energy and entropy - sites.science.oregonstate.edu

Energy and EntropyPhysics 423

David Roundy

Spring 2012Draft 1/30/2012

Contents

Contents i

1 Monday: What kind of beast is it? 11.1 Thermodynamic variables (Chapter 1 ) . . . . . . . . . . 1

Activity: Thermodynamic variable cards (4 min) . . . . . . . . . . . . . . . 1Required homework 1.1: steam-tables . . . . . . . . . . . . . . . . . . 2

Activity: Dimensions (5 min?) . . . . . . . . . . . . . . . . . . . . . . . . . . 2Activity: Conjugate pairs (10 min?) . . . . . . . . . . . . . . . . . . . . . . 2Activity: Intensive vs. extensive (10-15 min) . . . . . . . . . . . . . . . . . . 2

Required homework 1.2: extensive-internal-energy . . . . . . . . . . . 2Activity: State variables and functions (5 min?) . . . . . . . . . . . . . . . . 2Activity: Conserved quantities (??? min) . . . . . . . . . . . . . . . . . . . . 3Thermodynamic equilibrium . . . . . . . . . . . . . 3Lecture: (??? min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Equations of state . . . . . . . . . . . . . . . 3Lecture: (3-4 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

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2 Tuesday: 4Activity: Math pretest (15 min) . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1 Partial derivatives as change of variables . . . . . . . . . 4Lecture: (???) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Activity: Paramagnetism (??? min) . . . . . . . . . . . . . . . . . . . . . . 4Activity: Ideal gas bulk modulus (??? min) . . . . . . . . . . . . . . . . . . 4

2.2 Total differential . . . . . . . . . . . . . . . . 5Lecture: (11 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5BWBQ: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Lecture: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6BWBQ: (15 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Exact vs. inexact differentials . . . . . . . . . . . . 7BWBQ: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Lecture: (5 min?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Required homework 1.3: total-differentials . . . . . . . . . . . . . . . 72.3 Chain rules . . . . . . . . . . . . . . . . . 7

Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Exam: Math pretest . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Wednesday: 12Handout: Changes of variables . . . . . . . . . . . . . . . . . . . . . . 12

Homework 1 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Problem 1.1 Steam tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Problem 1.2 Extensive internal energy . . . . . . . . . . . . . . . . . . . . . 14Problem 1.3 Total differentials . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 Thursday: 154.1 Mixed partial derivatives . . . . . . . . . . . . . . 15

Activity: BWBQ: Mixed partials (30-40 min?) . . . . . . . . . . . . . . . . . 15Activity: Mixed partials vanishing (??? min) . . . . . . . . . . . . . . . . . 15

Handout: Mixed partial derivatives . . . . . . . . . . . . . . . . . . . 15Required homework 2.1: a-maxwell-relation . . . . . . . . . . . . . . 19Required homework 2.2: adiabatic-susceptibility . . . . . . . . . . . . 19Required homework 2.3: summation-notation . . . . . . . . . . . . . 19Required homework 2.4: homogeneous-function-theorem . . . . . . . 19

Monotonicity and invertibility . . . . . . . . . . . . 19Activity: Monotonicity lecture/discussion (10 min) . . . . . . . . . . . . . . 19Legendre transform . . . . . . . . . . . . . . . 20Lecture: (??? min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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5 Friday: Lagrange multipliers for minimization 23Homework 2 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Problem 2.1 A Maxwell relation . . . . . . . . . . . . . . . . . . . . . . . . 23Problem 2.2 Adiabatic susceptibility . . . . . . . . . . . . . . . . . . . . . . 23Problem 2.3 Summation notation . . . . . . . . . . . . . . . . . . . . . . . . 24Problem 2.4 Euler’s homogeneous function theorem . . . . . . . . . . . . . . 24

6 Monday: Lab 1: Heat and Temperature 26Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Activity: Energy/Heat equivalence (15 min now, 10 min later) . . . . . . . . 26SWBQ: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26SWBQ: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26SWBQ: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26SWBQ: (5 min?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Dulong-Petit Law . . . . . . . . . . . . . . . . 27Lecture: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7 Tuesday: First and Second Laws 28Activity: Name-the-experiment pretest (10 min) . . . . . . . . . . . . . . . . 28System and surroundings . . . . . . . . . . . . . . 28Lecture: (2-3 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28BWBQ: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

7.1 First Law . . . . . . . . . . . . . . . . . . 28Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28SWBQ: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

7.2 Second Law and Entropy . . . . . . . . . . . . . . 29SWBQ: (1 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Lecture: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Fast and slow . . . . . . . . . . . . . . . . . 30Lecture: (7-12 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

7.3 The thermodynamic identity (5.1.2.2 ) . . . . . . . . . . 30Lecture: (8-15 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Activity: Name the experiment (20 min) . . . . . . . . . . . . . . . . . . . . 30

7.4 Heat capacity . . . . . . . . . . . . . . . . . 31SWBQ: (5 min (skipped in 2011)) . . . . . . . . . . . . . . . . . . . . . . . . 31Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31BWBQ: (10 min (skipped in 2011)) . . . . . . . . . . . . . . . . . . . . . . . 32Lecture: (5 min (skipped in 2011)) . . . . . . . . . . . . . . . . . . . . . . . 32Activity: Entropy change of cooling coffee (20-30 min (skipped 2010)) . . . . 32

Required homework 3.1: adiabatic-ideal-gas . . . . . . . . . . . . . . 33Required homework 3.2: bottle-in-bottle . . . . . . . . . . . . . . . . 33Exam: Pretest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

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Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

8 Wednesday: Second Law lab 35Activity: Melting ice (45 min) . . . . . . . . . . . . . . . . . . . . . . . . . . 35Lab 1 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Question 1.1 Plot your data I . . . . . . . . . . . . . . . . . . . . . . . . . . 36Question 1.2 Plot your data II . . . . . . . . . . . . . . . . . . . . . . . . . . 36Question 1.3 Specific heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Question 1.4 Latent heat of fusion . . . . . . . . . . . . . . . . . . . . . . . . 37Question 1.5 Entropy for a temperature change . . . . . . . . . . . . . . . . 37

9 Thursday: Heat and work 38Activity: Expanding Gas Quiz (20-45 min) . . . . . . . . . . . . . . . . . . . 38

Handout: Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Question 1 Free expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39Activity: Name the experiment with changing entropy (20 min) . . . . . . . 40

9.1 Work . . . . . . . . . . . . . . . . . . . 40Lecture: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40BWBQ: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40Activity: Using pV plots (20-30 min) . . . . . . . . . . . . . . . . . . . . . . 40Activity: Using TS plots (15-20 min) . . . . . . . . . . . . . . . . . . . . . . 41Activity: Name the experiment for a rubber band (15 min) . . . . . . . . . . 41

10 Friday: 4210.1 Engines and Fridges . . . . . . . . . . . . . . . 42

Other views of the Second Law . . . . . . . . . . . . 42Lecture: (??? min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Carnot efficiency . . . . . . . . . . . . . . . . 42Lecture: (??? min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Activity: Carnot efficiency (??? min) . . . . . . . . . . . . . . . . . . . . . . 44Activity: Big money (15 min?) . . . . . . . . . . . . . . . . . . . . . . . . . 44Lab 2 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Question 2.1 Mass of ice remaining . . . . . . . . . . . . . . . . . . . . . . . 46Question 2.2 Final temperature of water . . . . . . . . . . . . . . . . . . . . 46Question 2.3 Change in entropy of water . . . . . . . . . . . . . . . . . . . . 46Question 2.4 Change in entropy of ice . . . . . . . . . . . . . . . . . . . . . . 47Question 2.5 Net change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Question 2.6 Mass of ice remaining . . . . . . . . . . . . . . . . . . . . . . . 47Question 2.7 Final temperature . . . . . . . . . . . . . . . . . . . . . . . . . 47Question 2.8 Errors? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Homework 3 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48Problem 3.1 Adiabatic compression . . . . . . . . . . . . . . . . . . . . . . . 48

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Problem 3.2 A bottle in a bottle . . . . . . . . . . . . . . . . . . . . . . . . 48Practice homework 4.1: power-from-ocean . . . . . . . . . . . . . . . 49Required homework 4.2: power-plant-river . . . . . . . . . . . . . . . 49Required homework 4.3: heat-pump . . . . . . . . . . . . . . . . . . . 49

11 Monday: 50Monotonicity and invertibility . . . . . . . . . . . . 50Activity: Monotonicity lecture/discussion (10 min) . . . . . . . . . . . . . . 50

11.1 Thermodynamic potentials (Chapter 7 ) . . . . . . . . . . 51Legendre transform . . . . . . . . . . . . . . . 51Activity: Legendre transform revisited (15 min?) . . . . . . . . . . . . . . . 51Activity: Understanding the potentials (20 min) . . . . . . . . . . . . . . . . 52Lecture: (5 min?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Required homework 4.4: gibbs-free-energy . . . . . . . . . . . . . . . 5311.2 Maxwell relations . . . . . . . . . . . . . . . . 53

Lecture: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53BWBQ: (15-25 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Activity: Name the experiment with Maxwell relations (10 min) . . . . . . . 54

12 Tuesday: Lab 2: rubber band 55Activity: Rubber band lab . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Prelab 3 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Question 3.1 Tension vs. temperature . . . . . . . . . . . . . . . . . . . . . . 56Question 3.2 Isothermal stretch . . . . . . . . . . . . . . . . . . . . . . . . . 56Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Required homework 4.5: free-expansion . . . . . . . . . . . . . . . . . 58

13 Wednesday: 59Activity: Name another experiment with Maxwell relations (10-20 min) . . . 59Homework 4 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Problem 4.1 Power from the ocean (practice) . . . . . . . . . . . . . . . . . . 59Problem 4.2 Power plant on a river . . . . . . . . . . . . . . . . . . . . . . . 60Problem 4.3 Heat pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60Problem 4.4 Using the Gibbs free energy . . . . . . . . . . . . . . . . . . . . 61Problem 4.5 Free expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

14 Thursday: Thermodynamics practice 62Activity: Simple cycle (40-110 min) . . . . . . . . . . . . . . . . . . . . . . . 62Activity: Temperature change of dissolving salt (40 min?) . . . . . . . . . . 62Activity: Never, sometimes or always true (60 min?) . . . . . . . . . . . . . 63

Handout: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

15 Friday: Thermodynamics practice 66

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Activity: Black body thermodynamics (30 min) . . . . . . . . . . . . . . . . 66Activity: Applying the second law . . . . . . . . . . . . . . . . . . . . . . . . 66

Required homework 5.1: maine-entropy-2nd-law-spontaneous-metal . 67Required homework 5.2: isothermal-adiabatic-compressibility . . . . . 67

Lab 3 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Question 3.3 Tension vs. temperature . . . . . . . . . . . . . . . . . . . . . . 69Question 3.4 Tension vs. length . . . . . . . . . . . . . . . . . . . . . . . . . 70Question 3.5

(∂S∂L

)T

vs. length . . . . . . . . . . . . . . . . . . . . . . . . . . 70Question 3.6 Isothermal stretch . . . . . . . . . . . . . . . . . . . . . . . . . 70Question 3.7 Adiabatic stretch . . . . . . . . . . . . . . . . . . . . . . . . . . 70

16 Monday: Statistical approach 71Lecture: (6 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

16.1 Fairness function (Chapter 6 ) . . . . . . . . . . . . 72Lecture: (13 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72Activity: Combining probabilities (??? min) . . . . . . . . . . . . . . . . . . 72Lecture: (??? min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73Activity: Demonstrating extensivity (??? min) . . . . . . . . . . . . . . . . 73

17 Tuesday: Optimizing the fairness 74Activity: Students as molecules (20 min) . . . . . . . . . . . . . . . . . . . . 74

17.1 Least bias lagrangian . . . . . . . . . . . . . . . 74Lecture: (10 min?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

17.2 Weighted averages . . . . . . . . . . . . . . . 75Lecture: (20 min?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

17.3 Probabilities of microstates (Chapter 11 ) . . . . . . . . . 75Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75SWBQ: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77SWBQ: (3 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Lecture: (7 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Required homework 5.3: boltzmann-ratio . . . . . . . . . . . . . . . . 77Challenge homework 5.4: plastic-rod . . . . . . . . . . . . . . . . . . 77

18 Wednesday: From statistics to thermodynamics 7818.1 Thermodynamic properties from the Boltzmann factor . . . . . 78

Lecture: (5 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78Activity: Solving for maximum fairness (10 min) . . . . . . . . . . . . . . . 78Lecture: (15 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79BWBQ: (skipped this) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80BWBQ: (skip this?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80Activity: Entropy of microcanonical ensemble (20 min) . . . . . . . . . . . . 81Homework 5 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81Problem 5.1 Hot metal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

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Problem 5.2 Isothermal and adiabatic compressibility . . . . . . . . . . . . . 81Problem 5.3 Boltzmann ratio . . . . . . . . . . . . . . . . . . . . . . . . . . 82Problem 5.4 A plastic rod (challenge) . . . . . . . . . . . . . . . . . . . . . . 82

Required homework 6.1: rubber-band-model . . . . . . . . . . . . . . 83Required homework 6.2: rubber-meets-road . . . . . . . . . . . . . . 83

19 Thursday: Statistical mechanics of air 8419.1 Quantum spectra . . . . . . . . . . . . . . . . 84

Lecture: (10 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8419.2 Diatomic gas . . . . . . . . . . . . . . . . . 85

Lecture: (20 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Activity: Diatomic molecule from quantum up (1 hour 30 min without wrap-up) 87

20 Friday: 8820.1 Diatomic gas wrapup . . . . . . . . . . . . . . . 88

Lecture: (20 min) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88Homework 6 due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92Problem 6.1 A rubber band model . . . . . . . . . . . . . . . . . . . . . . . 93Problem 6.2 The rubber meets the road . . . . . . . . . . . . . . . . . . . . 93

20.2 Third law . . . . . . . . . . . . . . . . . . 94Ice rules . . . . . . . . . . . . . . . . . . 95

Handout: Measurement of entropy of water . . . . . . . . . . . . . . 95Handout: Pauling ice rules . . . . . . . . . . . . . . . . . . . . . . . . 98Handout: Entropy of water revisited . . . . . . . . . . . . . . . . . . 103

Activity: Concept diagram (10 min) . . . . . . . . . . . . . . . . . . . . . . 109Exam: Survey and post-test . . . . . . . . . . . . . . . . . . . . . . . 109

Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111Question 4 Lab 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Question 5 Lab 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Question 6 Lab 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Question 7 Class as a whole . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

21 Monday: Final exam 114Exam: Final exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

Problem 1 Masses on a piston . . . . . . . . . . . . . . . . . . . . . . . . . . 115Problem 2 Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . 116Problem 3 Two processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117Problem 4 Hanging Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118Problem 5 Insulated room . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119Problem 6 Soap bubble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

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Class schedule 121

Index 123

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1 Monday: What kind of beast is it?

Current board:

1.1 Thermodynamic variables (Chapter 1 )

Activity: Thermodynamic variable cards (4 min) Give each student a set of 3× 5cards (or larger?), and ask them to write the name of each variable at the top of the linedside.

Write down any thermodynamic variables that you can think of. If students seem puzzled,ask them to write down any properties of solids, liquids or gasses that they might measure,or values that might be helpful in predicting these properties.

I’ll write down all these thermodynamic variables, and talk through any that seem par-ticularly interesting.

As we go through the variables that students have come up with, I’ll give them the math-ematical notation that we’ll be using in this class, and they’ll add it on the blank side. Ineach of the following activities, we’ll ask students to divide up their cards, so each one mustgo into either one pile or the other. When we’ve established the correct answers, studentswill write them on the lined side, to make sure they know what each thing is.

Add to this list any important variables the students may have missed, in particular, wewant heat and work to be here, as well as the standard p, V , T , S, U , N , M , and density. Imight also like to add number density n, specific heat or heat capacity, and maybe coefficientof thermal expansion or isothermal compressibility.

“Are there any additional properties you might add if we were talking about a rubber bandinstead, looking at it as a one-dimensional system?”

“How do we measure this?”We should now have τ and L as well.Here are some extra questions that may be worth asking: (but maybe not right

now)

a) What is temperature?

b) How do we measure temperature?

c) How do we measure pressure?

Physics 423 1 Monday 4/18/2012

d) Would pV = (NkBT )2 be a possible equation of state?

e) How do we measure entropy?

f) How many of the variables T , S, p and V are independent?

Required homework 1.1: steam-tables

Activity: Dimensions (5 min?) “What are the dimensions of each of the thermody-namic quantities?”

Activity: Conjugate pairs (10 min?) If I have a solid object (say a stack of papers,or a book), and I put another object on top of it, what will happen to the height of the firstobject?

It will get shorter. Why? Is this always true, or will it in some cases get taller? Heightand force in the previous example are a conjugate pair.

In your groups, try to divide our stack of thermodynamic variables up into analogousconjugate pairs. For each thing you can ask “What would I need to apply to change it?”

Talk over their pairings. In cases where there isn’t a partner available, ask students if wemight be missing a state variable that would be its conjugate partner.

Talk over which way the pairings go... if you increase pressure, what happens to volume.Will this always be the case? If you increase temperature, what happens to other variables?

Pairings: pV , TS, τL, ~M · ~B, ~E · ~P . There are several variables that don’t show uphere. Q and W won’t, although they relate to some of these conjugate pairs. Similarly, thosestate variables that are simply products or ratios of other state variables don’t partake inthese pairs.

Activity: Intensive vs. extensive (10-15 min) Ideally here I’d like to have two cupsof water (the same amount), and ask what would happen to each of those thermodynamicquantities, if instead of asking about one cup of water, I put the two together and asked abouta cup with twice as much water. Hopefully this will clarify the “doubling” that we mean.

Break our thermodynamic variables into categories based on how they change if you havetwice as much stuff.

Once students have split things up, give the definitions of intensive and extensive.Extensive: U , V , N , M , p, S, L (sort of: Q, W ).Intensive: T , n, ρ, τ .

Required homework 1.2: extensive-internal-energy

Activity: State variables and functions (5 min?) A state variable is one of the thingsthat I would need to tell you in order for you to reproduce my experiment. There may beredundant state variables.

A state function is something that could be measured (or possibly computed) if you knowthe state of a system (i.e. a sufficient set of state variables.


This may be more clear if we use an example from classical mechanics. We can considera single point particle, which is moving in some external gravitational field. We can split upthe following variables: ~r, ~v, ~a, ~p, ~F , m, KE, PE, W .

“Right down on your small whiteboard which of these variables are state variables, statefunctions (but not state variables), and which are neither.”

State variables: ~r, ~v, ~p, m.State functions: ~a, ~F , KE, PE.Neither: W .

Categorize the thermodynamic variables that we brainstormed earlier.State variables: V , N , M , p, T , n, ρ, S, τ , L.State functions (but not state variables): U , C, α.Neither: Q, W .

Activity: Conserved quantities (??? min) Which of the thermodynamic quantitiesare conserved?

Thermodynamic equilibrium

Lecture: (??? min) Most state functions are only well-defined when a system is inthermodynamic equilibrium. e.g. if I throw sodium metal into a cup of water, and ask whatthe temperature of the cup is, you won’t be able to give me a good answer, since there aretemperature variations, pressure variations, etc. However, if I ask what the volume of thecup is, you’ll have no problems.

Equations of state

Lecture: (3-4 min) Many of our state variables are redundant: a few are needed todefine the state, and the rest can be computed or measured based on those. An equation ofstate is how we express this computation.

An equation of state is an equation that relates a set of mutually-dependent thermody-namic state variables, such as pressure-volume or tension-length. The most famous is theideal gas equation pV = NkBT .

Physics 423 3 Tuesday 4/19/2012

2 Tuesday:

Current board:

Activity: Math pretest (15 min)

2.1 Partial derivatives as change of variables

Lecture: (???) We should say something on this topic.

Activity: Paramagnetism (??? min) We have the following equations of state for thetotal magnetization M , and the entropy S:

M = NµeµBkBT − e−

µBkBT

eµBkBT + e

− µBkBT

(2.1)

S = NkB

{ln 2 + ln

(eµBkBT + e

− µBkBT

)+

µB

kBT

eµBkBT − e−

µBkBT

eµBkBT + e

− µBkBT

}(2.2)

Solve for the magnetic susceptibility, which is defined as:

χB =

(∂M

∂B

)T

(2.3)

Also solve for the same derivative, taken with the entropy S held constant:(∂M

∂B

)S

(2.4)

Why does this come out as zero?

Activity: Ideal gas bulk modulus (??? min) We have the following equations ofstate for a monatomic ideal gas. The first is the famous ideal gas law. The second is trueonly for a monatomic ideal gas. The third is the Sackur-Tetrode equation, which is true forany ideal gas.

pV = NkBT (2.5)


U =3

2NkBT (2.6)

S = NkB

{ln

[V

N

(mU

3πNh2

) 32

]+

5

2

}(2.7)

From these, solve for the following partial derivatives:

B = −V(∂p

∂V

)T

(2.8)

BS = −V(∂p

∂V

)S

(2.9)

The former is the isothermal bulk modulus...

2.2 Total differential

Lecture: (11 min) Introduce total differentials here.

f(x, y) (2.10)

df =

(∂f

∂x

)y

dx+

(∂f

∂y

)x

dy (2.11)

This may look pretty weird at first, but if you think of the differential dx as a small changein x, then this is just a nice notation for the first terms in a Tailor expansion of f(x, y),which is

f(x′, y′) = f(x, y) +

(∂f

∂x

)y

(x′ − x) +

(∂f

∂y

)x

(y′ − y) +1

2!

(∂2f

∂x2

)y

(x′ − x)2 + · · ·

(2.12)

f(x′, y′)− f(x, y) =

(∂f

∂x

)y

(x′ − x) +

(∂f

∂y

)x

(y′ − y) +1

2!

(∂2f

∂x2

)y

(x′ − x)2 + · · · (2.13)

∆f =

(∂f

∂x

)y

∆x+

(∂f

∂y

)x

∆y + · · · (2.14)

Now if we take the limit that ∆x is small and ∆y is small, the higher-order terms vanish,and we are left with

df =

(∂f

∂x

)y

dx+

(∂f

∂y

)x

dy (2.15)

Note also, that we can study the total differential of functions with any number ofarguments, so if we had f(a, b, c, e), we could write

df =

(∂f

∂a

)b,c,e

da+

(∂f

∂b

)a,c,e

db+

(∂f

∂c

)a,b,e

dc+

(∂f

∂e

)a,b,c

de (2.16)


One exciting thing about total derivatives in thermodynamics is that the derivatives(∂f∂x

)y

and(∂f∂y

)x

are often observable state variables themselves! Thus we often interpret a

total differential in order to find definitions for the terms.

