external pressure is given by P external = RPT /V 2- However, if the expansion is carried out reversibly, the system pressure is always greater than this value. By comparing the areas in the indicator diagrams of Figure 2.16, it is seen that
\^reversibl\ — \^irreversibl\
By contrast, for the compression step,
Irreversible \ — irreversible
The reversible work is the lower bound for the compression work and the upper bound for the expansion work. This result for the expansion work can be generalized to an important statement that holds for all forms of work: The maximum work that can he extracted from a process between the same initial and final states is that obtained under reversible conditions.
Although the preceding statement is true, it suggests that it would be optimal to operate an automobile engine under conditions in which the pressure inside the cylin¬ ders differs only infinitesimally from the external atmospheric pressure. This is clearly not possible. A practical engine must generate torque on the drive shaft, and this can only occur if the cylinder pressure is appreciably greater than the external pressure. Similarly, a battery is only useful if one can extract a sizable rather than an infinitesimal current. To operate such devices under useful irreversible conditions, the work output is less than the theoretically possible limit set by the reversible process.3
(2.23)
(2.24)
Determining AC/ and Introducing Enthalpy, a New State Function
Measuring the energy taken up or released in a chemical reaction is of particular inter¬ est to chemists. How can the A U for a thermodynamic process be measured? This will be the topic of Chapter 4. However, this topic is briefly discussed here in order to enable you to carry out calculations on ideal gas systems in the end-of-chapter
3For a more detailed discussion of irreversible work, see D. Kivelson and I. Oppenheim, “Work in Irreversible Expansions,” Journal of Chemical Education 43 (1966): 233.
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2.10 CALCULATING q, w, A U, AND A H FOR PROCESSES INVOLVING IDEAL GASES 35
problems. The first law states that A U = q + w. Imagine that the process is carried out under constant volume conditions and that nonexpansion work is not possible. Because under these conditions w = — / P external^ = 0,
A U = qv (2.25)
Equation (2.25) states that AU can be experimentally determined by measuring the heat flow between the system and surroundings in a constant volume process.
Chemical reactions are generally carried out under constant pressure rather than constant volume conditions. It would be useful to have an energy state function that has a relationship analogous to Equation (2.25), but at constant pressure conditions. Under constant pressure conditions, we can write
dU = dqp - P externally = J qP - P dV (2.26)
Integrating this expression between the initial and final states:
i
j PdV = qP - P{Vf
~vd
= qP- (pfVf - PiVi)
(2.27)
Note that in order to evaluate the integral involving P, we must know the functional relationship P(V), which in this case is Pt = Pf = P where P is constant. Rearranging the last equation, we obtain
(Uf + PfVf) - (Ui + PjYi) = qp (2.28)
Because P, V , and U are all state functions, U + PV is a state function. This new state function is called enthalpy and is given the symbol H.
H = U + PV (2.29)
As is the case for U, H has the units of energy, and it is an extensive property. As shown in Equation (2.30), AH for a process involving only P-V work can be determined by measuring the heat flow between the system and surroundings at constant pressure:
AH = qP (2.30)
This equation is the constant pressure analogue of Equation (2.25). Because chemical reactions are much more frequently carried out at constant P than constant V, the energy change measured experimentally by monitoring the heat flow is AH rather than AU. When we classify a reaction as being exothermic or endothermic, we are talking about AH, not AU.
Calculating q, w, A U, and AH for Processes Involving Ideal Gases
In this section we discuss how AU and AIT, as well as q and w, can be calculated from the initial and final state variables if the path between the initial and final state is known. The problems at the end of this chapter ask you to calculate q, w, AU, and AH for simple and multistep processes. Because an equation of state is often needed to carry out such calculations, the system will generally be an ideal gas. Using an ideal gas as a surrogate for more complex systems has the significant advantage that the mathematics is simplified, allowing one to concentrate on the process rather than the manipulation of equations and the evaluation of integrals.
What does one need to know to calculate AU? The following discussion is restricted to processes that do not involve chemical reactions or changes in phase. Because U is a state function, AU is independent of the path between the initial and final states. To describe a fixed amount of an ideal gas (i.e., n is constant), the values of two of the variables P, V, and T must be known. Is this also true for AU for processes involving ideal gases? To answer this question, consider the expansion of an ideal gas
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
from an initial state V\,T\ to a final state V2,T2. We first assume that H is a function of both V and T. Is this assumption correct? Because ideal gas atoms or molecules do not interact with one another, H will not depend on the distance between the atoms or mol¬ ecules. Therefore, H is not a function of V , and we conclude that A H must be a function of T only for an ideal gas, A U = At/ (T).
We also know that for a temperature range over which Cv is constant,
A U = qv = Cv(Tf - Tt) (2.31)
Is this equation only valid for constant V? Because H is a function of only T for an ideal gas, Equation (2.31) is also valid for processes involving ideal gases in which V is not constant. Therefore, if one knows Cv, Th and T2, A H can be calculated, regardless of the path between the initial and final states.
How many variables are required to define A H for an ideal gas? We write
AH = A U(T) + A (PV) = A U(T) + A (nRT) = A H(T) (2.32)
We see that AH is also a function of only T for an ideal gas. In analogy to Equation (2.31),
A H = qP = CP(T f - Tt) (2.33)
Because AH is a function of only T for an ideal gas, Equation (2.33) holds for all processes involving ideal gases, whether P is constant or not, as long as it is reasonable to assume that CP is constant. Therefore, if the initial and final temperatures are known or can be calculated, and if Cv and CP are known, A H and AH can be calculated regardless of the path for processes involving ideal gases using Equations (2.31) and (2.33), as long as no chemical reactions or phase changes occur. Because H and H are state functions, the previous statement is true for both reversible and irreversible processes. Recall that for an ideal gas CP — Cv = nR , so that if one of Cv and CP is known, the other can be readily determined.
We next note that the first law links q , w, and AH. If any two of these quantities are known, the first law can be used to calculate the third. In calculating work, often only expansion work takes place. In this case one always proceeds from the equation
w = - J P external dV (2-34)
This integral can only be evaluated if the functional relationship between Pexternai and V is known. A frequently encountered case is Pexternal = constant, such that
W = — P external ( V f ~ V i) (2-35)
Because P external ^ the work considered in Equation (2.35) is for an irreversible process.
A second frequently encountered case is that the system and external pressure differ only by an infinitesimal amount. In this case, it is sufficiently accurate to write P external = P, and the process is reversible:
w = - j ff dV (2.36)
This integral can only be evaluated if T is known as a function of V. The most com¬ monly encountered case is an isothermal process, in which T is constant. As was seen in Section 2.2, for this case
fdV V f
w = -nRT / — = —nRT In — (2.37)
J V Vi
In solving thermodynamic problems, it is very helpful to understand the process thor¬ oughly before starting the calculation, because it is often possible to obtain the value of one or more of q , w, AH, and AH without a calculation. For example, AH = AH = 0 for an isothermal process because AH and AH depend only on T. For an adiabatic process, q = 0 by definition. If only expansion work is possible, w = 0 for a constant volume process. These guidelines are illustrated in the following two example problems.
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2.10 CALCULATING q, w, A U, AND A H FOR PROCESSES INVOLVING IDEAL GASES 37
EXAMPLE PROBLEM 2.5
A system containing 2.50 mol of an ideal gas for which CV m = 20.79 J mol-1 K-1 is taken through the cycle in the following diagram in the direction indicated by the arrows. The curved path corresponds to PV = nRT , where T = T\ = T2.
a. Calculate q , w, A U, and A H for each segment and for the cycle assuming that the heat capacity is independent of temperature.
b. Calculate q , w, A U, and A H for each segment and for the cycle in which the direction of each process is reversed.
Solution
We begin by asking whether we can evaluate q , w, A U, or A H for any of the segments without any calculations. Because the path between states 1 and 3 is isothermal, A U and A H are zero for this segment. Therefore, from the first law, q^i = — w3_>i. For this reason, we only need to calculate one of these two quan¬ tities. Because AV = 0 along the path between states 2 and 3, w2^ 3 = 0. Therefore, AU2^2 — ^2^3- Again, we only need to calculate one of these two quantities. Because the total process is cyclic, the change in any state function is zero. Therefore, A U = A H = 0 for the cycle, no matter which direction is cho¬ sen. We now deal with each segment individually.
Segment 1 — > 2
The values of n , P\ and and P2 and V2 are known. Therefore, T\ and T2 can be cal¬ culated using the ideal gas law. We use these temperatures to calculate AU as follows:
ftCy m
At/1^2 = nCVtm(T2 - r,) = — ^ (P2V2 - PXVX)
nR
20.79 JmoF1 K"1 0.08314 L bar K_1 mo F1
X (16.6 bar X 25.0 L - 16.6bar X 1.00 L)
= 99.6 kJ
The process takes place at constant pressure, so
, , 105 Nm~2
P external w 2 ^ I ) 16.6bar X
X (25.0 X 10“3m3 - 1.00 X 10“3m3)
= — 39.8kJ
Using the first law,
q = AU - w = 99.6 kJ + 39.8 kJ = 139.4 kJ We next calculate T2 :
T 2 =
P2V2
nR
_ 16.6 bar X 25.QL _
2.50mol X 0.083 14 L bar K-1 mol-1
= 2.00 X 103 K
We next calculate T2 = T] and then A H\^,2.
T fiVi 16.6bar X 1.00L
1 nR 2.50mol X 0.08314LbarK"1moF1 = At/^2 + A(PV) = At/^2 + nR(T2 - Tx) = 99.6 X 103J + 2.5 mol X 8.314Jmol-1K_1 X (2000 K - 79.9 K) = 139.4kJ
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Segment 2^3
As previously noted, w = 0, and
A(/2^3 = #2^3 Cv(T3 ~ T2)
= 2.50 mol X 20.79JmoF1K“1(79.9K - 2000 K)
= -99.6 kJ
The numerical result is equal in magnitude, but opposite in sign to A £7^2 because T3 = T\. For the same reason, AH2^3 = —A H3^2.
Segment 3 — » 1
For this segment, AU3^\ = 0 and A H3^>\ = 0 as noted earlier, and w3^i = —q3^i. Because this is a reversible isothermal compression,
Vi , ,
w3—>i = -nRT In— = -2.50mol X 8.314 Jmol-1 K_1 X 79.9K
^3
1.00 X 10“3m3
X In - ; - T
25.0 X 10“3m3
= 5.35 kJ
The results for the individual segments and for the cycle in the indicated direction are given in the following table. If the cycle is traversed in the reverse fashion, the magni¬ tudes of all quantities in the table remain the same, but all signs change.
Path
1- >2
2— >3 3 — » 1 Cycle
Q (kJ) 139.4 -99.6 -5.35 34.5
AU (kJ) 99.6 -99.6 0 0
w( kJ) -39.8 0
5.35
-34.5
AH (kJ)
139.4 - 139.4 0 0
EXAMPLE PROBLEM 2.6
In this example, 2.50 mol of an ideal gas with m = 12.47 J mol-1 K-1 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure of the gas are 325 K and 2.50 bar, respectively. The final pressure is 1.25 bar. Calculate the final temperature, q , w, A U, and AH.
Solution
Because the process is adiabatic, q = 0, and AU = w. Therefore, AU = nCy m(T f ~ Tt) - — P external ( V f Vi)
Using the ideal gas law,
'Tf T
nCv,m (T f - Ti) = —nRP ,
external
pf Pi
Tf\ nCyjn +
nRPexternal | _ nRP external
- I = 7 ,| nCv m +
C +
v,m 1
pf )
RP external \
Pi
Tf = Ti\
Pi
C +
v-'v,m 1
RP,
external
12,
= 325 K X
7
, , 8.314 J mol-1 K_1 X 1.00 bar \
.47 J mol1 K 1 + - \
2.50 bar
12.47 J mol-1 K_1 +
8.314 J mol- 1 K-i X 1.00 bar
= 268 K
1.25 bar
7
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2.1 1 THE REVERSIBLE ADIABATIC EXPANSION AND COMPRESSION OF AN IDEAL GAS 39
We calculate A U = w from
A U = nCv,m (T f - T,) = 2.5 mol X 12.47 J moF'K-1 X (268 K - 325 K)
= -1.78 kJ
Because the temperature falls in the expansion, the internal energy and enthalpy decreases: AH = AU + A (PV) = A U + nR(T2 - 7’1)
= -1.78 X 103 J + 2.5 mol X 8.314 J mol-1K-1 X (268 K - 325 K) = -2.96 kJ
The Reversible Adiabatic Expansion and Compression of an Ideal Gas
The adiabatic expansion and compression of gases is an important meteorological process. For example, the cooling of a cloud as it moves upward in the atmosphere can be modeled as an adiabatic process because the heat transfer between the cloud and the rest of the atmosphere is slow on the timescale of its upward motion.
Consider the adiabatic expansion of an ideal gas. Because q = 0, the first law takes the form
A U W Or CydT P external^V
For a reversible adiabatic process, P = Pexternab and
dV dT dV
CydT = —nRT or, equivalently, Cy — = —nR
Integrating both sides of this equation between the initial and final states,
dT
' dV
Cy — TlR I
V T Jv V
(2.38)
(2.39)
(2.40)
If Cy is constant over the temperature interval Tf- Th then
T V
Cv\ny=-nR\ny (2.41)
Because CP — Cy = nR for an ideal gas, Equation (2.41) can be written in the form
= -{y - 1 ) In (^y^j or, equivalently, y = (2.42)
where y = CP^m/Cv^m. Substituting Tf/Tt = PfV f/ PjVi in the previous equation, we obtain
W = PfV} (2.43)
for the adiabatic reversible expansion or compression of an ideal gas. Note that our der¬ ivation is only applicable to a reversible process, because we have assumed that
P — P external •
Reversible adiabatic compression of a gas leads to heating, and reversible adiabatic expansion leads to cooling. Adiabatic and isothermal expansion and compression are compared in Figure 2.17, in which two systems containing 1 mol of an ideal gas have the same volume at P = 1 atm. One system undergoes adiabatic compression or expansion, and the other undergoes isothermal compression or expansion. Under isothermal conditions, heat flows out of the system as it is compressed to P > 1 atm, and heat flows into the system as it is expanded to P < 1 atm to keep T constant. Because no heat flows into or out of the system under adiabatic conditions, its temper¬ ature increases in compression and decreases in expansion. Because T > Tisothermai for a compression starting at 1 atm, P adiabatic > pisothermal f«r a given volume of
FIGURE 2.17
Two systems containing 1 mol of N2 have the same P and V values at 1 atm. The red curve corresponds to reversible expansion and compression under adiabatic condi¬ tions. The blue curve corresponds to reversible expansion and compression under isothermal conditions.
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40
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
(a) (b) (c)
FIGURE 2.18
(a) A system has the translational energy levels shown before undergoing an adia¬ batic compression, (b) After the compres¬ sion, the energy levels are shifted upward but the occupation probability of the lev¬ els shown on the horizontal axis is unchanged, (c) Subsequent cooling to the original temperature at constant V restores the energy to its original value by decreas¬ ing the probability of occupying higher energy states. Each circle corresponds to a probability of 0. 10.
Vocabulary
cyclic path
degrees of freedom
enthalpy
exact differential
first law of thermodynamics
heat
heat capacity
the gas. Similarly, in a reversible adiabatic expansion originating at 1 atm, p adiabatic < P isothermal for a given volume of the gas.
I EXAMPLE PROBLEM 2.7
A cloud mass moving across the ocean at an altitude of 2000. m encounters a coastal mountain range. As it rises to a height of 3500. m to pass over the mountains, it undergoes an adiabatic expansion. The pressure at 2000. m and 3500. m is 0.802 and 0.602 atm, respectively. If the initial temperature of the cloud mass is 288 K, what is the cloud temperature as it passes over the mountains? Assume that CPm for air is 28.86 J K-1 mol-1 and that air obeys the ideal gas law. If you are on the mountain, should you expect rain or snow?
Solution
-(y ~ !)ln( y.
(r - !)lnl 2-“ ) = ~(y - l)ln ( -= I - (y - l)ln I —L
TiPf
Cpj
Cp,m R
Ti
- 1
/
C
P,m
Cp,m ~~ R 28.86 JK 1 mol-1
28.86 JK 1 mol-1 - 8.314 J K 1 mol-1
28.86 JK 1 mol
-l
X In
0.802 atm 0.602 atm
28.86 JK_1 mol-1 - 8.314 J K_1 mol-1 - 0.0826
Tf = 0.9207 Tt = 265 K You can expect snow.
It is instructive to consider an adiabatic compression or expansion from a micro¬ scopic point of view. In an adiabatic compression, the energy levels are all raised, but the probability that a given level is accessed is unchanged. This behavior is observed in contrast to that shown in Figure 2.5 because in an adiabatic compression, T and there¬ fore U increases. For more details, see R. E. Dickerson, Molecular Thermodynamics , W.A. Benjamin, Menlo Park, 1969. If the gas is subsequently cooled at constant V, the energy levels remain unchanged, but the probability that higher energy states are popu¬ lated decreases as shown in Figure 2.18c.
indicator diagram
internal energy
irreversible
isobaric
isochoric
isothermal
path
path function quasi-static process reversible state function work
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NUMERICAL PROBLEMS 41
Conceptual Problems
Q2.1 Electrical current is passed through a resistor immersed in a liquid in an adiabatic container. The tempera¬ ture of the liquid is varied by 1°C. The system consists solely of the liquid. Does heat or work flow across the boundary between the system and surroundings? Justify your answer. Q2.2 Two ideal gas systems undergo reversible expansion under different conditions starting from the same P and V. At the end of the expansion, the two systems have the same volume.
The pressure in the system that has undergone adiabatic expan¬ sion is lower than in the system that has undergone isothermal expansion. Explain this result without using equations.
Q2.3 You have a liquid and its gaseous form in equilibrium in a piston and cylinder assembly in a constant temperature bath. Give an example of a reversible process.
Q2.4 Describe how reversible and irreversible expansions differ by discussing the degree to which equilibrium is main¬ tained between the system and the surroundings.
Q2.5 For a constant pressure process, A H = qP. Does it follow that qP is a state function? Explain.
Q2.6 A cup of water at 278 K (the system) is placed in a microwave oven and the oven is turned on for 1 minute during which the water begins to boil. State whether each of q , w, and A U is positive, negative, or zero.
Q2.7 In the experiments shown in Figure 2.4a and 2.4b, surroundings ^ 0, but A T surroundings ^ 0. Explain how this is possible.
Q2.8 What is wrong with the following statement? Burns caused by steam at 100°C can be more severe than those caused by water at 100°C because steam contains more heat than water. Rewrite the sentence to convey the same informa¬ tion in a correct way.
Q2.9 Why is it incorrect to speak of the heat or work asso¬ ciated with a system?
Q2.10 You have a liquid and its gaseous form in equilib¬ rium in a piston and cylinder assembly in a constant tempera¬ ture bath. Give an example of an irreversible process.
Q2.ll What is wrong with the following statement? Because the well-insulated house stored a lot of heat, the temperature didn’t fall much when the furnace failed. Rewrite the sentence to convey the same information in a correct way.
Q2.12 Explain how a mass of water in the surroundings can be used to determine q for a process. Calculate q if the tempera¬ ture of a 1 .00 kg water bath in the surroundings increases by 1.25°C. Assume that the surroundings are at a constant pressure.
Q2.13 A chemical reaction occurs in a constant volume enclosure separated from the surroundings by diathermal walls. Can you say whether the temperature of the surround¬ ings increases, decreases, or remains the same in this process? Explain.
Q2.14 Explain the relationship between the terms exact differential and state function.
Q2.15 In the experiment shown in Figure 2.4b, the weight drops in a very short time. How will the temperature of the water change with time?
Q2.16 Discuss the following statement: If the temperature of the system increased, heat must have been added to it. Q2.17 Discuss the following statement: Heating an object causes its temperature to increase.
Q2.18 An ideal gas is expanded reversibly and adiabati- cally. Decide which of q, w, A U, and AH are positive, nega¬ tive, or zero.
Q2.19 An ideal gas is expanded reversibly and isother- mally. Decide which of q , w, At/, and AH are positive, nega¬ tive, or zero.
Q2.20 An ideal gas is expanded adiabatically into a vac¬ uum. Decide which of q, w, At/, and AH are positive, nega¬ tive, or zero.
Q2.21 A bowling ball (a) rolls across a table, and (b) falls on the floor. Is the work associated with each part of this process positive, negative, or zero?
Q2.22 A perfectly insulating box is partially filled with water in which an electrical resistor is immersed. An external electrical generator converts the change in potential energy of a mass m that falls by a vertical distance h into electrical energy that is dissipated in the resistor. What value do q , w, and A U have if the system is defined as the resistor and the water? Everything else is in the surroundings.