BWBQ: (5 min) Interpret the following total differential

dR = adB + Cde (2.17)

to find expressions for a and C.Interpret the following total differential

dH = TdS + V dp (2.18)

to find expressions for the temperature T and the volume V .

Lecture: (5 min) Generally, common derivative sense applies to total differentials...

product rule

f = gh (2.19)

df = gdh+ hdg (2.20)

chain rule

f(x, y) = g(h) (2.21)

df =dg

dhdh (2.22)

BWBQ: (15 min) Given f(x, y) = ln (x2 + y2), find df .Find dF when:

F = −∑i

Pi lnPi (2.23)

Find dF if

F = U − TS (2.24)

dU = TdS − pdV (2.25)


Exact vs. inexact differentials

BWBQ: (10 min) Given an expression for a total differential

df = ydx+ xdy (2.26)

what is(∂f∂x

)y

and(∂f∂y

)x? What is f(x, y)?

“When you are given a total differential, you should be able to interpret it as two or moreequations involving partial derivatives. This usually often the first step in making use of atotal differential. ”

Can you solve the same problem for

dg =

(ln

(x

y

)+ 1

)dx− x

ydy? (2.27)

Can you solve the same problem for

dh = ydx− xdy? (2.28)

Lecture: (5 min?) In the above example, dh is an inexact differential , while df and dgare a exact differentials. I’ll try to always write inexact differentials with a d, to make itclear that it is not the total differential of a function! Inexact differentials are useful, butnot nearly so useful as exact differentials.

Generally, inexact differentials refer to processes, while exact differentials refer to states.e.g. work is an inexact differential for non-conservative force fields.

Required homework 1.3: total-differentials

2.3 Chain rules

Lecture: (10 min)

Inverting a partial derivative Write on board:(∂a

∂b

)c

=1(∂b∂a

)c

Ordinary chain rule for total derivatives

dx

dy

dy

dz=dx

dz(2.29)


Ordinary chain rule for partial derivatives Write on board:(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)d

=

(∂a∂c

)d(

∂b∂c

)d

(2.30)

Cyclic chain rule for partial derivatives Write on board:(∂a

∂b

)c

= −(∂c∂b

)a(

∂c∂a

)b

(∂a

∂b

)c

(∂c

∂b

)a

(∂c

∂a

)b

= −1 (2.31)

This is particularly handy when you’ve got something that is being held constant that youhappen to know a derivative of, or if you’re taking a derivative of something you’d ratherhold constant.

Note: in most problems, it only helps to use the cyclic chain rule once. If you use it asecond time, you’re starting to go in circles, and may wish to go back to square one.


Math pretest

Question 1 The Gibbs Free Energy, G, is a function of the independent variables tem-perature T and pressure p, i.e., it can be written as G(T, p). The total differential of G canbe written as dG = −SdT + V dp, where S is the entropy and V is the volume.

a) Interpret the above equation in order to determine an expression for the entropy S.

b) From the total differential dG, obtain a different thermodynamic derivative that isequal to (

∂S

∂p

)T

Question 2 Z is a function of the two variables x and y, i.e. Z = Z(x, y). Consider thefollowing two expressions:

α = − 1

Z

∂Z

∂x(2.32)

β =1

Z

∂Z

∂y(2.33)

Show that in general ∂α∂y

+ ∂β∂x

= 0. Explain your reasoning.


Question 3 A curve has been traced out on the z-y graph below, and has been labeledPath A. Consider the integral

I1 ≡∮A,clockwise

zdy (2.34)

where the integral is taken around Path A starting at point P , proceeding clockwise untilreaching point P again.

a) Is the integral I1 positive, negative, zero or is there not enough information to decide?Please explain your reasoning.

0

Path A

y

z

P

Refer again to the graph in part (a). We define H as a function of the independentvariables z and y; i.e. H = H(z, y). Consider the integral


~∇H · d~r (2.35)

where the integral is taken around Path A starting at point P , proceeding clockwiseuntil reaching point P again. (Note: d~r = dyy + dzz)

b) Is the integral I2 positive, negative, zero or is there not enough information to decide?Please explain your reasoning.


Question 4 This Pressure-Volume (p-V ) diagram represents a system consisting a fixedamount of ideal gas that undergoes two different quasistatic processes in going from state Ato state B:

0 Volume

Pressu

re

State

State

A

B

Process #1

Process #2

[In these questions, W represents the energy the system loses by working during a process;Q represents the energy the system gains by heating during a process.]

a) Is W for Process #1 greater than, less than, or equal to that for Process #2? Explain.

b) Is Q for Process #1 greater than, less than, or equal to that for Process #2? Explainyour answer.

c) Which would produce the largest change in the total energy (kinetic plus potential) ofall the atoms in the gas: Process #1, Process #2, or both processes produce the samechange?


3 Wednesday:

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

Physics 423 12 Wednesday 4/20/2012

We have the following equations of state for a monatomic ideal gas. The first is the famousideal gas law. The second is the internal energy of a monatomic ideal gas. The third is theSackur-Tetrode equation for entropy, which is true for any ideal gas.

pV = NkBT

U =3

2NkBT

S = NkB

{ln

[V

N

(mU

3πNh2

) 32

]+

5

2

}

From the above equations, find the isothermal bulk modulus, which you can think of themas an intensive spring constant for a three-dimensional material.

bT = −V(∂p

∂V

)T

Now find the adiabatic bulk modulus, which involves holding the entropy constant:

bS = −V(∂p

∂V

)S

We have the following equations of state for the total magnetization M , and the entropy:


µBkBT

eµBkBT + e

− µBkBT

S = NkB

{ln 2 + ln

(eµBkBT + e

− µBkBT

)+

µB

kBT

eµBkBT − e−

µBkBT

eµBkBT + e

− µBkBT

}

Solve for the magnetic susceptibility, which is defined as:

χB =

(∂M

∂B

)T


∂B

)S



Energy and Entropy Homework 1Due Wednesday 4/20

Problem 1.1 Steam tables Use the NIST web site “Thermophysical Properties of FluidSystems” to answer the following questions.

a) What is the change in entropy of 100 g of water at atmospheric pressure, when it istaken from room temperature to 500 K (which is about 440◦F, so it’s acheivable inyour oven)?

b) How would this answer differ if we were considering 100 g of benzene?1

c) What is the change in internal energy of water for the same process?

d) What is the change in internal energy of benzene for the same process?

e) What are the change in internal energy and entropy of water when it boils, at oneatmosphere?

f) What are the change in internal energy and entropy of benzene when it boils at oneatmosphere?

Problem 1.2 Extensive internal energy Consider a system which has an internalenergy defined by:

U = γV αSβ (3.1)

where α, β and γ are constants. The internal energy is an extensive quantity. What con-straint does this place on the values α and β may have?

Problem 1.3 Total differentials Find the total differential of R where. . .

a) R(B,C) = B2 + C2

b) R(B,C) = BC

c) R(B,C) = eB2+C2

d) R(B,C) = eS(B,C)

e) R(B,C) = ST , where S = S(B,C) and T = T (B,C)

End of Energy and Entropy Homework 1Due Wednesday 4/20

1Benzene is a significant component of gasoline, and is a common carcinogenic solvent.

Physics 423 14 Thursday 4/21/2012

4 Thursday:

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

4.1 Mixed partial derivatives

Activity: BWBQ: Mixed partials (30-40 min?) Compate(∂(∂V∂T

)p

∂p

)T

and

∂(∂V∂p

)T

∂T

p

(4.1)

Is one greater than the other, are they equal, or are they both zero?Do worksheet on mixed partial derivatives. Be sure to give them plenty of time for this!Wrapup: (

∂(∂V∂T

)p

∂p

)T

=

∂(∂V∂p

)T

∂T

p

(4.2)

=∂2V

∂p∂T(4.3)

=∂2V

∂T∂p(4.4)

It may look confusing, or it may look boring, but the point is that it’s true.

Activity: Mixed partials vanishing (??? min) This activity was originally a home-work problem

Create a function for which the mixed second partial derivatives are in fact both zero.Be as general as possible. Identify specifically the constraint that forces both to be zero.


Mixed partial derivatives

In this activity we will explore what it means for a function to have nonzero and equal mixedsecond partials by reasoning about the situation graphically.

Temp

eratur

e

Pressure

Volume

V1

V2

V3

p1

p2

p3

T1

T3

T3

a

b

c

Figure 4.1: The pV T surface for an ideal gas with fixed N .


Partial Derivatives and Material Properties

©2006 University of Maine Physics Education Research Laboratory, Orono, ME

V2

V3

T1

V1

T2 T3

P3

P2

P1

Figure 2. V-T projection of an ideal gas P-V-T surface with fixed n.

P3P2P1

V2

V3

V1

T3

T2

T1

Figure 3. V-P projection of an ideal gas P-V-T surface with fixed n.Physics 423 Thursday 4/21/2012

A With two different colors of ink, draw and identify the following features on the surfaceat point a. For each expression, explain in words what these features represent graphically.(

∂V

∂T

)p

(4.5)

(∂V

∂p

)T

(4.6)

Do you expect the value of either of these two derivatives to change for different valuesof pressure or temperature? Explain.

B Identify on the diagram the same two derivatives given in part A, only now at points band c instead of at a.

Do each of these two derivatives change as you move along the variable with respect towhich they are differentiated? If so, how?

Do each of these two derivatives change as you move along the variable that is heldconstant during differentiation? If so, how?

C Consider the graph of the projection of the ideal gas pV T surface onto the V -T planeon the next page. Identify

(∂V∂T

)p

at temperature T2 for the three different values of pressure

depicted.Does

(∂V∂T

)p

at temperature T2 increase, decrease, or remain the same as you go from

lower to higher gas pressures? Explain what this means physically. Does it make sense toyou? Is this consistent with your response in part B?

D Consider the projection onto the V -p plane. Identify(∂V∂p

)T

at pressure p2 for the three

different values of temperature depicted.

Does(∂V∂p

)T

at pressure P2 increase, decrease, or remain the same as you go from lower

to higher temperatures? Explain what this means physically. Does it make sense to you? Isthis consistent with your response in part B?

E Compare how, if at all, the two derivatives in parts C and D change (i.e., compare thesigns of the changes). Can you say anything about the magnitude of the change at thispoint?

F Provide a graphical interpretation of a mixed second partial derivative. (Hint: includethe concept of slope in your interpretation.) Be sure to include an explanation as to whymixed second partials are in general not equal to zero.


Required homework 2.1: a-maxwell-relation

Required homework 2.2: adiabatic-susceptibility

Required homework 2.3: summation-notation

Required homework 2.4: homogeneous-function-theorem

Monotonicity and invertibility

Activity: Monotonicity lecture/discussion (10 min) Ask groups to discuss howmany variables are needed to define the state of a given system. It depends on what you arewilling to allow to change. V and T would be sufficient, but you might also include N , andmight want to include even more if you have more than one sort of molecule.

p

V

good(x)bad(x)

Since 2 (or 3) variables fully define the state of a given sys-tem, we know that every other state variable could be writtenas a function of those two variables. “But which two variablesare those?”

We can reason about this pretty easily. “Which state vari-ables can I directly control?” T , S, p, V . But which of thesecould be the two variables that define the state of our system?Let us consider a simple plot of p versus V at constant tem-perature. “Which of these curves are physically reasonable?”

Talk about creating a box full of some material, say a gasor liquid. I could choose to fix the volume of the box, oralternatively I could give it a piston on top and exert a fixed force on it, thus determiningthe pressure. “Should it matter which approach I use?” This tells us that for any givenpressure, there must be only one volume, and for any given volume, there can only be onepressure. Thus our curve must be monotonic. “How would our answer change if we were toplot p versus V at fixed pressure?” (

∂p

∂V

)T

≤ 0 (4.7)(∂p

∂V

)S

≤ 0 (4.8)

Now let’s think about temperature and entropy. The same argument applies, indicating thattemperature and entropy must be fixed. Note that although it’s hard to “fix” the entropy,it’s easy to change entropy by heating up a system. And to keep it from changing, we onlyneed insulate the system and avoid doing anything irreversible. So the same argument showsthat temperature and entropy are monotonically related. The sign we can work out fromthe property of temperature that hot things heat up cold things, and the thermodynamicdefinition of entropy change. (

∂T

∂S

)p

≥ 0 (4.9)


(∂T

∂S

)V

≥ 0 (4.10)

We will also see that the heat capacity is always positive, since it relates to precisely thesederivatives.

Note that monotonicity does not indicate proportionality!

Legendre transform

Lecture: (??? min) Today we’ll be talking about Legendre transforms, but I’d like tobegin with a physical motivation for why we might want to use them.

Let’s talk about the bulk modulus, which is a measure of how hard it is to compresssomething:

K = −V ∂p

∂V(4.11)

A high bulk modulus means that a material is very stiff, and it requires a very high pressureto change the volume by a given fraction. The factor of volume in front is chosen to makethe bulk modulus an intensive quantity, and the negative sign is chosen to make it a positivequantity. The dimensions of bulk modulus are those of pressure.

But what should be held constant? We have two reasonable choices, we could either holdthe temperature fixed, or we could hold then entropy fixed.

KT = −V(∂p

∂V

)T

(4.12)

KS = −V(∂p

∂V

)S

(4.13)

The former is the isothermal compressibility, while the latter is the adiabatic compressibility.I’ll talk in a few minutes about how we would measure these things.

First, let’s talk a little bit about what the pressure p is. One way to define it is to usethe thermodynamic relation

dU = TdS − pdV (4.14)

p = −(∂U

∂V

)S

(4.15)

Using this definition of p, we can quite easily find the adiabatic compressibility, if we knowU as a function of S and V .

KS = −V(∂p

∂V

)S

(4.16)(∂p

∂V

)S

= −

(∂(∂U∂V

)S

∂V

)S

(4.17)


=

(∂2U

∂V 2

)S

(4.18)

So if we know U as a function of S and V , we’re golden. Let’s see how things work for theisothermal compressibility.

KT = −V(∂p

∂V

)T

(4.19)(∂p

∂V

)S

= −

(∂(∂U∂V

)S

∂V

)T

(4.20)

Now we’ve got something that looks a lot nastier, it’s not like a normal second derivative.The reason is that we’re talking about something at fixed temperature, but U isn’t naturallya function of temperature, but of entropy! So what we do is to look for something thatis a function of temperature. The approach to find somthing like this is called a Legendretransform. In this case, we want to switch from entropy to temperature, so we do

F = U − TS (4.21)

“In your groups, find dF .”

F = U − TS (4.22)

dF = dU − TdS − SdT (4.23)

= TdS − pdV − TdS − SdT (4.24)

= −SdT − pdV (4.25)

So this gives us a new thermodynamic function, which we call the Helmholtz free energy ,which is naturally a function of temperature and volume. It also gives us a new definitionfor pressure

p = −(∂F

∂V

)T

(4.26)

KT = −V(∂p

∂V

)T

(4.27)(∂p

∂V

)T

= −

(∂(∂F∂V

)T

∂V

)T

(4.28)

= −(∂2F

∂V 2

)T

(4.29)

So the question naturally arises, “What is the Helmholtz free energy?” To find oneanswer, we can look at the meaning of pdV which shows up in the total differential of bothU and F . This term represents the tiny amount work done as the volume is changing by


a tiny amount. We can let the volume change by a larger amount by integrating. If weintegrate while holding the entropy fixed (this is an adiabatic process), we find that the totalwork done is equal to the change in the internal energy. This is the conservation of energy.If we instead change the volume while holding the temperature fixed, the work done is equalto the change in free energy. That’s where the “free” part of its name comes in: it’s theamount of energy that is free for doing work.

Back to the bulk moduli. We have two bulk moduli, adiabatic and isothermal. What dothey mean, or how would we measure them? One way to measure a bulk modulus (somewhatindirectly) would be to measure the speed of sound (heavier things accelerate more slowly).You also need to know the density, but the bulk modulus provides the “springiness” thatdetermines the speed of sound. But which bulk modulus determines this? In practice, itis the adiabatic bulk modulus that matters, because the thermal conductivity is sufficientlylow that it can be neglected on time (and length) scales corresponding to sound waves.

If you wanted to measure the isothermal bulk modulus, you’d use a static measurement,in which you applied a fixed pressure and measured the change in volume. Doing this slowlywould ensure that the temperature didn’t change as a result of applying the pressure.

The following are properties of a Legendre transform, where you could replace S and Twith p and V if you renamed F . (

∂U

∂S

)V

= T (4.30)(∂F

∂T

)V

= S (4.31)(∂2U

∂S2

)V

=

(∂T

∂S

)V

(4.32)(∂2F

∂T 2

)V

=

(∂S

∂T

)V

(4.33)(∂2U

∂S2

)V

(∂2F

∂T 2

)V

= 1 (4.34)

Physics 423 22 Friday 4/22/2012

5 Friday: Lagrange multipliers forminimization

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

Energy and Entropy Homework 2Due Friday 4/22

Problem 2.1 A Maxwell relation A useful consequence of the First Law is that dU =TdS − pdV , where dU is the change in internal energy of a system and S is the systementropy. Use the equality of mixed partial derivatives to obtain a relationship betweencertain derivatives of T and p. Explain the relationship in words. Can you think of a use forthis relationship?

Problem 2.2 Adiabatic susceptibility Spend no more than a half an hour on thefollowing problem. You will be graded based on making a reasonable amount of progressdone, and on the correctness of your intermediate answer. We have the following equationsof state for the total magnetization M , and the entropy:


µBkBT

eµBkBT + e

− µBkBT

S = NkB

{ln 2 + ln

(eµBkBT + e

− µBkBT

)+

µB

kBT

eµBkBT − e−

µBkBT

eµBkBT + e

− µBkBT

}Solve for the magnetic susceptibility, which is defined as:

χB =

(∂M

∂B

)T


∂B

)S



Problem 2.3 Summation notation Towards the end of this course, we will be usingsummation notation quite a bit. Write x defined below without summation notation. Pleasewrite x in terms of only lower-case variables (in other words, eliminate A).

a)

x =5∑

n=0

2n (5.1)

b)

x =3∑

n=1

3∑m=1

m

n(5.2)

c)

x =∞∑i=0

1

2i(5.3)

d)

A =1∑i=0

bi (5.4)

x =1∑i=0

ciA

(5.5)

Problem 2.4 Euler’s homogeneous function theorem

a) Consider an extensive function X, which is a function of two extensive variables, Band D:

X = X(B,D)

What does this tell us about X(λB, λD)?

b) If we define

a ≡(∂X

∂B

)D

c ≡(∂X

∂D

)B

What is the total differential of X, expressed in terms of a, B, c and D?


c) Are a and c intensive, extensive or neither?

d) What are

a(λB, λD) =

c(λB, λD) =

e) Take a (total) derivative with respect to λ of both sides of your equation from part (a)for X(λB, λD). What does this tell us about X(B,D) in terms of a, B, c and D?

End of Energy and Entropy Homework 2Due Friday 4/22


6 Monday: Lab 1: Heat andTemperature

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

Lecture: (10 min) This class is called Energy and Entropy. You’ve already learned quitea bit about energy, but entropy is probably quite new to you.

Thermodynamics is a field that involves making experimental measurements of bulksubstances, and using theory to connect those measurements with other measurements. Wewill begin this course by making measurements of how much energy is needed to heat upwater (and ice) by a certain amount.

Activity: Energy/Heat equivalence (15 min now, 10 min later) We will use aresistive heating element to heat up some ice and water, measuring the current and thevoltage to find the power. We will measure its change in temperature to work out the heatcapacity of water and the latent heat of fusion of ice.

SWBQ: (5 min) What is temperature? What kind of a thing is temperature?

SWBQ: (5 min) What is energy? What kind of a thing is energy?

SWBQ: (5 min) What is heat? What kind of a thing is heat?

Lecture: (10 min) Now that you all have some data being collected, let’s talk abouthow you will be analyzing it. Many thermodynamic measurements are measurements ofderivatives. The temperature you can measure directly, and the power of the heater (which


is the energy dissipated into the water per unit time) we can work out directly. From thetwo of those, we will work out the heat capacityWrite on board:

“Cp =

(dQ

∂T

)p

”

The heat (called Q) is the amount of energy which is thermally transfered, just likework is the amount of energy which is mechanically transfered. I write the derivative funnybecause Q is not a function of T . The p subscript just means we’re keeping the pressureconstant.

SWBQ: (5 min?) What is entropy? What kind of a thing is entropy?

Another quantity we’ll be looking at is the entropy. We’ll spend much of this coursetalking about what entropy “really is”, but for now, just know that you can measure entropyby measuring heat and integrating:Write on board:

∆S =

∫dQreversible

T

Dulong-Petit Law

Lecture: (5 min) In 1819, shortly after Dalton had introduced the concept of atomicweight in 1808, Dulong and Petit observed that if they measured the specific heat per unitmass of a variety of solids, and divided by the atomic weights of those solids, the resultingper-atom specific heat was essentially constant. This is the Dulong-Petit law, although wehave since given a name to that constant, which is 3R or 3kB, depending on whether therelative atomic mass (atomic weight) or the absolute atomic mass is used.

This law isn’t precisely true, and isn’t always true, and is never true at low temperatures.But it captures some physics that we will later call the equipartition theorem. We will writeDulong-Petit’s law as:

Cp = 3NkB (6.1)

where N is the total number of atoms.


7 Tuesday: First and Second Laws

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

Activity: Name-the-experiment pretest (10 min)

System and surroundings

Lecture: (2-3 min) We distinguish in thermodynamics between a system and its sur-roundings . The system is the thing we’re measuring or predicting the properties of, andthe surroundings is everything else. Which is which will depend on what experiment we’redoing.

BWBQ: (10 min) In yesterday’s lab, what composed the system, and what composedits surroundings?

a) In the first stage, when we were melting the ice, we could say that the ice was thesystem, and the water, heater, thermometer and the rest of the world were the sur-roundings.

b) In the second bit, the water was the system, and everything else was the surroundings.

c) But from another perspective, the system was the water and the heater, and the ther-mometer (and the cup), since all these objects were being heated. Our measurement ofQ really measured the heat transferred to the whole combined system. We just hopedthat the heat capacities of the extraneous parts was negligible.

7.1 First Law

Lecture: (10 min) The first law of thermodynamics simply states that energy is con-served (or is a substance, to use Aristotle’s terminology). But it is useful to look at those


two non-state variables work and heat. Both are changes in energy of a system, so we canwrite the first law as

∆U = Q+W (7.1)

where U is the internal energy of the system, Q is the energy added to the system by heating,and W is the work done by the system (or the energy removed from the system by working).