Q2.23 A student gets up from her chair and pushes a stack of books across the table. They fall to the floor. Is the work asso¬ ciated with each part of this process positive, negative, or zero? Q2.24 Explain why ethene has a higher value for Cym at 800 K than CO.
Q2.25 Explain why CPm is a function of temperature for ethane, but not for argon in a temperature range in which electronic excitations do not occur.
Q2.26 What is the difference between a quasi-static process and a reversible process?
Numerical Problems
Problem numbers in red indicate that the solution to the prob¬ lem is given in the Student’s Solutions Manual.
P2.1 A 3.75 mole sample of an ideal gas with CV m = 3R/2 initially at a temperature Tt = 298 K and Pt = 1.00 bar is enclosed in an adiabatic piston and cylinder assembly. The
gas is compressed by placing a 725 kg mass on the piston of diameter 25.4 cm. Calculate the work done in this process and the distance that the piston travels. Assume that the mass of the piston is negligible.
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42
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.2 The temperature of 1.75 moles of an ideal gas increases from 10.2°C to 48.6°C as the gas is compressed adi- abatically. Calculate q, w, A U, and A H for this process, assuming that m = 3R/2.
P2.3 A 2.50 mole sample of an ideal gas, for which Cy m = 3R/2, is subjected to two successive changes in state: (1) From 25.0°C and 125 X 103 Pa, the gas is expanded isothermally against a constant pressure of 15.2 X 103 Pa to twice the initial volume. (2) At the end of the previous process, the gas is cooled at constant volume from 25.0°C to -29.0°C. Calculate q , w, A U, and A H for each of the stages. Also calculate q , w, A U, and AH for the complete process.
P2.4 A hiker caught in a thunderstorm loses heat when her clothing becomes wet. She is packing emergency rations that if completely metabolized will release 35 kJ of heat per gram of rations consumed. How much rations must the hiker consume to avoid a reduction in body temperature of 2.5 K as a result of heat loss? Assume the heat capacity of the body equals that of water and that the hiker weighs 51 kg. State any additional assumptions.
P2.5 Count Rumford observed that using cannon boring machinery a single horse could heat 1 1.6 kg of ice water (T = 273 K) to T = 355 K in 2.5 hours. Assuming the same rate of work, how high could a horse raise a 225 kg weight in 2.5 minutes? Assume the heat capacity of water is 4.18 J K-1 g-1.
P2.6 A 1.50 mole sample of an ideal gas at 28.5°C expands isothermally from an initial volume of 22.5 dm3 to a final vol¬ ume of 75.5 dm3. Calculate w for this process (a) for expan¬ sion against a constant external pressure of 1.00 X 105 Pa, and (b) for a reversible expansion.
P2.7 Calculate q, w, A U, and AH if 2.25 mol of an ideal gas withCy m = 3R/2 undergoes a reversible adiabatic expansion from an initial volume Vt = 5.50 m3 to a final vol¬ ume Vf = 25.0 m3. The initial temperature is 275 K.
P2.8 Calculate w for the adiabatic expansion of 2.50 mol of an ideal gas at an initial pressure of 2.25 bar from an initial temper¬ ature of 450. K to a final temperature of 300. K. Write an expression for the work done in the isothermal reversible expan¬ sion of the gas at 300. K from an initial pressure of 2.25 bar. What value of the final pressure would give the same value of w as the first part of this problem? Assume that CPm = 5R/2.
P2.9 At 298 K and 1 bar pressure, the density of water is 0.9970 g cm-3, andCpjW = 75.3 J K-1 mol-1. The change in volume with temperature is given by AV = Vinitiail3AT where /3, the coefficient of thermal expansion, is 2.07 X 10-4 K-1. If the temperature of 325 g of water is increased by 25.5 K, calculate w, q , AH, and A U.
P2.10 A muscle fiber contracts by 3.5 cm and in doing so lifts a weight. Calculate the work performed by the fiber. Assume the muscle fiber obeys Hooke’s law F = —k x with a force constant k of 750. N m_1.
P2.ll A cylindrical vessel with rigid adiabatic walls is sep¬ arated into two parts by a frictionless adiabatic piston. Each part contains 45.0 L of an ideal monatomic gas with Cv,m = 3/?/2. Initially, T, = 300. K and P, = 1.75 X 105 Pa in each part. Heat is slowly introduced into the left part using an electrical heater until the piston has moved sufficiently to
the right to result in a final pressure Pf = 4.00 bar in the right part. Consider the compression of the gas in the right part to be a reversible process.
a. Calculate the work done on the right part in this process and the final temperature in the right part.
b. Calculate the final temperature in the left part and the amount of heat that flowed into this part.
P2.12 In the reversible adiabatic expansion of 1.75 mol of an ideal gas from an initial temperature of 27.0°C, the work done on the surroundings is 1300. J. If Cy m = 3R/2, calcu¬ late q, w, A U, and AH.
P2.13 A system consisting of 82.5 g of liquid water at 300. K is heated using an immersion heater at a constant pressure of 1.00 bar. If a current of 1.75 A passes through the 25.0 ohm resistor for 100. s, what is the final temperature of the water?
P2.14 A 1.25 mole sample of an ideal gas is expanded from 320. K and an initial pressure of 3.10 bar to a final pressure of 1.00 bar, and CP^m = 5R/2. Calculate w for the following two cases:
a. The expansion is isothermal and reversible.
b. The expansion is adiabatic and reversible.
Without resorting to equations, explain why the result to part (b) is greater than or less than the result to part (a).
P2.15 A bottle at 325 K contains an ideal gas at a pressure of 162.5 X 103 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against Pexternai = 120.0 X 103 Pa, and some gas is expelled from the bottle in the process. When P = Pexternab the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 325 K. What is the final pressure in the bottle for a monatomic gas, for which CVm = 3R/2, and a diatomic gas, for which CV m = 5R/21
P2.16 A 2.25 mole sample of an ideal gas with CV m = 3R/2 initially at 310. K and 1.25 X 105 Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is 3.10 X 106 Pa. Calculate the final temperature of the gas. Calculate q, w, A U, and AH for this process.
P2.17 A vessel containing 1.50 mol of an ideal gas with Pt = 1.00 bar and CPm = 5R/2 is in thermal contact with a water bath. Treat the vessel, gas, and water bath as being in thermal equilibrium, initially at 298 K, and as separated by adiabatic walls from the rest of the universe. The vessel, gas, and water bath have an average heat capacity of CP = 2450. J K-1. The gas is compressed reversibly to Pf = 20.5 bar. What is the temperature of the system after thermal equilibrium has been established?
P2.18 An ideal gas undergoes an expansion from the initial state described by Ph Vh T to a final state described by Pf, Vf, Tin (a) a process at the constant external pressure Pf, and (b) in a reversible process. Derive expressions for the largest mass that can be lifted through a height h in the surroundings in these processes.
P2.19 An ideal gas described by Tt = 275 K, Pt = 1.10 bar, and Vi = 10.0 L is heated at constant volume until P =
10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.10 bar. It is then restored to its original state by the extraction of heat at constant pressure. Depict this closed-cycle
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NUMERICAL PROBLEMS
43
process in a P-V diagram. Calculate w for each step and for the total process. What values for w would you calculate if the cycle were traversed in the opposite direction?
P2.20 In an adiabatic compression of one mole of an ideal gas with CyjW = 5R/2, the temperature rises from 278 K to 450. K. Calculate q, w, A H, and A U.
P2.21 The heat capacity of solid lead oxide is given by
T
C p m = 44.35 + 1.47 X 10-3 — in units of J K-1 mol-1
K
Calculate the change in enthalpy of 1.75 mol of PbO(s) if it is cooled from 825 K to 375 K at constant pressure.
P2.22 A 2.25 mole sample of carbon dioxide, for which Cp m = 37.1 J K-1 mol-1 at 298 K, is expanded reversibly and adiabatically from a volume of 4.50 L and a temperature of 298 K to a final volume of 32.5 L. Calculate the final tem¬ perature, q, w, A/7, and A U. Assume that CPjn is constant over the temperature interval.
P2.23 A 1.75 mole sample of an ideal gas for which P = 2.50 bar and T = 335 K is expanded adiabatically against an external pressure of 0.225 bar until the final pres¬ sure is 0.225 bar. Calculate the final temperature, q , w, AH, and A U for (a) Cy?m = 3R/2, and (b) Cv,m = 5R/2.
P2.24 A 3.50 mole sample of N2 in a state defined by Tt = 250. K and Vt = 3.25 L undergoes an isothermal reversible expansion until V f = 35.5 L Calculate w, assuming (a) that the gas is described by the ideal gas law, and (b) that the gas is described by the van der Waals equation of state. What is the percent error in using the ideal gas law instead of the van der Waals equation? The van der Waals parameters for N2 are listed in Table 7.4.
P2.25 A major league pitcher throws a baseball with a speed of 162 kilometers per hour. If the baseball weighs 235 grams and its heat capacity is 1.7 J g-1 K-1, calculate the temperature rise of the ball when it is stopped by the catcher’s mitt. Assume no heat is transferred to the catcher’s mitt and that the catcher’s arm does not recoil when he or she catches the ball. Also assume that the kinetic energy of the ball is completely converted into thermal energy.
P2.26 A 2.50 mol sample of an ideal gas for which Cy m = 3R/2 undergoes the following two-step process:
(1) From an initial state of the gas described by T = 13.1°C and P = 1.75 X 105 Pa, the gas undergoes an isothermal expansion against a constant external pressure of 3.75 X 104 Pa until the volume has doubled. (2) Subsequently, the gas is cooled at constant volume. The temperature falls to - 23.6°C. Calculate q, w, A U, and AH for each step and for the overall process.
P2.27 A 2.35 mole sample of an ideal gas, for which CV m = 3R/2, initially at 27.0°C and 1.75 X 106 Pa, undergoes a two- stage transformation. For each of the stages described in the fol¬ lowing list, calculate the final pressure, as well as q, w, A U, and AH. Also calculate q, w, A U, and AH for the complete process.
a. The gas is expanded isothermally and reversibly until the volume triples.
b. Beginning at the end of the first stage, the temperature is raised to 105°C at constant volume.
P2.28 A 3.50 mole sample of an ideal gas with Cv,m = 3R/2 is expanded adiabatically against a constant external pres¬ sure of 1.45 bar. The initial temperature and pressure are Tt = 310. K and Pt = 15.2 bar. The final pressure is Pf = 1.45 bar. Calculate q, w, A U, and AH for the process.
P2.29 A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Assume the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Tt = 280. K The final pressure in the tire is Pf = 3.75 bar. Calculate the final temperature of the air in the tire. Assume that CyjW = 5R/2.
P2.30 For 1.25 mol of an ideal gas, P external = P =
350. X 103 Pa. The temperature is changed from 135°C to 21.2°C, and m = 3R/2. Calculate q, w, A U, and AH.
P2.31 Suppose an adult is encased in a thermally insulating barrier so that all the heat evolved by metabolism of food¬ stuffs is retained by the body. What is her temperature increase after 2.5 hours? Assume the heat capacity of the body is 4.18 J g-1 K_1 and that the heat produced by metabo¬ lism is 9.4 kJ kg-1hr-1.
P2.32 Consider the isothermal expansion of 2.35 mol of an ideal gas at 415 K from an initial pressure of 18.0 bar to a final pressure of 1.75 bar. Describe the process that will result in the greatest amount of work being done by the sys¬ tem with P extemai — 1-75 bar, and calculate w. Describe the process that will result in the least amount of work being done by the system with P external — 1-75 bar, and calculate w. What is the least amount of work done without restric¬ tions on the external pressure?
P2.33 An automobile tire contains air at 225 X 103 Pa at 25.0°C. The stem valve is removed and the air is allowed to expand adiabatically against the constant external pressure of one bar until P = P external- For aifi Cv,m = 5R/2. Calculate the final temperature. Assume ideal gas behavior.
P2.34 One mole of an ideal gas is subjected to the following changes. Calculate the change in temperature for each case if Cym = 3R/2.
a. q = —425 J, w = 185 J
b. q = 315. J, w = -315 J
c. q = 0, w = 225 J
P2.35 Consider the adiabatic expansion of 0.500 mol of an ideal monatomic gas with CVm = 3R/2. The initial state is described by P = 6.25 bar and T = 300. K.
a. Calculate the final temperature if the gas undergoes a reversible adiabatic expansion to a final pressure of P = 1.25 bar.
b. Calculate the final temperature if the same gas undergoes an adiabatic expansion against an extemai pressure of
P = 1.25 bar to a final pressure P = 1.25 bar.
Explain the difference in your results for parts (a) and (b).
P2.36 A pellet of Zn of mass 3 1 .2 g is dropped into a flask containing dilute H2S04 at a pressure of P = 1.00 bar and a temperature of T = 300. K. What is the reaction that occurs? Calculate w for the process.
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44
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.37 Calculate A H and A U for the transformation of
2.50 mol of an ideal gas from 19.0°C and 1.00 atm to 550.°C and
T
19.5 atm if CPm = 20.9 + 0.042 — in units of J K-1 mol-1.
K
P2.38 A 1.75 mole sample of an ideal gas for which CyjW = 20.8 J K-1 mol-1 is heated from an initial tempera¬ ture of 21.2°C to a final temperature of 380.°C at constant volume. Calculate q, w, A U, and A H for this process.
P2.39 An ideal gas undergoes a single-stage expansion against a constant external pressure P external — Pf at constant temperature from T, Ph Vt, to T, Pf, Vf.
a. What is the largest mass m that can be lifted through the height h in this expansion?
b. The system is restored to its initial state in a single- state compression. What is the smallest mass m' that must fall through the height h to restore the system to its initial state?
c. If h = 15.5 cm, Pi = 1.75 X 106Pa, Pf = 1.25 X 106Pa, T = 280. K, and n = 2.25 mol, calculate the values of the masses in parts (a) and (b).
P2.40 The formalism of the Young’s modulus is sometimes used to calculate the reversible work involved in extending or compressing an elastic material. Assume a force F is applied to an elastic rod of cross-sectional area A0 and length L0. As a result of this force the rod changes in length by AL. The Young’s modulus E is defined as
_ tensile stress _ F/ a0 _ FLq tensile strain AL/l0
a. Relate k in Hooke’s Law to the Young’s modulus expres¬ sion just given.
b. Using your result in part (a) show that the magnitude of the reversible work involved in changing the length L0 of an elastic cylinder of cross-sectional area A0 by A L is
- 1 lmy
w
2\Lj
EAftLt
OMT
P2.41 The Young’s modulus (see Problem P2.40) of muscle fiber is approximately 2.80 X 107 Pa. A muscle fiber 3.25 cm in length and 0.125 cm in diameter is suspended with a mass M hanging at its end. Calculate the mass required to extend the length of the fiber by 10%.
P2.42 DNA can be modeled as an elastic rod that can be twisted or bent. Suppose a DNA molecule of length L is bent such that it lies on the arc of a circle of radius Rc. The
reversible work involved in bending DNA without twisting is B T j
wbend = — 2 w^ere B *s the bending force constant. The 2 Rc
DNA in a nucleosome particle is about 680. A in length. Nucleosomal DNA is bent around a protein complex called the histone octamer into a circle of radius 55 A. Calculate the reversible work involved in bending the DNA around the his¬ tone octamer if the force constant B = 2.00 X 10-28 J m.
P2.43 A 1.75 mole sample of an ideal gas is compressed isothermally from 62.0 L to 19.0 L using a constant external pressure of 2.80 atm. Calculate q , w, A U, and A H.
P2.44 Assume the following simplified dependence of the pressure in a ventricle of the human heart as a function of the volume of blood pumped.
PS9 the systolic pressure, is 120. mm Hg, corresponding to 0.158 atm. Pj, the diastolic pressure, is 80.0 mm Hg, corre¬ sponding to 0.105 atm. If the volume of blood pumped in one heartbeat is 75.0 cm3, calculate the work done in a heartbeat.
Web-Based Simulations, Animations, and Problems
W2.1 A simulation is carried out in which an ideal gas is heated under constant pressure or constant volume conditions. The quantities AV (or AP), w, A U, and AT are determined as a function of the heat input. The heat taken up by the gas under constant P or V is calculated and compared with A U and AH.
W2.2 The reversible isothermal compression and expansion of an ideal gas is simulated for different values of T. The work w is calculated from the T and V values obtained in the simulation. The heat q and the number of moles of gas in the system are calculated from the results.
W2.3 The reversible isobaric compression and expansion of an ideal gas is simulated for different values of pressure gas as heat flows to/from the surroundings. The quantities q , w, and A U are calculated from the AT and AV values obtained in the simulation.
W2.4 The isochoric heating and cooling of an ideal gas is simulated for different values of volume. The number of moles of gas and A U are calculated from the constant V value and from the T and P values obtained in the simulation.
W2.5 Reversible cyclic processes are simulated in which the cycle is either rectangular or triangular on a P-V plot. For each segment and for the cycle, A U, q , and w are determined. For a given cycle type, the ratio of work done on the sur¬ roundings to the heat absorbed from the surroundings is determined for different P and V values.
W2.6 The reversible adiabatic heating and cooling of an ideal gas is simulated for different values of the initial tem¬ perature. The quantity y = CPm/Cv^m as well as CPm and Cym are determined from the P, V values of the simulation;
A U and A U are calculated from the V, T , and P values obtained in the simulation.
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The Importance of State Functions: Internal Energy and Enthalpy
I he mathematical properties of state functions are utilized to express the infinitesimal quantities dU and dH as exact differentials. By doing so, expressions can be derived that relate the change of U with T and V and the change in H with T and P to experimentally accessible quantities such as the heat capacity and the coefficient of thermal expansion. Although both U and H are functions of any two of the variables P, V, and T, the dependence of U and H on temperature is generally far greater than the dependence on P or V. As a result, for most processes involving gases, liquids, and solids, U and H can be regarded as functions of T only. An exception to this state¬ ment is the cooling on the isenthalpic expansion of real gases, which is com¬ mercially used in the liquefaction of N2, O2, He, and Ar.
3.1
The Mathematical Properties of State Functions
3.2
The Dependence of U on V and T
3.3
Does the Internal Energy Depend More Strongly on V or 77
3.4
The Variation of Enthalpy with Temperature at Constant Pressure
3.5
How Are CP and Cv Related?
3.6
The Variation of Enthalpy with Pressure at Constant Temperature
3.7
The Joule-Thomson Experiment
3.8
Liquefying Gases Using an Isenthalpic Expansion
The Mathematical Properties of State Functions
In Chapter 2 we demonstrated that U and H are state functions and that w and q are path functions. We also discussed how to calculate changes in these quantities for an ideal gas. In this chapter, the path independence of state functions is exploited to derive rela¬ tionships with which A U and A H can be calculated as functions of P, V, and T for real gases, liquids, and solids. In doing so, we develop the formal aspects of thermodynam¬ ics. We will show that the formal structure of thermodynamics provides a powerful aid in linking theory and experiment. However, before these topics are discussed, the math¬ ematical properties of state functions need to be outlined.
The thermodynamic state functions of interest here are defined by two variables from the set P, V , and T. In formulating changes in state functions, we will make extensive use of partial derivatives, which are reviewed in the Math Supplement (Appendix A). The fol¬ lowing discussion does not apply to path functions such as w and q because a functional
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45
46 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
FIGURE 3.1
Starting at the point labeled z on the hill, a person first moves in the positive x direction and then along the y direction. If dx and dy are sufficiently small, the change in height dz is given by
relationship such as Equation (3.1) does not exist for path-dependent functions. Consider 1 mole of an ideal gas for which
P = f(V,T)
RT
(3.1)
Note that P can be written as a function of the two variables V and T. The change in P resulting from a change in V or T is proportional to the following partial derivatives:
= hmAy^0
- limAr^0
P(V + A V,T) - P(V,T)
Ay
p{vj + a t) - p{vj)
A T
_RT
“y^
R
V
(3.2)
The subscript T in {dP/dV)T indicates that T is being held constant in the differen¬ tiation with respect to V. The partial derivatives in Equation (3.2) allow one to determine how a function changes when the variables change. For example, what is the change in P if the values of T and V both change? In this case, P changes to P + dP where
dP
I dT + v
I dV
T
(3.3)
^ + ^ Consider the following practical illustration of Equation (3.3). A person is on a hill and
\dx Jy \dy Jx has determined his or her altitude above sea level. How much will the altitude (denoted
by z) change if the person moves a small distance east (denoted by x) and north (denoted by y)l The change in z as the person moves east is the slope of the hill in that direction, ( dz/dx)y , multiplied by the distance dx that he or she moves. A similar expression can be written for the change in altitude as the person moves north. Therefore, the total change in altitude is the sum of these two changes or
dz
dy
These changes in the height z as the person moves first along the x direction and then along the y direction are illustrated in Figure 3.1. Because the slope of the hill is a nonlinear function of x and y9 this expression for dz is only valid for small changes dx and dy. Otherwise, higher order derivatives need to be considered.
Second or higher derivatives with respect to either variable can also be taken. The mixed second partial derivatives are of particular interest. Consider the mixed partial derivatives of P:
f _d_fdP\ \ _ d2P
VTTVaV/r/v- dTW
R_
V2
f d f dP\ \ _ irP VdV \df)v)T ~~ dVdT
(3.4)
For all state functions /and for our specific case of P, the order in which the function is differentiated does not affect the outcome. For this reason,
fd_ (df(V,T)\ \ = / 3 /df(V,T)\ \ \dT V dV JtJv \dV\ dT JvJt
(3.5)
Because Equation (3.5) is only satisfied by state functions/, it can be used to deter¬ mine if a function / is a state function. If / is a state function, one can write A / = f/ df = f finai — f initiai. This equation states that / can be expressed as an infinitesimal quantity df that when integrated depends only on the initial and final states; df is called an exact differential. An example of a state function and its exact differential is U and dU = dq — P external dV.
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3.1 THE MATHEMATICAL PROPERTIES OF STATE FUNCTIONS 47
EXAMPLE PROBLEM 3.1
a. Calculate
for the function /(x, y ) = yex + xy + xlny.
b. Determine if /(x, y) is a state function of the variables x and y.
c. If /(x, y) is a state function of the variables x and y, what is the total differential df!
Solution
a. ( ^ ) = yex + y + In y.
dx 2
= yex9
i — ex + 1 H — ,
\ ty Jx y
b. Because we have shown that
— I = ex + x + -
dy
tl_
ay2
x
^2
= ex + 1 + -
y y
f(x,y) is a state function of the variables x and y. Generalizing this result, any well-behaved function that can be expressed in analytical form is a state function.
c. The total differential is given by
df
I dx +
y
dy
= (yex + y + lny)dx +
+ x H - jdy
y;
Two other important results from differential calculus will be used frequently. Consider a function z — f(x9 y) that can be rearranged to x = g(y, z) or y = h(x, z). For example, if P = nRT /V, then V = nRT/P and T = PV/nR. In this case
The cyclic rule will also be used:
(3.6)
(3.7)
It is called the cyclic rule because x, y, and z in the three terms follow the orders x, y, z\ y, z, x; and z, x, y. Equations (3.6) and (3.7) can be used to reformulate Equation (3.3) shown next below:
dP
I dT + v
dV
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Suppose this expression needs to be evaluated for a specific substance, such as N2 gas. What quantities must be measured in the laboratory in order to obtain numerical values for ( 8P/dT)v and ( SP/dV)T ? Using Equations (3.6) and (3.7),
dP_
dT
v
dP_
dV
T
and
(3.8)
where (3 and k are the readily measured isobaric volumetric thermal expansion coefficient and the isothermal compressibility, respectively, defined by
P =
V
p
and
{-)
V\dPj T
(3.9)
Both ( dV/dT)P and ( dV/dP)T can be measured by determining the change in volume of the system when the pressure or temperature is varied, while keeping the second variable constant.
The minus sign in the equation for k is chosen so that values of the isothermal compressibility are positive. For small changes in T and P, Equations (3.9) can be written in the more compact form: V(T2) = V(T1)(1 + (3[T2 ~ 7^]) and V(P2) = V(Pi)(l - k[P2 ~ Pi]). Values for (3 and k for selected solids and liq¬ uids are shown in Tables 3.1 and 3.2, respectively.
Equation (3.8) is an example of how seemingly abstract partial derivatives can be directly linked to experimentally determined quantities using the mathematical properties of state functions. Using the definitions of [3 and k, Equation (3.3) can be written in the form
j3 1
dP = — dT - dV
k kV
(3.10)
which can be integrated to give
if V f
f (3 f 1 B 1 Vf
A P = -dT - / — dV « ~(Tf - TA - - In— (3.11)
J K J KV K v f lJ K Vi
Ti Vi
TABLE 3.1 Volumetric Thermal Expansion Coefficient for Solids and Liquids at 298 K
Element
106 f3/(K~l)
Element or Compound
io4^/(k-‘;
Ag(s)
51.6
mi)
1.81
Al(.v)
69.3
CCl 4(Z)
11.4
Au(i)
42.6
CH3COCH3(/)
14.6
Cu(.v)
49.5
CH3OH(/)
14.9
Fe(.s)
36.9
C2H5OH(/)
11.2
Mg(i)
78.3
c6h5ch3(/)
10.5
SICs)
7.5
c6h6(/)
11.4
W(j)
13.8
h2o(/)
2.04
Zn(.v)
90.6
H20(.v)
1.66
Sources: Benenson, W., Harris, J. W., Stocker, H., and Lutz, H. Handbook of Physics. New York: Springer, 2002; Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002; Blachnik, R., ed. D’Ans Lax Taschenbuch fiir Chemiker und Physiker. 4th ed. Berlin: Springer, 1998.
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3.1 THE MATHEMATICAL PROPERTIES OF STATE FUNCTIONS 49
1 TABLE 3.2
Isothermal Compressibility at 298 K
Substance
T
&
"S'
o
Substance
106 k/ bar-1
Al(.v)
1.33
Br2(/)
64
Si02(^)
2.57
C2H5OH(Z)
110
Ni (s)
0.513
C6H5OH(/)
61
TiO 2(s)
0.56
c6h6(Z)
94
Na (s)
13.4
CC14(Z)
103
Cu (s)
0.702
CH3COCH3(Z)
125
C (graphite)
0.156
CH3OH(Z)
120
Mn (s)
0.716
CS2(Z)
92.7
Co (s)
0.525
h20(Z)
45.9
Au (s)
0.563
H g(l)
3.91
Pb (s)
2.37
SiCl4(Z)
165
F e(5)
0.56
TiCU(Z)
89
Ge(j)
1.38
Sources: Benenson, W., Harris, J. W., Stocker, H., and Lutz, H. Handbook of Physics. New York: Springer, 2002; Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton FL: CRC Press, 2002; Blachnik, R., ed. D’Ans Lax Taschenbuch fur Chemiker und Physiker. 4th ed. Berlin: Springer, 1998.
The second expression in Equation (3.11) holds if A T and AV are small enough that (3 and k are constant over the range of integration. Example Problem 3.2 shows a useful application of this equation.
EXAMPLE PROBLEM 3.2
You have accidentally arrived at the end of the range of an ethanol-in-glass ther¬ mometer so that the entire volume of the glass capillary is filled. By how much will the pressure in the capillary increase if the temperature is increased by another 10.0°C? pglass = 2.00 X 10~5(°C)-\ pethanol = 11.2 X 10_4(°C)_1, and Kethanoi = 11.0 X 10 5(bar) 1 . Will the thermometer survive your experiment?
Solution
Using Equation (3.11),
ethanol
A P =
/fiethanol lrr, /
—dT~j
dT - / — dV kV
ethanol A ^ 1 ^ f
- A T - In —
K K v ;
P ethanol A ^ 1, L( 1 + Pglass^T) (3ethanoi 1 ViPglass^T
- AT - In - ~ - AT -
K K Vi K K Vi
(P ethanol P glass')
AT
(11.2 - 0.200) X 10“4(°C)_1 11.0 X 10_5(bar)“
X 10.0°C = 100. bar
In this calculation, we have used the relations V(T2) = V(T{)(\ + /3[T2 ln(l + x) ~ x if v « 1.
The glass is unlikely to withstand such a large increase in pressure.
7^]) and
J
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50
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
The Dependence of U on V and T
In this section, the fact that dU is an exact differential is used to establish how U varies with T and V. For a given amount of a pure substance or a mixture of fixed composition, U is determined by any two of the three variables P, V , and T. One could choose other combinations of variables to discuss changes in U. However, the following discussion will demonstrate that it is particularly convenient to choose the variables T and V. Because U is a state function, an infinitesimal change in U can be written as
dU
dT + v
I dV
T
(3.12)
This expression says that if the state variables change from T, V to T + dT , V + dV, the change in U, dU , can be determined in the following way. We determine the slopes of U{T,V) with respect to T and V and evaluate them at T, V. Next, these slopes are multiplied by the increments dT and dV, respectively, and the two terms are added. As long as dT and dV are infinitesimal quantities, higher order deriva¬ tives can be neglected.
How can numerical values for ( 3U/dT)v and ( dU/dV)T be obtained? In the follow¬ ing, we only consider P-V work. Combining Equation (3.12) and the differential expression of the first law,
fdU\ fdU\
Jq ~ PexternaldV = dT + ( — dV (3.13)
The symbol dq is used for an infinitesimal amount of heat as a reminder that heat is not a state function. We first consider processes at constant volume for which dV = 0, so that Equation (3.13) becomes
dqy
(3.14)
Note that in the previous equation, d qv is the product of a state function and an exact differential. Therefore, d qv behaves like a state function, but only because the path (constant V) is specified. The quantity dq is not a state function.
Although the quantity ( dU/dT)v looks very abstract, it can be readily measured. For example, imagine immersing a container with rigid diathermal walls in a water bath, where the contents of the container are the system. A process such as a chem¬ ical reaction is carried out in the container and the heat flow to the surroundings is measured. If heat flow dqv occurs, a temperature increase or decrease dT is observed in the system and the water bath surroundings. Both of these quantities can be measured. Their ratio, dqv/dT , is a special form of the heat capacity dis¬ cussed in Section 2.5:
Jqy _ fdu\ _
dT WvV V
(3.15)
where dqv/dT corresponds to a constant volume path and is called the heat capacity at constant volume.
The quantity Cv is extensive and depends on the size of the system, whereas Cv m is an intensive quantity. As discussed in Section 2.5, Cv m is different for different sub¬ stances under the same conditions. Observations show that Cv m is always positive for a single-phase, pure substance or for a mixture of fixed composition, as long as no chem¬ ical reactions or phase changes take place in the system. For processes subject to these constraints, U increases monotonically with T.
With the definition of Cv, we now have a way to experimentally determine changes in U with T at constant V for systems of pure substances or for mixtures of constant composition in the absence of chemical reactions or phase changes. After Cv has been
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3.2 THE DEPENDENCE OF U ON V AND T
51
determined as a function of T as discussed in Section 2.5, the following integral is numerically evaluated:
A Uv =
(3.16)
Over a limited temperature range, Cv m can often be regarded as a constant. If this is the case, Equation (3.16) simplifies to
A Uy
1 2
1
CydT — CyAT — nCym\T
which can be written in a different form to explicitly relate qv and A U :
/
i
■d'qy —
(3.17)
(3.18)
Although dq is not an exact differential, the integral has a unique value if the path is defined, as it is in this case (constant volume). Equation (3.18) shows that A U for an arbitrary process in a closed system in which only P—V work occurs can be determined by measuring q under constant volume conditions. As discussed in Chapter 4, the technique of bomb calorimetry uses this approach to determine A U for chemical reactions.
Next consider the dependence of U on V at constant T, or (dU/dV)T. This quantity has the units of J/m3 = (J/m)/m2 =kgms_2/m2 = force/area = pressure and is called the internal pressure. To explicitly evaluate the internal pressure for different substances, a result will be used that is derived in the discussion of the second law of thermodynamics in Section 5.12:
Using this equation, the total differential of the internal energy can be written as
dU = dUv + dUT = CydT +
dV
(3.20)
In this equation, the symbols dUy and dUj have been used, where the subscript indi¬ cates which variable is constant. Equation (3.20) is an important result that applies to systems containing gases, liquids, or solids in a single phase (or mixed phases at a con¬ stant composition) if no chemical reactions or phase changes occur. The advantage of writing dU in the form given by Equation (3.20) over that in Equation (3.12) is that [dU/dV)T can be evaluated in terms of the system variables P, V , and T and their deriv¬ atives, all of which are experimentally accessible.
Once [dU/dV)T and ( 3U/dT)v are known, these quantities can be used to deter¬ mine dU. Because U is a state function, the path taken between the initial and final states is unimportant. Three different paths are shown in Figure 3.2, and dU is the same for these and any other paths connecting V» Tt and Vf, Tf. To simplify the calculation, the path chosen consists of two segments, in which only one of the variables changes in a given path segment. An example of such a path is Vf, Tt —>Vf,Ti—>Vf,T y. Because T is constant in the first segment,
dU = dUT
dV
Because V is constant in the second segment, dU = dUv = CydT. Finally, the total change in U is the sum of the changes in the two segments, dUtotai = dUv + dUT.
FIGURE 3.2
Because U is a state function, all paths connecting Vb Tt and Vf, Tf are equally valid in calculating A U. Therefore, a specification of the path is irrelevant.
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52 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Does the Internal Energy Depend More Strongly on V or 77
Chapter 2 demonstrated that U is a function of T alone for an ideal gas. However, this statement is not true for real gases, liquids, and solids for which the change in U with V must be considered. In this section, we ask if the temperature or the volume depend¬ ence of U is most important in determining A U for a process of interest. To answer this question, systems consisting of an ideal gas, a real gas, a liquid, and a solid are consid¬ ered separately. Example Problem 3.3 shows that Equation (3.19) leads to a simple result for a system consisting of an ideal gas.
/ -
EXAMPLE PROBLEM 3.3
Evaluate ( dU/dV)T for an ideal gas and modify Equation (3.20) accordingly for the specific case of an ideal gas.
Solution
(dU\ / dP_\ Jd[nRT/V]
\dVjT \dTjv V dT
nRT
V
- P = 0
Therefore, dU = CydT , showing that for an ideal gas, U is a function of T only.
Example Problem 3.3 shows that U is only a function of T for an ideal gas. Specifically, U is not a function of V. This result is understandable in terms of the potential function of Figure 1.10. Because ideal gas molecules do not attract or repel one another, no energy is required to change their average distance of separation (increase or decrease V):
Tf
A U= Jcv(T)dT (3.21)
t,
Recall that because U is only a function of T, Equation (3.21) holds for an ideal gas even if V is not constant.
Next consider the variation of U with T and V for a real gas. The experimental determination of ( dU/dV)T was carried out by James Joule using an apparatus con¬ sisting of two glass flasks separated by a stopcock, all of which were immersed in a water bath. An idealized view of the experiment is shown in Figure 3.3. As a valve between the volumes is opened, a gas initially in volume A expands to completely fill the volume A + B. In interpreting the results of this experiment, it is important to understand where the boundary between the system and surroundings lies. Here, the decision was made to place the system boundary so that it includes all the gas. Initially, the boundary lies totally within VA, but it moves during the expansion so that it continues to include all gas molecules. With this choice, the volume of the system changes from before the expansion to VA + VB after the expansion has taken place.
The first law of thermodynamics [Equation (3.13)] states that
FIGURE 3.3
Schematic depiction of the Joule experi¬ ment to determine (dU/dV)T. Two spheri¬ cal vessels, A and B, are separated by a valve. Both vessels are immersed in a water bath, the temperature of which is monitored. The initial pressure in each vessel is indicated.
'd'tf P external
dT +
I dV
T
However, all the gas is contained in the system; therefore, P external = 0 because a vac¬ uum cannot exert a pressure. Therefore Equation (3.13) becomes
dq
dT +
(3.22)
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3.3 DOES THE INTERNAL ENERGY DEPEND MORE STRONGLY ON V OR 77
53
To within experimental accuracy, Joule found that dT surroundings = 0- Because the water bath and the system are in thermal equilibrium, dT = dT surroundings = 0. With this observation, Joule concluded that jfq = 0. Therefore, Equation (3.22) becomes
(3.23)
Because dV A 0, Joule concluded that (dU /dV)T = 0. Joule’s experiment was not defin¬ itive because the experimental sensitivity was limited, as shown in Example Problem 3.4.
EXAMPLE PROBLEM 3.4
In Joule’s experiment to determine ( dU/dV)T , the heat capacities of the gas and the water bath surroundings were related by C surroundings/ C system ~ 1000. If the precision with which the temperature of the surroundings could be measured is ±0.006°C, what is the minimum detectable change in the temperature of the gas?
Solution
View the experimental apparatus as two interacting systems in a rigid adiabatic enclosure. The first is the volume within vessels A and B, and the second is the water bath and the vessels. Because the two interacting systems are isolated from the rest of the universe,
*7 — C water bath^' water bath ^ gas gas ~ ^
A Tgas = A Twaterbath = -1000 X (±0.006°C) = T6°C
^ gas
In this calculation, A T gas is the temperature change that the expanded gas undergoes to reach thermal equilibrium with the water bath, which is the negative of the tempera¬ ture change during the expansion.
Because the minimum detectable value of A T gas is rather large, this apparatus is clearly not suited for measuring small changes in the temperature of the gas in an expansion. J
More sensitive experiments were carried out by Joule in collaboration with William Thomson (Lord Kelvin). These experiments, which are discussed in Section 3.8, demonstrate that ( dU/dV)T is small, but nonzero for real gases.
Example Problem 3.3 has shown that (dU/dV)T = 0 for an ideal gas. We next calcu¬ late (dU/dV)T and A UT = fy™,.f(dU/dV)TdVm for a real gas, in which the van der Waals equation of state is used to describe the gas, as illustrated in Example Problem 3.5.
[ EXA
EXAMPLE PROBLEM 3.5
For a gas described by the van der Waals equation of state,
P = nRT /(V — nb) — an2 /V2. Use this equation to complete these tasks:
a. Calculate (dU/dV)T using (dU/dV)T = T(dP/dT)v - P
b. Derive an expression for the change in internal energy, A Uj = fy.f(dU/dV)TdV, in compressing a van der Waals gas from an initial molar volume V* to a final molar volume Vj at constant temperature.
Solution
a. T
(— ) - P = T\
\dTJv
nRT V - nb
n2a V2 .
nRT
dT
nRT
nRT
- P = - -
r y V — nb
- P
n2a
n2a
V — nb V — nb
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
b. AUr =
Vf Vf ~
n ^ dU \ f n a
— 1 dV = I -
dV J:
1
1
dV = na\ -
' f
U Vi
Note that A UT is zero if the attractive part of the intermolecular potential is zero.
J
Example Problem 3.5 demonstrates that in general (dU/dV)T ^ 0, and that A UT can be calculated if the equation of state of the real gas is known. This allows the rela¬ tive importance of A UT = fy.f (dU/dV)TdV and A Uv = Jj.f CydT to be determined in a process in which both T and V change, as shown in Example Problem 3.6.
EXAMPLE PROBLEM 3.6
One mole of N2 gas undergoes a change from an initial state described by T = 200. K and Pj = 5.00 bar to a final state described by T = 400. K and Pf = 20.0 bar. Treat N2 as a van der Waals gas with the parameters a = 0.137 Pa m mol and b = 3.87 X 10“5 m3 mol-1. We use the path N2 (g, T = 200. K ,P = 5.00 bar)
N2(g, T = 200. K ,P = 20.0 bar) N2(g, T = 400. K, P = 20.0 bar), keeping in mind that all paths will give the same answer for A U of the overall process.
a. Calculate A UT = fy.f ( dU/dV)TdV using the result of Example Problem 3.5. Note that Vt = 3.28 X 10-3 m3 and Vf = 7.88 X 10-4 m3 at 200. K, as calcu¬ lated using the van der Waals equation of state.
b. Calculate A Uv = n Cy mdT using the following relationship for Cy m in this temperature range:
Cy,m
J K_1 mol-1
= 22.50 - 1.187 X 10“2— + 2.3968 X 10“5— r - 1.0176 X 10“8^ K K2 K3
The ratios Tn /Kn ensure that Cy m has the correct units,
c. Compare the two contributions to A U. Can A Uj be neglected relative to A Uyl
Solution
a. Using the result of Example Problem 3.5,
A UT — n2a
X
1
1
V V t
v m, i y m, j
= 0.137 Pam0 1
3.28 X 10-3 m3 7.88 X 10-4 m3
= -132 J
b. A Uy — n I Cy m dT
400.
200.
22.50 - 1.187 X 10“2— + 2.3968 X 10“5— r
K
K
V
-1.0176 X 10“8— r K3
Ml - I J
= (4.50 - 0.712 + 0.447 - 0.0610)kJ = 4.17 kJ
c. A Uj is 3.2% of A Uy for this case. In this example, and for most processes, At/7- 1 can be neglected relative to A Uy for real gases.