It is more convenient mathematically, however, to have a framework for talking aboutinfinitesimal changes in the total energy...Write on board:

dU = dQ+dW

SWBQ: (5 min) If I stretch a rubber band and snap it shut repeatedly, what will happento its temperature? If students look puzzled, suggest they assume that as long as you dothis reasonably quickly, the air doesn’t get heated up much (nor the fingers).

7.2 Second Law and Entropy

SWBQ: (1 min) If you drop a hot chunk of metal into a cup of water, which way willenergy be transfered by heating? What is the rule that governs this?

Why can’t two objects that are at the same temperature spontaneously change temper-ature?

Lecture: (5 min) The second law of thermodynamics clarifies this rule, and extends itto cases where there might be other things going on, e.g. in the case of a refrigerator.

The second law involves the change in entropy, which I defined for you yesterday:

∆S =

∫dQreversible

T(7.2)

The Second Law of Thermodynamics simply states that for any possible process, thechange in entropy of a system plus its surroundings is either positive or zero.Write on board:

∆Ssystem + ∆Ssurroundings ≥ 0

In 2011, I ad libbed quite a bit after this.


Fast and slow

Lecture: (7-12 min)

a) Fast vs. slow

b) quasistatic vs. reversible vs. irreversible vs. spontaneous

c) adiabatic and its various meanings (including isentropic)

7.3 The thermodynamic identity (5.1.2.2 )

Lecture: (8-15 min) The internal energy is clearly a state function, and thus its differ-ential must be an exact differential.

dU = ? (7.3)

= dQ−dW (7.4)

= dQ− pdV only when change is quasistatic (7.5)

I spent some time on work being pdV .What is this dQ? As it turns out, we can define a state function S called entropy and so

long as a process is done reversibly

dQ = TdS only when change is quasistatic (7.6)

so finally we find out thatWrite on board:

dU = TdS − pdV

The fact that the T in this equation is actually the physical temperature was originally anexperimental observation. At this point, the entropy S is just some weird heat-related statefunction.

Activity: Name the experiment (20 min) “Name the experiment!” For each of thederivatives below, describe and draw a picture of the experiment that you would perform inorder to measure it. This is a big-white-board activity.(∂T

∂V

)S

(∂V

∂p

)T

(∂L

∂τ

)T

(∂U

∂T

)V

(∂V

∂T

)p

(∂L

∂T

)τ

(∂V

∂p

)S

(∂L

∂τ

)T

(∂U

∂p

)S

This first name-the-experiment activity is intended to use things that can be measured directly,or using the first law, provided one knows how to hold entropy constant.

In 2011, I had groups report on how they did their measurements, which took 6 of my 20minutes, but I think it was worthwhile. I might be tempted to put more groups on something

like(∂U∂p

)S

.


7.4 Heat capacity

SWBQ: (5 min (skipped in 2011)) Monday, we introduced the first law, and thethermodynamic identity in which we introduced a new state variable called the entropy S:

dU = dQ−dW (7.7)

= TdS − pdV (7.8)

Write down one thing we know about entropy from these two equations. Please try to becreative, so I can see more than one answer.

dQQS = TdS (7.9)

∆S =

∫dQQS

T(7.10)

T =

(∂U

∂S

)V

(7.11)

Any suggestions how we might go about measuring the entropy?

Lecture: (10 min) As we learned last week, heat capacity is amount of energy requiredto raise the temperature of an object by a small amount.

C ∼ dQ

∂T(7.12)

dQ = CdT At constant what? (7.13)

If we hole the volume constant, then we can see from the first law that

dU = dQ− pdV (7.14)

since dV = 0 for a constant-volume process,

CV =

(∂U

∂T

)V

(7.15)

But we didn’t measure CV on Monday, since we didn’t hold the volume of the waterconstant. Instead we measured Cp, but what is that? To distinguish between different sortsof heat capacities, we need to specify the sort of path used. So, for instance, we could write

dQ = TdS (7.16)

dQ = CαdT+?dα (7.17)

TdS = CαdT+?dα (7.18)

dS =CαTdT +

?

Tdα (7.19)


Cα = T

(∂S

∂T

)α

(7.20)

This may look like an overly-tricky derivative, so let’s go through the first law and checkthat we got it right in a few cases. I’ll do the CV case. We already know that

dU = dQ− pdV (7.21)

CV =

(∂U

∂T

)V

(7.22)

=

(∂U

∂S

)V

(∂S

∂T

)V

(7.23)

= T

(∂S

∂T

)V

(7.24)

where the second step just uses the ordinary chain rule.

BWBQ: (10 min (skipped in 2011)) Students work out the heat capacity Cp from thefirst law. If it looks necessary, tell them to use the enthalpy, but hopefully they’ll rememberthis from Tuesday.

dU = dQ− pdV (7.25)

H = U + pV (7.26)

dH = dQ+ V dp (7.27)(∂H

∂T

)p

= Cp (7.28)

Cp =

(∂H

∂S

)p

(∂S

∂T

)p

(7.29)

= T

(∂S

∂T

)p

(7.30)

Lecture: (5 min (skipped in 2011)) We can find a change in entropy from the heatcapacity quite easily:

∆S =

∫1

TdQQS (7.31)

=

∫C(T )

TdT (7.32)

Activity: Entropy change of cooling coffee (20-30 min (skipped 2010)) Studentswork out the change in entropy of a cup of coffee cooling from 100◦C to room temperature.Let them also work out the change in entropy of the room.


Required homework 3.1: adiabatic-ideal-gas

Required homework 3.2: bottle-in-bottle


Pretest

Question 1 Describe an experiment that would measure(∂τ

∂L

)T

where τ is the tension in a rubber band, L is its length, and T is its temperature. Be explicitabout how you would perform the measurement, and draw a sketch of your apparatus.

Question 2 Describe an experiment that would measure(∂L

∂T

)τ

where τ is the tension in a rubber band, L is its length, and T is its temperature. Be explicitabout how you would perform the measurement, and draw a sketch of your apparatus.


8 Wednesday: Second Law lab

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW

∆Ssystem+∆Ssurroundings ≥ 0

dU = TdS − pdV

Activity: Melting ice (45 min) Let’s talk about entropy changes in irreversible pro-cesses.

Let’s try doing this as a quickie lab! We’ll reuse the same materials as for Lab 1. We’llmass some water (possibly hot water) and measure its initial temperature. Then we add some0◦C ice (in cube form, so it can be removed), and mass the total to find the mass of the ice.Groups will solve to find out how much ice will be left, if any, and the final temperature oftheir ice water. They will also work out the change in entropy of the ice and the initial waterfor the process. By the time they’re done with the calculation, things should have equilibrated.They can then remove the ice cube to see how much of it has melted.

What happens when I add a 50g ice cube (at 0◦C) to 100g of room temperature water inan insulated container (but at constant pressure)? What is the change in entropy? What isthe final temperature? How much ice is left? What is the change in entropy of the ice? Ofthe water?

The heat capacity of liquid water is Cp = 4.18 J/g/K and is roughly constant over thistemperature range, and that the latent heat of fusion (a.k.a. the enthalpy of fusion) of iceis 333 J/g.

In 2010, (when the students only computed the answers without a lab portion) this couldhave taken more than the 20 minutes it took, as students didn’t have time to even look at theentropy, but only were able to figure out how much ice was melted. I told them the answer(that entropy increases again), but expect to address it again today (the next day).


Lab 1: Energy and temperatureIn this lab, we will be measuring how much energy it takes to melt ice and heat water.

Materials:

• Styrofoam cup

• Heating element

• Scale

• 2 digital multimeters

• Temperature guage

• Ice and water

V

A

The setup

You will put some mass of ice (about 50g) and ice-cold water (about 150g) into your styrofoamcup. Use the scale to record the mass of the ice and water as you add them to the cup. Finally,add your ice-cold heating element and thermometer through the lid of the cup.

Collect data

We will be measuring the temperature of the water and the power dissipated in the heatingelement (which is just a resistor). Thus we can find out how much energy was added to thewater, and how this changes the temperature. In order to keep the temperature measurementreasonable, we will need to periodically stir the cup and heat it moderately slowly.

You will be collecting temperature data using the computer, so before you turn on theheater, you should make sure the computer is taking data. Turn on the heater, and writedown the time you do so as well as the current and voltage, from which you can find thepower dissipated in the resistor. If the current or voltage changes during the course of theexperiment, take note of the new values—and the time.

Question 1.1 Plot your data I Plot the temperature versus total energy added to thesystem (which you can call Q). To do this, you will need to integrate the power. Discussthis curve and any interesting features you notice on it.

Question 1.2 Plot your data II Plot the heat capacity versus temperature. This willbe a bit trickier. You can find the heat capacity from the previous plot by looking at theslope.

Cp =

(∂Q

∂T

)p

(8.1)

This is what is called the heat capacity, which is the amount of energy needed to change thetemperature by a given amount. The p subscript means that your measurement was madeat constant pressure. This heat capacity is actually the total heat capacity of everythingyou put in the calorimeter, which includes the resistor and thermometer.


Question 1.3 Specific heat From your plot of Cp(T ), work out the heat capacity per unitmass of water. You may assume the effect of the resistor and thermometer are negligible.How does your answer compare with the prediction of the Dulong-Petit law?

Question 1.4 Latent heat of fusion

a) What did the temperature do while the ice was melting? How much energy was requiredto melt the ice in your calorimeter? How much energy was required per unit mass?per molecule?

b) The change in entropy is easy to measure for a reversible isothermal process (such asthe slow melting of ice), it is just

∆S =Q

T(8.2)

where Q is the energy thermally added to the system and T is the temperature inKelvin. What is was change in the entropy of the ice you melted? What was thechange in entropy per molecule? What was the change in entropy per molecule dividedby Boltzmann’s constant?

Question 1.5 Entropy for a temperature change Choose two temperatures that yourwater reached (after the ice melted), and find the change in the entropy of your water. Thischange is given by

∆S =

∫dQ

T(8.3)

=

∫ tf

ti

P (t)

T (t)dt (8.4)

where P (t) is the heater power as a function of time and T (t) is the temperature, also as afunction of time.


9 Thursday: Heat and work

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

Activity: Expanding Gas Quiz (20-45 min) Let’s start the class with a little quizabout expanding gasses, which is basically an exercise in using the first law.

In 2010, this took 45 minutes total, including 30 minutes of discussion and questions.Students were shocked by the increase in entropy in the irreversible free expansion, and didn’tlike the fact that entropy isn’t conserved.


Quiz

Question 1 Free expansion Consider the two processes described below.

Process #1 Five moles of an ideal gas are initially confined in a one-liter cylinder witha movable piston, at a temperature of 300 K. Slowly the gas expands against the movablepiston, while the cylinder is in contact with a thermal reservoir at 300 K. The temperatureof the gas remains constant at 300 K while the volume increases to two liters.

Process #2 A thin plastic sheet divides an insulated two-liter container in half. Fivemoles of the same ideal gas are confined to one half of the container, at a temperature of300 K. The other half of the container is a vacuum. The plastic divider is suddenly removedand the gas expands to fill the container. Because it is a free expansion of an ideal gas (nowork is done on or by the gas), the final temperature of the gas is also 300 K.

Process #3 The same cylinder as in process #1 is thermally insulated and then allowedto slowly expand, starting at 300 K, to twice its original size (two liters).

#1 Isothermal expansion #2 Free expansion #3 Adiabatic expansion

300K 300K 300K

300K 300K ? K

a) Are ∆Sisothermal, ∆Sfree and ∆Sadiabatic, the change in entropy of the gas for each process,positive, negative, or zero? Please explain your reasoning.

b) Is ∆Sisothermal greater than, less than, or equal to ∆Sfree? How do each of these comparewith ∆Sadiabatic? Please explain.

c) Are ∆Ssurr-isothermal, ∆Ssurr-free and ∆Ssurr-adiab, the change in entropy of the surround-ings for each process, positive, negative, or zero? Please explain.


Activity: Name the experiment with changing entropy (20 min) “Name theexperiment!” For each of the derivatives below, describe the experiment that you wouldperform in order to measure it.(

∂S

∂V

)T

(∂S

∂T

)V

(∂S

∂p

)T

(∂S

∂T

)p

(9.1)

In this activity, we’ll look again at what changing entropy means, and how we might measurethat change in entropy.

9.1 Work

Lecture: (5 min) What work is can depend on the system you are looking at.

a) For a 3-D thing, −pdV

b) For a 2-D thing, σdA

c) For a 1-D thing, τdL

d) For a dielectric material, − ~E · d~P , for a paramagnet − ~B · d ~M etc...

BWBQ: (10 min) For each of the following pictures, is work done by or on the system?i.e. does it lose or gain energy by working?

a) A gas in a piston that is being compressed. Does it matter if it’s a liquid instead?

b) A balloon in a box (of vacuum) that is popped. What if the air is the system? Whatif the balloon & the air together form the system?

c) A gas in a piston that is expanded.

d) Ice in a piston that is being compressed.

e) A rubber band that is being stretched.

f) A soap bubble that is being blown.

g) A piece of iron that is being magnetized.

h) A rubber band that is snapped shut in a vacuum.

Activity: Using pV plots (20-30 min) We’ll begin the day by looking at processesthat happen at constant pressure, volume, temperature or entropy. This touches on thename-the-experiment discussion we had yesterday.


0 Volume

Pressu

re

We talked about measurements such as(∂τ

∂L

)S

(9.2)(∂L

∂T

)τ

(9.3)

(9.4)

In these measurements, we measure how a state variable changes aswe vary another one. Another sort of measurement involves integralsrather than derivatives, and measures finite changes. To discuss this,it’s common to use what are called pV diagrams. For instance, considerthe following square.

“What does this describe?”“Is p a function of V ?”In your groups, work out the following questions:What is the net work done after one cycle of this process? How much work was done at

each step?What is the net heat transfer over one cycle of this process? For each step?

0 S

T

Activity: Using TS plots (15-20 min) Now let’s look at anothercycle. Let’s consider The following figure, which looks similar, but isnow a plot of T vs. S.

What is this cycle? How would you go about running a cycle likethis?

What is the net heat transfer over one cycle of this process? Howmuch was transfered on each step?

What is the net work done after one cycle of this process? How muchwork was done at each step?

Activity: Name the experiment for a rubber band (15 min)“Name the experiment!” For each of the derivatives I write on the board,describe the experiment that you would perform in order to measure it for a rubber band.(

∂τ

∂L

)T

(∂τ

∂L

)S

(∂τ

∂T

)L

(∂τ

∂T

)S

(∂S

∂L

)T

(∂L

∂τ

)T

(∂L

∂τ

)S

(9.5)(∂L

∂T

)τ

(∂L

∂T

)S

(∂T

∂L

)S

(∂S

∂τ

)T

(∂T

∂τ

)S

(∂S

∂T

)τ

(∂S

∂T

)L

(9.6)

In class 2010, we ended up spending most of our wrap-up talking about what it means to holdentropy fixed. Students were also confused as to whether constant Q meands constant T .


10 Friday:

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

10.1 Engines and Fridges

Other views of the Second Law

Lecture: (??? min) We’ve already looked at the second law of thermodynamics:

∆Ssystem + ∆Ssurroundings ≥ 0 (10.1)

As it turns out there are a couple of other equivalent ways to state this law. The Kelvinformulation states that No process is possible in which the sole result is the absorption ofheat from a reservoir and its complete conversion into work. The Clausius formulationstates that No process is possible whose sole result is the transfer of heat from a body oflower temperature to a body of higher temperature. These formulations make it clear that ifyou could violate the Second Law, you could become filthy rich.

Tc

Th

Qc

Qh

W

Carnot efficiency

Lecture: (??? min) A heat engine is a device that converts heatinto work. I will diagram heat engines as displayed in this picture. Theheat engine contains several parts.

• At the top and bottom are hot and cold heat sinks . The operator ofthe engine has to keep these two heat sinks at fixed temperature, whichmeans burning fuel to warm up the hot sink, and using something likea radiator to keep the cold sink cool.


• In the middle of the picture is the engine itself, which will contain somesort of a working substance that is (most likely) alternately heated andcooled.

• There is some amount of heat Qh transfered from the hot sink to theengine, and some other amount of heat Qc (both taken to be positivenumbers) transfered from the engine to the cold sink. By the first law,the difference between these must be the work.

W = Qh −Qc

It may seem like heat engines (and steam engines) are a bit old-fashioned,but about 80% (according to wikipedia) of electric power in the worldis generated by steam turbines1—which are simple heat engines. Soit’s not really a 19th century application, although it was pretty wellunderstood in the 19th century.

Tc

Th

Qc

Qh

W

Efficiency The efficiency in general is what you get out divided bywhat you put in. In this case, what we have to put in is the heat addedto the hot resevoir Qh, and what we get out is the work W , so

η =W

Qh

(10.2)

=Qh −Qc

Qh

(10.3)

= 1− Qc

Qh

(10.4)

So clearly, we’d like to minimize the amount of heat sent to the coldsink. This also has environmental advantages.

If each step is done reversibly, then we could run a heat engine inreverse, and have a refridgerator. Thus we put do work on the fridge,and cool off the cold sink, while warming up the hot sink.

Tc

Th

Qc

Qh

W

Qc

Qh

The efficiency of any reversible heat engine must be the same as theefficiency of any reversible refridgerator. We can see this by using a heatengine to drive a refridgerator. By choice the work done by the engine isthe same as the work done on the fridge. If the efficiencies of the fridgeand engine differ, then there will be a net transfer of heat either fromhot sink to cold sink or from cold sink to hot sink. The former wouldbe reasonable and natural, but the latter would be crazy, which meansthat the fridge cannot be more efficient than the engine. However, if

1This consists of all coal-burning plants and nuclear power plants, and I’m not sure what else... almostcertainly geothermal.


both fridge and engine are reversible, then if the fridge is less efficientthan the engine, then we could run the thing in reverse, and get thecrazy situation again happening, in which nothing changes except thatheat is transferred from a cold place to a hot place... and that just isn’t natural! “So we canonly conclude that every possible reversible heat engine must have the same efficiency!”

Activity: Carnot efficiency (??? min) For a reversible engine, the net entropy changeof system plus surroundings must be zero over each cycle. Since the entropy change of thesystem is always zero, even for an irreversible engine, we only need consider the entropychange of the surroundings.

“In your groups, on big white boards, use these properties to work out the efficiency of areversible heat engine.”

This could be done (more quickly) as lecture, in which case the following is what I’d do:

∆Sh = −Qh

Th(10.5)

∆Sc =Qc

Tc(10.6)

∆Sh + ∆Sc = 0 (10.7)

=Qc

Tc− Qh

Th(10.8)

Qc

Tc=Qh

Th(10.9)

Qc

Qh

=TcTh

(10.10)

ηcarnot = 1− TcTh

(10.11)

This is the Carnot efficiency , and is acheived by any reversible heat engine. This efficiencyalso is the upper bound on the efficiency of any heat engine. Note that this efficiencyapplies equally well to traditional cyclic heat engines, steam turbines, photovoltaics, andthermopower devices. There is also a corresponding limit on the efficiency of refridgerationdevices, which has the same broad applicability.

Activity: Big money (15 min?) “Supposing you are able to violate the Second Law inthe following way, work out a design for a device that would allow you to make big money.In each case, I will give you . ”

a) You develop a special filter with nanopores that allow air molecules to pass through inone direction, but not the other direction.

b) Using metamaterials you create a thin film that allows blue light to pass through it inone direction, but not the other.


c) You develop a nano-door that allows only high-energy molecules to pass through itin one direction, but allows only low-energy molecules to pass through in the otherdirection—and requires no energy input.


Lab 2: Second Law

Materials: styrofoam cup, scale, thermometer, ice and water

The setup You will put around 200g of water in your cup, massing it with the scale as youadd it. You will be given a goal for the initial temperature for the water used by your group,which will be somewhere between room temperature and boiling. Measure the temperatureof the water after it has been in the cup long enough to equilibrate with the cup. Then addaround 100g of 0◦ ice to the cup and record its mass. Cover the cup and wait for the ice tomelt. While you wait, work in your groups on the following questions using your big whiteboards.

The heat capacity of liquid water is Cp = 4.18 J/g/K and is roughly constant over thetemperature range we will be using, and that the enthalpy of fusion (a.k.a. latent heat) ofice is 334 J/g.

Group analysis

Question 2.1 Mass of ice remaining Work out the mass of ice (if any) that will remainafter the cup has reached thermal equilibrium.

Question 2.2 Final temperature of water Work out the final temperature of the wa-ter/ice mixture.

Question 2.3 Change in entropy of water Work out the change in entropy of the waterthat happened as it cooled down.


Question 2.4 Change in entropy of ice Work out the change in entropy of the ice asit melted and (possibly) warmed up.

Question 2.5 Net change What is the net change of entropy for this entire adiabaticprocess?

Compare with experiment Look inside your cup once you’ve done all the above work.

Question 2.6 Mass of ice remaining Is there any ice left? If so, carefully fish it outand measure the mass of the remainder to find out how much ice was left.

Question 2.7 Final temperature Measure the final temperature of the water in thecup.

Question 2.8 Errors? Comment on the magnitude and sources of errors in your experi-ment and/or prediction. What could you do to reduce or estimate these errors?



Problem 3.1 Adiabatic compression Consider the isothermal expansion of a simpleideal gas. The internal energy is given by

U = CvT (10.12)

where you may take Cv to be a constant—although for a polyatomic gas such as oxygen ornitrogen, it is temperature-dependent. The ideal gas law

pV = NkBT (10.13)

determines the relationship between p, V and T . You may take the number of molecules Nto be constant.

a) Use the first law to relate the inexact differential for work to the exact differential dTfor an adiabatic process.

b) Find the total differential dT where T is a function T (p, V ).

c) In the previous two sections, we found two formulas involving dT . Use the additionaldefinition of work dW = −pdV to solve for the relationship between p, dp, V and dVfor an adiabatic process.

d) Integrate the above differential equation to find a relationship between the initial andfinal pressure and volume for an adiabatic process.

Problem 3.2 A bottle in a bottle The internal energy of helium gas at temperatureT is to a very good approximation given by

U =3

2NkBT (10.14)

Consider a very irreversible process in which a small bottle of helium is placed inside a largebottle, which otherwise contains vacuum. The inner bottle contains a slow leak, so that thehelium leaks into the outer bottle. The inner bottle contains one tenth the volume of theouter bottle, which is insulated. What is the change in temperature when this process iscomplete? How much of the helium will remain in the small bottle?



Practice homework 4.1: power-from-ocean

Required homework 4.2: power-plant-river

Required homework 4.3: heat-pump


11 Monday:

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

Monotonicity and invertibility

Activity: Monotonicity lecture/discussion (10 min) Ask groups to discuss howmany variables are needed to define the state of a given system. It depends on what you arewilling to allow to change. V and T would be sufficient, but you might also include N , andmight want to include even more if you have more than one sort of molecule.

p

V

good(x)bad(x)

Since 2 (or 3) variables fully define the state of a given sys-tem, we know that every other state variable could be writtenas a function of those two variables. “But which two variablesare those?”

We can reason about this pretty easily. “Which state vari-ables can I directly control?” T , S, p, V . But which of thesecould be the two variables that define the state of our system?Let us consider a simple plot of p versus V at constant tem-perature. “Which of these curves are physically reasonable?”

Talk about creating a box full of some material, say a gasor liquid. I could choose to fix the volume of the box, oralternatively I could give it a piston on top and exert a fixed force on it, thus determiningthe pressure. “Should it matter which approach I use?” This tells us that for any givenpressure, there must be only one volume, and for any given volume, there can only be onepressure. Thus our curve must be monotonic. “How would our answer change if we were toplot p versus V at fixed pressure?” (

∂p

∂V

)T

≤ 0 (11.1)


(∂p

∂V

)S

≤ 0 (11.2)

Now let’s think about temperature and entropy. The same argument applies, indicating thattemperature and entropy must be fixed. Note that although it’s hard to “fix” the entropy,it’s easy to change entropy by heating up a system. And to keep it from changing, we onlyneed insulate the system and avoid doing anything irreversible. So the same argument showsthat temperature and entropy are monotonically related. The sign we can work out fromthe property of temperature that hot things heat up cold things, and the thermodynamicdefinition of entropy change. (

∂T

∂S

)p

≥ 0 (11.3)(∂T

∂S

)V

≥ 0 (11.4)

We will also see that the heat capacity is always positive, since it relates to precisely thesederivatives.

Note that monotonicity does not indicate proportionality!

11.1 Thermodynamic potentials (Chapter 7 )

Legendre transform

Activity: Legendre transform revisited (15 min?) Recall that:

dU = TdS − pdV (11.5)

As you may also recall, we worked out three other thermodynamic potentials using Legendretransforms:

H = U + pV (11.6)

F = U − TS (11.7)

G = U + pV − TS (11.8)

On your whiteboards, work out dH, dU and dG.Write on board:

dF = −SdT − pdV

Write on board:dH = TdS + V dp

Write on board:dG = −SdT + V dp


Activity: Understanding the potentials (20 min) Students work out in groups whatthe changes in H and F are, in processes that hold one or the other of their natural variablesconstant. Groups that are ahead can be asked to consider the Gibbs free energy.∫

constant p

dH =

∫TdS (11.9)

∆H = Qconstant p (11.10)

What is ∫constant T

dF =

∫pdV (11.11)

∆F = Wconstant T (11.12)

Without proof

K = e−β∆G (11.13)

Lecture: (5 min?) This is a possible wrap-up for the previous activity.What is the enthalpy , Helmholtz free energy and Gibbs free energy?This is easiest to see in a couple of simple cases. Suppose you want to know how much

energy is needed to melt some ice or boil some water at constant pressure? For one thing,you’d need to know the change in internal energy U , but that’s not all. Since the volumechanges, you also will need to push aside some air, and that’s going to require some work...in fact, it’ll require p∆V of work. So the total energy we’ll need to boil the water or meltthe ice will have to account for that work as well. In one case, it’ll be easier than we think,and in the other case it’ll be harder. The change in enthalpy will give us the answer in eithercase.

Now suppose we have have some steam and we want to use it to do some work. Howmuch work can we get out of it? Obviously, the internal energy is relevant here, since energymust be conserved. But suppose we want to achieve this work at constant temperature (sowe won’t have to insulate things)? If we’re keeping things at constant temperature, thenthere is heating going on... so the amount of work can’t be the same as the change in internalenergy. How much work can we do? The answer is given by the change in Helmholtz freeenergy.

Finally we get to the Gibbs free energy. As you might gather, it’s helpful when you wantto keep both temperature and pressure constant. You might wonder what you could possiblybe changing, if you keep both pressure and temperature constant! One possibility is thatyou could be doing a phase transition, like melting ice at zero centigrade and atmosphericpressure. This is possible because the Gibbs free energies of ice and water are the same atthat temperature and pressure. Another possibility is that you’re undergoing a chemicalreaction... which is another sort of a phase transition. Reactions or phase transitions arereversible (or “in equilibrium”) when the change in Gibbs free energy is zero. If it’s negative,they happen spontaneously, and if it’s positive they don’t happen at all.


Required homework 4.4: gibbs-free-energy

11.2 Maxwell relations

Lecture: (5 min) Last week, we learned that mixed partial derivatives are the same,regardless of the order in which we take the derivative, so(

∂(∂f∂x

)y

∂y

)x

=

∂(∂f∂y

)x

∂x

y

(11.14)

∂2f

∂x∂y=

∂2f

∂y∂x(11.15)

In thermodynamics, partial derivatives are often physical quantities, things we can measure.In such a case, their derivatives also may be measurable—and important—quantities.

For instance, we already know that

dU = TdS − pdV (11.16)

dF = −SdT − pdV (11.17)

dH = TdS + V dp (11.18)

dG = −SdT + V dp (11.19)

where T is the temperature, p is the pressure, V is the volume, U is the internal energy , Sis the entropy , F is the Helmholtz free energy , H is the enthalpy , and G is the Gibbs freeenergy .

BWBQ: (15-25 min) How are the following thermodynamic derivatives related? (Eachgroup gets one or two)

a) (∂T

∂p

)V

and

(∂S

∂V

)T

b) (∂T

∂p

)S

and

(∂S

∂V

)p

c) (∂T

∂V

)p

and

(∂p

∂S

)T

In 2010 class, I turned some of these derivatives upside down, which made the problemconsiderably more challenging, which I think was useful.

In 2011, I have the derivatives already pre-turned-upside-down.


Activity: Name the experiment with Maxwell relations (10 min) “Name theexperiment!” For each of the derivatives I write on the board, describe the experiment thatyou would perform in order to measure it.(

∂S

∂V

)T

(∂S

∂p

)T

(11.20)

For extra fun... (∂τ

∂S

)L

(∂τ

∂S

)T

(∂L

∂S

)T

(∂L

∂S

)τ

(∂S

∂τ

)L

(∂S

∂L

)τ

(11.21)


12 Tuesday: Lab 2: rubber band

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c(

∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp

Activity: Rubber band lab See Lab 15.Note: students will probably want to take data separated by 5◦C or perhaps 10◦C.


Prelab 3: Thermodynamics of a rubber bandPlease answer the following questions before beginning the experiment. Think carefullybefore making your predictions, but don’t worry if you aren’t sure what will happen. Inevery case, give a reason for your prediction—guessing is perfectly acceptable, but if you areonly guessing, just say so.

Question 3.1 Tension vs. temperature Will the tension increase or decrease withincreasing temperature? Sketch a graph of what you expect to observe. Be sure to includezero on the graph, and try to sketch the expected results to scale.

Question 3.2 Isothermal stretch Pick a temperature and a range of lengths for whichyou (expect to) have clean data.1

a) If you stretch a rubber band, holding it fixed temperature, what will be the change inits internal energy? its entropy? Be quantitative and be explicit about whether theywill increase or decrease.

b) What do you expect will be the sign of the work W? of the heat transfered Q? Explainwhat this means in terms of an actual observation (i.e. are you doing work or is therubber band, and is it heating its environment or vice versa).

c) If the rubber band is reversibly isothermally stretched, what will the magnitude of Wand Q be? Which is larger, and does this make sense to you? How do the magnitudesof W and Q compare with the magnitude of the change in internal energy?

1I understand that you will probably have to guess on many of these. Use intuition where you can, andsee if you can come up for a justification for reasonable values. The units of entropy are J/K, and all otherquantities have dimensions of energy.


Lecture: (10 min) About Wednesday’s homework. . .On solving for Cp(T ), one of the challenging problems was dealing with the melting

portion of the time, when dT/dt = 0. Most of you gathered that Cp was either undefined orinfinite. It can be seen to be infinite, from the definition of Cp on the board, which meansthat

∫CpdT = Q, which is finite and positive.

On the last problem, more of you had trouble. First, as in the previous problem, youneeded to be sure to use T in Kelvins. If it ever makes a difference, you always need to useKelvins, as the zero value of Celsius is arbitrary.

Secondly, many of you struggled with actually doing the integral.

∆S =

∫ t2

t1

P (t)

T (t)dt (12.1)

= P

∫ t2

t1

1

T (t)dt (12.2)

At this point, there are two reasonable options. One is to simply do the integral numerically.After all, your data is already in a spreadsheet.

∆S =

∫ t2

t1

1

T (t)dt (12.3)

= P∑i

1

Ti∆t (12.4)

= P∑i

1

Ti(1s) (12.5)

This you can just plug into the spreadsheet and get the answer. This works especially wellif your data doesn’t match a very nice analytic form. In most of your data, however, thetemperature was quite linear, so you could write

T (t) = T0 +mt (12.6)

∆S = P

∫ t2

t1

1

T (t)dt (12.7)

= P

∫ t2

t1

1

T0 +mtdt (12.8)

(12.9)

But this starts looking pretty nasty. A nicer approach would be to write

dT = mdt (12.10)

∆S = P

∫ T2

T1

1

T

dT

m(12.11)

=P

m

∫ T2

T1

1

TdT (12.12)


=P

mlnT |T2

T1(12.13)

= Cp lnT |T2

T1(12.14)

I should also note that this problem could be approached more directly, although myhomework question didn’t lead you in this direction.

∆S =

∫dQ

T(12.15)

=

∫ T2

T1

Cp(T )dT

T(12.16)

= Cp

∫ T2

T1

dT

T(12.17)

Finally, I’d like to point out that Cp is not generally a constant. If we had had a chanceto go to higher temperatures, we would have seen Cp dropping off (i.e. slope would havedecreased a bit). Some of you may have been able to see this as a subtle drop in the slopeof your curve.

Required homework 4.5: free-expansion


13 Wednesday:

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c(

∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp

Activity: Name another experiment with Maxwell relations (10-20 min) “Nameanother experiment!” For each of the following derivatives, we can already come up witha simple experiment. Use a Maxwell relation to come up with yet another way to measurethis. (

∂τ

∂L

)T

(∂τ

∂T

)L

(∂V

∂T

)p

(∂T

∂p

)V

(∂T

∂V

)S

(∂S

∂T

)L

(13.1)


Problem 4.1 Power from the ocean (practice) It has been proposed to use the thermalgradient of the ocean to drive a heat engine. Supoose that at a certain location the watertemperature is 22oC at the ocean surface and 4oC at the ocean floor.

a) What is the maximum possible efficiency of an engine operating between these twotemperatures?

b) If the engine is to produce 1 GW of electrical power, what minimum volume of wa-ter must be processed every second? Note that the heat capacity of water Cp =4.2 Jg−1K−1 and the density of water is 1 g cm−3, and both are roughly constant overthis temperature range.


Problem 4.2 Power plant on a river At a power plant that produces 1 GW (109watts)of electricity, the steam turbines take in steam at a temperature of 500oC, and the wasteenergy is expelled into the environment at 20oC.

a) What is the maximum possible efficiency of this plant?

b) Suppose you arrange the power plant to expel its waste energy into a chilly mountainriver at 15oC. Roughly how much money can you make in a year by installing yourimporved hardware, if you sell the addtional electricity for 5 cents per kilowatt-hour?

c) At what rate will the plant expel waste energy into this river?

d) Assume the river’s flow rate is 100m3/s. By how much will the temperature of theriver increase?

e) To avoid this “thermal pollution” of the river the plant could instead be cooled byevaporation of river water. This is more expensive, but it is environmentally preferable.At what rate must the water evaporate? What fraction of the river must be evaporated?

Tc

Th

Qc

Qh

W

Problem 4.3 Heat pump A heat pump is a refridgerator (or airconditioner) run backwards, so that it cools the outside air (or ground)and warms your house. We will call Qh the amount of heat delivered toyour home, and W the amount of electrical energy used by the pump.

a) Define a coefficient of performance γ for a heat pump, which (likethe efficiency of a heat engine) is the ratio of “what you get out”to “what you put in.”

b) Use the second law of thermodynamics to find an equation forthe coefficient of performance of an ideal (reversible) heat pump,when the temperature inside the house is Th and the temperatureoutside the house is Tc. What is the efficiency in the limit asTc � Th?

c) Discuss your result in the limit where the indoor and outdoortemperatures are close, i.e. Th − Tc � Tc. Does it make sense?

d) What is the ideal coefficient of performance of a heat pump whenthe indoor temperature is 70◦F and the outdoor temperature is50◦F? How does it change when the outdoor temperature drops to30◦F?


Problem 4.4 Using the Gibbs free energy You are given the following Gibbs freeenergy:

G = −kTN ln

(aT 5/2

p

)where a is a constant (whose dimensions make the argument of the logarithm dimensionless).

a) Compute the entropy.

b) Work out the heat capacity at constant pressure Cp.

c) Find the connection among V, p,N, and T , which is called the equation of state.

d) Compute the internal energy U .

Problem 4.5 Free expansion The internal energy is of any ideal gas can be written as

U = U(T,N) (13.2)

meaning that the internal energy depends only on the number of particles and the temper-ature, but not the volume.1 The ideal gas law

pV = NkBT (13.3)

defines the relationship between p, V and T . You may take the number of molecules N tobe constant. Consider the free adiabatic expansion of an ideal gas to twice its volume. “Freeexpansion” means that no work is done, but also that the process is also neither quasistaticnor reversible.

a) What is the change in temperature of the gas?

b) What is the change in entropy of the gas? How do you know this?


1This relationship happens to be linear at low temperatures, where “low” is defined relative to the energyof the excited states of the molecules or atoms.


14 Thursday: Thermodynamics practice

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c(

∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp

Activity: Simple cycle (40-110 min) Consider the following simple cycle for a monatomicideal gas.

U =3

2NkBT (14.1)

pV = NkBT (14.2)

We start at V0, p0 and T0. We first allow the gas to expand to twice its original volume at fixedtemperature. We then cool it at fixed pressure until it returns to its original volume. Finally,we heat it up at fixed volume until it returns to the original pressure and temperature.

What is the net work done? How much work for each leg of the cycle?What is the net energy transfered by heating? For each leg of the cycle?What is the efficiency of this engine? How does it compare to the Carnot efficiency?

Why is it less efficient?What is the change in entropy on each leg? The net change in entropy?In 2011, this took a total of 120 minutes, spread between two class periods. It felt like a

very long time, but students in both groups that went very quickly or groups that went veryslowly felt that it was quite helpful.

Activity: Temperature change of dissolving salt (40 min?) The enthalpy of dis-solution (dissolving) of NaCl in water is 4 × 103J/mol. The specific heat cp of water is4.2 J/g/K. What is the change in temperature of a cup holding 100 g of water (moderatelywell insulated), if you dissolve 30 g of salt in it?


What happened to the entropy of the isolated salt plus water system? Can we predictthe value of the entropy change from the information given? Could we measure the changein entropy? How?

Can we do this experiment?What assumptions did we make? We neglected the heat capacity of the salt, and any

change it would have on the heat capacity of the water.

∆Hdis =4× 103J/mol · 30g

23g/mol + 35.4g/mol(14.3)

= 2000J (14.4)

∆T ≈ Q

Cp(14.5)

=2000J

4.3J/g/K · 100g(14.6)

≈ 5K (14.7)

Activity: Never, sometimes or always true (60 min?) “For each of the scenariosin the handout, work out in your groups whether it is never true, sometimes true or alwaystrue. But more importantly, provide reasoning for your answer, expressed in terms of thelaws or principles of thermodynamics! ”

Students will naturally expect that 1b is always true, neglecting the choice to keep pressureconstant. I needed to talk with each group, and have them write down the First Law, in orderto realize that they don’t know the sign or magnitude of the work, and thus don’t know whetherthe internal energy goes up or down.

Students continue to be unclear as to which quantities are state functions, and also arevague on the process of using a reversible path with the same starting and ending points todetermine the change in a state function for an irreversible process.


For each of the following statements, work out in your group whether it is never true,sometimes true or always true. Explain why your answer is correct, being specific aboutwhat fundamental laws are responsible. If it is only sometimes true, try to give examples ofeach case.

Hot object in cool water If I put a hot object (of any material) into a tub of colderwater without changing its pressure:

a) The temperature of the object will decrease.

b) The internal energy of the object will decrease.

c) The entropy of the object will decrease.

d) The volume of the object will decrease.

e) The entropy of the object plus the entropy of the water will increase.

Identical objects Consider two identical objects A and B, which could be of any material,and are of the same mass, but are subject to different conditions (pressure, temperature).

a) If A has higher volume than B, then A has higher entropy.

b) If A has higher temperature than B, then A has higher internal energy.

c) If A has higher temperature than B, but they have the same volume, then A has higherinternal energy.

d) If A has higher temperature than B, but they have the same pressure, then A hashigher internal energy.

e) If A has higher temperature than B, but they have the same pressure, then A hashigher enthalpy. (H = U + pV )

f) If A has higher temperature than B, but they have the same pressure, then A hashigher entropy.


Bag of hot air? Consider an insulated spherical bag full containing a fluid—which couldbe either liquid or solid. The bag is very strong, so it cannot be stretched, and also happensto be an excellent thermal insulator.

a) If I sit on the bag, it will no longer be spherical.

b) If I sit on the bag, the pressure of the bagfull of fluid will increase.

c) If I sit on the bag, the volume of the bagfull of fluid will increase.

d) If I sit on the bag, the temperature of the bagfull of fluid will increase.

e) If I sit on the bag, the entropy of the bagfull of fluid will increase.

f) If I sit on the bag, the internal energy of the bagfull of fluid will increase.

g) If I sit on the bag, the enthalpy of the bagfull of fluid will increase.

h) If I sit on the bag, the Helmholtz free energy of the bagfull of fluid will increase.

Aluminum balloon? Consider a similar spherical bag that is an excellent thermal con-ductor. In this problem, assume that my bottom is at room temperature—perhaps becauseI’m wearing asbestos undergarments.

a) If I sit on the bag and wait a while, it will no longer be spherical.

b) If I sit on the bag and wait a while, the pressure of the bagfull of fluid will increase.

c) If I sit on the bag and wait a while, the volume of the bagfull of fluid will increase.

d) If I sit on the bag and wait a while, the temperature of the bagfull of fluid will increase.

e) If I sit on the bag and wait a while, the entropy of the bagfull of fluid will increase.

f) If I sit on the bag and wait a while, the internal energy of the bagfull of fluid willincrease.

g) If I sit on the bag and wait a while, the enthalpy of the bagfull of fluid will increase.

h) If I sit on the bag and wait a while, the Helmholtz free energy of the bagfull of fluidwill increase.


15 Friday: Thermodynamics practice

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c(

∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp

insulation

T

1

2

T1 2

Q

Q

Activity: Black body thermodynamics (30 min) Let’s considerthe following diagram. Here we’ve got two objects in an insulated en-vironment, surrounded by vacuum, so the only way they can exchangeenergy is through electromagnetic radiation. Further, let’s assume thateach object is perfectly black, so that it absorbs any radiation incidenton it.

• Suppose T1 > T2, what can we say about Q1 and Q2 on the basisof the Second Law?

• Suppose T1 = T2, what can we say about Q1 and Q2?

• Now let us suppose that we replace the right-hand object with onethat only absorbs two thirds of incident radiation and reflects theother one third. How does that change our previous answers?

Wrap-up Note that our results must hold even if we put a filter that reflects all but onefrequency in between the two objects, which means that at every frequency, the absorptionmust be proportional to the emmision, with the same proportionality constant!

Activity: Applying the second law Students work out which processes are reversible,which are irreversible, and which are impossible.

Hopefully from their first lab, they’ll have a value for the differences in entropy per moleof liquid water at several temperatures and of ice.


• What happens if you throw an ice cube into water at room temperature? Is the reversepossible?

• What happens if you drop a room-temperature rubber band into ice-cold water? Isthe reverse possible?

• You leave a cup of water in a sealed room and some of it evaporates.

• You drop a stone into a bucket of water.

Required homework 5.1: maine-entropy-2nd-law-spontaneous-metal

Required homework 5.2: isothermal-adiabatic-compressibility


Lab 3: Thermodynamics of a rubber bandIn this lab, we will measure the tension of a rubber band stretched to a fixed length, as afunction of temperature. By measuring this at several similar lengths, we will be able todetermine the change in internal energy U and entropy S for a rubber band that is stretchedat fixed temperature.

Materials:

• One rubber band

• Tube

• Stopper with hook

• Vernier force guage

• Vernier temperature guage

• Several clamps

• Boiling water

• Ice

• Pan

Background

The inexact differential of work for a system such as a rubber band that is stretched in justone direction is

d−W = −τdL. (15.1)

This follows naturally from the definition of work as force dotted with distance—providedone takes into account the sign convention for tension, which is opposite that of pressure.Thus the thermodynamic identity is:

dU = TdS + τdL (15.2)

We could use the thermodynamic identity directly, but since we are working at constantT , it is more helpful to consider the Helmholtz free energy

F ≡ U − TS (15.3)

The total differential of F is

dF = τdL− SdT (15.4)

from which we can extract a Maxwell relation:

∂2F

∂L∂T=

∂2F

∂T∂L(15.5)(

∂

∂L

(∂F

∂T

)L

)T

=

(∂

∂T

(∂F

∂L

)T

)L

(15.6)

−(∂S

∂L

)T

=

(∂τ

∂T

)L

(15.7)


By measuring the variation of tension with temperature at fixed length, we can find out howentropy will change when the length is changed at fixed temperature! At the same time,a measurement of the tension will tell us how the free energy will vary under the sameisothermal stretch:

τ =

(∂F

∂L

)T

(15.8)

Thus, provided we know τ and ∂τ/∂TL for our rubber band, as a function of the length, wecan integrate to find ∆S, ∆F and ∆U for an isothermal stretch. .

The setup

You will stretch your rubber band between a force me-ter and a hook in the stopper in the bottom of a pipe.If you attach the rubber band to the force meter bymeans of a chain of paper clips, then you can ensurethat when the pipe is filled with water, the rubber bandis completely immersed.

You will need to insert a thermometer probe intothe top of the pipe in order to measure the tempera-ture of your water—and thus the temperature of yourrubber band.