The calculations in Example Problems 3.5 and 3.6 show that to a good approxima¬ tion A Uj = fy.f (dU /dV)jdV ~ 0 for real gases under most conditions. Therefore, it
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3.4 THE VARIATION OF ENTHALPY WITH TEMPERATURE AT CONSTANT PRESSURE
55
is sufficiently accurate to consider U as a function of T only \U = U(T)] for real gases in processes that do not involve unusually high gas densities.
Having discussed ideal and real gases, what can be said about the relative magni¬ tude of A Up = fy/ ( dU/dV)TdV and A Uv = fr/ CydT for processes involving liq¬ uids and solids? From experiments, it is known that the density of liquids and solids varies only slightly with the external pressure over the range in which these two forms of matter are stable. This conclusion is not valid for extremely high pressure conditions such as those in the interior of planets and stars. However, it is safe to say that dV for a solid or liquid is very small in most processes. Therefore,
v2
= » (f » 0 (3.24)
Vi
because AV ~ 0. This result is valid even if ( dU/dV)T is large.
The conclusion that can be drawn from this section is as follows. Under most condi¬ tions encountered by chemists in the laboratory, U can be regarded as a function of T alone for all substances. The following equations give a good approximation even if V is not constant in the process under consideration:
U(Tf,Vf)
U {T j, Vt) = A U =
(3.25)
Note that Equation (3.25) is only applicable to a process in which there is no change in the phase of the system, such as vaporization or fusion, and in which there are no chem¬ ical reactions. Changes in U that arise from these processes will be discussed in Chapters 4 and 8.
The Variation of Enthalpy with Temperature at Constant Pressure
As for U, H can be defined as a function of any two of the three variables P, V , and T. It was convenient to choose U to be a function of T and V because this choice led to the identity A U = qy. Using a similar reasoning, we choose H to be a function of T and P. How does H vary with P and 77 The variation of H with T at constant P is dis¬ cussed next, and a discussion of the variation of H with P at constant T is deferred to Section 3.6.
Consider the constant pressure process shown schematically in Figure 3.4. For this process defined by P = P external*
dU = dqP - PdV
(3.26)
Although the integral of dq is in general path dependent, it has a unique value in this case because the path is specified, namely, P = Pexternai = constant. Integrating both sides of Equation (3.26),
/
i
dU
i
or Uf ~ Ut = qP- P(Vf ~ Vt) (3.27)
Because P = Pf = Pi9 this equation can be rewritten as
(Uf + PfVf) ~ ( Ui + PiVi) = qp or A H = qP (3.28)
The preceding equation shows that the value of AH can be determined for an arbitrary process at constant P in a closed system in which only P-V work occurs by simply meas¬ uring qP , the heat transferred between the system and surroundings in a constant pressure
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P = P
' external '
Initial state Final state
FIGURE 3.4
The initial and final states are shown for an undefined process that takes place at constant pressure.
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
process. Note the similarity between Equations (3.28) and (3.18). For an arbitrary process in a closed system in which there is no work other than P-V work, A U = qy if the process takes place at constant V, and A H = qP if the process takes place at constant P. These two equations are the basis for the fundamental experimental techniques of bomb calorimetry and constant pressure calorimetry discussed in Chapter 4.
A useful application of Equation (3.28) is in experimentally determining A H and AT/ of fusion and vaporization for a given substance. Fusion (solid — » liquid) and vaporization (liquid — » gas) occur at a constant temperature if the system is held at a constant pressure and heat flows across the system-surroundings boundary. In both of these phase transitions, attractive interactions between the molecules of the system must be overcome. Therefore, q > 0 in both cases and CP — > oo. Because A H = qP, A Hfusion and AH vaporization can be determined by measuring the heat needed to effect the transition at constant pressure. Because AH = A U + A (PV), at constant P,
vaporization vaporization ~ P A VVa porization ^ 0 (3.29)
The change in volume upon vaporization is AV vaporization = Vgas — Viiquij » 0; therefore, A U vaporization < A H vaporization. An analogous expression to Equation (3.29) can be written relating A U fusion and AH fusion. Note that AV fusion is much smaller than AV vaporization and can be either positive or negative. Therefore, A U fusion ~ fusion- The thermodynamics of fusion and vaporization will be discussed in more detail in Chapter 8.
Because H is a state function, dH is an exact differential, allowing us to link ( 3H/dT)P to a measurable quantity. In analogy to the preceding discussion for dU , dH is written in the form
dH
I dT +
p
(3.30)
Because dP = 0 at constant P, and dH = dqP from Equation (3.28), Equation (3.30) becomes
<€fqP —
dT
p
(3.31)
Equation (3.31) allows the heat capacity at constant pressure CP to be defined in a fashion analogous to Cv in Equation (3.15):
CP -
dqP
dT
(3.32)
Although this equation looks abstract, CP is a readily measurable quantity. To measure it, one need only measure the heat flow to or from the surroundings for a constant pres¬ sure process together with the resulting temperature change in the limit in which dT and q approach zero and form the ratio lim (zfq/dT)P.
dT — >0
As was the case for Cy, Cp is an extensive property of the system and varies from substance to substance. The temperature dependence of CP must be known in order to calculate the change in H with T. For a constant pressure process in which there is no change in the phase of the system and no chemical reactions,
AHP =
j CP(T)dT
Ti
(3.33)
If the temperature interval is small enough, it can usually be assumed that CP is con¬ stant. In that case,
AHP = CPAT = nCP^mAT (3.34)
The calculation of AH for chemical reactions and changes in phase will be discussed in Chapters 4 and 8.
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3.5 HOW ARE CP AND Cv RELATED? 57
EXAMPLE PROBLEM 3.7
A 143.0 g sample of C(s) in the form of graphite is heated from 300. to 600. K at a constant pressure. Over this temperature range, CP m has been determined to be
j1 j^2 j-3
12.19 + 0.1126 - 1.947 X 10“4— r + 1.919 X 10“7^
K K2 K3
jA
7.800 X 10-11 —7 K4
Calculate AH and qP. How large is the relative error in AH if we neglect the temperature-dependent terms in CP m and assume that CP m maintains its value at 300. K throughout the temperature interval?
J K-1mol-1
Solution
lfr
m I
~MJ
AH = — CPm(T)dT
600.
143.0 g
J
T T2 \
-12.19 + 0.1126 - 1.947 X 10“4 — r + 1.919 '
12.00 g mol 1 mol
300.
K
rj-i 3 rj-, 4
X 10-% - 7.800 X 10-1%
K
K
d —
K
143.0
12.00
X
-12.19- + 0.0563— r - 6.49 X 10-5— r + 4.798
K
K
K
X 10_8^t - 1.56 X 10_ 11 —
K4
K
600.
J = 46.9 kJ
J300.
From Equation (3.28), AH = qP.
If we had assumed CP m = 8.617 J mol-1 K-1, which is the calculated value at 300. K, AH = 143.0 g/ 12.00 g mol-1 X 8.617 JK_1 mol-1 X [600. K - 300. K] = 30.8 kJ. The relative error is 100 X (30.8 kJ - 46.9 kJ)/46.9 kJ = -34.3%. In this case, it is not reasonable to assume that CP m is independent of temperature.
How Are CP and Cv Related?
To this point, two separate heat capacities, CP and Cv , have been defined. How are these quantities related? To answer this question, the differential form of the first law is written as
dq - Cv dT + dV + Pexternal dV (3.35)
Consider a process that proceeds at constant pressure for which P = Pexternal • In this case, Equation (3.35) becomes
dqP — Cy dT +
d_U\
dVjj
dV + PdV
(3.36)
Because <dqP = CPdT ,
CP — Cy +
= Cy + T
~) (~) + ^ + dvJAffrjp \8T J p v
dU\
dvj7
+ p
dV\
dTj,
(a®,
(3.37)
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
To obtain Equation (3.37), both sides of Equation (3.36) have been divided by dT , and the ratio dV/dT has been converted to a partial derivative at constant P. Equation (3.19) has been used in the last step. Using Equation (3.9) and the cyclic rule, one can simplify Equation (3.37) to
Cp — Cy + Tl
dp_
dT
= cv
p
P
Cp = Cv + TV — or
K
= c
V, m
TV m ——
m K
(3.38)
Equation (3.38) provides another example of the usefulness of the formal theory of thermodynamics in linking seemingly abstract partial derivatives with experimentally available data. The difference between CP m and Cy m can be determined at a given tem¬ perature knowing only the molar volume, the isobaric volumetric thermal expansion coefficient, and the isothermal compressibility.
Equation (3.38) is next applied to ideal and real gases, as well as liquids and solids, in the absence of phase changes and chemical reactions. Because /3 and k are always positive for real and ideal gases, CP — Cv > 0 for these substances. First, CP — Cv is calculated for an ideal gas, and then it is calculated for liquids and solids. For an ideal gas, (dU/dV)T = 0 as shown in Example Problem 3.3, and fdP\ fdV\ fnR\fnR\
i — I ( — I = 7i — II — I = nR so that Equation (3.37) becomes
\dTjy\dTjp \V
CP Cy — nR
(3.39)
This result was stated without derivation in Section 2.4. The partial derivative (dV/dT)P = V (3 is much smaller for liquids and solids than for gases. Therefore, generally
Cy ^5 >
(3.40)
so that CP ~ Cy for a liquid or a solid. As shown earlier in Example Problem 3.2, it is not feasible to carry out heating experiments for liquids and solids at constant volume because of the large pressure increase that occurs. Therefore, tabulated heat capacities for liquids and solids list CP m rather than Cy m.
The Variation of Enthalpy with Pressure at Constant Temperature
In the previous section, we learned how H changes with T at constant P. To calculate how H changes as both P and T change, ( dH/dP)T must be calculated. The partial derivative ( dH/dP)T is less straightforward to determine in an experiment than (dH/dT ) P. As will be seen, for many processes involving changes in both P and T, ( dH/dT)PdT » ( dH/dP)TdP and the pressure dependence of H can be neglected relative to its temperature dependence. However, the knowledge that ( dH/dP)T is not zero is essential for understanding the operation of a refrigerator and the liquefaction of gases. The following discussion is applicable to gases, liquids, and solids.
Given the definition H = U + PV, we begin by writing dH as
dH = dU + P dV + V dP (3.41)
Substituting the differential forms of dU and dH ,
rdH\ fdU\
— ap = at + —
CpdT + ( — dP = Cy dT + — dV + PdV + V dP
dP J p \dV J p
— Cy dT +
dU\
— + P
dV T
dV + V dP
(3.42)
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3.6 THE VARIATION OF ENTHALPY WITH PRESSURE AT CONSTANT TEMPERATURE
For isothermal processes, dT = 0, and Equation (3.42) can be rearranged to
-) =
dP J j \dV Jt
Using Equation (3.19) for (dU/dV)T,
« = t(e) ('«:') + v
\dP Jt \dTjy \dPjT
= V- = V(1 - TP) (3.44)
The second formulation of Equation (3.44) is obtained through application of the cyclic rule [Equation (3.7)]. This equation is applicable to all systems containing pure sub¬ stances or mixtures at a fixed composition, provided that no phase changes or chemical reactions take place. The quantity ( 3H/dP)T is evaluated for an ideal gas in Example Problem 3.8.
dV_\
dP) 7
+ V
(3.43)
[ EXAMPLE PROBLEM 3.8
Evaluate ( 3H/dP)T for an ideal gas.
Solution
(< BP/dT)v = (■ d[nRT/V]/dT)v = nR/V and (dV/dP)T = ( d[nRT / P]/ dP)T = — nRT/P 2 for an ideal gas. Therefore,
nRT \
f-) =*(-) t^) + v =
\dPjT \dT Jv\dP JT V
nRT nRT
rs I + v = - + V = o
2 / P nRT
This result could have been derived directly from the definition H = U + PV. For an ideal gas, U = U(T) only and PV = nRT . Therefore, H = H(T) for an ideal gas and ( 3H/dP)T = 0. _ J
Because Example Problem 3.8 shows that H is only a function of T for an ideal gas,
AH
CP(T)dT = n Cpm(T)dT
Ti
(3.45)
for an ideal gas. Because H is only a function of T, Equation (3.45) holds for an ideal gas even if P is not constant. This result is also understandable in terms of the potential function of Figure 1.10. Because ideal gas molecules do not attract or repel one another, no energy is required to change their average distance of separation (increase or decrease P).
Equation (3.44) is next applied to several types of systems. We have seen that (3H/dP)T = 0 for an ideal gas. For liquids and solids, 1 » T f3 for T < 1000 K as can be seen from the data in Table 3.1. Therefore, for liquids and solids, (■ 3H/dP)T ~ V to a good approximation, and dH can be written as
dH « CpdT + V dP (3.46)
for systems that consist only of liquids or solids.
| EXAMPLE PROBLEM 3.9
Calculate the change in enthalpy when 124 g of liquid methanol initially at 1.00 bar and 298 K undergoes a change of state to 2.50 bar and 425 K. The density of liquid methanol under these conditions is 0.791 g cm-3, and Cp m for liquid methanol is 81.1 JK-1 mol-1.
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60
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Solution
Because H is a state function, any path between the initial and final states will give the same A H. We choose the path methanol (/, 1.00 bar, 298 K) — > methanol (/, 1.00 bar, 425 K) — > methanol (/, 2.50 bar, 425 K). The first step is isothermal, and the second step is isobaric. The total change in H is
AH
nCP,m (Tf - Tt) + V{Pf ~ Pt)
= 81.1 J K-1 mol-1 X - - - r X (425 K - 298 K)
32.04 g mol-1
124 g 10“6 m3
+ - 5 — t X - r— x (2.50 bar - 1.00 bar
0.791 g cm-3 cm3
= 39.9 X 103 J + 23.5 J « 39.9 kJ
X
IQ5 Pa bar
Note that the contribution to AH from the change in T is far greater than that from change in P.
tl thej
Example Problem 3.9 shows that because molar volumes of liquids and solids are small, H changes much more rapidly with T than with P. Under most conditions, H can be assumed to be a function of T only for solids and liquids. Exceptions to this rule are encountered in geophysical or astrophysical applications, for which extremely large pressure changes can occur.
The following conclusion can be drawn from this section: under most conditions encountered by chemists in the laboratory, H can be regarded as a function of T alone for liquids and solids. It is a good approximation to write
H(Tf,Pf)
H(ThPi) = A H= CPdT = n / CP,m dT
(3.47)
even if P is not constant in the process under consideration. The dependence of H on P for real gases is discussed in Section 3.8 and Section 3.9 in the context of the Joule- Thomson experiment.
Note that Equation (3.47) is only applicable to a process in which there is no change in the phase of the system, such as vaporization or fusion, and in which there are no chemical reactions. Changes in H that arise from chemical reactions and changes in phase will be discussed in Chapters 4 and 8.
Having dealt with solids, liquids, and ideal gases, we are left with real gases. For real gases, ( 3H/dP)T and ( dU/dV)T are small, but still have a considerable effect on the properties of the gases upon expansion or compression. Conventional technology for the liquefaction of gases and for the operation of refrigerators is based on the fact that ( 8H/dP)T and (dU/dV)T are not zero for real gases. To derive a useful formula for calculating ( 8H/dP)T for a real gas, the Joule-Thomson experiment is discussed first in the next section.
The Joule-Thomson Experiment
If the valve on a cylinder of compressed N2 at 298 K is opened fully, it will become covered with frost, demonstrating that the temperature of the valve is lowered below the freezing point of H20. A similar experiment with a cylinder of H2 leads to a consid¬ erable increase in temperature and, potentially, an explosion. How can these effects be understood? To explain them, we discuss the Joule-Thomson experiment.
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3.7 THE JOULE-THOMSON EXPERIMENT 61
FIGURE 3.5
In the Joule-Thomson experiment, a gas is forced through a porous plug using a pis¬ ton and cylinder mechanism. The pistons move to maintain a constant pressure in each region. There is an appreciable pres¬ sure drop across the plug, and the temper¬ ature change of the gas is measured. The upper and lower figures show the initial and final states, respectively. As shown in the text, if the piston and cylinder assem¬ bly forms an adiabatic wall between the system (the gases on both sides of the plug) and the surroundings, the expansion is isenthalpic.
The Joule-Thomson experiment shown in Figure 3.5 can be viewed as an improved version of the Joule experiment because it allows ( dU/dV)T to be measured with a much higher sensitivity than in the Joule experiment. In this experiment, gas flows from the high-pressure cylinder on the left to the low-pressure cylinder on the right through a porous plug in an insulated pipe. The pistons move to keep the pressure unchanged in each region until all the gas has been transferred to the region to the right of the porous plug. If N2 is used in the expansion process (Pi > P2), it is found that T2 < Tx\ in other words, the gas is cooled as it expands. What is the origin of this effect? Consider an amount of gas equal to the initial volume V\ as it passes through the apparatus from left to right. The total work in this expansion process is the sum of the work performed on each side of the plug separately by the moving pistons:
^ ^ left ^ right
V2
Jp2dV= PlVl - P2V2 0
(3.48)
Because the pipe is insulated, q = 0, and
A U = U2 - Ui = w = PiVi - P2V2 (3.49)
This equation can be rearranged to
U2 + P2V2 = Ui + P\V i or H2 = Hi (3.50)
Note that the enthalpy is constant in the expansion; the expansion is isenthalpic. For the conditions of the experiment using N2, both dT and dP are negative, so (dT /dP) H > 0. The experimentally determined limiting ratio of A T to A P at constant enthalpy is known as the Joule-Thomson coefficient:
lxj-T = lim (^-] = (— ) (3.51)
T ap^o\ApJh \dPjH
If fJij-r is positive, the conditions are such that the attractive part of the potential dom¬ inates, and if /jlj-t is negative, the repulsive part of the potential dominates. Using experimentally determined values of ( 3H/dP)T can be calculated. For an isen¬
thalpic process,
dH = CpdT +
I dP = 0
T
(3.52)
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62 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
TABLE 3.3 Joule-Thomson
Coefficients for Selected
Substances at 273 K and 1 atm
Gas
lxj-T (K/MPa)
Ar
3.66
c6h14
-0.39
ch4
4.38
co2
10.9
h2
-0.34
He
-0.62
n2
2.15
Ne
-0.30
nh3
28.2
o2
2.69
Source: Linstrom, P. J., and Mallard, W. G.. eds. NIST Chemistry Webbook: NIST Standard Reference Database Number 69. Gaithersburg, MD: National Institute of Standards and Technology. Retrieved from http://webbook.nist.gov.
Dividing through by dP and making the condition dH = 0 explicit,
CP
+
0
(dH\
giving )t = ~CplX]~T (3,53)
Equation (3.53) states that ( dH/dP)T can be calculated using the measurement of material-dependent properties CP and fij-p • Because /jlj-t is not zero for a real gas, the pressure dependence of H for an expansion or compression process for which the pressure change is large cannot be neglected. Note that ( 8H/dP)T can be positive or negative, depending on the value of /ulj- t at the P and T of interest.
If /jlj-t is known from experiment, (8U/dV)T can be calculated as shown in Example Problem 3.10. This has the advantage that a calculation of ( 3U/dV)p based on measure¬ ments of Cp, fi j-T and the isothermal compressibility k is much more accurate than a measurement based on the Joule experiment. Values of iij-p are shown for selected gases in Table 3.3. Keep in mind that fi j-T is a function of P and A P, so the values listed in the table are only valid for a small pressure decrease originating at 1 atm pressure.
EXAMPLE PROBLEM 3.10
Using Equation (3.43), ( 3H/dP)T = [_(dU/dV)T + P](dV/dP)T + V, derive an expression giving (dU/dV)T entirely in terms of measurable quantities for a gas.
Solution
V
Cpiij-r + y kV
In this equation, k is the isothermal compressibility defined in Equation (3.9).
[ EXA
EXAMPLE PROBLEM 3.11
Using Equation (3.43),
fdH_"
dP
show that fij-p = 0 for an ideal gas.
Solution
dU ,
— + P
dV
Pj-T ~
j_ / a//
CP \dP
1
dU
CP _\dV Jp \dP J j
^1 +v
dP
dV
— I — I + p —
dV
dP
+ V
J_
Cp
J_
Cp L
. dV .
0 + P[ — + V
dP
d[nRT/P]
dP
+ V
_1
CpL
nRT
+ V
= 0
In this calculation, we have used the result that ( dU/dV)T = 0 for an ideal gas.
J
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3.8 LIQUEFYING GASES USING AN ISENTHALPIC EXPANSION 63
Example Problem 3.11 shows that for an ideal gas, /jlj-t is zero. It can be shown that for a van der Waals gas in the limit of zero pressure
O Liquefying Gases Using an Isenthalpic -0 Expansion
For real gases, the Joule-Thomson coefficient [jlj-t can take on either negative or positive values in different regions of P-T space. If /jlj-t is positive, a decrease in pressure leads to a cooling of the gas; if it is negative, the expansion of the gas leads to a heating. Figure 3.6 shows the variation of /jlj-t with T and P for N2 and H2. All along the solid curve, /jlj-t = 0. To the left of each curve, /ulj-t is positive, and to the right, it is negative. The temperature for which [Aj-t = 0 is referred to as the inversion temperature. If the expansion conditions are kept in the region in which jaj-j is positive, A T can be made sufficiently large as A P decreases in the expansion to liquefy the gas. Note that Equation (3.54) predicts that the inversion temperature for a van der Waals gas is independent of P, which is not in agreement with experiment.
The results in Figure 3.6 are in accord with the observation that a high-pressure (100 < P < 500 atm) expansion of N2 at 300 K leads to cooling and that similar con¬ ditions for H2 lead to heating. To cool H2 in an expansion, it must first be precooled below 200 K, and the pressure must be less than 160 atm. He and H2 are heated in an isenthalpic expansion at 300 K for P < 200 atm.
The Joule-Thomson effect can be used to liquefy gases such as N2, as shown in Figure 3.7. The gas at atmospheric pressure is first compressed to a value of 50 atm to 200 atm, which leads to a substantial increase in its temperature. It is cooled and subse¬ quently passed through a heat exchanger in which the gas temperature decreases to a value within ~50 K of the boiling point. At the exit of the heat exchanger, the gas expands through a nozzle to a final pressure of 1 atm in an isenthalpic expansion. The cooling that occurs because fij-r > 0 results in liquefaction. The gas that boils away passes back through the heat exchanger in the opposite direction than the gas to be liq¬ uefied is passing. The two gas streams are separated, but in good thermal contact. In this process, the gas to be liquefied is effectively precooled, enabling a single-stage expansion to achieve liquefaction.
100 200 300 400 500 Pressure /atm
FIGURE 3.6
All along the curves in the figure,
/jl j-T = 0, and \a j-T is positive to the left of the curves and negative to the right.
To experience cooling upon expansion at 100. atm, T must lie between 50. K and 150. K for H2. The corresponding temper¬ atures for N2 are 100. K and 650. K.
FIGURE 3.7
Schematic depiction of the liquefaction of a gas using an isenthalpic Joule-Thomson expansion. Heat is extracted from the gas exiting from the compressor. It is further cooled in the countercurrent heat exchanger before expanding through a nozzle. Because its temperature is suffi¬ ciently low at the exit to the countercur¬ rent heat exchanger, liquefaction occurs.
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64 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Vocabulary
cyclic rule
exact differential
heat capacity at constant pressure
heat capacity at constant volume
internal pressure isenthalpic
isobaric volumetric thermal expansion coefficient
isothermal compressibility Joule-Thomson coefficient Joule-Thomson experiment partial derivatives
Concept Problems
Q3.1 The heat capacity CPm is less than Cv m for H20(7) near 4°C. Explain this result.
Q3.2 What is the physical basis for the experimental result that U is a function of V at constant T for a real gas? Under what conditions will U decrease as V increases?
Q3.3 Why didn’t Joule change his experiment to make c surroundings/ C system ~ 0.001 to increase the sensitivity of the apparatus?
Q3.4 Why does the relation CP > Cv always hold for a gas? Can Cp < Cv be valid for a liquid?
Q3.5 Why can qv be equated with a state function if q is not a state function?
Q3.6 Explain without using equations why ( 3H/dP)T is generally small for a real gas.
Q3.7 Why is it reasonable to write dH ~ CPdT + VdP for a liquid or solid sample?
Q3.8 Refer to Figure 1.10 and explain why (dU/dV)p is generally small for a real gas.
Q3.9 Can a gas be liquefied through an isenthalpic expan¬ sion if /jlj-t = 0?
Q3.10 Why is qy = AU only for a constant volume process? Is this formula valid if work other than P-V work is possible?
Q3.ll Classify the following variables and functions as intensive or extensive: T, P, V, q, w9 U, H.
Q3.12 Why are q and w not state functions?
Q3.13 Why is the equation AH = fJ/CP(T) dT = n fr/C p m ( T ) dT valid for an ideal gas even if P is not con- stant in the process? Is this equation also valid for a real gas? Why or why not?
Q3.14 What is the relationship between a state function and an exact differential?
Q3.15 Is the following statement always, never, or some¬ times valid? Explain your reasoning: A/7 is only defined for a constant pressure process.
Q3.16 Is the following statement always, never, or sometimes valid? Explain your reasoning: a thermodynamic process is completely defined by the initial and final states of the system.
Q3.17 Is the following statement always, never, or sometimes valid? Explain your reasoning: q = 0 for a cyclic process. Q3.18 The molar volume of H20(/) decreases with increas¬ ing temperature near 4°C. Can you explain this behavior using a molecular level model?
Q3.19 Why was the following qualification made in Section 3.7? Note that Equation (3.47) is only applicable to a process in which there is no change in the phase of the sys¬ tem, such as vaporization or fusion, and in which there are no chemical reactions.
Q3.20 Is the expression A Uy = J^CydT = nf^CyifndT only valid for an ideal gas if V is constant?
Numerical Problems
Problem numbers in red indicate that the solution to the prob¬ lem is given in the Student’s Solutions Manual.
P3.1 Obtain an expression for the isothermal compressibil¬ ity k = — l/V (dV/dP)T for a van der Waals gas.
P3.2 Use the result of Problem P3.26 to show that ( dCv/dV)T for the van der Waals gas is zero.
P3.3 The molar heat capacity CP m of S02(g) is described by the following equation over the range 300 K < T < 1700 K:
CP m T T2
— — = 3.093 + 6.967 X 10“3 - 45.81 X 10“7— r
R K K2
T3
+ 1.035 X 10“9— r K3
In this equation, T is the absolute temperature in kelvin. The ratios 7 n/Kn ensure that CPm has the correct dimension. Assuming ideal gas behavior, calculate q, w , A U, and A H if 1.50 moles of S02(g) is heated from 22.5°C to 1 140.°C at a constant pressure of 1 bar. Explain the sign of w.
P3.4 Use the relation (dU/dV)T = T(dP/dT)v - P and the cyclic rule to obtain an expression for the internal pres¬ sure, (dU/dV)T, in terms of P, j3, T, and k.
P3.5 A mass of 34.05 g of H20(s) at 273 K is dropped into 185 g of H20(/) at 310. K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equi¬ librium has been reached. Assume that CP m for H20 is con¬ stant at its values for 298 K throughout the temperature range of interest.
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NUMERICAL PROBLEMS 65
P3.6 A vessel is filled completely with liquid water and sealed at 13.56°C and a pressure of 1.00 bar. What is the pressure if the temperature of the system is raised to 82. 0°C? Under these conditions , f3water = 2.04 X 10_4K_1, vessel = 1-42 X 10-4 K_1, and Kwater = 4.59 X 10_5bar_1. P3.7 Integrate the expression (3 = l/V (dV/dT)P assuming that [3 is independent of temperature. By doing so, obtain an expression for V as a function of T and /3 at constant P.
P3.8 A mass of 32.0 g of H20(g) at 373 K is flowed into 295 g of H20(/) at 310. K and 1 atm. Calculate the final tem¬ perature of the system once equilibrium has been reached. Assume that CPm for H20 is constant at its values for 298 K throughout the temperature range of interest. Describe the state of the system.
P3.9 Because (dH/dP)T = —Cp/jlj-T, the change in enthalpy of a gas expanded at constant temperature can be calculated. To do so, the functional dependence of [ij-j on P must be known. Treating Ar as a van der Waals gas, calculate A H when 1 mole of Ar is expanded from 325 bar to 1.75 bar at 375 K. Assume that /jlj-T is independent of pressure and is given by ix j-T = [(2 a/RT) - b]/CPm, and CPm = 5R/2 for Ar. What value would A H have if the gas exhibited ideal gas behavior?
P3.10 Derive the following expression for calculating the isothermal change in the constant volume heat capacity: (dCv/dV)T = T(d2P/dT2)v.
P3.ll A 75.0 g piece of gold at 650. K is dropped into 180. g of H20(/) at 310. K in an insulated container at 1 bar pres¬ sure. Calculate the temperature of the system once equilib¬ rium has been reached. Assume that CP m for Au and H20 is constant at their values for 298 K throughout the temperature range of interest.
P3.12 Calculate w, q, AH, and A U for the process in which 1.75 moles of water undergoes the transition H20(/, 373 K) — » H20(g, 610. K) at 1 bar of pressure. The volume of liquid water at 373 K is 1.89 X 10-5 m3 mol-1 and the molar volume of steam at 373 K and 610. K is 3.03 and 5.06 X 10-2 m3 mol-1, respectively. For steam, CP m can be considered constant over the temperature interval of interest at 33.58 J mol-1 K-1.
P3.13 Equation (3.38), CP = Cv + TV((32/k), links CP and Cv with [3 and k. Use this equation to evaluate CP-CV for an ideal gas.
P3.14 Use the result of Problem P3.26 to derive a formula for (dCv/dV)T for a gas that obeys the Redlich-Kwong equa¬ tion of state,
P =
RT
1
ym - b Vt y miy m + h)
P3.15 The function /(x, y) is given by f(x, y ) = xy sin 5x + x2 Vy In y + 3e~2x cos y. Determine
dx2
ay2
yjx
and [ — I —
dx \dy
.Is
dy\dx)y)x \dx\dy
X/ y
Obtain an expression for the total differential df.
P3.16 The Joule coefficient is defined by (dT /dV)u =
(1 /Cy)[P ~ T(dP/dT)v]. Calculate the Joule coefficient for an ideal gas and for a van der Waals gas.
P3.17 Using the result of Equation (3.8), ( dP/dT)v = f3 / k, express /3 as a function of k and Vm for an ideal gas, and /3 as a function of b, k, and Vm for a van der Waals gas.
P3.18 Show that the expression (dU/dV )T =
T ( dP/dT )v — P can be written in the form
P3.19 Derive an expression for the internal pressure of a gas that obeys the Bethelot equation of state,
RT a
P = -
V — b TV2
V m u i V m
P3.20 Because U is a state function, (d/dV ( dU/dT)v)T = (d/dT ( dU/dV)T)y . Using this relationship, show that (dCv/dV)T = 0 for an ideal gas.
P3.21 Starting with the van der Waals equation of state, find an expression for the total differential dP in terms of dV and dT. By calculating the mixed partial derivatives (d(dP/dV)T/dT)v and ( d(dP/dT)v/dV)T , determine if dP is an exact differential.
P3.22 Use ( dU/dV)T = {/3T - kP)/k to calculate (dU/dV)T for an ideal gas.
P3.23 Derive the following relation,
f du\ _ _ 3 a _
\dvm)T 2Vrvm(vm + b )
for the internal pressure of a gas that obeys the Redlich-Kwong equation of state,
_ RT _ a _ 1
Vm-b Vr y m(y m + b)
P3.24 A differential dz — f(x,y)dx + g(x,y)dy is exact if the integral f f(x, y)dx + f g(x, y)dy is independent
of the path. Demonstrate that the differential dz = 2 xydx + x2dy is exact by integrating dz along the paths (1,1) -> (1,8) -> (6,8) and (1,1) -> (1,3) -> (4,3)
(4,8) — > (6,8). The first number in each set of parentheses is the x coordinate, and the second number is the y coordinate. P3.25 Show that dp/p = ~(3dT + KdP where p is the density p = m/V. Assume that the mass m is constant.
P3.26 For a gas that obeys the equation of state
RT , x Vm = — + B(T)
derive the result
= B(T) ~ T
dB(T)
dT
P3.27 Because Vis a state function, ( d(dV/dT)P/dP)T =
( d(dV/dP)T/dT)P . Using this relationship, show that the isothermal compressibility and isobaric expansion coefficient are related by (d/3/dP)T = —(di</dT)P.
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66
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
P3.28 Use the relation
Cp,m C
V,m
V
the cyclic rule, and the van der Waals equation of state to derive an equation for CP m - Cy m in terms of Vm, T, and the gas constants R , a , and b.
P3.29 For the equation of state Vm = RT/P + B(T), show that
(dCP,m\ _ d2B(T)
V dP )t ~~ ~T dT2
[Hint: Use Equation (3.44) and the property of state functions with respect to the order of differentiation in mixed second derivatives.]
P3.30 Starting with /3 = (1 /V)(dV/dT)P, show that p = — (1 /p)(dp/dT)P, where p is the density.
P3.31 This problem will give you practice in using the cyclic rule. Use the ideal gas law to obtain the three func¬ tions P = /(V, T), V = g(P9 T), and T = h(P , V). Show that the cyclic rule (dP/dV)T (dV/dT)P(dT/dP)v = ~ 1 is obeyed.
P3.32 Regard the enthalpy as a function of T and P. Use the cyclic rule to obtain the expression
Cr
= J8H\ /fd_T\
dP JT/ \dpj
' H
P3.33 Using the chain rule for differentiation, show that the isobaric expansion coefficient expressed in terms of density is given by (5 = —l/p(dp/dT)P.
P3.34 Derive the equation (dP/dV)T = —1/(kV) from basic equations and definitions.
P3.35 Derive the equation ( dH/dT)v = Cy + U/3//<:from basic equations and definitions.
dU\
P3.36 For an ideal gas, I — I and I — I =0. Prove
dV J7
dH\
dP / 7
that Cv and CP are independent of volume and pressure.
= J*L\ fd-C\
P3.37 Prove that C
P3.38 Show that
P3.39 Show that
v
dCy dV ( dCy
dV J
= T
Waals gas.
V dV
dT )L d2p dT2 Jv
= 0 for an ideal and for a van der
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Thermochemistry
I hermochemistry is the branch of thermodynamics that investigates the heat flow into or out of a reaction system and deduces the energy stored in chemical bonds. As reactants are converted into products, energy can either be taken up by the system or released to the surroundings. For a reaction that takes place at constant volume, the heat that flows to or out of the system is equal to A U for the reaction. For a reaction that takes place at constant pressure, the heat that flows to or out of the system is equal to AH for the reaction. The enthalpy of formation is defined as the heat flow into or out of the system in a reaction between pure elements that leads to the formation of 1 mol of product. Because H is a state function, the reaction enthalpy can be written as the enthalpies of for¬ mation of the products minus those of the reactants. This property allows AH and A U for a reaction to be calculated for many reactions without carrying out an experiment.
Energy Stored in Chemical Bonds Is Released or Taken Up in Chemical Reactions
Internal Energy and Enthalpy Changes Associated with Chemical Reactions
Hess's Law Is Based on Enthalpy Being a State Function
The Temperature Dependence of Reaction Enthalpies
The Experimental Determination of AU and AH for Chemical Reactions
Supplemental: Differential Scanning Calorimetry
Energy Stored in Chemical Bonds Is Released or Taken Up in Chemical Reactions
A significant amount of the internal energy or enthalpy of a molecule is stored in the form of chemical bonds. As reactants are transformed to products in a chemical reac¬ tion, energy can be released or taken up as bonds are made or broken, respectively. For example, consider a reaction in which N 2(g) and H2(g) dissociate into atoms, and the atoms recombine to form NH3(g). The enthalpy changes associated with individ¬ ual steps and with the overall reaction 1/2 N 2(g) + 3/2 H2(g) - > NH3(g) are
shown in Figure 4.1. Note that large enthalpy changes are associated with the individ¬ ual steps but the enthalpy change in the overall reaction is much smaller.
The change in enthalpy or internal energy resulting from chemical reactions appears in the surroundings in the form of a temperature increase or decrease resulting from heat flow and/or in the form of expansion or nonexpansion work. For example, the combustion of gasoline in an automobile engine can be used to do expansion work on the surroundings. Nonexpansion electrical work is possible if the chemical
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67
68 CHAPTER 4 Thermochemistry
FIGURE 4.1
Enthalpy changes are shown for individual steps in the overall reaction
m + 3H(gf)
1/2 N2(g) + 3/2 H2(g) - * NH3(g).
314 X 103J
NH(g) + 2H(g)
390 X 103J
1124 X 103J
NH2(g) + H(g)
466 X 103J
-45.9 X 103J
K N2 (g) + % H2(gr)
NH3(g)
reaction is carried out in an electrochemical cell. In Chapters 6 and 1 1, the extraction of nonexpansion work from chemical reactions will be discussed. In this chapter, the focus is on using measurements of heat flow to determine changes in V and H due to chemical reactions.
Internal Energy and Enthalpy Changes Associated with Chemical Reactions
In the previous chapters, we discussed how A U and A H are calculated from work and heat flow between the system and the surroundings for processes that do not involve phase changes or chemical reactions. In this section, this discussion is extended to reaction systems.
Imagine that a reaction involving a stoichiometric mixture of reactants (the system) is carried out in a constant pressure reaction vessel with diathermal walls immersed in a water bath (the surroundings). If the temperature of the water bath increases, heat has flowed from the system (the contents of the reaction vessel) to the surroundings (the water bath and the vessel). In this case, we say that the reaction is exothermic. If the temperature of the water bath decreases, the heat has flowed from the surroundings to the system, and we say that the reaction is endothermic.
Consider the reaction in Equation (4.1):
Fe30 4(s) + 4 H2(g)
3 Fe(s) + 4 H20(Z)
(4.1)
■>
Note that the phase (solid, liquid, or gas) for each reactant and product has been speci¬ fied because U and H are different for each phase. This reaction will only proceed at a measurable rate at elevated temperatures. However, as we show later, it is useful to tab¬ ulate values for AH for reactions at a pressure of 1 bar and a specified temperature, generally 298.15 K. The pressure value of 1 bar defines a standard state, and changes in H and U at the standard pressure of 1 bar are indicated by a superscript ° as in A H° and A U°. The standard state for gases is a hypothetical state in which the gas behaves ideally at a pressure of 1 bar. For most gases, deviations from ideal behavior
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4.2 INTERNAL ENERGY AND ENTHALPY CHANGES ASSOCIATED WITH CHEMICAL REACTIONS
are very small. The enthalpy of reaction, A HR, at specific values of T and P is defined as the heat exchanged between the system and the surroundings as the reactants are transformed into products at conditions of constant T and P. By convention, heat flow¬ ing into the system is given a positive sign. A HR is, therefore, a negative quantity for an exothermic reaction and a positive quantity for an endothermic reaction. The standard enthalpy of reaction, A H°R, refers to one mole of the specified reaction at a pressure of 1 bar, and unless indicated otherwise, to T = 298.15 K.
How can the reaction enthalpy and internal energy be determined? We proceed in the following way. The reaction is carried out at 1 bar pressure, and the tempera¬ ture change A T that occurs in a finite size water bath, initially at 298.15 K, is meas¬ ured. The water bath is large enough that AT is small. If AT is negative as a result of the reaction, the bath is heated to return it, the reaction vessel, and the system to 298.15 K using an electrical heater. By doing so, we ensure that the initial and final states are the same and therefore the measured AH is equal to A Hr. The electrical work done on the heater that restores the temperature of the water bath and the system to 298.15 K is equal to A H°R. If the temperature of the water bath increases as a result of the reaction, the electrical work done on a heater in the water bath at 298.15 K that increases its temperature and that of the system by AT in a separate experiment is measured. In this case, A H°R is equal to the negative of the electrical work done on the heater.
Although an experimental method for determining A H°R has been described, to tabulate the reaction enthalpies for all possible chemical reactions would be a monu¬ mental undertaking. Fortunately, A H°R can be calculated from tabulated enthalpy values for individual reactants and products. This is advantageous because there are far fewer reactants and products than there are reactions among them. Consider A H°R for the reaction of Equation (4.1) at T = 298.15 K and P = 1 bar. These values for P and T are chosen because thermodynamic values are tabulated for these values. However, A HR at other values of P and T can be calculated as discussed in Chapters 2 and 3. In principle, we could express A H°R in terms of the individual enthalpies of reactants and products:
A /_/ O _ 11 o _ 770
LAI 1 R n products n reactants
= 3H°m(Fe,s) + 4H°m(H20,I) - H°m(Fe304,s) - 4H°m(H2,g) (4.2)
The m subscripts refer to molar quantities. Although Equation (4.2) is correct, it does not provide a useful way to calculate A H%. There is no experimental way to determine the absolute enthalpy for any element or compound because there is no unique refer¬ ence zero against which individual enthalpies can be measured. Only AH and A U, as opposed to H and U, can be determined in an experiment.
Equation (4.2) can be transformed into a more useful form by introducing the enthalpy of formation. The standard enthalpy of formation, A H°f, is defined as the enthalpy change of the reaction in which the only reaction product is 1 mol of the species of interest, and only pure elements in their most stable state of aggregation under the stan¬ dard state conditions appear as reactants. We refer to these species as being in their standard reference state. For example, the standard reference state of water and carbon at 298.15 K are H20(/) and solid carbon in the form of graphite. Note that with this defi¬ nition, A Hf = 0 for an element in its standard reference state because the reactants and products are identical.