Collect data

During the experiment you will pour water of variousdifferent temperatures into the pipe in order to heat upyour rubber band, and record the tension as a functionof length. When changing to a different temperature, you will need to empty your pipe andrefill it. Each time, you should measure the temperature at the top and bottom to ensurethat the water is well mixed.

Conclusions and Questions

Please answer the following questions. As always, show your work. In each case, discusswhether your result agrees with your predictions. If it disagrees, please attempt to explainwhat was wrong with your reasoning.

Question 3.3 Tension vs. temperature Plot the tension versus temperature for eachof your lengths on the same plot.


Question 3.4 Tension vs. length Plot the tension versus length for a few tempera-tures.

Question 3.5(∂S∂L

)T

vs. length Plot(∂S∂L

)T

versus length for the same set of temperaturesyou chose for Question 3.4.

Question 3.6 Isothermal stretch Pick a temperature and a range of lengths for whichyou have clean data.

a) What is the change in free energy for an isothermal stretch at this temperature fromthe smallest length to the largest?

b) What is the change in entropy?

c) What is the change in the internal energy?

d) What is Q?

e) What is W?

f) What is |Q/W |?

Question 3.7 Adiabatic stretch What additional experiments would you need to per-form in order to answer Question 3.6 for a stretch that is adiabatic rather than isothermal?


16 Monday: Statistical approach

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c(

∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp

Lecture: Topics for the day:

a) Law of large numbers (some things become easier)

b) We can’t keep track of what everything is doing, nor do we want to.

c) Probabilities of quantum eigenstates .

d) Probabilities of “single-particle states”.

e) Fairness function and maximizing fairness.

Lecture: (6 min) Statistical mechanics is the theoretical counterpart of thermodynamics.It’s how we can predict thermodynamic quantities from first principles, or use thermody-namic measurements to extract microscopic properties.

From quantum mechanics, you know that given a Hamiltonian describing a system, youcan solve for all the possible eigenstates and their energies. But how can you know which ofthose states a given system will be in? And given that state, how can you predict the resultof interactions of the system with its surroundings, when you don’t know the hamiltonianor eigenstates of the surroundings? These are the questions that are answered by statisticalmechanics.


16.1 Fairness function (Chapter 6 )

Lecture: (13 min) The primary quantity in statistical mechanics is the probability Piof finding the system in eigenstate i. The approach we are going to use is to state that theprobabilities are those which maximize the fairness (or minimize the bias). So we need todefine a fairness function F that we can maximize. First, let’s talk about some propertiesthe fairness function should satisfy.

a) it should be continuous

b) it should be symmetric

F(P1, P2, P3, . . .) = F(P3, P2, P1, . . .) (16.1)

c) it should be minimum when P1, P2, . . . = 0 and P1 = 1

F(1, 0, 0, . . .) = minimum (16.2)

d) it should be maximum when P1 = P2 = P3 = · · ·

F(P, P, P, . . .) = maximum (16.3)

e) “Addition rule” if I have two uncorrelated systems, then their fairness should add(extensivity!!!). This corresponds to the following, but I won’t give it to them untilthey’ve worked out the probabilities together.

F(PA, PB) + F(P1, P2, P3) = F(PAP1, PAP2, PAP3, PBP1, PBP2, PBP3) (16.4)

There aren’t many functions which satisfies all these rules!

Activity: Combining probabilities (??? min) The following is a simple exercise toget students thinking about probabilities and how they combine.

Consider two systems A and B, which you can think of as dice or quantum mechanicalsystems with several eigenstates. Each system has a set of possible states i. If the two areuncorrelated, then we can describe their separate probabilities as PA

i and PBj . What is the

probability Pij of finding A in state3 i and B in state j?

Pij = PAi P

Bj (16.5)

Suppose the two systems are correlated, and I know that the probability of finding thesystems in states i and j respectively is Pij for any states i and j. How would I find theprobability of finding A in state n, regardless of the state of B?

PAn =

∑j

Pnj (16.6)

Write on board:

Fairness = −kall states∑

i

Pi lnPi


Lecture: (??? min) Go through the rules above and demonstrate that this is continuous,symmetric, minimum when maximally unfair and maximum when maximally fair.

Activity: Demonstrating extensivity (??? min) Students will demonstrate that thefairness function is additive (and thus extensive), in the case of combining two uncorrelatedsystems with separate probabilities.

FA = −k∑i

Pi lnPi (16.7)

FB = −k∑i

Pi lnPi (16.8)

FAB = −k∑ij

PiPj ln (PiPj) (16.9)

= −k∑ij

PiPj (lnPi + lnPj) (16.10)

= −k

(∑ij

PiPj lnPi

)− k

(∑ij

PiPj lnPj

)(16.11)

= −k

(∑i

Pi lnPi

)(∑j

Pj

)− k

(∑j

Pj lnPj

)(∑i

Pi

)(16.12)

= −k

(∑i

Pi lnPi

)− k

(∑j

Pj lnPj

)(16.13)

= FA + FB (16.14)


17 Tuesday: Optimizing the fairness

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp


i

Pi lnPi

Activity: Students as molecules (20 min) Students, who represent molecules, willeach get 1 ordinary die, and the class will be split up into either six or nine sections (perhapstables?) Students will all start at a single table, and we’ll have the students at each tablecompute their own table’s contribution to the “fairness” of the room. 1

When so instructed, each student will roll his or her die. If they get a 1-4, they will goto the North, East, South or West respectively, otherwise they will remain stationary. We’llrecompute the fairness, and repeat.

They can think of the different tables as different quantum eigenstates, and of themselvesas distinguishable particles.

17.1 Least bias lagrangian

Lecture: I introduced the fairness function on Monday, and yesterday you made somemeasurements of how the fairness function changed as you randomly modified a polymer.We said on Monday that the fairness was maximized, and today we’ll look at how we goabout maximizing it.

1The fairness of the room is defined as the following, but let them work this out if possible...

F = N∑i

fi ln fi (17.1)

where fi is the fraction of students at each table and N is the total number of students.


SWBQ: How would you find the maximum of a function?

Lecture: (10 min?) Usually, (analytically) we maximize functions by setting theirderivatives equal to zero. So we could maximize the fairness by

∂F∂Pi

= 0 (17.2)

“Using the formula for the fairness function, what can this tell us about Pi?” The answeris that the probabilities should all be zero. _ There is a problem with this, which is thatPi can’t take just any values, because these are probabilities, and they have to add up toone. This is a constraint, and to solve a constrained maximization, we use the method ofLagrange multipliers. We first define a Lagrangian2:

L = F + α

(1−

∑i

Pi

)(17.3)

Note that since the added term should be zero, we haven’t changed the thing we want tomaximize. Now we maximize this in the same way, but we’ve got some extra terms thatshow up in our derivatives. We could, by the way, obtain our constraint by maximizing overα (the Lagrange multiplier) as well as the probabilities Pi.

17.2 Weighted averages

Lecture: (20 min?) Most thermodynamic quantities can be expressed as weighted av-erages over all possible eigenstates (or microstates). For instance, we the internal energy isgiven by:

U =∑i

PiEi (17.4)

Note that this will probably not be an eigenvalue of the energy, but that’s okay. Theenergy eigenvalues are so close for the total energy of a macroscopic object that we couldn’tdistinguish them anyhow.

17.3 Probabilities of microstates (Chapter 11 )

2The term Lagrangian means different things in different fields. In this case, we aren’t using the normalPhysics meaning for Lagrangian, but rather the definition from optimization theory, since we are optimizing.


Lecture: (10 min) I’d like to now summarize what I hope you all figured out yesterdayabout the probabilities that come from maximizing the fairness. Except that this time I’lluse an internal energy constraint instead of a length constraint.

L = −kB∑i

Pi lnPi + αkB

(1−

∑i

Pi

)+ βkB

(U −

∑i

PiEi

)(17.5)

where α and β are the two Lagrange multipliers. We want to maximize this, so we set itsderivatives to zero:

∂L∂Pi

= 0 (17.6)

= −kB (lnPi + 1)− kBα− βkBEi (17.7)

lnPi = −1− α− βEi (17.8)

At this point, it is convenient to invoke the normalization constraint...∑i

Pi = 1 (17.9)

1 =∑i

e−1−α−βEi (17.10)

1 = e−1−α∑i

e−βEi (17.11)

e1+α =∑i

e−βEi (17.12)

(17.13)

Write on board:

Z ≡all states∑

i

e−βεi

Write on board:

Pi =e−βεi

Z

Pi =Boltzmann factor

partition function(17.14)

At this point, we haven’t yet solved for β, and to do so, we’d need to invoke the internalenergy constraint:

U =∑i

EiPi (17.15)

U =

∑iEie

−βEi

Z(17.16)


SWBQ: (10 min) Introduce two non-interacting identical systems with N eigenstateseach, and talk about whether things will be extensive or intensive. “How many eigenstatesdoes the combined system have?” Students answer N or 2N . Using spin 1 system, ask forwhat the eigenstates of two spin-1 particles are. Soon we realize that it’s N (2) = N2, whichmeans it’s neither extensive nor intensive.

SWBQ: (3 min) What are the energy eigenstates of the combined system, if the energy

eigenstates of the small systems are E(1)i ?

Lecture: (7 min) Talk about the partition function, and work out what the partitionfunction of the combined system is, showing that it’s the product of the two separate systems.Then motivate it being exponential, if you have a whole bunch of non-interacting systems.

Note, however, that the log of the partition function is extensive! It will turn out to bea thermodynamic state variable, as we will see tomorrow.

Required homework 5.3: boltzmann-ratio

Challenge homework 5.4: plastic-rod


18 Wednesday: From statistics tothermodynamics

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp


i

Pi lnPi

Z ≡all states∑

i

e−βεi

Pi =e−βεi

Z

18.1 Thermodynamic properties from the Boltzmann

factor

Lecture: (5 min) Let’s talk a bit about fairness. We used the fairness to find theprobabilities of being in the various eigenstates, by assuming that the “fairest” distributionwould prevail. If you bring two separate systems together and allow them to equilibrate,then you would expect that the net fairness would either remain the same or would increase.This sounds a little like entropy in the second law, in that the net entropy of system plussurroundings can increase or stay the same, but cannot decrease. The maximum value of thefairness for a given system (which is the value it will have in equilibrium) is its entropy.

Activity: Solving for maximum fairness (10 min) Let’s look the maximum value ofthe fairness (a.k.a. entropy), which is

F = −k∑i

Pi lnPi (18.1)

U =∑i

PiEi (18.2)


Pi =e−βEi

Z(18.3)

“On your big white boards, solve for the fairness as a function of U , β and Z. i.e. eliminatethe sum over i.”

Fmax = −kB∑i

Pi lnPi (18.4)

= −kB∑i

e−βEi

Zln

(e−βEi

Z

)(18.5)

= −kB∑i

e−βEi

Z(−βEi − lnZ) (18.6)

= kBβ∑i

Eie−βEi

Z+ kB

∑i

lnZe−βEi

Z(18.7)

Fmax = kBβU + kB lnZ (18.8)

At this point, we may want to solve for U again, to get yet another relationship for U :

U =Fmax

kBβ− 1

βlnZ (18.9)

“We saw before that lnZ was extensive, so we can now conclude that β is intensive.From which it is also clear that entropy is extensive (which we already knew). ”

Lecture: (15 min) Since we believe that

S = Fmax (18.10)

Let’s see what else we can extract from this equation for U , which is Equation 18.9. We alsoknow that

dU = TdS − pdV (18.11)

T =

(∂U

∂S

)V

(18.12)

Since we have an equation for U in terms of S, we just need to figure out how to hold Vconstant, and we’ll know what T is! “What does it mean to hold V constant? It hasn’tshown up in any of our statistical equations?” If we change the volume, we change theenergy eigenvalues, so if we hold V constant (and in general, do no work) then the energyeigenvalues are fixed. So this derivative should be possible.

U =S

kBβ− 1

βlnZ (18.13)


T =

(∂U

∂S

)V

(18.14)

=1

kβ− S

kβ2

(∂β

∂S

)V

+lnZ

β2

(∂β

∂S

)V

− 1

Zβ

(∂Z

∂S

)V

(18.15)

=1

kβ− S

kβ2

(∂β

∂S

)V

+lnZ

β2

(∂β

∂S

)V

− 1

Zβ

(∂Z

∂β

)V

(∂β

∂S

)V

(18.16)

=1

kβ+

1

β

(− S

kβ+

lnZ

β− 1

Z

(∂Z

∂β

)V

)(∂β

∂S

)V

(18.17)

=1

kβ+

1

β

(− S

kβ+

lnZ

β+ U

)(∂β

∂S

)V

(18.18)

=1

kβ(18.19)

β =1

kT(18.20)

Then we can easily see now from Equation 18.9 that

U = TS − kT lnZ (18.21)

−kT lnZ = U − TS (18.22)

F = −kBT lnZ (18.23)

So it turns out that the log of the partition function just about gives us the Helmholtz freeenergy! ¨

BWBQ: (skipped this) Check whether this S is indeed extensive as expected.

S2 =N∑i=1

N∑j=1

P(1)i P

(1)j ln

(P

(1)i P

(1)j

)(18.24)

=N∑i=1

N∑j=1

P(1)i P

(1)j

(lnP

(1)i + lnP

(1)j

)(18.25)

= 2

(N∑i=1

P(1)i

)N∑j=1

P(1)j lnP

(1)j (18.26)

= 2N∑j=1

P(1)j lnP

(1)j (18.27)

= 2S1 (18.28)

BWBQ: (skip this?) Work out an expression for S in terms of Z and β (or T ).


Activity: Entropy of microcanonical ensemble (20 min) Consider a system withW eigenstates, all of which have the same energy. What is the entropy?

S = −kBW∑i=1

1

Wln

(1

W

)(18.29)

= kB lnW (18.30)

“This is the result that is on Boltzmann’s tombstone, and is sometimes taken as a definitionof the entropy, although it is less general than the one that we’re using. ”


Problem 5.1 Hot metal A 1 cm3 cube of hot metal is thrown into the ocean; severalhours pass.

a) During this time does the entropy of the metal increase, decrease, remain the same, oris this not determinable with the given information? Explain your reasoning.

b) Does the entropy of ocean increase, decrease, remain the same, or is this not deter-minable with the given information? Explain your reasoning.

c) Does the entropy of the metal plus the ocean increase, decrease, remain the same, oris this not determinable with the given information? Explain your reasoning.

Problem 5.2 Isothermal and adiabatic compressibility The isothermal compress-ibility is defined as

KT = − 1

V

(∂V

∂p

)T

(18.31)

KT is be found by measuring the fractional change in volume when the the pressure is slightlychanged with the temperature held constant. In contrast, the adiabatic compressibility isdefined as

KS = − 1

V

(∂V

∂p

)S

(18.32)

and is measured by making a slight change in pressure without allowing for any heat transfer.This is the compressibility, for instance, that would directly affect the speed of sound. Showthat

KT

KS

=CpCV

(18.33)

Where the heat capacities at constant pressure and volume are given by

Cp = T

(∂S

∂T

)p

(18.34)


CV = T

(∂S

∂T

)V

(18.35)

Problem 5.3 Boltzmann ratio At low temperatures, diatomic molecule can be welldescribed as a rigid rotor. The Hamiltonian of such a system is simply proportional to thesquare of the angular momentum

H =1

2IL2 (18.36)

and the energy eigenvalues are

Elm = h2 l(l + 1)

2I(18.37)

a) What is the energy of the ground state and the first and second excited states of theH2 molecule?

b) At room temperature, what is the relative probability of finding a hydrogen moleculein the l = 0 state versus finding it in any one of the l = 1 states?i.e. what is Pl=0,m=0/ (Pl=1,m=−1 + Pl=1,m=0 + Pl=1,m=1)

c) At what temperature is the value of this ratio 1?

d) At room temperature, what is the probability of finding a hydrogen molecule in anyone of the l = 2 states versus that of finding it in the ground state?i.e. what is Pl=0,m=0/ (Pl=2,m=−2 + Pl=2,m=−1 + · · ·+ Pl=2,m=2)

Problem 5.4 A plastic rod (challenge) When stretched to a length L the tension forceτ in a plastic rod at temperature T is given by its Equation of State

τ = aT 2(L− Lo)

where a is a positive constant and Lo is the rod’s unstretched length. For an unstretchedrod (i.e. L = Lo) the specific heat at constant length is CL = bT where b is a constant.Knowing the internal energy at To, Lo (i.e. U(To, Lo)) find the internal energy U(Tf , Lf ) atsome other temperature Tf and length Lf .

a) Write U = U(T, L) and take the total derivative dU .

b) Show that the partial derivative (∂U/∂L)T = −aT 2(L− Lo).

c) To integrate the resulting differential equation Line Integrate dU very carefully in theT, L plane, keeping in mind that CL = bT holds only at L = Lo.



Required homework 6.1: rubber-band-model

Required homework 6.2: rubber-meets-road


19 Thursday: Statistical mechanics of air

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp


i

Pi lnPi

Z ≡all states∑

i

e−βεi

Pi =e−βεi

Z

19.1 Quantum spectra

Lecture: (10 min) I’m going to quickly review and introduce the energy eigenvalues forsome simple quantum mechanical problems.

For each of the following, I’d like to sketch out the potential, then sketch the wavefunctionsand the spacing of the energy levels.

The first problem you handled was a particle in an infinite square well potential:

H =−h2

2m

∂2

∂x2(19.1)

En =h2π2n2

2mL2(19.2)

where n ≥ 1. We could solve the same problem in three dimensions, and we would have:

H =−h2

2m∇2 (19.3)

Enxnynz =h2π2

(n2x + n2

y + n2z

)L2

(19.4)


The next moderately simple problem is the rigid rotator. In this case, the only energy inthe Hamiltonian is the angular kinetic energy:

H =−h2

2IL2 (19.5)

Elm =h2l(l + 1)

2I(19.6)

Finally, we have a simple problem that hasn’t yet come up in the paradigms, which is thesimple harmonic oscillator. In this case we have both kinetic and potential energy:

H =−h2

2m

∂2

∂x2+mω2

0

2x2 (19.7)

En =

(n+

1

2

)hω0 (19.8)

Of course, you also studied the hydrogen atom, but its solution is less general than thosewe’ve listed here. Any diatomic molecule behaves like a rigid rotator and a simple harmonicoscillator, and like a particle in a box, too!

19.2 Diatomic gas

Lecture: (20 min) Let’s consider a diatomic ideal gas, such as nitrogen. In this case,the energy levels of a single molecule are given by:

E(1)nxnynznvlm

=h2π2

(n2x + n2

y + n2z

)2mL2

+h2l(l + 1)

2I+

(n+

1

2

)hω0 (19.9)

That’s an awful lot of quantum numbers, and that’s just one molecule, and we’re neglectingany possible electronic excited states!

How does this change when we’ve got N molecules all confined in the same box? We’vealready talked about how energies relate when we combine systems:

Etot =N∑i

E(1)i (19.10)

where I’ve left out all the quantum numbers, since there are so very many.If we want to know the internal energy, we’ll need to sum over every possible state, with

the probability of that particular state. To do this, we’ll need to know the partition function,so let’s start with that.

Z =all states∑

e−βEthis state (19.11)

=∑

nx1ny1nz1nv1 l1m1,nx2ny2nz2nv2 l2m2,···

e−β(E

(1)nx1ny1nz1nv1 l1m1

+E(2)nx2ny2nz2nv2 l2m2

+···)

(19.12)


. . . except that this isn’t quite right. We can’t distinguish between the different molecules...when we swap two of them in the sum, we’re really talking about the same state! We canfix this double-counting by multiplying by an N !, which gives us:

Z =1

N !

∑nx1ny1nz1nv1 l1m1,nx2ny2nz2nv2 l2m2,···

e−β(E


+E(2)nx2ny2nz2nv2 l2m2

+···)

(19.13)

=1

N !


e−βE(1)

nx1ny1nz1nv1 l1m1e−βE(2)

nx2ny2nz2nv2 l2m2 · · · (19.14)

=1

N !

∑nx1ny1nz1nv1 l1m1

e−βE(1)

nx1ny1nz1nv1 l1m1

∑nx2ny2nz2nv2 l2m2

e−βE(2)

nx2ny2nz2nv2 l2m2

· · ·(19.15)

=1

N !

∑nxnynznvlm

e−βE(1)

nxnynznvlm

N

(19.16)

=1

N !

∑nxnynznvlm

e−β(h2π2(n2

x+n2y+n2

z)2mL2 +

h2l(l+1)2I

+(n+ 12)hω0

)N

(19.17)

=1

N !

∑nxnynznvlm

e−βh2π2(n2

x+n2y+n2

z)2mL2 e−β

h2l(l+1)2I e−β(n+ 1

2)hω0

N

(19.18)

=1

N !

∑nxnynz

e−βh2π2(n2

x+n2y+n2

z)2mL2

∑nv

e−βh2l(l+1)

2I

∑lm

e−β(n+ 12)hω0

N

(19.19)

=1

N !

∑nxnynz

e−βh2π2(n2

x+n2y+n2

z)2mL2

N (∑nv

e−βh2l(l+1)

2I

)N (∑lm

e−β(n+ 12)hω0

)N

(19.20)

Now, if we were computing the internal energy U , we’d be able to do something very similar:

U =∑i

PiEi (19.21)

=1

N !


P...Etot (19.22)

=1

N !


P...

(E


+ E(2)nx2ny2nz2nv2 l2m2

+ · · ·)

(19.23)

=N

N !


P...E(1)nx1ny1nz1nv1 l1m1

(19.24)


= N∑

nxnynznvlm

PnxnynznvlmE(1)nxnynznvlm

(19.25)

= N∑

nxnynznvlm

Pnxnynznvlm

(h2π2

(n2x + n2

y + n2z

)2mL2

+h2l(l + 1)

2I+

(n+

1

2

)hω0

)(19.26)

= N

∑nxnynz

Pnxnynzh2π2

(n2x + n2

y + n2z

)2mL2

+∑lm

Plmh2l(l + 1)

2I+∑nv

Pnv

(n+

1

2

)hω0

(19.27)

So you can see, if we can work out the average translational kinetic energy, rotational energyand vibrational energy of a molecule in this gas, then we’ll easily have the total internalenergy of this system just by adding everything up and multiplying by N .

Activity: Diatomic molecule from quantum up (1 hour 30 min without wrap-up) I’m going to divide you into groups. Each group will have a separate task, so thathopefully when we’re done as a class we’ll have an answer for the total internal energy of adiatomic ideal gas.

These sums are pretty challenging, so I’ll ask each group to consider one of two dis-tinct limits: the low-temperature limit and the high-temperature limit. The low- and high-temperature limits have a different meaning for each term in the energy.

For the translational kinetic energy, the limits will be defined by

βh2π2

2mL2� 1 (19.28)

or the reverse. For the rotational energy, it will be

βh2

2I� 1 (19.29)

and for the vibrational energy, it will be

βhω0 � 1 (19.30)

These energy scales have large gaps, so the translational energy may be in the high-temperaturelimit while the other two are in the low-temperature limit, for instance.

You should have five groups, so I’d like each group to do one term in the energy, ineither the low-temperature limit or the high-temperature limit, but let’s not do the low-temperature limit of the translational kinetic energy. The reason we avoid that particularlimit is that we’ve ignored the possibility that two of our molecules may be in precisely thesame state. You know the Pauli exclusion principle for electrons (and fermions in general),which means we could have a problem. For bosons (all particles that aren’t fermions) thereis another problem, which I also won’t go into. As long as our temperature is high enoughthat the probability of any given state is pretty low, we won’t run into these troubles (whichcould give us things like a Bose-Einstein condensate).


20 Friday:

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp


i

Pi lnPi

Z ≡all states∑

i

e−βεi

Pi =e−βεi

Z

20.1 Diatomic gas wrapup

Lecture: (20 min) Yesterday we worked out the internal energy per molecule of a di-atomic gas associated with translational kinetic energy, rotational kinetic energy, and vibra-tional energy (which has both a kinetic and potential component). For each case (excepttranslation), you considered both the low- and high- temperature limits.

In each case you had sums that looked like∑i

somethinge−βEi (20.1)

and β(E1 − E0) was either large or small.For the high-temperature limits, you needed to convert summations into integrals, which

was a reasonable approximation because the change of the thing being summed (summand?)was small as you changed the quantum numbers by one, so treating it as a continuum wasokay.

For the low-temperature limit, you had an easier scenario, as all the Boltzmann factorswere all very small compared with the ground state. So you could just truncate the sumafter a few terms.


Translation at high T

U

N=∑

nxnynz

h2π2(n2x + n2

y + n2z

)2mL2

1

Ze−β

h2π2(n2x+n2

y+n2z)

2mL2 (20.2)

≈∫ ∞

0

∫ ∞0

∫ ∞0

h2π2(n2x + n2

y + n2z

)2mL2

1

Ze−β

h2π2(n2x+n2

y+n2z)

2mL2 dnxdnydnz (20.3)

=h2π2

2mL2

1

Z

∫ ∞0

∫ ∞0

∫ ∞0

(n2x + n2

y + n2z

)e−β

h2π2(n2x+n2

y+n2z)

2mL2 dnxdnydnz (20.4)

ux =√β

hπ√2mL

nx (20.5)

U

N=

h2π2

2mL2

1

Z

(√2mL

√kBT

hπ

)3 ∫ ∞0

∫ ∞0

∫ ∞0

(u2x + u2

y + u2z

)e−(u2

x+u2y+u2

z)duxduyduz

(20.6)

=

√2mL(kBT )

32

hπ

1

Z

∫ ∞0

∫ ∞0

∫ ∞0

(u2x + u2

y + u2z

)e−(u2

x+u2y+u2

z)duxduyduz (20.7)

At this point we’ve extracted the physics from the integral. It’s clearly not zero, and it alsoisn’t infinite, so it’s just some number that we can work out later. But we still need Z...

Z =∑

nxnynz

e−βh2π2(n2

x+n2y+n2

z)2mL2 (20.8)

≈∫ ∞

0

∫ ∞0

∫ ∞0

e−βh2π2(n2

x+n2y+n2

z)2mL2 dnxdnydnz (20.9)

ux =

√β

2m

hπ

Lnx (20.10)

Z =

(√2mkBTL

hπ

)∫ ∞0

∫ ∞0

∫ ∞0

e−(u2x+u2

y+u2z)duxduyduz (20.11)

=

(√2mkBTL

hπ

)(∫ ∞0

e−u2

du

)3

(20.12)

Once again, we’ve extracted the physics from the integral, leaving a dry, dimensionless husk.In this case, I cleaned that husk up a bit, so it’ll be a bit more compact. Putting thesetogether (with a minimum of simplification, we get:

U

N=

√2mL(kBT )

32

hπ3(∫∞

0u2e−u

2du)(∫∞

0e−u

2du)2(√

2mkBTLhπ

) (∫∞0e−u2du

)3(20.13)

= kBT3

(∫ ∞0

u2e−u2

du

)(∫ ∞0

e−u2

du

)(20.14)


=3

2kBT (20.15)

Rotation at low T I’ll skip over this, as it isn’t very exciting (and it’s taking a longtime to LATEX this). I’ll just mention that it drops exponentially to zero at low temperature.This is a universal property of systems with a “gap,” which is to say, with a finite energydifference between the ground state and the first excited state.

Rotation at high T

Z =∞∑l=0

l∑m=−l

e−βh2l(l+1)

2I (20.16)

=∞∑l=0

(2l + 1)e−βh2l(l+1)

2I (20.17)

≈∫ ∞

0

(2l + 1)e−βh2l(l+1)

2I dl (20.18)

As it turns out, we can relatively easily do this integral. However, the “+1” terms areinsignificant, since l is integrated up to infinity, and the large l contribution dominates. Sowe can:

Z ≈∫ ∞

0

2le−βh2l2

2I dl (20.19)

u =hl√

2IkBT(20.20)

Z =8IkBT

h2

∫ ∞0

ue−u2

u (20.21)

(20.22)

The integral is easy to do, but there’s no urgent need to do so: we have already taken thephysics out of the integral.

U

N≈ 1

Z

∫ ∞0

2lh2l2

2Ie−β

h2l2

2I dl (20.23)

=1

Z2h2

2I

(4IkBT

h2

)2 ∫ ∞0

u3e−u2

du (20.24)

=2 h

2

2I

(4IkBTh2

)2 ∫∞0u3e−u

2du

8IkBTh2

∫∞0ue−u2u

(20.25)

= 4kBT

∫∞0u3e−u

2du∫∞

0ue−u2u

(20.26)

= kT (20.27)


Here we can see much of the essential physics by recognizing that the energy is proportionalto kBT without performing the integrals. I probably have mistakes somewhere in the above,but the final answer is correct.

Harmonic oscillator at high T For the harmonic oscillator, I’ll demonstrate a differentapproach, since I think showing the same approach for the third time in a row is a bit boring.

Z =∞∑n=0

e−β(n+ 12)hω0 (20.28)

= e−β12hω0

∞∑n=0

e−βnhω0 (20.29)

= e−β12hω0

∞∑n=0

(e−βhω0

)n(20.30)

This is just a harmonic series, so we can solve it using the standard trick, where I’ll call theseries s:

s ≡∞∑n=0

(e−βhω0

)n(20.31)

e−βhω0s = e−βhω0

∞∑n=0

(e−βhω0

)n(20.32)

=∞∑n=1

(e−βhω0

)n(20.33)

=

(∞∑n=0

(e−βhω0

)n)− 1 (20.34)

e−βhω0s = s− 1 (20.35)

1 =(1− e−βhω0

)s (20.36)

s =1

1− e−βhω0(20.37)

Z =e−β

12hω0

1− e−βhω0(20.38)

So that’s nice. Of course, we still want to find the energy. To do this, we can employ yetanother trick—although it’s not so hard to do in the high-temperature limit the same waywe solved the previous problem. We can recognize that

U =∑i

PiEi (20.39)

=∑i

Eie−βEi

Z(20.40)


=1

Z

∑i

Eie−βEi (20.41)

Z =∑i

e−βEi (20.42)(∂Z

∂β

)Ei

=∑i

−Eie−βEi (20.43)

U = −

(∂Z∂β

)Ei

Z(20.44)

So once we have the partition function, we could just take a derivative to find the internalenergy. So for the simple harmonic oscillator, we have:(

∂Z

∂β

)Ei

= −1

2hω0Z −

e−β12hω0

(1− e−βhω0)2 e−βhω0hω0 (20.45)

= −1

2hω0Z −

1

1− e−βhω0e−βhω0hω0Z (20.46)

= −1

2hω0Z −

hω0

eβhω0 − 1Z (20.47)

U

N=

1

2hω0 +

hω0

eβhω0 − 1(20.48)

This gives us an exact solution for the internal energy of a simple harmonic oscillator, butwe still haven’t found the high-temperature limit. To find that, we have to take βhω0 � 1.In this case, we can just use a simple Taylor’s expansion approach:

U

N=

1

2hω0 +

hω0

eβhω0 − 1(20.49)

≈ 1

2hω0 +

hω0

1 + βhω0 − 1(20.50)

=1

2hω0 +

hω0

βhω0

(20.51)

=1

2hω0 + kBT (20.52)

≈ kBT (20.53)

This tells us that like the rotational energy, the vibrational energy approaches kBT permolecule at high temperatures.

There is a general rule which occurs in the classical limit (which is the high-temperaturelimit), that any quadratic term in the energy ends up providing 1

2kBT to the internal energy.

This is called the equipartition theorem. Since there are three translational degrees of freedom(v2 in each direction), the kinetic energy gives us 3

2kBT . There are two ways to rotate a

diatomic molecule, which gives us an additional kBT . And finally, the vibration has bothkinetic and potential energy, which each provide half of the final kBT .



Problem 6.1 A rubber band model Consider a model in which rubber is composed ofsegments of fixed length a, each of which can be oriented in any of the six cardinal directions(±~x, ±~y or ±~z). We will assume that the rubber band is aligned in the ~x direction, and

there is an energetic penalty ε for segments to be in any of the ~y or ~z directions, but thatthere is no interaction involving the orientations of neighboring segments.1

E~x = 0 (20.54)

E−~x = 0 (20.55)

Eanything else = ε (20.56)

a) Using these energies, write down the relative probabilities for each alignment using theBoltzmann factor and their relative energies.

b) Find the mean displacement in the ~x direction of a molecule composed of N segmentsbased on the above probabilities. Why does it have this value?

c) If we pull on the molecule with tension τ in the +~x direction, we will need to adjust theBoltzmann factors to include terms such as±βτa. How does this affect the probabilitiesyou worked out in part (a) of this problem?

d) What is the mean displacement in the ~x direction L of a molecule composed of Nsegments under tension τ?

e) What is the entropy S of this system, as a function of β and τ? Keep in mind thatthe entropy is given by

S = −kBall states∑

i

Pi lnPi

f) What is the internal energy U of this system, as a function of β and τ?

g) What is the Helmholtz free energy of this system, F ≡ U − TS?

Problem 6.2 The rubber meets the road In the previous problem, you solved for theGibbs free energy of a model polymer. In this problem, you will use “reasonable” values forthe parameters a, ε and N and compare with your results from lab 2.

You may assume that the number of links in a polymer molecule are N = 109, thelength of each link in the polymer is a = 1 nm, and the energy of vertically-oriented linksis ε = −10−21 J. In addition, you will need to assume that the rubber band is made up of

1This model is actually considerably more accurate (taking ε = 0) for long linear polymers such as DNAin solution.


a number of rubber molecules Nwide which are connected in parallel, so the tension is Nwide

times the value for a single molecule, and the internal energy, etc. is Nwide times the valuefor a single molecule.

a) Find a value for Nwide that gives a reasonable (within an order of magnitude) predictionfor the change in tension in your rubber band when it is stretched by a given amount.You will need to measure displacements from the length that gives zero tension—whichis not when the rubber band itself has zero length.

b) Plot the predicted length L at room temperature versus the tension τ , over the rangeof tensions you measured experimentally. Also include on your plot (possibly with anoffset) your experimental data for this temperature. Comment on your agreement ordisagreement.

c) For a specific tension, plot the predicted length versus temperature, along with exper-imental data points.

d) For a specific length, plot your predicted tension versus temperature. This will prob-ably require a computer and some time.2 Plot also your experimental data.

e) Let us consider the isothermal stretch you analyzed in the rubber band lab. Youcomputed from your measurement the work done, the heat transfered, and the changein internal energy and entropy. Using the model for the same change in tension andat the same temperature, what are Q, W , ∆U and ∆S?

f) Discuss the agreements and disagreements between the theory and the model. Suggestspecific problems and possible improvements in the model.


20.2 Third law

Lecture:

The third law of thermodynamics The entropy of any perfect crystal at zero temper-ature is zero.

S = −kB∑i

Pi lnPi (20.57)

“When is this zero?” It’s equivalent to saying that any perfect crystal at zero temperature isin a non-degenerate ground state. It doesn’t sound so deep when looked at from a stat-mechviewpoint, but when you look at it from a thermodynamic standpoint, it’s close to amazing.It means that if you...

2If you can only solve for a few data points, then just plot a few data points.


• Start with solid oxygen and graphite at zero temperature.

• Gradually warm them up.

• Allow the graphite to slowly burn, producing CO2. We have to do this at just the righttemperature, so that the process is reversible.

• Slowly cool down the resulting CO2 to zero temperature.

If we integrate

∆S =

∫dQreversible

T(20.58)

we find no change in the entropy! This means that we really can tabulate the absoluteentropy of any material we like. All we have to do to figure out the zero of entropy is coolit down to zero Kelvin.

Ice rules


Physics 423 Friday 5/13/2012

2680 LINUS PAULING Vol. 57

summary 1. The velocity constant for the &acetone al-

cohol decomposition in the presence of dilute sodium hydroxide has been measured at 5' intervals from 0 to 50".

2. The constancy of ratio of velocity constant to sodium hydroxide concentration has been confirmed over a limited concentration range a t 25'.

3. The energy of activation, calculated from the Arrhenius equation for a series of temperature intervals, has been shown to be a function of temperaturl: well outside the limits of error. E,,, increases consistently from a value of 15,850 cal. at 5" to 17,250 a t 32.5 and then decreases by about 400 cal. a t 45'.

4. On the addition of 18.5% of methyl alcohol Eact increases by over 1700 cal. but the general character of the EaCt-T curve remains little changed to 30".

5. The corresponding B values from the integrated Arrhenius equation, In k = 2.3 B - (Eact/ RT), both in water and the methyl alcohol solution parallel these Ea,, values and furnish experimental evidence for abandoning the unfortunate term "temperature independent constant" for this quantity.

6. The data show that the collision theory is inadequate and that the entropy of activation is an important quantity in considering solution reactions. NEW YORK, N. Y. RECEIVED AUGUST 15, 1935

[CONTRIBUTION FROM THE GATES CEEMICAL LABORATORY, CALIFORNIA INSTITUTE OF TECHNOLOGY, N O . 5061

The Structure and Entropy of Ice and of Other Crystals with Some Randomness of Atomic Arrangement

BY LINUS PAULING

Investigations of the entropy of substances a t low temperatures have produced very important information regarding the structure of crystals, the work of Giauque and his collaborators being particularly noteworthy. For example, the observed entropy of crystalline hydrogen shows that even a t very low temperatures the molecules of orthohydrogen in the crystal are rotating about as freely as in the gas;' subsequent to this dis- covery the phenomenon of rotation of molecules in crystals was found to be not uncommon. Also the entropy values of carbon monoxide2 and nitrous oxide3 show that in crystals of these substances the molecules are not uniquely oriented, but have instead a choice between two orientations, presumably the opposed orientations CO and OC or NNO and ONN along fixed axes. It is pointed out in this note that the observed entropy of ice at low temperatures provides strong support for a particular structure of ice, and thus gives an answer to a question which has been extensively discussed during the past few years.

It has been generally recognized since the dis- (1) W. F. Oiauque and H. L. Johnston, TIIIS JOURNAL, 60, 3221

(1928); L. Pauling, Phys. Rev., 86, 430 (1930). (2) J. 0. Clayton and W. F. Giauque, Tms JOURNAL, 64, 2610

(1932). (3) R. W. Blue and W. F. Giauque, ibid. , 67, 991 (1936); K.

Clusius, Z . Elcklrochem., 40, 99 (1934).

covery of the hydrogen bond4 that the unusual properties of water and ice (high melting and boiling points, low density, association, high dielectric constants, etc.) owe their existence to hydrogen bonds between water molecules. The arrangement of oxygen atoms (but not of hydrogen atoms) in crystals of ice is known from x-ray studies;6 it is not a close-packed arrangement (as of sulfur atoms in the high-temperature form of hydrogen sulfide), but a very open one, like that of the silicon atoms in high-tridymite. Each oxygen atom in ice is tetrahedrally surrounded by four other oxygen atoms a t the distance 2.76 A., and it has been assumed that it is bonded to these atoms by hydrogen bonds, the number of hydrogen atoms being just that required to place one hydrogen atom between each pair of oxygen atoms. (Similarly in high-tridymite there is an oxygen atom between each pair of silicon atoms; we might say that each silicon atom is attached to four others by oxygen bonds.)

The question now arises as to whether a given hydrogen atom is midway between the two oxygen

(4) W. M. Latimer and W. H. Rodebusb, THIS JOURNAL, 42, 1419 (1920).

(5) D. M. Dennison, Phys. Reo., 17, 20 (1921); W. H. Bragg, Puoc. Phys. SOC. (London), 34, 98 (1922); U'. H. Barnes, Proc. Roy. SOC. (London), Al26, 670 (1929).


Dec., 1935 THE STRUCTURE AND ENTROPY OF ICE 2681

atoms it connects or closer to one than to the other. The answer to this is that it is closer to one than to the other. In the gas molecule the 0-H distance is 0.95 A., and the magnitudes of the changes in properties from steam to ice are not sufficiently great. to permit us to assume that this distance is increased to 1.38 8. In ice each hydrogen atom is about 0.95 A. from one oxygen atom and 1.81 4.. from another.

We now ask whether each hydrogen atom (or rather hydrogen nucleus) has a choice of two positions along its oxygen-oxygen axis, independent of the positions of the other hydrogen nuclei. The answer is in the negative; for we know that the concentration of (OH)-and (H30)+ ions in water is very small, and we expect the situation to be essentially unchanged in ice. Hence the hydrogen nuclei will assume positions such that each oxygen atom (with a t most very few exceptions) will have two hydrogen atoms attached to it.

Let us now make the following assumptions (to be supported later by a discussion of the entropy) regarding the structure of ice.

(1) In ice each oxygen atom has two hydrogen atoms attached to it a t distances of about 0.95 8., forming a water molecule, the HOH angle being about 105' as in the gas molecule.

(2) Each water molecule is oriented so that its two hydrogen at oms are directed approximately toward two of the four oxygen atoms which surround it tetrahedrally, forming hydrogen bonds.

(3) The orientations of adjacent water molecules are such that only one hydrogen atom lies approximately along each oxygen-oxygen axis. (4) Under ordinary conditions the interaction

of non-adjacent molecules is not such as to appreciably stabilize any one of the many configurations satisfying the preceding conditions with reference to the others.

Thus we assurne that an ice crystal can exist in any one of a large number of configurations,B each corresponding to certain orientations of the water molecules. The crystal can change from one configuration to another by rotation of some of the molecules or by the motion of some of the hydrogen nuclei, each moving a distance of about 0.86 8. from a potential minimum 0.95 8. from

(6 ) One of the large number of configurations is represented by the structure of ice suggested by J, D. Bernal and R. H. Fowler, J . Chcm. Phys , 1, 515 (1'333); these authors also suggested that at temperatures just below the melting point (but not at lower temperatures) the molecular arrangement might be partially or largely irregular.

one oxygen atom to another one 0.95 A. from an adjacent oxygen atom. It is probable that both processes occur. As an example of a change from one to another configuration, we may consider one of the puckered rings of six oxygen atoms occurring in ice

\H 0 - H 0 /O H-0

H

\O H

0-H O/H

\H 0 do O/H

H \O H-0

Change from one of the two cyclic arrangements of hydrogen nuclei to the other is permitted by our postulates. The fact that a t temperatures above about 200A. the dielectric constant of ice is of the order of magnitude of that of water shows that the molecules can orient themselves with considerable freedom, the crystal changing under the influence of the electric field from unpolarized to polarized configurations satisfying the above conditions. On cooling the crystal to low temperatures i t freezes into some one of the possible configurations; but i t does not go over (in a reasonable period of time) to a perfect crystal with no randomness of molecular orientation. It will have at very low temperatures the entropy L In W, in which W is the number of configurations accessible to the crystal.

Let us now calculate W, using two different methods.

There are N molecules in a mole of ice. A given molecule can orient itself in six ways satisfying condition 2. However, the chance that the ad- jacen t molecules will permit a given orientation is

inasmuch as each adjacent molecule has two hydrogen-occupied and two unoccupied tetrahedral directions, making the chance that a given direction is available for each hydrogen of the original molecule I / t 1 and the chance that both can be located in accordance with the given orientation 1/4. The total number of configurations for N molecules is thus W = (6/4)'" = ( 3 / 2 ) N .

The same result is given by the following equivalent argument. Ignoring condition 1, there are 22N configurations with hydrogen bonds between adjacent oxygen atoms, each hydrogen nucleus having the choice of two positions, one near one oxygen atom and the other near the other. Some of these are ruled out by condition 1. Let us now consider a given oxygen atom and the four sur- rounding hydrogen atoms. There are sixteen arrangements of the four hydrogen nuclei. Of


2682 LINUS PAULING VOl. 57

these ten are ruled out by condition l - o n e gives (H40)++, four give (HsO)+, four (OH)- and one 0--. This condition for each oxygen atom thus permits only 6/la = 3//8 of the configurations, the total number of permitted configurations being accordingly W = 22"(3/8)N = ( 3 / . J N .

The contribution of this lack of regularity to the entropy of ice is thus R in 3//z = 0.805 E. U. The observed entropy discrepancy of ice at low temperatures is 0.87 E. U., obtained by subtracting the entropy difference of ice at very low temperatures and water vapor at standard conditions, for which the value 44.23 E. U. has been calculated from thermal data by Giauque and Ashley,? from the spectroscopic value 45.101 E. U. for the entropy of water vapor given by Gordon.8 The agreement in the experimental and theoretical entropy vaSues provides strong support of the postulated structure of icess

The structure of ice is seen to be of a type intermediate between that of carbon monoxide and nitrous oxide, in which each molecule can assume either one of two orientations essentially inde- pendently of the orientations of the other molecules in the crystal, and that of a perfect molecular crystal, in which the position and orientation of each molecule are uniquely determined by the other molecules. In ice the orientation of a given molecule is dependent on the orientations of its four immediate neighbors, but not directly on the orientations of the more distant molecules.