We next illustrate how reaction enthalpies can be expressed in terms of formation enthalpies. The only compounds that are produced or consumed in the reaction Fe30 4(s) + 4 H2(g) - > 3 Fe(s) + 4 H20(Z) are Fe304(s) and H20(Z). All ele¬
ments that appear in the reaction are in their standard reference states. The formation reactions for the compounds at 298.15 K and 1 bar are
H2(£) + ^02(g) - > H20(Z)
A H% = A H°f (H20, /) = H°m( H20, /) - H°m( H2, g) - | H°m( 02, g) (4.3)
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70
CHAPTER 4 Thermochemistry
3 Fe(s) + 2 02(g) - > Fe304(s)
A H°r
paA + pbB IpqC + vd D
-vA\H°f ^-vbAHI b
vc AH^ q + vD A H°f D
Elements in standard reference state
FIGURE 4.2
Equation (4.8) follows from the fact that A H for both paths is the same because they connect the same initial and final states.
A H°r = A//y(Fe304, s) = H°m( Fe304, s) ~ 3H^(Fe, s) ~ 2H°m(02 , g) (4.4)
If Equation (4.2) is rewritten in terms of the enthalpies of formation, a simple equation for the reaction enthalpy is obtained:
A H°r = 4A//y(H20, /) - A//y(Fe304, s)
(4.5)
Note that elements in their standard reference state do not appear in this equation because A Hf = 0 for these species. This result can be generalized to any chemical transformation
(4.6)
Vp^A + + . . . - > PXX + J'yY + . . .
which we write in the form
o =
(4.7)
The X( refer to all species that appear in the overall equation. The unitless stoichiometric coefficients V{ are positive for products and negative for reactants. The enthalpy change associated with this reaction is
(4.8)
The rationale behind Equation (4.8) can also be depicted as shown in Figure 4.2. Two paths are considered between the reactants A and B and the products C and D in the reaction vaA +
vbB - > vcC + vdD. The first of these is a direct path for which A H° = A H% In the
second path, A and B are first broken down into their elements, each in its standard refer¬ ence state. Subsequently, the elements are combined to form C and D. The enthalpy change for the second route is A H% = 'LivlAH°fproducts - A,\v,\AHfreactants = 'LvlAH°f i. Because H is a state function, the enthalpy change is the same for both paths. This is stated in mathematical form in Equation (4.8).
Writing A H°R in terms of formation enthalpies is a great simplification over compil¬ ing measured values of reaction enthalpies. Standard formation enthalpies for atoms and inorganic compounds at 298.15 K are listed in Table 4.1, and standard formation enthalpies for organic compounds are listed in Table 4.2 (Appendix B, Data Tables).
Another thermochemical convention is introduced at this point in order to calculate enthalpy changes involving electrolyte solutions. The solution reaction that occurs when a salt such as NaCl is dissolved in water is
NaCl(s) - » Na +{aq) + C \~{aq)
Because it is not possible to form only positive or negative ions in solution, the measured enthalpy of solution of an electrolyte is the sum of the enthalpies of all anions and cations formed. To be able to tabulate values for enthalpies of formation of individual ions, the enthalpy for the following reaction is set equal to zero at P = 1 bar for all temperatures:
1/2 H2(g) - > H +{aq) + e_ (metal electrode)
In other words, solution enthalpies of formation of ions are measured relative to that for H +{aq). The thermodynamics of electrolyte solutions will be discussed in detail in Chapter 10.
As the previous discussion shows, only the A Hf of each reactant and product is needed to calculate A H°R. Each A Hf is a difference in enthalpy between the compound and its con¬ stituent elements, rather than an absolute enthalpy. However, there is a convention that allows absolute enthalpies to be specified using the experimentally determined values of the A Hf of compounds. In this convention, the absolute enthalpy of each pure element in its standard reference state is set equal to zero. With this convention, the absolute molar enthalpy of any chemical species in its standard reference state H°m is equal to A Hf for that species. To demonstrate this convention, the reaction in Equation (4.4) is considered:
A H°r = A//y(Fe304, s) = H°m( Ee304, s) - 3iC(Fe, s) ~ 2H°m(02, g ) (4.4)
A H°r
vpA + pbB |vcC + vqD
-vAAH°f b
vc A c +pd A ~Tft □
Elements in standard reference state
FIGURE 4.2
Equation (4.8) follows from the fact that A H for both paths is the same because they connect the same initial and final states.
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4.3 HESS'S LAW IS BASED ON ENTHALPY BEING A STATE FUNCTION
71
Setting H^ = 0 for each element in its standard reference state,
AH}(Fq304,s) = H°m(Fe304,s) -3X0-2X0= H°m(Fe304,s) (4.9)
The value of A H°R for any reaction involving compounds and elements is unchanged by this convention. In fact, one could choose a different number for the absolute enthalpy of each pure element in its standard reference state, and it would still not change the value of A H°R. However, it is much more convenient (and easier to remember) if one sets H°m = 0 for all elements in their standard reference state. This convention will be used again in Chapter 6 when the chemical potential is discussed.
Hess's Law Is Based on Enthalpy Being a State Function
As discussed in the previous section, it is extremely useful to have tabulated values of A Hf for chemical compounds at one fixed combination of P and T. Tables 4.1 and 4.2 list this data for 1 bar and 298.15 K. With access to these values of A Hf, A H°R can be calculated for all reactions among these elements and compounds at 1 bar and 298.15 K.
But how is A Hf determined? Consider the formation reaction for C2H 5(g):
2 C (graphite) + 3 H2(g) - * C2H6(g) (4.10)
Graphite is the standard reference state for carbon at 298.15 K and 1 bar because it is slightly more stable than diamond under these conditions. However, it is unlikely that one would obtain only ethane if the reaction were carried out as written. Given this experimental hindrance, how can A Hf for ethane be determined? To determine A Hf for ethane, we take advantage of the fact that A H is path independent. In this context, path independence means that the enthalpy change for any sequence of reactions that sum to the same overall reaction is identical. This statement is known as Hess’s law. Therefore, one is free to choose any sequence of reactions that leads to the desired outcome. Combustion reactions are well suited for these purposes because in general they proceed rapidly, go to completion, and produce only a few products. To determine A Hf for ethane, one can carry out the following com¬ bustion reactions:
C2H6(g) + 7/2 02(g) •
- * 2 C02(g) + 3H20(Z)
\HJ
(4.11)
C (graphite) + 02(g) -
— > co2(g)
A H°u
(4.12)
H2(g) + 1/2 02(g)
-* H20(Z)
A H°IU
(4.13)
These reactions are combined in the following way to obtain the desired reaction:
2 X [C {graphite) + 02(g) - > C02(g)] 2 A H°n (4.14)
2 C02(g) + 3 H20(Z) - > C2H6(g) + 7/2 02(g) —AH] (4.15)
3 X [H2(g) + 1/2 02(g) - > H2Q(/)] _ 3A H]u _ (4.16)
2C {graphite) + 3H2(g) - > C2H6(g) 2A H°n - AH] + ?>AH°m
We emphasize again that it is not necessary for these reactions to be carried out at 298.15 K. The reaction vessel is immersed in a water bath at 298.15 K and the com¬ bustion reaction is initiated. If the temperature in the vessel rises during the course of the reaction, the heat flow that restores the system and surroundings to 298.15 K after completion of the reaction is measured, allowing A H°R to be determined at 298.15 K.
Several points should be made about enthalpy changes in relation to balanced over¬ all equations describing chemical reactions. First, because H is an extensive function, multiplying all stoichiometric coefficients with any number changes A H°R by the same factor. Therefore, it is important to know which set of stoichiometric coefficients has
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been assumed if a numerical value of A H°R is given. Second, because the units of A Hj for all compounds in the reaction are kJ mol-1, the units of the reaction enthalpy A H°R are also kJ mol-1. One might pose the question “per mole of what?” given that all the stoichiometric coefficients may differ from each other and from one. The answer to this question is per mole of the reaction as written. Doubling all the stoichiometric coeffi¬ cients doubles A H°R.
EXAMPLE PROBLEM 4.1
The average bond enthalpy of the O — H bond in water is defined as one-half of the
enthalpy change for the reaction H20(g) - > 2H(g) + 0(g). The formation
enthalpies, A Hf, for H(g) and 0(g) are 218.0 and 249.2 kJ mol-1, respectively, at 298.15 K, and A Hf for H20(g) is -241.8 kJ mol-1 at the same temperature.
a. Use this information to determine the average bond enthalpy of the O — H bond in water at 298.15 K.
b. Determine the average bond energy AU of the O — H bond in water at 298.15 K. Assume ideal gas behavior.
Solution
a. We consider the sequence
H20(g) - > H2(g) + 1/2 02(g) A H° = 241.8 kJ mol-1
H2(g) - > 2 H(g) A H° = 2 X 218.0 kJ mol"1
1 /2 02(g) - * 0(g) AH° = 249.2 kJ mol"1
H20(g) - > 2 H(g) + 0(g) A H° = 927.0 kJ mol"1
This is the enthalpy change associated with breaking both O — H bonds under standard conditions. We conclude that the average bond enthalpy of the O — H bond in water is
— X 927.0 kJ mol-1 = 463.5 kJ mol- 1 . We emphasize that this is the average value
because the values of A H for the transformations H20(g) - > H(g) + OH(g) and
OH(g) - > 0(g) + H(g) differ.
b. A U° = A H° - A(PV) = A H° - AnRT
= 927.0 kJ mol-1 - 2 X 8.314 J mol-1K-1 X 298.15 K = 922.0 kJmol-1
The average value for A U° for the O — H bond in water is ^ X 922.0 kJmol-1 = 461 .0 kJmol-1 . The bond energy and the bond enthalpy are nearly identical.
Example Problem 4.1 shows how bond energies can be calculated from reaction enthalpies. The value of a bond energy is of particular importance for chemists in esti¬ mating the thermal stability of a compound as well as its stability with respect to reac¬ tions with other molecules. Values of bond energies tabulated in the format of the periodic table together with the electronegativities are shown in Table 4.3 [Kildahl, N. K. “Bond Energy Data Summarized.” Journal of Chemical Education. 72 (1995): 423]. The value of the single bond energy, AUa_b> for a combination A-B not listed in the table can be estimated using the empirical relationship due to Linus Pauling in Equation (4.17):
A£/A-b = VA Ua.a X AE/B-b + 96.48(*a - *b)2 (4.17)
where \a and Xb are the electronegativities of atoms A and B.
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4.4 THE TEMPERATURE DEPENDENCE OF REACTION ENTHALPIES
73
The Temperature Dependence of Reaction Enthalpies
Suppose that we plan to carry out a reaction that is mildly exothermic at 298.15 K at another temperature. Is the reaction endothermic or exothermic at the second tempera¬ ture? To answer this question, it is necessary to determine A H°R at the second tempera¬ ture. We assume that no phase changes occur in the temperature interval of interest. The enthalpy for each reactant and product at temperature T is related to the value at 298.15 K by Equation (4.18), which accounts for the energy supplied in order to heat the substance to the new temperature at constant pressure:
T
H°t = #298.15 k + J Cp(T')dT' (4.18)
298.15 K
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The prime in the integral indicates a “dummy variable” that is otherwise identical to the temperature. This notation is needed because T appears in the upper limit of the inte¬ gral. In Equation (4.18), #298.15 k is the absolute enthalpy at 1 bar and 298.15 K. However, because there are no unique values for absolute enthalpies, it is useful to com¬ bine similar equations for all reactants and products with the appropriate stoichiometic coefficients to obtain the following equation for the reaction enthalpy at temperature T:
T
A H°rj = A H°R'29SA5K + j A CP(T')dT' (4.19)
298.15 K
where
A Cp(T') = ]>>,Cp/n (4.20)
i
Recall that in our notation, A H°R or A Hf without an explicit temperature value implies that T = 298.15 K. In Equation (4.20), the sum is over all reactants and products, including both elements and compounds. A calculation of A H°R at an elevated tempera¬ ture is shown in Example Problem 4.2.
[ EXA
EXAMPLE PROBLEM 4.2
Calculate A H°R 145o k for the reaction 1/2 H2(g) + l/2Cl2(g) - *■ HCl(g) and
1 bar pressure given that A///(HCl,g) = —92.3 kJ mol-1 at 298.15 K and that
Cp,m(H 2>g) —
cPj.ci2,g) -
CP)m(HCl,g) =
/ j T2
29.064 - 0.8363 X 10“3 — + 20.111 X 10“7^
V K K2
(3I.695 + 10.143 X 10“3- - 40.373 X 10“7^r
V K K2.
f 28.165 + 1.809 X 10“3- + 15.464 X 10“7^)
V K K2/
^JKT'moF1
K_1mol_1
JK^'moF1
over this temperature range. The ratios T /K and T2/ K2 appear in these equations in order to have the right units for the heat capacity.
Solution
1450
ACpCO
A#)?,1450K = A#^298.15K + J A CP(T)dT
298.15
T T2
28.165 + 1.809 X 10-3— + 15.464 X 10-7— r
K
K
- -( 29.064 - 0.8363 X 10-3- + 20.111 X 10~7^
K K2
1
r-2 \
-- 31.695 + 10.143 X 10"3 - 40.373 X 10"7— r
K
K-
JK_1mor
= -2.215 - 2.844 X 10“3- + 25.595 X 10“7— r Jr'mof1
K
K
A#^i45o^ = -92. 3 kJ mol 1
1450
298.15
-2.215 - 2.844 X 10-3— + 25.595 X 10-7 — r ) X d — J mol
K
K"
K
= — 92.3 kJ mol-1 - 2.836kJmol_1 = -95.1kJmoF1
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4.5 THE EXPERIMENTAL DETERMINATION OF AD AND A H FOR CHEMICAL REACTIONS
75
In this particular case, the change in the reaction enthalpy with T is not large. This is I the case because A CP(T) is small and not because an individual CPi(T) is small. J
The Experimental Determination of A U and A H for Chemical Reactions
For chemical reactions, A U and A H are generally determined through experiment. In this section, we discuss how these experiments are carried out. If some or all of the reactants or products are volatile, it is necessary to contain the reaction mixture for which A U and A H are being measured. Such an experiment can be carried out in a bomb calorimeter, shown schematically in Figure 4.3. In a bomb calorimeter, the reaction is carried out at constant volume. The motivation for doing so is that if dV = 0, At/ = qy. Therefore, a measure¬ ment of the heat flow normalized to 1 mole of the specified reaction provides a direct measurement of A Ur. Bomb calorimetry is restricted to reaction mixtures containing gases because it is impractical to carry out chemical reactions at constant volume for systems consisting solely of liquids and solids, as shown in Example Problem 3.2. In the following, we describe how A U r and A HR are determined for an experiment in which a single liquid or solid reactant undergoes combustion in an excess of 02(g).
The bomb calorimeter is a good illustration of how one can define the system and surroundings to simplify the analysis of an experiment. The system is defined as the contents of a stainless steel thick-walled pressure vessel, the pressure vessel itself, and the inner water bath. Given this definition of the system, the surroundings consist of the container holding the inner water bath, the outer water bath, and the rest of the uni¬ verse. The outer water bath encloses the inner bath and, through a heating coil, its tem¬ perature is always held at the temperature of the inner bath. Therefore, no heat flow will occur between the system and surroundings, and q = 0. Because the combustion experiment takes place at constant volume, w = 0. Therefore, A£7 = 0. These conditions describe an isolated system of finite size that is not coupled to the rest of the universe. We are only interested in one part of this system, namely, the reaction mixture.
What are the individual components that make up AU1 Consider the system as consisting of three subsystems: the reactants in the calorimeter, the calorimeter vessel, and the inner water bath. These three subsystems are separated by rigid diathermal walls and are in thermal equilibrium. Energy is redistributed among the subsystems as reactants are converted to products, the temperature of the inner water bath changes, and the temperature of the calorimeter changes.
A ms A mH0
— At/ combustion X Cp,m( H20) X AT + Ccalorimeter x A T = 0(4.21)
Ms MHl0
In Equation (4.21), A T is the change in the temperature of the three subsystems. The mass of water in the inner bath, mHl0\ its molecular weight, MHiq\ its heat capacity, Cp m(H20); the mass of the sample, ms\ and its molecular weight, Ms, are known. At/ combustion is defined per mole of the combustion reaction, but because the reaction includes exactly 1 mole of reactant, the factor ms/Ms in Equation (4.21) is appropri¬ ate. We wish to measure A U combustion- However, to determine A U combustion, the heat capacity of the calorimeter, Ccaiorimeter , must first be determined by carrying out a reaction for which A U r is already known, as illustrated in Example Problem 4.3. To be more specific, we consider a combustion reaction between a compound and an excess of 02.
FIGURE 4.3
Schematic diagram of a bomb calorimeter. The liquid or solid reactant is placed in a cup suspended in the thick- walled steel bomb, which is filled with 02 gas. The vessel is immersed in an inner water bath, and its temperature is monitored. The diathermal container is immersed in an outer water bath (not shown) whose temperature is maintained at the same value as the inner bath through a heating coil. By doing so, there is no heat exchange between the inner water bath and the rest of the universe.
EXAMPLE PROBLEM 4.3
When 0.972 g of cyclohexane undergoes complete combustion in a bomb calorimeter, A T of the inner water bath is 2.98°C. For cyclohexane, A U combustion is -3913 kJ mol-1. Given this result, what is the value for A Ucombustion for the combustion of benzene if A T is 2.36°C when 0.857 g of benzene undergoes complete combustion in the same calorimeter? The mass of the water in the inner bath is 1.812 X 103 g, and the CPm of water is 75.3 J K-1 mol-1.
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76 CHAPTER 4 Thermochemistry
Solution
To calculate the calorimeter constant through the combustion of cyclohexane, we write Equation (4.21) in the following form:
C calorimeter
0.972g
ms .
~ww-
mH20
combustion ~ A/f ^
M
h2o
84. 16 g mol
-l
X 3913 XlCr Jmol-1 —
A T
o 1.812 X 10 g , ,
t - - V 7 ^ Q T m/'U t 1/ 1 N
1 8. 02 g mol
-X75.3 Jmol-1K-1X2.98°C
2.98°C
= 7.59 X 103J(°C)_1
In calculating A U(
combustion
for benzene, we use the value for Ccaiorimeter:
A U,
combustion
Ms( mH2o
= -zr(^7^cp,m(K20)kT + c , r V
ms\M
h2o 78.12gmoF1
calorimeter
at
X
0.857 g = -3.26 X 106 Jmol-1
1.812 X lCrg , .
- 1 X 75.3 J mol-1 K-1 X 2.36°C
18.02 g mol-1
+ 7.59 X 103J(°C)_1 X 2.36°C
J
Once kU combustion has been determined, combustion can be determined using the following equation:
FIGURE 4.4
Schematic diagram of a constant pressure calorimeter suitable for measuring the enthalpy of solution of a salt in water.
combustion combustion A (PE) (4.22)
For reactions involving only solids and liquids, A U » A (PV) and A H ~ A U. If some of the reactants or products are gases, the small change in the temperature that is measured in a calorimetric experiment can generally be ignored and A (PV) = MnRT) = A nRT
combustion combustion AnRT (4.23)
where An is the change in the number of moles of gas in the overall reaction. For the first reaction of Example Problem 4.3,
C6H12(/) + 9 02(g) - > 6 C02(g) + 6 H20(Z) (4.24)
and An = -3. Note that at T= 298.15 K, the most stable form of cyclohexane and water
is a liquid.
AH combustion combustion 3 RT 3913 X 10 kJmol
-3 X 8.314 J K-1mol-1 X 298.15 K = -3920 X 103 Jmol-1 (4.25)
For this reaction, A Hcombustion and A Ucombustion differ by only 0.2%. Note that because the contents of the bomb calorimeter are not at 1 bar pressure, AJJ comhustion rather than AU combustion *s measured. The difference is small, but it can be calculated.
If the reaction under study does not involve gases or highly volatile liquids, there is no need to operate under constant volume conditions. It is preferable to carry out the reaction at constant P using a constant pressure calorimeter. AIT is directly deter¬ mined because AH = qP. A vacuum-insulated vessel with a loosely fitting stopper as shown in Figure 4.4 is adequate for many purposes and can be treated as an isolated composite system. Equation (4.21) takes the following form for constant pressure calorimetry involving the solution of a salt in water:
mKnjt mH2o
AH° = -dissolution + tX CPim{ H20) Ar + CcalorimeterAT = 0 (4.26)
Msalt MH20
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4.6 DIFFERENTIAL SCANNING CALORIMETRY 77
AH solution is defined per mole of the solution reaction, but because the reaction includes exactly 1 mole of reactant, the factor msait/Msait in Equation (4.21) is appro¬ priate. Because A(PV) is negligibly small for the solution of a salt in a solvent, AU solution = AH solution- The solution must be stirred to ensure that equilibrium is attained before AT is measured.