It should be pointed out that a finite residual entropy calculated for a substance from experimental data obtained at temperatures extending down to a certain temperature, with extrapolation below that point, may arise either from failure of the experimenter to obtain thermodynamic equilibrium in his calorimetric measurements or from error in the extrapolation. Measurements made under ideal conditions and extended to sufficiently low temperatures would presumably lead to zero entropy for any system. O-H-

crystals will be found to have residual entropy a t very low temperatures as a result of some randomness of atomic arrangement. It is probable, however, that experimental verification of the residual entropy would be difficult for most of the cases mentioned below.

Diaspore, "02, contains groups OH0 with hydrogen bonds.1° As in water, we expect the hydrogen nucleus to be closer to one oxygen atom than to the other. Inasmuch as there is no restriction on the choice of the two positions, each oxygen atom forming only one hydrogen bond, this would lead to a residual entropy of R In 2 per mole of AlH02. The same residual entropy is expected per HOz group for other crystals containing this group (staurolite).

Hydrogen bonds may not always lead to residual entropy. For example, the crystal lepidocro- cite,l' FeOOH, contains infinite strings of oxygen atoms joined by hydrogen bonds, oHoHoHoHoH. If we assume that the choice of two positions by a hydrogen nucleus is restricted by the requirement that each oxygen have one and only one attached hydrogen, then there are two accessible configurations per string. This lack of definiteness does not lead to an appreciable residual entropy for a large crystal.

In the double molecules of formic acid, with the O-H-

structure12 HC< O-H- >CH and in other mono-

carboxylic acids forming double molecules, we expect that the interaction within a carboxyl group is strong enough to permit only the two configurations in which one hydrogen nucleus is attached to each carboxyl group to be stable a t low temperatures. This would lead to a residual entropy of '/z R In 2 per RCOOH.

By a similar argument we calculate for 0-oxalic acid (anhydrous), which contain finite strings13 of molecules

.c-c<-H->c-c< O-H- >C-.e,,

O-H- O-H-

the residual entropy R In 2 per mole of (COOH)z, whereas a-oxalic acid, containing layers with the dealized structure18 in which there are infinite sequences of carboxyl groups attached by hydro-

Other Crystals with Residual Entropy Our knowledge Of the structure of Crystals per-

Other mits the prediction to be made that (7) W. F. Giauque and M. F. Ashley, PAYS. Rev., 4% 81 (1933). (8) A. R. Gordon, J. Chem. Phys. , 2, 65 (1934). (9) The existence of this residual entropy of ice a t very low tem-

peratures was discovered by Giauque and Ashley (ref. 7) , who pre- liminarily ascribed it to the pefsistance of rotation of ortho-water molecules (comprising of the total) about their electric-moment 336 (1934). axes, giving an entropy of %/a R In 2 = 1.03 E. U.

gen bonds, would have no appreciable residual (10) F. J. Ewing, J . Chem. Phys., 3, 203 (1935). (11) F. J. Ewing, {bid., 3,420 (1935). (12) L. Pauling and L. 0. Brockway, Proc. Naf. Acad. Sci., 20,

(13) S. B. Hendricks, Z. Krist., 91, 48 (1935).


Dec., 1935 ”HE STRUCTURE AND ENTROPY OF ICE 2683

ted apparently at random among the positions provided, of which they occupy the fraction 1 - 6 .

H/O” “\‘H dH/o’yo \H In the cubic tungsten bronzes,16 Na,WOa, the sodium ions occupy the fraction x of the avail- I able positions. These crystals

o / y o would presumably show residual entropy. The experimental verification of this entropy of mixing of the ions W6+ and We+ as prediction by heat capacity measurements and the well as the entropy of random distribution of the study of the equilibrium between the two forms sodium ions. A somewhat similar case is pro- might be practicable. vided by the hexagonal form of silver iodide,”

At present i t is not possible to predict with con- in which at room temperature each silver atom fidence whether the hydrogen nucleus between has the choice of four or five positions a few tenths two fluorine atoms connected by a hydrogen bond of an hgstri jm apart. At liquid air tempera- has the choice of two positions (as for oxygen tures, however, most of the atoms settle into one atoms) or not. The FHF distance of 2.35 8. is position, so that no residual entropy would be somewhat greater than twice the H-F distance in shown. HF, 0.91 k., although the difference (29%) is not Many crystals show an uncertainty in structure so great as for OH0 and H-0 (45%). The ques- similar to that of CO and NNO or of solid solution would be answered by a determination of the tions. For potassium cyanate the x-ray datala residual entropy of potassium hydrogen fluoride, indicate that each cyanate ion has the choice of KHF2, or of some other crystal containing the two orientations, NCO and OCN, which would HF2 group; the residual entropy would be R In 2 lead to the residual entropy R In 2. In spinel, if the potential function for the hydrogen nucleus MgA204, and other crystals with the spinel has two minima rather than one. The same re- structure, the bivalent and trivalent cations are sidual entropy 1vVould also be shown by KHzF3, distributed with considerable randomness among KH3F4, etc. If crystalline hydrofluoric acid the available positions,1s leading to a residual contains cyclic molecules (HF)6, with the structure entropy corresponding to a crystalhe solution.

In muscovite,20 KA12(AlSi~)Olo(OH)2, one alumi- F, the two-minimum function and the num and three silicon atoms per formula are dis- F

tributed, presumably with considerable random- restriction that one hydrogen is attached to each ness, among the available tetrahedral positions. fluorine would lead to a residual entropy of ‘/e R The same phenomenon, leading to residual en- In 2 per mole of HF. tropy, without doubt occurs in many alumino-

Hydrogen bonds between unlike atoms, as in silicates. NH4F, would no4 lead to residual entropy. Crystals of cadmium bromidez1 and nickel

Residual entropy may also result from the bromide22 Prepxed in certain ways show a type multiplicity of rstable positions for atoms other Of randomness which does not lead to mY aPPreci- than hydrogen, In y-&Oa and v-FezOa the able residual entropy, provided that the Crystals oxygen ions are i3rranged in a close-packed frame- are not extremely small. This randomness of work, which provides nine positions for every eight structure (“Wechselstmktur~” alternating layer cations. The x..ray datal4 indicate that the cat- StmCtUre) COnSiStS in a choice between two posi- ions are distributed essentially at random among tions for each layer of a layer structure, leading to these positions, leading to a residual entropy of an entropy of k In 2 for each layer, which remains 3/2 R In In

agonal arrangemsent, the iron atoms being distribu-

O\C/O O\C/O I ‘c

CI I o/c\o d b

H ~ ~ ~ H

H ~ ~ ~ H

= ‘L.558 E. u. per mole of RzO3. (16) G. Hsgg, ibid. , Bag, 192 (1935). (17) L. Helmholz, J . Chem. P h y s . , 3, 740 (1935). (18) S. B. Hendricks and L. Pauling, THIS JOURNAL, 47, 2904

(1925). (19) T. F. W. Barth and E. Posnjak, J . Washington Acad. Sci.,

21, 255 (1931); F. Machatschki, Z. Kuist., 80, 416 (1031). (20) L. Pauling, Proc. N a f . Acad. Sci., 16, 123 (1930); W. W.

Jackson and J. West, Z. Kvist . , 76, 211 (1930).

Fe1-6S~ the atoms have a hex-

(14) G. Hagg and G . SBderhohn, 2. physik. Chem., 819, 88 (1935); G. Hagg, i b id . , B29, 95 (1935); E. J. W. Verwey, Z. Kris f . , 91, 65 (1935); E. Kordes, ibzd., 91, 193 (1935).

(1933) (15) G HAyg and I. Sucksdorff, 2. p h y s i k . Chem., B22, 444 (21) J. M. Bijvoet and W, Nieuwenkamp, ibid. , 86, 466 (1933).

(22) J. A. A . Ketelaar, i b id . , 88, 26 (1934).


2684 LINUS PAULING AND L. 0. BROCKWAY Vol. 57

inappreciable, inasmuch as the number of layers in a crystal (other than an extremely small crystal) is very small compared with the number of atoms.

In this connection i t might be mentioned that there exists the possibility that ice may crystallize with such an alternating layer structure. The oxygen-atom arrangement assigned to ice corresponds to superimposing double oxygen layers in the sequence ABAB-(A a t 00, B at '/s 2 / ~ , C a t z / / 3 of a hexagonal net). The sequence ABC- ABCABC-- would also lead to an arrangement (diamond) such that each oxygen atom is surrounded by four others arranged tetrahedrally, which is indeed, so far as I can see, just as satis- factory as the reported arrangement. There is no good evidence that such a cubic modification of ice has been observed. However, the arbitrariness of orientation which we have found to exist for the water molecules in ice suggests that there may also be an arbitrariness in the sequence of double oxygen layers, with configurations such as ABABABCBCB- occurring. Such an alternating layer structure would have hexagonal symmetry, might develop faces a t angles corresponding to the axial ratio c/a = 1.63, and would not be distinguishable so far as residual entropy is concerned from a crystal with fixed oxygen atom

arrangement. The x-ray data show that the sequence of layers is not completely random, the structure being essentially ABABAB-; i t is possible, however, that a change in the sequence, corresponding to twinning on the basal plane, occurs occasionally.

I am indebted to Professor W. F. Giauque for discussing the question of the structure and entropy of ice, as well as related questions, with me.

Summary

It is suggested that ice consists of water molecules arranged so that each is surrounded by four others, each molecule being oriented in such a way as to direct its two hydrogen atoms toward two of the four neighbors, forming hydrogen bonds. The orientations are further restricted by the requirement that only one hydrogen atom lie near each 0-0 axis. There are ( 3 / 2 ) N such configurations for N molecules, leading to a residual entropy of R In 3/2 = O.SO5 E. U., in good agreement with the experimental value 0.S7 E. U.

The structure and entropy of other crystals showing randomness of atom arrangement are discussed. PASADENA, CALIF. RECEIVED SEPTEMBER 24,1935

[ CONTRIBUTJON FROM GATES CHEMICAL LABORATORY, CALIFORNIA INSTITUTE OF TECHNOLOGY, KO. 5071

The Radial Distribution Method of Interpretation of Electron Diffraction Photographs of Gas Molecules

B Y LINUS PAULING AND L. 0. BROCKWAY

Introduction The only method of interpretation of electron

diffraction photographs of gas molecules which has been used to any great extent is the so-called visual method, involving the correlation of apparent maxima and minima on the photographs with maxima and minima on simplified theoretical curves calculated for various models of the molecule under consideration. This method of interpretation, originally developed by Wierl, ' has been thoroughly tested by Pauling and Brockway2 who have shown it to yield values of interatomic distances accurate to within about 1 yo (estimated probable error). The main disadvantage of the method is that i t does not involve a straightfor-

(1) R Wierl Ann Phys ik , 8, 521 (19311, 13, 453 (1932) ( 2 ) L Pauling and L 0 Brockway, J Chem Phis , 2, 867 (1934)

ward process of determining the structure of a molecule from the analysis of experimental results, but consists instead in the testing (and rejection or acceptance) of any structures which may be formulated, a tedious calculation being required for each structure.

We have developed a new method of interpretation of the photographs which does not suffer from this disadvantage. This radial distribution method, which is closely related to the method of interpretation of x-ray diffraction data developed by Zernike and Prim3 for the study of the structure of liquids and applied by Warren and Ging- rich4 to crystals, consists in the calculation (from

( 3 ) F. Zernike and J. A. Pn'ns, 2. Phyvik, 41, 184 (1927); see also P. Debye and H. Menke, Ergab . Tech. Rontgenkunde, Akad. Verlagsges., Leipzig, Vol. 11, 1931. (4) B. E. Warren and N. S. Gingrich, P h y s . REI. , 46, 368 (1934).


1144 W. F. GIAUQUE AND J. W. STOUT VOl. 58

Jenkins4 and Bell and The latter authors have shown in benzene solutions of con- centrations from M = 0.01 to M = 1.5 trichloroacetic acid is completely associated. We assume, therefore, that under the conditions of our measurements all the acids were entirely present as dimers.

The three infra-red absorption bands which can be investigated in this solvent correspond to the frequencies vl, vz and vs of the previous paper. The first of these ( V I ) (which has been omitted in Fig. 1 to conserve space) is mainly due to the C-H valence vibration, but is also in part made up of overtone and combination vibrations from other parts of the molecule. Thus i t occurs even in trichloroacetic acid, but considerably displaced and much weaker. The remaining two bands are to be attributed to vibrations within the ring formed by the union of the two single molecules. Attempts to measure some of these bands in the single molecules at high temperatures were un- successful on account of decomposition.

It may be seen by reference to Table I that the band vz is shifted toward higher frequencies by groups recognized by the organic chemist as electronegative, while i t is shifted to lower frequencies by groups electropositive in the same sense. The uniform shift in the chloroacetic acids discovered by Bennett and Daniels is confirmed. The effect of the weight of the groups may be seen by considering the monohalogen acids.

(4) Bury and Jenkins, J . Chem. Soc., 688 (1934). (5) Bell and Arnold, i b i d . , 1432 (1935).

Here the shift in the band decreases as the weight of the halogen increases until in iodoacetic acid i t occurs in the same position as acetic. This suggests that there may be two opposing factors operative, one a weight factor and the other perhaps connected with the electron affinity or polari- zability of the substituted group. It is noticeable that there is a correlation between the strength of the acid in aqueous solution and the position of this band. This is perhaps to be associated with changes in the contribution of ionic states to the energy of the molecule.

The remaining band, v3, is shifted in each case in the opposite direction, but the displacement is smaller. This behavior was observed also in the investigation on the effect of association.2 It does not seem possible in the light of our present knowledge to give an exact interpretation of these band shifts although they seem too uniform to be fortuitous.

The author is grateful to Dr. Farrington Daniels for helpful suggestions and for continued interest in this problem.

Summary

The infra-red absorption spectra of acetic acid and eight substituted acetic acids have been measured in carbon tetrachloride solution. Cer - tain uniform shifts in the absorption bands have been observed and correlated with other measurements. MADISON, WISCONSIN

1.

RECENED MAY 5, 1936

[CONTRIBUTION FROM THE CHEMICAL LABORATORY OF THE UNIVERSITY OF CALIFORNIA]

The Entropy of Water and the Third Law of Thermodynamics. The Heat Capacity of Ice from 15 to 273OK.

BY W. F. GIAUQUE AND J. w. STOUT

It has long been known to those interested in the accurate application of the third law of thermodynamics that measured entropy changes in reactions involving water did not agree k t h those calculated from low temperature heat capacity data. In early comparisons the inaccuracy of the available data seemed sufficient to explain the disagreement, but even after more accurate experiments were performed discrepancies still remained. The investigations of Wiebe, Johns-

ton, Overstreet, and one of us1 on the entropies of hydrogen chloride, hydrogen, oxygen and, chlo- rine, combined with very accurate determinations of the heats of reactions

Hz + '/SO* = HzO and 2HC1 4- 1/202 = HzO + Cln

(1) (a) Giauque and Wiebe, (HCI), THIS JOURNAL, 60, 101 (1928); (b) Giauque and Johnston, (Hz), ibid., 10, 3221 (1928); ( c ) Giauque and Johnston, (a), i b i d . , 61, 2300 (1929); (d) Giauque, (Hs), ibid., 68, 4816 (1930); (e) Giauque and Overstreet, (HC1, CIz), ibid., 64, 1731 (1932).


July, 1936 THE HEAT CAPACITY OF ICE 1145

made by Rossini2 and various equilibrium data, including those relating to the reactions3

HgO + Hz = Hg + Hz0 Hg + ' / O z HgO

indicated that the C,d In T for water did not give the correct entropy. That this was so be- came a certainty when Giauque and Ashley4 calculated the entropy of gaseous water from its band spectrum and showed that an entropy discrepancy of about one calorie per degree per mole existed. They presumed this to be due to false equilibrium in ice at low temperatures.

Water is a substance of such importance that we considered further experimental investigation to be desirable not only to check the above discrepancy but especially to see whether slow cooling or other conditions favorable to the attainment of equilibrium could alter the experimental result.

Apparatus.-In order to prevent strains in the resistance thermometer when the water was frozen a double-walled calorimeter, Fig. 1, was constructed. The outside wall was of copper, 0.5 mm. thick, 4.4 cm. 0. d., and 9 cm. long. The inside copper wall, 0.5 mm. thick, was tapered, being 3.8 cm. 0. d. a t the bottom and 4.0 cm. 0. d. a t the upper end. The top of the inner container was made from a thin copper sheet, 0.2 mm. thick, which prevented the trans- mission of strains to the resistance thermometer. The neck for filling the calorimeter was in the center of this sheet. A series of thin circular slotted vanes of copper were soldered to the inner container, and the assembly forced inside the outer tube. A heavy copper plate, 1 mm. thick inside the inner wall, and 2 mm. thick between the walls, served as the bottom of both tubes. The thermocouple was soldered into tube D by means of Rose's metal.

A resistance thermometer-heater of No. 40 double silk covered gold wire containing about 0.1% silver was wound on the outside of the calorimeter. The resistance was 310 ohms at 290°K. and dropped to about 17 ohms at G0K. The resistance thermometer was calibrated during the measurements by means of copper-constantan thermocouple No. 16 which had been compared with a hydrogen gas thermometer.6 However one of the five parallel constantan wires in the thermocouple had accidentally been broken since the original calibration. This wire was discarded and after the completion of the measurements the thermocouple was compared with the oxygen and hydrogen vapor pressure thermometers.6b The thermocouple was also checked against the melting point, 54.39"K. and higher transition point, 43.76'K., of oxygen.'' On

(2) (a) (HzO), Bur. Standards J . Research, 6, 1 (1931); (b) (HCI),

(3) See summary by Eastman, Circular 6125, U. S. Dept. of

(4) Giauque and Ashley, Phys. Rev. , 4S, 81 (1933). ( 5 ) (a) Giauque, Buffington and Schulze, THIS JOURXAL, 49, 2343

(1927): (h) Giauque, Johnston and Kelley, i b id . , 49, 2867 (1927).

J T

ibid.,9, 883 (1932).

Comm., Bur. of Mines (1929).

the basis of these comparisons a small correction to the original calibration was readily made.

Helium gas was introduced into the space between the two walls by means of a German silver tube, A. A similar German silver tube was soldered by means of Wood's metal into the cap, B. The sample, C, was transferred through this tube into the calorimeter, and helium gas at one atmosphere pressure admitted. The German silver tube was then heated and removed from the cap, leaving the hole sealed with Wood's metal. After the measurements on the full calorimeter had been completed, the calorimeter was heated to the melting point of the Wood's metal (72°C.) and the water completely pumped out without dismantling the apparatus. The heat capacity of the empty calorimeter was then

The remainder of the heat capacity apparatus, the method of making the measurements and calculations, and accuracy consid- erations were similar to those previously described.'".'

Purification of Water,- Distilled water from the laboratory still was transferred into the vacuum- tight purification apparatus constructed from Pyrex glass. The apparatus was evacuated to remove dis- solved gases, and flushed out several times with helium gas. The water was distilled into a receiving bulb, the first fraction being discarded The calorimeter had previously been attached to the purification system and evacuated. When sufficient water had collected in the receiving bulb, it was transferred into

I measured.

D Fig. 1.-Calorimeter

the calorimeter. Helium gas a t one atmosphere pressure was admitted to the calorimeter which was then sealed off as described above.

A series of short heat capacity measurements were made in the temperature region immediately below the melting point in order to determine the pre-melting effect due to liquid-soluble solid-insoluble impurity. From these measurements it was calculated that the mole fraction of impurity was three parts in a million,

The Heat Capacity of Ice.-The results of the heat capacity measurements are given in Table I.

The data are shown in Fig. 2. In the calculations one 1 5 O calorie was taken

as equal to 4.1832 international joules. The calorimeter contained 72.348 g. of ice.

In order to allow time for the establishment of an equilibrium state in the solid, the ice was cooled


1116 W. F. GIAUQUE AND J. W. STOUT Vol. 5s

very slowly. The following are temperatures reached at various times after the ice was frozen : 0 hours, 273.1'; 12 hours, 246'; 37 hours, 203'; GO hours, 180'; 84 hours, 168'; 92 hours, 156'; 108 hours, 105'; 120 hours, 91'; 156 hours, 90'. The sample was then cooled from 90 to 68' in about three hours and the heat capacity measurements of series I taken. Next the calorimeter was cooled to the temperatures of liquid hydrogen at the following rate: 0 hours, 85'; 1.5 hours, 72'; 2 hours, 60'; 2.5 hours, 56'; 7 hours, 39'; (liquid hydrogen evaporated) 17.5 hours, 49' ; 22 hours, 50' (more liquid hydrogen added) ; 23 hours, 41'; 27 hours, 13'. The measurements of series I1 were then made.

I--- - I

01 / I I I I

0 80 160 240 Temperature, OK.

Fig. 2.-Heat capacity in calories per degree per mole of ice.

During this series of measurements which extended from 15'K. to the melting point, and covered a period of eighty hours, the calorimeter was under constant observation. To make certain that no unusual thermal situation was present in the solid near the melting point, the heat of fusion was determined a t the end of the above series of measurements. The value obtained, 1436 cal./mole, agrees well with that which has been chosen for the entropy calculation.

TABLE I HEAT CAPACITY OF ICE

(Molecular weight, 18.0156) OOC. = 273.10"K. T,OR. AT C, cal./deg./mole Series

16.43 1.403 0.303 I1 18.37 1.729 .410 I1 20.78 2.964 .528 I1 24.20 3.815 .700 I1 28.05 3.596 .883 I1 31.64 3.578 1.OG5 I1 35.46 4 073 1.251 I1 39 62 4 242 1.440 IT

43.96 48.52 52.98 57.66 62.63 G7.83 70.61 73.01 75.60 78.51 79.98 81.44 82.42 83.72 83.94 86.66 87.25 89.20 91.32 91.93 94.93 95.85 97.37 99.57

100.69 104.69 110.13 115.84 121.74 127.54 133.50 139.48 145.43 151.43 157.48 163.52 169.42 175.36 181.25 187.20 192.96 199.11 205.32 211.56 217.97 224.36 230.08 236.19 242.40 249.31 256.17 262.81 267.77

4.469 1.641 i1 4.571 1.837 i1 4.361 2.014 i1 5.041 2.203 11 5.228 2.418 11 4.910 2.612 11 5.403 2.723 1 5.737 2.821 11 4 I 638 2.922 1 4.991 3.016 11 4.133 3.070 1 5.538 3.115 111 4.860 3.163 I v 5.438 3.191 11 3.765 3.199 1 4.893 3.286 111 4.756 3.336 I V 5.557 3,389 11 4.394 3.488 111 4.651 3.532 IV 5.233 3.649 11 4.649 3.660 111 6.234 3.724 IV 4.778 3.814 11 4.980 3.832 111 5.497 3.985 11 5.373 4.136 11 6.031 4.315 11 5.908 4.489 11 5.813 4. fi55 11 6.005 4.808 11 5.952 4,978 11 5.928 5.135 11 6.240 5.306 11 5.837 5.466 11 5.851 5 . G63 11 5.908 5.996 5.678 5.983 5.658 6.133 6.309 6.554 6.200 6.935 6.068 6.101 6.795 6.903 6.591 6.303 4,465

5.842 (i ,007 6.185 6.359 A . 530 6.710 6.935 7.119 7.326 7.519 7.711 7.887 8.048 8.295 8.526 8.732 8.909

11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11

In the heat capacity measurements between 85 and lOO'K., the attainment of temperature equilibrium in the solid was much less rapid than at other temperatures. This observation is of considerable interest and some of its implications will be discussed below.