EXAMPLE PROBLEM 4.4
The enthalpy of solution for the reaction
Na2S04(.?) H2°(;) > 2Na +(aq) + SO j~{aq)
is determined in a constant pressure calorimeter. The calorimeter constant was deter¬ mined to be 342.5 J K-1. When 1.423 g of Na2S04 is dissolved in 100.34 g of H20(/), AT = 0.031 K. Calculate AH°soiution for Na2S04 from these data. Compare your result with that calculated using the standard enthalpies of formation in Table 4.1 (Appendix B, Data Tables) and in Chapter 10 in Table 10.1.
Solution
A HI
solution
Mjahf
msalt\M Hl0
cV,m(H2o)Ar + c
I
AT
100.34 g
142.04 gmoF1
calorimeter
X 75.3 J K-1 mol-1 X 0.031 K
X
18.02 g mol 1 + 342.5 JK"1 X 0.031 K
1.423 g = -2.4 X 103 JmoF1
We next calculate A H°soiution using the data tables.
solution = 2AHf(Na+,aq) + A H °f(S024~ ,aq) - A//?(Na2S04,s)
= 2 X (-240.1 kJmoF1) - 909.3k.lmoF1 + 1387.1 kJmoF1 = —2. 4 kJmoF1
The agreement between the calculated and experimental results is good.
SUPPLEMENTAL
Differential Scanning Calorimetry
Differential scanning calorimetry (DSC) is a form of constant pressure calorimetry that is well suited to routine laboratory tests in pharmaceutical and material sciences. It is also used to study chemical changes such as polymer cross-linking, melting, and unfolding of protein molecules in which heat is absorbed or released in the transition. The experimen¬ tal apparatus for such an experiment is shown schematically in Figure 4.5. The word differential appears in the name of the technique because the uptake of heat is measured relative to that for a reference material, and scanning refers to the fact that the tempera¬ ture of the sample is varied, usually linearly with time.
The temperature of the enclosure TE is increased linearly with time using a power supply. Heat flows from the enclosure through the disk to the sample because of the tem¬ perature gradient generated by the heater. Because the sample and reference are equidis¬ tant from the enclosure, the heat flow to each sample is the same. The reference material is chosen such that its melting point is not in the range of that of the samples.
We consider a simplified one-dimensional model of the heat flow in the DSC in Figure 4.6 following the treatment of Hohne, Hemminger, and Flammersheim in Differential Scanning Calorimetry , 2nd Edition, Berlin: Springer, 2003. The electrical current through the resistive heater increases the temperature of the calorimeter enclosure TE to a value greater than that of the sample and reference, Ts, and TR. The
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78 CHAPTER 4 Thermochemistry
FIGURE 4.5
A heat flux differential scanning calorimeter consists of an insulated massive enclosure and lid that are heated to the temperature TE using a resistive heater. A support disk in good thermal contact with the enclosure supports the sample and reference materials. The temperatures of the sample and reference are measured with a thermocouple. In practice, the reference is usually an empty sample pan.
FIGURE 4.6
A T is shown as a function of time for an exothermic process occuring in the sample. For this example, it is assumed that the heat capacities of the sample and reference are constant over the temperature range shown. The heat associated with the process on interest is proportional to the blue area. The green area arises from the difference in heat capacities of sample and reference. The zero line is obtained without material in the crucibles.
Lid
Power supply
heat flow per unit time from the enclosure to the sample and reference are designated by <&Es and respectively, which typically have the units of J g_1 s~l.
Assume that a process such as melting occurs in the sample and not in the refer¬ ence. The heat flow per unit time associated with the process is given by 0(0- It is time dependent because, using melting as an example, heat flow associated with the phase change begins at the onset of melting and ceases when the sample is completely in the liquid state. This additional heat flow changes the sample temperature by dTs and consequently both TE — Ts and the heat flow rate to the sample <&Es change. In the experiment A T(t) = T s(t) — T R(t) is measured. The change in the heat flow to the sample resulting from the process is
dT s (t)
Cs ^ = ®Es(t) -<&(*) (4.27)
where Cs is the constant pressure heat capacity of the sample. Equation (4.27) relates the sample temperature to the heat flow generated by the process of interest. If the process is exothermic, 0(0 < 0 and if it is endothermic, 0(0 > 0. We rewrite Equation (4.27) to explicitly contain the experimentally accessible function A 7)7).
dTR(t) dATjt) S dt S dt
0(0
(4.28)
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4.6 DIFFERENTIAL SCANNING CALORIMETRY
79
The corresponding equation for the reference in which only heating occurs is
dTR (?)
Cr — — — = ®ER(t) (4.29)
The essence of DSC is that a difference between the heat flux to the sample and reference is measured. We therefore subtract Equation (4.29) from Equation (4.28) and obtain
dTR(t) dAT(t)
®Es(t) ~ ® er (0 = (Cs ~ Cr) ^ ^ Cs — f <5(0 (4.30)
The quantities <$>ES and <&ER are directly proportional to the temperature differences Te — Ts and TE — TR and are inversely proportional to the thermal resistance Rthermal between the heated enclosure and the sample and reference. The latter are equal because of the symmetry of the calorimeter. The thermal resistance Rthermal is an intrinsic prop¬ erty of the calorimeter and can be obtained through calibration of the instrument.
^£5 -
Tv - T c
R
and <&ER =
Tf ~ Tn
thermal
R
thermal
Therefore,
*(t)
AT(t)
R,
thermal
{CS ~ CR)
dTR (?) dt
C<
dAT(t)
dt
(4.31)
(4.32)
Equation (4.32) links the heat flow per unit mass generated by the process of interest, <E>(f), to the measured quantity A T(t). Typically TR is increased linearly with time,
TR(t) = T0 + at
and Equation (4.32) simplifies to
$(?)
A T{t)
R
thermal
a(CS ~ Cr) ~ CS
dAT(t)
dt
(4.33)
We see that A T(t) is not simply related to <l>(r), which is to be determined in the experiment. The second term in Equation (4.33) arises because the heat capacities of the sample and reference are not equal. Additionally, the heat capacity of the sample changes as it undergoes a phase change.
The third term in Equation (4.33) is proportional to d AT (t)/dt and has the effect of broadening A T(t) relative to <[>(/) and shifting it to longer times. A schematic picture of a DSC scan in this model is shown in Figure 4.6.
Because CRa = ®ER(t) from Equation (4.29) and AT are proportional as shown in Equation (4.31), a graph of CR versus time has the same shape as shown in Figure 4.6. An analogous equation can be written for the sample, so that
®es ~ ®er ~ a(Cs - CR)
Therefore, a graph of (Cs ~ Cr) versus time also has the same shape as Figure 4.6.
The goal of the experiment is to determine the heat absorbed or evolved in the process per unit mass, which is given by
rh
qP = AH = / $(t)dt Jt\
R
thermal J 1 1
A T(t)dt - [ (~a(cs - CR))dt
" thermal Jt\
(4.34)
In Equation (4.34), the baseline contribution to the integral has been subtracted as it has no relevance for the process of interest.
Interpreting DSC curves in terms of heat capacities must be done with caution as illustrated for a melting transition in Figure 4.7. At the melting temperature, the enthalpy rises abruptly as discussed in Chapter 8 and as shown in Figure 4.7a. The heat capacity is the derivative of the enthalpy with respect to temperature and has the form shown in Figure 4.7b. As discussed in Section 2.5, the heat capacity of the liquid is higher than that of the solid. However, a measured DSC curve has the shape shown in Figure 4.7d rather than in Figure 4.7b.
There are two reasons for this discrepancy. Heat is taken up by the sample during the melting transition, but the sample temperature does not change. However, the
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(b) Ttrue
FIGURE 4.7
(a) AH is shown as a function of the true temperature for a melting process.
(b) The heat capacity becomes infinite at the transition temperature because the temperature remains constant as heat flows into the sample, (c) The heat capac¬ ity curve in (b) is distorted because the measured temperature increases while the sample temperature remains constant dur¬ ing the melting transition, (d) Further dis¬ tortions to the hypothetical scan (c) and a shift of the peak to higher temperature occur because of the thermal inertia of the calorimeter.
80 CHAPTER 4 Thermochemistry
"7~i 7m T2
FIGURE 4.8
The apparent heat capacity of a protein undergoing a reversible, thermal denatura- tion within the temperature interval T\~T2 is shown. The temperature at the heat capacity maximum is to a good approxima¬ tion the melting temperature Tm, defined as the temperature at which half the total pro¬ tein has been denatured. The excess heat capacity A CP = CP(T ) - Cp(T)is com¬ posed of two parts; the intrinsic excess heat capacity SC'p[ and the transition excess heat capacity 8CPS. See the text.
sample crucible temperature continues to increase as heat flows to the sample. Therefore the apparent heat capacity would have the shape shown in Figure 4.7c if there were no resistances to heat flow in the calorimeter. The thermal resistances pres¬ ent in the calorimeter smear out the curve shown in Figure 4.7c and shift it to higher temperatures as shown in Figure 4.7d. Clearly, the temperature dependence of the apparent heat capacity obtained directly from a DSC scan is significantly different from the temperature dependence of the true heat capacity. Deconvolution procedures must be undertaken to obtain accurate heat capacities. Because the heat absorbed or evolved in the process per unit mass is proportional to the area under the DSC scan, it is less affected by the instrument distortions than the heat capacity.
DSC is the most direct method for determining the energetics of biological macromol¬ ecules undergoing conformational transitions, which is important in understanding their biological activity. In particular, DSC has been used to determine the temperature range over which proteins undergo the conformational changes associated with reversible denaturation, a process in which a protein unfolds. The considerations for a melting process discussed earlier apply as well as for denaturation, although melting enthalpies are generally much larger than denaturation enthalpies. From the preceding discussion, when a protein solution is heated at a constant rate and at constant pressure, DSC reports the apparent heat capacity of the protein solution. Protein structures are stabilized by the coop¬ eration of numerous weak forces. Assume that a protein solution is heated from a tempera¬ ture T\ to a temperature T2. If the protein denatures within this temperature interval reversibly and cooperatively, the CP(T) versus T curve has the form shown in Figure 4.8.
The heat capacity of denaturation is defined as
A CdPen = C$(T) - Cp(T) (4.35)
which is the difference between the heat capacity of the denatured protein, CP{T ), and the heat capacity of the native (i.e., structured) protein, CP(T). The value of A CPen can be determined as shown in Figure 4.8. The value of A CPen is a positive quantity and for many globular proteins varies from 0.3 to 0.7 J K-1 g-1. The higher heat capacity of the protein in the denatured state can be understood in the following way. In the structured state, internal motions of the protein are characterized by coupled bending and rotations of several bonds that occur at frequencies on the order of kBT / h, where kB is Boltzmann’s constant and h is Planck’s constant. When the protein denatures, these “soft” vibrations are shifted to lower frequencies, higher frequency bond vibrations are excited, and the heat capacity increases. In addition to increased contributions from vibrational motions, when a protein structure is disrupted, nonpolar amino acid residues formerly isolated from solvent within the interior of the protein are exposed to water. Water molecules now order about these nonpolar amino acids, further increasing the heat capacity.
The excess heat capacity is defined as
A CP = CP{T) - Cp(T) (4.36)
where ACp is obtained at a given temperature from the difference between the point on the heat capacity curve CP{T ) and the linearly extrapolated value for the heat capacity of the native state CP{T)\ see Figure 4.8. The excess heat capacity is in turn composed of two parts:
A CP(T) = 8Cf + 8Clfs (4.37)
Within the interval T\-T2 the protein structure does not unfold suddenly. There occurs instead a gradual unfolding of the protein such that at any given temperature a fraction of the structured protein remains, and an ensemble of unfolded species is produced. The heat capacity and enthalpy change observed arise from this ensemble of physical forms of the partially unfolded protein, and as such these properties are averages over possible configurations of the protein. The component of the heat capacity that accounts for the sum of molecular species produced in the course of making a transition from the folded to the unfolded state is the intrinsic excess heat capacity or 8ClP. The second com¬ ponent of the heat capacity is called the transition excess heat capacity 8CPS. The transition excess heat capacity results from fluctuations of the system as the protein changes from different states in the course of denaturation.
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CONCEPTUAL PROBLEMS
81
The intrinsic and transition excess heat capacities are determined by first extrapo¬ lating the functions Cp(T ) and Cp(T) into the transition zone between T\ and T2. This is indicated in Figure 4.8 by the solid line connecting the heat capacity baselines above and below Tm. Once this baseline extrapolation is accomplished, 8Clp is the difference at a given temperature between the extrapolated baseline curve and Cp{T). The value of 8Cps is obtained from the difference between CP{T ) and the extrapo¬ lated baseline curve. The excess enthalpy of thermal denaturation is finally given by T2
A Hden — f 8CpsdT. In other words the excess enthalpy of thermal denaturation is the
Ti
area under the peak in Figure 4.8 above the extrapolated baseline curve.
Vocabulary
bomb calorimeter bond energy bond enthalpy
constant pressure calorimeter denaturation
differential scanning calorimetry
endothermic enthalpy of fusion enthalpy of reaction excess heat capacity exothermic Hess’s law
intrinsic excess heat capacity standard enthalpy of formation standard enthalpy of reaction standard reference state standard state
transition excess heat capacity
Conceptual Problems
Q4.1 In calculating A H°R at 285.15 K, only the A Hf of the compounds that take part in the reactions listed in Tables 4.1 and 4.2 (Appendix B, Data Tables) are needed. Is this statement also true if you want to calculate A H°R at 500. K?
Q4.2 What is the point of having an outer water bath in a bomb calorimeter (see Figure 4.3), especially if its tempera¬ ture is always equal to that of the inner water bath?
Q4.3 Is the following statement correct? If not rewrite it so that it is correct. The standard state of water is H20(g).
Q4.4 Does the enthalpy of formation of H20(/) change if the absolute enthalpies of H2(g) and 02(g) are set equal to 100. kJ mol-1 rather than to zero? Answer the same question for C02(g). Will A H°R for the reaction H20(/) +
C02(g) - > H2C03(/) change as a result of this change in
the enthalpy of formation of the elements?
Q4.5 Why are elements included in the sum in Equation (4. 14) when they are not included in calculating A H°R at 298 K?
Q4.6 Why are heat capacities of reactants and products required for calculations of A H°R at elevated temperatures?
Q4.7 Is the following statement correct? If not rewrite it so that it is correct. The superscript zero in A Hf means that the reactions conditions are 298.15 K.
Q4.8 Why is it valid to add the enthalpies of any sequence of reactions to obtain the enthalpy of the reaction that is the sum of the individual reactions?
Q4.9 In a calorimetric study, the temperature of the system rises to 325 K before returning to its initial temperature of
298 K. Why does this temperature rise not affect your measurement of A H°R at 298 K?
Q4.10 Is the following statement correct? If not rewrite it so that it is correct. Because the reaction
H2(g) + 02(g) - > H20(Z) is exothermic, the products are
at a higher temperature than the reactants.
Q4.ll The reactants in the reaction 2NO(g) + 02(g) - >
2N02(g) are initially at 298 K. Why is the reaction enthalpy the same if (a) the reaction is constantly kept at 298 K or (b) if the reaction temperature is not controlled and the heat flow to the surroundings is measured only after the tempera¬ ture of the products is returned to 298 K?
Q4.12 What is the advantage of a differential scanning calorimeter over a bomb calorimeter in determining the enthalpy of fusion of a series of samples?
Q4.13 You wish to measure the heat of solution of NaCl in water. Would the calorimetric technique of choice be at constant pressure or constant volume? Why?
Q4.14 Is the following statement correct? If not rewrite it so that it is correct. Because the enthalpy of formation of elements is zero, A Hf (0(g)) = 0.
Q4.15 If the A Hf for the chemical compounds involved in a reaction is available at a given temperature, how can A H°R be calculated at another temperature?
Q4.16 Is the following statement correct? If not rewrite it so that it is correct. If A H°R for a chemical reaction does not change appreciably with temperature, the heat capacities for reactants and products must be small.
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82
CHAPTER 4 Thermochemistry
Q4.17 Under what conditions are A H and A U for a reac¬ tion involving gases and/or liquids or solids identical?
Q4.18 Dogs cool off in hot weather by panting.
Write a chemical equation to describe this process and calculate A H°R.
Q4.19 Is A H for breaking the first C — H bond in methane equal to the average C — H bond enthalpy in this molecule? Explain your answer.
Q4.20 Humans cool off through perspiration. How does the effectiveness of this process depend on the relative humidity?
Numerical Problems
Problem numbers in red indicate that the solution to the problem is given in the Student's Solutions Manual.
P4.1 Given the data in Table 4.1 (Appendix B, Data Tables) and the following information, calculate the single bond enthalpies and energies for Si-F, Si-Cl, C-F, N-F, O-F, H-F:
Substance SiF 4(g) SiCl 4(g) CF 4(g) NF 3(g) OF 2{g) HF(g)
AHf (kJ mol-1) -1614.9 -657.0 -925 -125 -22 -271
P4.2 At 1000. K, A H°r = -123.77 kJ mol-1 for the reaction
N2(g) + 3 H2(g) - > 2 NH3(g), with C.Pm = 3.502 R, 3.4667?,
and 4.2177? for N 2(g), H2(g), and NH3(g), respectively. Calculate AHf of NH3(g) at 450. K from this information. Assume that the heat capacities are independent of temperature.
P4.3 A sample of K (s) of mass 2.740 g undergoes combustion in a constant volume calorimeter at 298.15 K.
The calorimeter constant is 1849 J K_1, and the measured temperature rise in the inner water bath containing 1450. g of water is 1.60 K. Calculate A Uf and A Hf for K20.
P4.4 Calculate A77j for NO (g) at 975 K, assuming that the heat capacities of reactants and products are constant over the temperature interval at their values at 298.15 K.
P4.5 The total surface area of the earth covered by ocean is 3.35 X 108 km2. Carbon is fixed in the oceans via photosyn¬ thesis performed by marine plants according to the reaction
6 C02(g) + 6H20(Z) - > C6H1206(s) + 6 02(g).
A lower range estimate of the mass of carbon fixed in the oceans is 44.5 metric tons/km2. Calculate the annual enthalpy change resulting from photosynthetic carbon fixation in the ocean given earlier. Assume P = 1 bar and T= 298 K.
P4.6 Derive a formula for AHR(T) for the reaction CO(g) +
1/2 02(g) - > C02(g) assuming that the heat capacities of
reactants and products do not change with temperature.
P4.7 Given the data in Table 4.3 and the data tables, calcu¬ late the bond enthalpy and energy of the following:
a. The C — H bond in CH4
b. The C — C single bond in C2H6
c. The C=C double bond in C2H4
Use your result from part (a) to solve parts (b) and (c).
P4.8 Use the following data at 298.15 K to complete this problem:
AHr (kJ mol"1)
1/2 H2(g) + 1/2 02(g) - * OH (g) 38.95
H2(g) + 1/2 02(g) - * H20(g) -241.814
H2(g) - * 2 H(g) 435.994
02(g) - * 2 O(g) 498.34
Calculate A H°R for
a. OH(g) - > H(g) + O(g)
b. H20(g) - * 2 H(g) + O(g)
c. H20(g) - » H(g) + OH(g)
Assuming ideal gas behavior, calculate AHR and A U°R for all three reactions.
P4.9 Calculate the standard enthalpy of formation of FeS2(+) at 600.°C from the following data at 298.15 K. Assume that the heat capacities are independent of temperature.
Substance Fe(s) FeS2(s) Fe2(>3(s) S (rhombic) S02(g) AHf(k] mol-1) -824.2 -296.81
Cp'jR 3.02 7.48 2.72
You are also given that for the reaction 2 FeS2(s) +11/2 02(g) - > Fe203(s) + 4 S02(g), AHR = -1655 kJ mol-1.
P4.10 The following data are a DSC scan of a solution of a T4 lysozyme mutant. From the data determine Tm. Determine also the excess heat capacity A CP at T = 308 K. Determine also the intrinsic 8Clp and transition SCps excess heat capaci¬ ties at T = 308 K. In your calculations use the extrapolated curves, shown as dotted lines in the DSC scan.
Temperature/K
P4.ll At 298 K, &H°r = 131.28 kJ mol-1 for the reaction
C (graphite) + H20(g) - > CO(g) + H 2(g), with CP, m = 8.53,
33.58, 29.12, and 28.82 J K-1 mol-1 for graphite, H20(g), CO(g), and H2(g), respectively. Calculate AH°R at 240°C from this information. Assume that the heat capacities are inde¬ pendent of temperature.
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NUMERICAL PROBLEMS 83
P4.12 Consider the reaction TiO20) + 2 C(graphite) +
2 Cl2(s) - > 2 CO (g) + TiCl4(/) for which A^298K =
—80. kJ mol-1. Given the following data at 25°C, (a) calculate A H°r at 135. 8°C, the boiling point of TiCl4, and (b) calculate A H°f for TiCl4(/) at 25°C:
Substance Ti02(s) Cl2(g) C(graphite) CO(g) TiCl4(/)
kH°f (kJ mol-1) -945 -110.5
Cp m(J K_1 mol-1) 55.06 33.91 8.53 29.12 145.2
Assume that the heat capacities are independent of temperature.
P4.13 Calculate A HR and MJ°R for the oxidation of benzene. Also calculate
A H\ ~ AU%
A H°r
P4.14 Several reactions and their standard reaction enthalpies at 298.15 K are given here:
A HI (kJ mol"1)
CaC2(i) + 2 H20(Z) - » Ca(OH)2(i) + C2H2(g) -127.9
Ca(.s) + 1/2 02(g) - » CaO(s) -635.1
CaO(s) + H20(Z) - > Ca(OH )2(s) -65.2
The standard enthalpies of combustion of graphite and C2H2(g) are -393.51 and -1299.58 kJ mol-1, respectively. Calculate the standard enthalpy of formation of CaC20) at 25 °C.
P4.15 Benzoic acid, 1.35 g, is reacted with oxygen in a constant volume calorimeter to form H20(/) and C02(g) at 298 K. The mass of the water in the inner bath is 1.55 X 103 g. The temperature of the calorimeter and its contents rises 2.76 K as a result of this reaction. Calculate the calorimeter constant.
P4.16 The total surface area of Asia consisting of forest, cultivated land, grass land, and desert is 4.46 X 107 km2. Every year, the mass of carbon fixed by photosynthesis by vegetation covering this land surface according to the reac¬ tion 6 C02(g) + 6H20(/) - > C^Hj^O^s) + 6 02(g) is
about 455. X 103 kg km-2. Calculate the annual enthalpy change resulting from photosynthetic carbon fixation over the land surface given earlier. Assume P = 1 bar and T = 298 K.
P4.17 Calculate A H°R and A U°R at 298. 15 K for the follow¬ ing reactions:
a. 4NH3(g) + 6 NO(g) - * 5N2(g) + 6H20(g)
b. 2 NO(g) + 02(g) - » 2 N02(g)
c. TiCl4(Z) + 2 H20(Z) - » Ti02(s) + 4HCl(g)
d. 2NaOH(ag) + H2S04(ag) - * Na2S04(ag) + 2 H20(Z)
Assume complete dissociation of NaOH, H2S04, and Na2S04
e. CH4(g) + H20(g) - > CO(g) + 3 H2(g)
f. CH3OH(g) + CO(g) - > CH3COOH(/)
P4.18 A sample of Na2S04(s) is dissolved in 225 g of water at 298 K such that the solution is 0.325 molar in Na2S04.
A temperature rise of 0.146°C is observed. The calorimeter
constant is 330. J K-1. Calculate the enthalpy of solution of Na2S04 in water at this concentration. Compare your result with that calculated using the data in Table 4.1 (Appendix B, Data Tables).
P4.19 Nitrogen is a vital component of proteins and nucleic acids, and thus is necessary for life. The atmosphere is com¬ posed of roughly 80% N2, but most organisms cannot directly utilize N2 for biosynthesis. Bacteria capable of “fixing” nitro¬ gen (i.e., converting N2 to a chemical form, such as NH3, which can be utilized in the biosynthesis of proteins and nucleic acids) are called diazotrophs. The ability of some plants like legumes to fix nitrogen is due to a symbiotic rela¬ tionship between the plant and nitrogen-fixing bacteria that live in the plant’s roots. Assume that the hypothetical reaction for fixing nitrogen biologically is
N 2(g) + 3H20 (0 - > 2NH 3(aq) + §02(g)
a. Calculate the standard enthalpy change for the biosynthetic fixation of nitrogen at T = 298 K. For NH3(ag), ammonia dissolved in aqueous solution, A Hf = —80.3 kJmol-1.
b. In some bacteria, glycine is produced from ammonia by the reaction
NH3(g)+2CH4(g) + |02(g)
- > NH2CH2COOH(s) + H20(Z)
Calculate the standard enthalpy change for the synthesis of glycine from ammonia. For glycine,
A H°f = -537.2 kJ mol-1. Assume T= 298 K.
c. Calculate the standard enthalpy change for the synthesis of glycine from nitrogen, water, oxygen, and methane.
P4.20 If 3.365 g of ethanol C2H5OH(/) is burned com¬ pletely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ.
a. Calculate A H°combustion f°r ethanol at 298.15 K.
b. Calculate A Hf of ethanol at 298.15 K.
P4.21 From the following data, calculate AHR 391 4 K for the reaction CH3COOH(g) + 2 O 2(g) > 2 H20(g) + 2 C02(g):
A HI (kJ mol"1)
CH3COOH(7) + 2 02(g) - > 2 H2O(0 + 2 C02(g) -871.5
H20(Z) - -> H20(g) 40.656
CH3COOH(Z) - > CH3COOH(g) 24.4
Values for A HR for the first two reactions are at 298.15 K, and for the third reaction at 391.4 K.
Substance CH3COOH (/) O 2(g) CO 2(g) H2Q(/) H 2Q(g) Cp,m/R 14.9 3.53 4.46 9.055 4.038
P4.22 A 0.1429 g sample of sucrose Ci2H220n is burned in a bomb calorimeter. In order to produce the same temperature rise in the calorimeter as the reaction,
2353 J must be expended.
a. Calculate A U and A H for the combustion of 1 mole of sucrose.
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84 CHAPTER 4 Thermochemistry
b. Using the data tables and your answer to (a), calculate A Hf for sucrose.
c. The rise in temperature of the calorimeter and its contents as a result of the reaction is 1.743 K. Calculate the heat capacity of the calorimeter and its contents.
P4.23 Calculate A H°R at 675 K for the reaction 4 NH3(g) +
6 NO(g) - > 5 N 2(g) + 6 H20(g) using the temperature
dependence of the heat capacities from the data tables. Compare your result with A H°R at 298.15. Is the difference large or small? Why?
P4.24 From the following data at 298.15 K as well as data in Table 4.1 (Appendix B, Data Tables), calculate the standard enthalpy of formation of H2S(g) and of FeS^s):
A H% (kj moP1)
Fe(s) + 2 H2S(g) - * FeS20) + 2 H 2(g) -137.0
H2S(g) + 3/2 02(g) - > H20(/) + S02(g) -562.0
P4.25 Using the protein DSC data in Problem P4.10, calcu¬ late the enthalpy change between T= 288 K and T= 318 K. Give your answer in units of kJ per mole. Assume the molec¬ ular weight of the protein is 14,000. grams. [Hint: You can perform the integration of the heat capacity by estimating the area under the DSC curve and above the dotted baseline in Problem P4.10. This can be done by dividing the area up into small rectangles and summing the areas of the rectangles. Comment on the accuracy of this method .]
P4.26 Given the following heat capacity data at 298 K, cal¬ culate A Hf of C02(g) at 525 K. Assume that the heat capaci¬ ties are independent of temperature.
Substance C (graphite) O 2(g) CO 2(g)
CP,m/i mor'K-' 8.52 28.8 37.1
P4.27 Calculate A H for the process in which Cl2(g) initially at 298.15 K at 1 bar is heated to 690. K at 1 bar. Use the tem¬ perature-dependent heat capacities in the data tables. How large is the relative error if the molar heat capacity is assumed to be constant at its value of 298.15 K over the temperature interval?
P4.28 From the following data at 298.15 K C, calculate the standard enthalpy of formation of FeO(T) and of Fe203(s):
A H% (kj mol"1)
Fe203(.y) + 3 C (graphite) - > 2 Fe(s) + 3 CO(g) 492.6
FeO (s) + C(graphite) - » Fe(s) + CO(g) 155.8
C (graphite) + 02(g) - * C02(g) -393.5 1
CO(g) + 1/2 02(g) - > C02(g) -282.98
P4.29 Calculate the average C — H bond enthalpy in methane using the data tables. Calculate the percent error in equating the average C — H bond energy in Table 4.3 with the bond enthalpy.
P4.30 Use the average bond energies in Table 4.3 to esti¬ mate A U for the reaction C2H4(g) + H2(g) - > C2H6(g).
Also calculate A U°R from the tabulated values of A Hf for reactant and products (Appendix B, Data Tables). Calculate the percent error in estimating A U°R from the average bond energies for this reaction.
P4.31 Use the tabulated values of the enthalpy of combus¬ tion of benzene and the enthalpies of formation of C02(g) and H20(/) to determine A Hf for benzene.
P4.32 Compare the heat evolved at constant pressure per mole of oxygen in the combustion of sucrose (Ci2H220n) and palmitic acid (Ci6H3202) with the combustion of a typi¬ cal protein, for which the empirical formula is C4 3H^ ^NO. Assume for the protein that the combustion yields N2(g), C02(g), and H20(/). Assume that the enthalpies for com¬ bustion of sucrose, palmitic acid, and a typical protein are 5647 kJ mol-1, 10,035 kJ mol-1, and 22.0 kJ g-1, respectively. Based on these calculations, determine the average heat evolved per mole of oxygen consumed, assuming combustion of equal moles of sucrose, palmitic acid, and protein.
P4.33 A camper stranded in snowy weather loses heat by wind convection. The camper is packing emergency rations consisting of 58% sucrose, 31% fat, and 11% protein by weight. Using the data provided in Problem P4.32 and assum¬ ing the fat content of the rations can be treated with palmitic acid data and the protein content similarly by the protein data in Problem P4.32, how much emergency rations must the camper consume in order to compensate for a reduction in body temperature of 3.5 K? Assume the heat capacity of the body equals that of water. Assume the camper weighs 67 kg. State any additional assumptions.
P4.34 In order to get in shape for mountain climbing, an avid hiker with a mass of 60. kg ascends the stairs in the world’s tallest structure, the 828 m tall Burj Khalifa in Dubai, United Arab Emirates. Assume that she eats energy bars on the way up and that her body is 25% efficient in converting the energy content of the bars into the work of climbing. How many energy bars does she have to eat if a single bar produces 1.08 X 103 kJ of energy upon metabolizing?
P4.35 We return to the 60. kg hiker of P4.34, who is climbing the 828 m tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body. She eats two energy bars and a single bar produces 1.08 X 103 kJ of energy upon metabolizing. Assume that the heat capacity of her body is equal to that for water. Calculate the increase in her temperature at the top of the structure. Is your result reasonable? Can you think of a mechanism by which her body might release energy to avoid a temperature increase?
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Entropy and the Second and Third Laws of Thermodynamics
D
I \eal-world processes have a natural direction of change. Heat flows from hotter bodies to colder bodies, and gases mix rather than separate. Entropy, designated by 5, is the state function that predicts the direction of natural, or spontaneous, change and entropy increases for a sponta¬ neous change in an isolated system. For a spontaneous change in a system interacting with its environment, the sum of the entropy of the system and that of the surroundings increases. In this chapter, we introduce entropy, derive the conditions for spontaneity, and show how 5 varies with the macroscopic variables P. V, and T.
IThe Universe Has a Natural Direction of Change
To this point, we have discussed q and w, as well as U and H. The first law of thermody¬ namics states that in any process, the total energy of the universe remains constant. However, it does not predict which of several possible energy conserving processes will occur. Consider the following two examples. A metal rod initially at a uniform tempera¬ ture could, in principle, undergo a spontaneous transformation in which one end becomes hot and the other end becomes cold without being in conflict with the first law, as long as the total energy of the rod remains constant. However, experience demonstrates that this does not occur. Similarly, an ideal gas that is uniformly distributed in a rigid adiabatic container could undergo a spontaneous transformation such that all of the gas moves to one-half of the container, leaving a vacuum in the other half. For an ideal gas, (i dU/dV)T = 0, therefore the energy of the initial and final states is the same. Neither of these transformations violates the first law of thermodynamics — and yet neither occurs.
Experience tells us that there is a natural direction of change in these two processes. A metal rod with a temperature gradient reaches a uniform temperature at some time after it has been isolated from a heat source. A gas confined to one-half of a container with a vacuum in the other half distributes itself uniformly throughout the container if a valve separating the two parts is opened. The transformations described in the previous paragraph are unnatural transformations. The word unnatural is
5.1
The Universe Has a Natural Direction of Change
5.2
Heat Engines and the Second Law of Thermodynamics
5.3
Introducing Entropy
5.4
Calculating Changes in Entropy
5.5
Using Entropy to Calculate the Natural Direction of a Process in an Isolated System
5.6
The Clausius Inequality
5.7
The Change of Entropy in the Surroundings and A Stofa/ =
AS + A S surrouncjijng5
5.8
Absolute Entropies and the Third Law of Thermodynamics
5.9
Standard States in Entropy Calculations
5.10
Entropy Changes in Chemical Reactions
5.11
(Supplemental) Energy Efficiency: Heat Pumps, Refrigerators, and Real Engines
5.12
(Supplemental) Using the Fact that 5 Is a State Function to Determine the Dependence of S on 1/and T
5.13
(Supplemental) The Dependence of S on T and P
5.14
(Supplemental) The Thermodynamic
Temperature Scale
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85
86
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
used to indicate that such an energy-conserving process can occur but is extremely unlikely. By contrast, the reverse processes, in which the temperature gradient along the rod disappears and the gas becomes distributed uniformly throughout the con¬ tainer, are natural transformations, also called spontaneous processes, which are extremely likely. Spontaneous does not mean that the process occurs immediately, but rather that it will occur with high probability if any barrier to the change is overcome. For example, the transformation of a piece of wood to C02 and H20 in the presence of oxygen is spontaneous, but it only occurs at elevated temperatures because an activa¬ tion energy barrier must be overcome for the reaction to proceed.
Our experience is sufficient to predict the direction of spontaneous change for the two examples cited, but can the direction of spontaneous change be predicted in less obvious cases? In this chapter, we show that there is a thermodynamic function called entropy that allows us to predict the direction of spontaneous change for a system in a given initial state. For example, assume that a reaction vessel contains a given number of moles of N2, H2, and NH3 at 600 K and at a total pressure of 280 bar. An iron cata¬ lyst is introduced that allows the mixture of gases to equilibrate according to 1/2 N2 + 3/2 H2 v NH3. What is the direction of spontaneous change, and what are the partial pressures of the three gases at equilibrium? The answer to this question is obtained by calculating the entropy change in the system and the surroundings.
Most students are initially uncomfortable when working with entropy because entropy is further removed from direct experience than energy, work, or heat. Historically, entropy was introduced by Clausius in 1850, several decades before entropy was understood at a microscopic level by Boltzmann. Boltzmann’s explanation of entropy will be presented in Chapter 32, and we briefly state his conclusions here. At the microscopic level, matter consists of atoms or molecules that have energetic degrees of freedom (i.e., translational, rotational, vibrational, and electronic), each of which is associated with discrete energy levels that can be calculated using quantum mechanics. Quantum mechanics also characterizes a molecule by a state associated with a set of quantum numbers and a molecular energy. Entropy is a measure of the number of quantum states accessible to a macroscopic system at a given energy. Quantitatively, S = k In W, where W is the number of states accessible to the system, and k = R/N A. As demonstrated later in this chapter, the entropy of an isolated system is maximized at equilibrium. Therefore, the approach to equilibrium can be envisioned as a process in which the system achieves the distribution of energy among molecules that corresponds to a maximum value of W and, correspondingly, to a maximum in S.
Heat Engines and the Second Law of Thermodynamics
5.2
The development of entropy presented here follows the historical route by which this state function was first introduced. The concept of entropy arose as 19th-century scien¬ tists attempted to maximize the work output of engines. An automobile engine operates in a cyclical process of fuel intake, compression, ignition and expansion, and exhaust, which occurs several thousand times per minute and is used to perform work on the sur¬ roundings. Because the work produced by such an engine is a result of the heat released in a combustion process, it is referred to as a heat engine. An idealized version of a heat engine is depicted in Figure 5.1. The system consists of a working substance (in this case an ideal gas) confined in a piston and cylinder assembly with diathermal walls. This assembly can be brought into contact with a hot reservoir at Thot or a cold reservoir at Tcoid. The expansion or contraction of the gas caused by changes in its temperature drives the piston in or out of the cylinder. This linear motion is converted to circular motion as shown in Figure 5.1, and the rotary motion is used to do work in the surroundings.
The efficiency of a heat engine is of particular interest in practical applications. Experience shows that work can be converted to heat with 100% efficiency. Consider an example from calorimetry discussed in Chapter 4, in which electrical work is done on a resistive heater immersed in a water bath. We observe that all of the electrical work done on the heater has been converted to heat, resulting in an increase in the temperature
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5.2 HEAT ENGINES AND THE SECOND LAW OF THERMODYNAMICS 87
FIGURE 5.1
A schematic depiction of a heat engine is shown. Changes in temperature of the working substance brought about by con¬ tacting the cylinder with hot or cold reser¬ voirs generate a linear motion that is mechanically converted to a rotary motion, which is used to do work.
of the water and the heater. What is the maximum theoretical efficiency of the reverse process, the conversion of heat to work? As shown later, it is less than 100%. There is a natural asymmetry in the efficiency of converting work to heat and converting heat to work. Thermodynamics provides an explanation for this asymmetry.
As discussed in Section 2.7, the maximum work output in an isothermal expansion occurs in a reversible process. For this reason, we next calculate the efficiency of a reversible heat engine, because the efficiency of a reversible engine is an upper bound to the efficiency of a real engine. This reversible engine converts heat into work by exploiting the spontaneous tendency of heat to flow from a hot reservoir to a cold reser¬ voir. It does work on the surroundings by operating in a cycle of reversible expansions and compressions of an ideal gas in a piston and cylinder assembly. We discuss auto¬ motive engines in Section 5.11.
The cycle for a reversible heat engine is shown in Figure 5.2 in a P-V diagram. The expansion and compression steps are designed so that the engine returns to its initial state after four steps. Recall from Section 2.7 that the area within the cycle equals the work done by the engine. As discussed later, four separate isothermal and adiabatic steps are needed to make the enclosed area in the cycle greater than zero. Beginning at point a , the first segment is a reversible isothermal expansion in which the gas absorbs heat from the reservoir at Thou and does work on the surroundings. In the second segment, the gas expands further, this time adiabatically. Work is also done on the surroundings in this step. At the end of the sec¬ ond segment, the gas has cooled to the temperature Tcoid. The third segment is an isother¬ mal compression in which the surroundings do work on the system and heat is absorbed by the cold reservoir. In the final segment, the gas is compressed to its initial volume, this time adiabatically. Work is done on the system in this segment, and the temperature returns to its initial value, Thot • In summary, heat is taken up by the engine in the first segment at Thou and released to the surroundings in the third segment at Tcoid- Work is done on the
Volume
FIGURE 5.2
A reversible Carnot cycle for a sample of an ideal gas working substance is shown on an indicator diagram. The cycle con¬ sists of two adiabatic and two isothermal segments. The arrows indicate the direc¬ tion in which the cycle is traversed. The insets show the volume of gas and the coupling to the reservoirs at the beginning of each successive segment of the cycle. The coloring of the contents of the cylin¬ der indicates the presence of the gas and not its temperature. The volume of the cylinder shown is that at the beginning of the appropriate segment.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
surroundings in the first two segments and on the system in the last two segments. An engine is only useful if net work is done on the surroundings, that is, if the magnitude of the work done in the first two steps is greater than the magnitude of the work done in the last two steps. The efficiency of the engine can be calculated by comparing the net work per cycle with the heat taken up by the engine from the hot reservoir.
Before carrying out this calculation, we discuss the rationale for the design of this reversible cycle in more detail. To avoid losing heat to the surroundings at temperatures between That and Tcoid , adiabatic segments 2 (b — > c) and 4 ( d — > a) are used to move the gas between these temperatures. To absorb heat only at Thot and release heat only at Tcol<b segments 1 (a — > b) and 3 (c — > d) must be isothermal. The reason for using alternating isothermal and adiabatic segments is that no two isotherms at different temperatures inter¬ sect, and no two adiabats starting from two different temperatures intersect. Therefore, it is impossible to create a closed cycle of nonzero area in an indicator diagram out of