'fo study possible effects due to rapid cooling. Physics 423 Friday 5/13/2012


the sample was cooled from 273.1 to 90' in four hours, The heat capacities in series I11 were then measured. Slow equilibrium was again en- countered] and the heat capacities were not significantly different from those previously obtained.

To determine whether the ice was being trans- formed from an unstable to an equilibrium condition at an appreciable rate, measurements extending over several hours each were made of the rate of temperature drift at 60, 72, 78 and 85'K. After correcting for the known heat interchange with the surroundings, the rate of evolution of heat due to internal changes was zero within the limits of experimental error. An amount of 0.0008 calorie per mole per minute could have been detected.

The calorimeter was allowed to stand for four days at temperatures between 60 and 80' before taking the measurements in series IV. Again the measurements were not significantly different from those previously obtained.

In Table I1 are listed the values of the heat capacity a t even temperatures as read off a smooth curve through the observations. These values are compared with those of previous investigators. A review of the measurements prior to 1913 has been given by Dickinson and Osborne.'j Of these the measurements of Nernst and co-workers are most important. The values used for comparison in Table I1 are calculated from an equation proposed by Nernst, Koref and Lindemann? as repre- senting these data. Pollitzers made measurements between 20 and 90'K. His measurements were plotted and a smooth curve drawn through them to give the values compared in Table 11. The very accurate work of Dickinson and Os- borne6 between 0 and -4O'C. has been calculated from an equation proposed by these authors. Maass and Waldbauerg and Barnes and Maasslo have measured the total change in heat content from various low temperatures to 25'C., and from the results have proposed equations for the specific heat of ice. Simon1' lists values for the heat capacity of ice from 1O0K. to the melting point.

(1915). (6) Dickinson and Osborne, Bull. U. S. Bur. Standards, l2, 49

(7) Nernst, Koref and tindemann. Silebcr. Berlin Akad. Wiss.,

(8) Pollitzer, 2. Elcktrochcm., 19, 513 (1913). (9) Maass and Waldbauer, THIS JOURNAL, 47, 1 (1925). (10) Barnes and Maass, Can. J. Research, 3, 205 (1930). (1 I ) Simon, "Hunilhlich der Physik," Vol. X, 1926, p, 363.

247 (1910).

Professor Sirnon12 informs us that, of the values listed in the "Handbuch der Physik," only the one at 10'R. is based on the results of his own measurements. The heat capacities at higher temperatures were calculated from the work of earlier observers. The experimental results of Simon1$ are in excellent agreement with the con- tinuation of the curve drawn through our measurements. We have included in Table I1 a value for the heat capacity of ice a t 10'K. picked from a smooth curve through Simon's data.

TABLE I1

(Molecular Weight, 18.0156) HEAT CAPACITY OF ICE

0°C. = 273.10'K. Values taken from smooth curve through observations

CP, cal./ Deviations previous results-This research, % deg./ Nernst Pollitzer D . and M. and B . and

T,"K. mole 1910 1913 0. 1915 W. 1925 M. 1830

10'8 0,066 20 .490 -13.3 30 .984 - 0.8 40 1.466 f13 .1 50 1.896 + 4 .1 60 2.304 + 2 . 8 70 2.701 + 8 . 5 80 3.075 + 4.4 + 6 . 6 90 3.448 + 1 . 3 - 0.4 -29.6

100 3.796 - 0 .6 -22.4 110 4.130 - 1.9 -16.8 120 4.434 - 2 . 3 -11.8 130 4.728 - 2 .4 - 7.6 140 4.993 - 1.8 - 3.8 150 5.265 - 1.5 - 0.8 160 5.550 - 1 . 4 + 1.3 170 5.845 - 1 .5 + 2 . 5 180 6.142 - 1 .5 + 3 . 4 190 6.438 - 1.5 + 3 .9 200 6.744 - 1 . 5 + 3 . 9 -2.1 210 7.073 - 1.7 + 3 . 3 -0 .9 220 7.391 - 1 . 7 f 2.6 - .I 230 7.701 - 1.3 -0.5 + 1.9 + .3 240 8.013 - 0.4 - . 2 + 0 .9 + . 2 250 8.326 + 1 . 3 + . 1 - 0 . 3 - . 4 260 8.642 + 6 . 1 + .3 - 1.6 -1.3 270 8.960 f 4 3 . 8 + . 5 - 3.2 -2.6

The Entropy of Water.-Values of the heat of fusion, heat capacity of the liquid and heat of

(12) Simon, personal communication. Measurements of the heat capacity of ice between Q and 13OK. were made in 1923 by Simon, but the results have not as yet been published. Since these values are of considerable interest in connection with the present investigation, we are, with Professor Simon's kind permission, presenting them here. HEAT CAPACITY OF ICE BETWEEN 9 AND 13'K. MEABUREMENTS

T, OK. 9 . 4 7 9.88- 10.46 11.35 l l . 5 j 0 12 10 12 .85 C, cal./deg. j

OF F. S I M O N

mole 0 . 0 5 6 0.063 o 075 0.096 0.102 o 118 n 141 a 0bt:iined in an independent second experiment. (1.7) Ctrlcilluted fromdataof Simon, see Ref. 12.


114s W. F. GIAUQUE AND J. W. STOUT lTol. 58

vaporization of water are so accurately known that further investigation was unnecessary.

The heat of fusion of ice has been accurately determined by a number of workers. The measurements prior to 1913 have been critically sum- marized by Dickinson, Harper and Osborne,14 who also made a number of measurements of the heat of fusion both by an electrical method and by the method of mixtures. Dickinson and Osborne6 measured the heat of fusion in an aneroid calorimeter, using electrical heating. The measurements in which energy was introduced electrically were recalculated by us on the basis of 1 int. joule = 4.1832 calories (15’). A weighted average of all the reported values yields 1435.7 cal./ mole with an estimated accuracy of *0.9 cal./ mole, for the heat of fusion.

Fiock16 has reviewed the measurements of the heat of vaporization of water and compared them with the results of determinations at the Bureau of Standards16 extending down to 50’. All measurements were converted into international joules. Of the data considered by Fiock, those of Grif- fiths, of Smith and of Henning contained measurements in the neighborhood of 25’C. Giving equal weight to the result of each of the above three observers and t o the value extrapolated from the Bureau of Standards measurements, and taking 1 calorie (15’) = 4.1832 int. joules, we obtained an average value of 10,499 * 3 (av. dev.) calories/mole for the heat of vaporization of water a t 25’.

The “I.C.T.” values for the heat capacity of liquid waterI7 were plotted against the logarithm of the absolute temperature and integrated graphically to obtain the entropy between 0 and 25’. The value for the vapor pressure a t 25’ was also obtained from the “I.C.T.”lS Using Berthe- lot’s equation of state and thermodynamics it can be shown that the entropy correctionla to the ideal gas state is almost negligible in this case The critical constants18 used were Tc = 647.1°K. and P, = 217.7 atm.

The entropy between 10 and 273.10’K. was obtained by graphical integration of the measured heat capacities. The entropy between 0 and

(14) Dickinson, Harper and Osborne, Bull. U. S. Bur. Slandards,

(15) Fiock, Bur. Standards J . &search, 6, 481 (1930). (16) (a) Osborne, Stimson and Fiock, ibid., 6, 411 (1930); (b)

(17) “International Critical Tables,” McGraw-Hill Book Co.,

(18) “I. C. T.,” Vol. 111, p. 211. (19) “I. C. T.,”Vol. I I I , p . 248.

10, 235 (1014).

Fiock and Ginnings, ibid., 8, 321 (1932).

New York, Vol. V, 1926, p. 113.

IO’K. was calculated by means of the Debye equation, using hv/k = 192. A summary of the entropy calculation is given in Table 111.

TABLE rrr CALCULATION OF ENTROPY OF U~ATER

O-1O0K., Debye function hv/k = 192 0.022

Fusion 1435.7/273.10 ,5,257 273.10-298.1OGK., graphical 1 ,580

10-273. 10°K., graphical 9.081

Vaporization 10499/298.10 35,220 Correction for gas imperfection O,OfL? Compression R In 2.3756/760 -6.88Ci

Cal./dcg./~iiole 44.28*0.05

The value of the entropy given in Table I11 may be compared with that calculated from spectroscopic data. Giauque and Ashley4 util- ized the preliminary molecular constants of water as given by Mecke and Baumann20 to determine the entropy of water. Later Gordon21 recaleu- lated the thermodynamic quantities for water using the revised moments of inertia of Freuden- berg and Mecke.22 He obtained So~gB.l = 45.10 cal./deg./mole. The difference between the spectroscopic and calorimetric values is 0.82 cal./deg. /mole.

The Problem of the False Equilibrium in Ice.-To account for the discrepancy between the calorimetric and spectroscopic values for the entropy of water Giauque and Ashley‘ offered an explanation based on the assumption that the ortho and para molecular states, which are known to exist in gaseous water, had persisted in the crystalline state at low temperatures. The situation was assumed to be similar to that which ac- counts for the entropy discrepancy in the case of solid The ortho water was assumed to have non-polar clockwise and counter clockwise rotations in ice, since the dielectric constant of ice at low temperatures corresponds to that of non-polar substances. This leads to a calculated discrepancy of 3/4R In 2 = 1.03 cal./ deg. /mole.

We have had many interesting private discus- sions with Professor Linus Pauling who has consistently objected to the ortho-para explanation. During the course of the present investigation Pauling2a offered an alternative explanation based

(20) Mecke and Baumann, (a) Nafuvwiss . . 20, 657 (1932); (b)

(21) Gordon, J . Chem. Phys. , 2,65 (1934). (22) Freudenberg and Mecke, Z. Phrsik, 81,465 (1933). (23) Pauling, (a) personal communication; (b) THIS JOURNAL, S I B

Phys . Z., 89, 833 (1932).

2680 (1935) Physics 423 Friday 5/13/2012


on the random orientation of hydrogen bonds in ice. Crystal structure investigations have shown that each oxygen atom is surrounded by four equivalent oxygen atoms. Although the positions of the hydrogen atoms have not been determined experimentally it seems reasonable to assume that they are located on lines joining adjacent oxygens. Since the oxygen-dxygen distance is considerably greater than twice the sepa- ration which is characteristic of an oxygen-hydrogen linkage, it has been assumed by Bernal and Fowlerz4 that each oxygen is joined to two close and two distant hydrogen atoms. At higher temperatures the bond directions of the close or distant hydrogen atoms of a given oxygen are a matter of chance. We quote from the paper of Ber- nal and Fowler. “Therefore it is quite conceivable and even likely that at temperatures just below the melting point the molecular arrangement is still partially or even largely irregular, though pre- serving at every point tetrahedral coordination and balanced dipoles. In that case ice would be crystalline only in the position of its molecules but glass-like in their orientation. Such a hypothesis may be still necessary to explain the dielectric constant and the absence of pyroelectricity.”

Pauling assumes further that when ice is cooled to low temperatures, i t fails to attain the ordered arrangement which would correspond to zero entropy. He shows that the discrepancy corresponding to the above lack of order would be R In 6/4 = 0.806 cal./deg./mole. This is in very close agreement with the experimentally determined discrepancy of 0.82 * 0.05 cal./deg./mole.

However, it should be pointed out that in two previous cases, carbon monoxidez5 and NNOZ6 where random molecular orientations led to entropy discrepancies, the experimental discrepancy was a few tenths of a unit lower than the calculated value due to partial attainment of an ordered state. If this were the case in ice, the ortho-para explanation might be correct. How- ever, we consider that the explanation advanced by Pauling is the more plausible as well as being in better agreement with the experimental results.

It is of interest to note that MacDougall and Giauquez7 investigated ice from 0.2 to 4’K.

(24) Bernal and Fowler, J . Chcm. Phys. , 1, 515 (1933). ( 2 5 ) Clayton and Giauque, Tars JOURNAL, 54,2610 (1932). (26) ( 8 ) Clusius, Z. Elektrochcm., 40, 99 (1934);

(27) hbacDougak1 and Giauque, ibid., IS, 1032 (1936).

(b) Blue and Giauque, THIS JOURNAL, 17,991 (1935).

and found no appreciable heat capacity in this region.

One of the purposes of the present work was to investigate the possibility of more complete attainment of equilibrium. We have mentioned above that various experiments, in which ice was cooled slowly or rapidly to low temperatures, or was allowed to stand for long periods of time a t low temperatures, resulted in heat capacities which were not appreciably different in the various series of measurements. At temperatures between 85 and 100’K. the attainment of thermal equilibrium in the solid was very much less rapid than at other temperatures. For this reason the heat capacity measurements in this region are somewhat less accurate than the others. This slow equilibrium presumably is due to the initial stages of excitation of some new degrees of freedom. From the value of the heat capacity this evidently is connected with motion of the hydrogen atoms. At temperatures below the region of slow equilibrium the dielectric constant2* is of the order of magnitude characteristic of a non- polar substance. At higher temperatures the dielectric constant rises rapidly and the orientation time of the dipole decreases. From this it would appear that the new degrees of freedom re- ferred to above are associated with the dipole orientation mechanism.

From the temperature drift experiments at 60, 72, 78 and 85OK. it appears that no great change in the heat content would be expected if ice were kept a t these temperatures for a considerable period of time.

We may mention in passing that low temperature heat capacity measurements on some of the high pressure forms of ice would undoubtedly lead to correct values for the entropy of water.

It is perhaps worth noting that while the theory of random bond orientation would lead to the same discrepancy, R In 3/2 = 0.806 cal./deg./ mole in both hydrogen and deuterium oxides, the theory of molecular rotation in ortho states of water would lead to different values for the calculated discrepancy. For hydrogen oxide the value is 3/4 R In 2 = 1.033, for deuterium oxide, 1/3 R In 2 = 0.459 cal./deg./mole. Although we believe the latter theory to be less plausible it will be of interest to make the comparison when measurements on deuterium oxide become available.

(28) “I. C. T.,” Vol VI , page 78. Physics 423 Friday 5/13/2012

Activity: Concept diagram (10 min) Students list concepts we’ve learned, and weform concept diagram on board.


Survey and post-test

Question 1 How would you perform a measurement of the quantity(∂S

∂τ

)T

(20.59)

for a rubber band without making a measurement of heat transfer (e.g. if you don’t haveany styrofoam). Describe the experiment and draw a sketch of your apparatus.

Question 2 Consider an atom with three energy states,

E0 = 0 (20.60)

E1 = E2 = ε (20.61)

What is the internal energy U at temperature T?


Question 3 A curve has been traced out on the z-y graph below, and has been labeledPath A. Consider the integral


zdy (20.62)

where the integral is taken around Path A starting at point P , proceeding clockwise untilreaching point P again.

a) Is the integral I1 positive, negative, zero or is there not enough information to decide?Please explain your reasoning.

0

Path A

y

z

P

Refer again to the graph in part (a). We define H as a function of the independentvariables z and y; i.e. H = H(z, y). Consider the integral


~∇H · d~r (20.63)

where the integral is taken around Path A starting at point P , proceeding clockwiseuntil reaching point P again. (Note: d~r = dyy + dzz)

b) Is the integral I2 positive, negative, zero or is there not enough information to decide?Please explain your reasoning.


Question 4 Lab 1 What did you learn from Lab 1, in which we heated up ice water?

Did you like this lab? Do you have any suggestions as to how we could improve it?

Question 5 Lab 2 What did you learn from Lab 2, in which we measured amount of iceremaining and final temperature of water?


Question 6 Lab 3 What did you learn from Lab 3, in which we measured the tension ofa rubber band?



Question 7 Class as a whole Do you like Thermal Physics? Why or why not?


21 Monday: Final exam

Current board:

(∂a

∂b

)c

=1(∂b∂a

)c

(∂a

∂b

)d

(∂b

∂c

)d

=

(∂a

∂c

)d

(∂a

∂b

)c

= −

(∂c∂b

)a(

∂c∂a

)b

“Cp =

(dQ

∂T

)p

”

∆S =

∫dQreversible

T

dU = dQ+dW


dU = TdS − pdV

dF = −SdT − pdV

dH = TdS + V dp

dG = −SdT + V dp


i

Pi lnPi

Z ≡all states∑

i

e−βεi

Pi =e−βεi

Z


Final exam

Problem 1 Masses on a piston An ideal gas is contained in a cylinder with a tightlyfitting piston. Several small masses are on the piston. (Neglect friction between the pistonand cylinder walls.) The cylinder is placed in an insulating jacket, and a large number ofmasses are then added to the piston.

Tell whether the pressure, temperature, and volume of the gas will increase, decrease, orremain the same. Explain.

additional

masses

insulating

jacket

Phys 423 115 Name

Problem 2 Gibbs free energy The Gibbs Free Energy, G, is a function of the indepen-dent variables T and p, i.e., it can be written as G(T, p). The total differential of G can bewritten as dG = −SdT + V dp.

a) Interpret the above equation in order to determine an expression for the entropy S.

b) From the total differential dG, obtain a different thermodynamic derivative that isequal to (

∂S

∂p

)T

Physics 423 116 Name

Problem 3 Two processes This Pressure-Volume (P -V ) diagram represents a systemconsisting a fixed amount of ideal gas that undergoes two different quasistatic processes ingoing from state A to state B:

0 Volume

Pressu

re

State

State

A

B

Process #1

Process #2

[In these questions, W represents the energy the system gains by working during a process;Q represents the energy the system gains by heating during a process.]

a) Is W for Process #1 greater than, less than, or equal to that for Process #2? Explain.

b) Is Q for Process #1 greater than, less than, or equal to that for Process #2? Explainyour answer.

c) Which would produce the largest change in the total energy (kinetic plus potential) ofall the atoms in the gas: Process #1, Process #2, or both processes produce the samechange?

Phys 423 117 Name

Problem 4 Hanging Chain The upper end of a hanging chain is fixed while the lowerend is attached to a mass M . The (massless) links of the chain are ellipses with major axesl + a and minor axes l − a, and can place themselves only with either the major axis or theminor axis vertical. The figure below shows a four-link chain in which the major axes of thefirst and fourth links and the minor axes of the second and third links are vertical. Assumethat the chain has N links and is in thermal equilibrium at temperature T .

a) What is the probability of finding the first link oriented with its major axis vertical?

b) What is this probability in the limit of high temperature?

c) What is the average length of the chain?

.


Problem 5 Insulated room An object is placed in a thermally insulated room thatcontains air. The object and the air in the room are initially at different temperatures. Theobject and the air in the room are allowed to exchange energy with each other, but the airin the room does not exchange energy with the rest of the world or with the insulating walls.

a) During this process, does the entropy of the object [Sobject] increase, decrease, remainthe same, or is this not determinable with the given information? Explain your answer.

b) During this process, does the entropy of the air in the room [Sair] increase, decrease,remain the same, or is this not determinable with the given information? Explain youranswer.

c) During this process, does the entropy of the object plus the entropy of the air in theroom [Sobject + Sair] increase, decrease, remain the same, or is this not determinablewith the given information? Explain your answer.

d) During this process, does the entropy of the universe [Suniverse] increase, decrease, re-main the same, or is this not determinable with the given information? Explain youranswer.

Phys 423 119 Name

Problem 6 Soap bubble Consider the reversible expansion (stretching) of a film ofsoapy water. The film’s internal energy is proportional to its area—this is why bubbles arespherical, since this minimizes their surface area.

The infinitesimal of work done by changing the area of a film is given by the surfacetension σ:

dW = σdA (21.1)

a) Show how the surface tension is related to the internal energy, and to the Helmholtzfree energy of the film. Briefly explain one of these relations in words.

b) Deduce expressions for the differentials of the internal energy and the Helmholtz freeenergy appropriate for this system (the film).

c) Deduce two different Maxwell relations describing such a film.

d) A student carefully measures the effect of temperature on surface tension of a givenbubble, and finds that over a wide range of temperatures and areas,(

∂σ

∂T

)A

= λ(T − T3) (21.2)

where λ is a constant with the value 2×10−4 Jm−2K−2, and T3 has the value of 600 K.What is the change in entropy when this bubble is isothermally expanded from an areaof 1 cm2 to an area of 2 cm2 at room temperature?

e) In the above scenario, does the bubble heat its surroundings, or vice versa? Explain.

f) If you can work out the change in internal energy U for the above process, do so. Ifnot, explain what additional information you would need.


Tentative schedule

Day Topic/subtopic Reading Homework due

1Monday

4/18

What kind of beast is it?Thermodynamic variables Chapter 1

Thermodynamic equilibriumEquations of state

2Tuesday

4/19

Partial derivatives as change of variablesTotal differential

Exact vs. inexact differentialsChain rules

3Wednesday

4/20Homework 1

4Thursday

4/21

Mixed partial derivativesMonotonicity and invertibilityLegendre transform

5Friday4/22

Lagrange multipliers for minimization Homework 2

6Monday

4/25

Lab 1: Heat and TemperatureDulong-Petit Law

7Tuesday

4/26

First and Second LawsSystem and surroundings

First LawSecond Law and Entropy

Fast and slowThe thermodynamic identity 5.1.2.2Heat capacity

8Wednesday

4/27Second Law lab Lab 1

9Thursday

4/28

Heat and workWork

10Friday4/29

Engines and FridgesOther views of the Second LawCarnot efficiency

Lab 2Homework 3

Phys 423 121 Name

11Monday

5/2

Monotonicity and invertibilityThermodynamic potentials Chapter 7

Legendre transformMaxwell relations

12Tuesday

5/3Lab 2: rubber band Prelab 3

13Wednesday

5/4Homework 4

14Thursday

5/5Thermodynamics practice

15Friday

5/6Thermodynamics practice Lab 3

16Monday

5/9

Statistical approachFairness function Chapter 6

17Tuesday

5/10

Optimizing the fairnessLeast bias lagrangianWeighted averagesProbabilities of microstates Chapter 11

18Wednesday

5/11

From statistics to thermodynamicsThermodynamic properties from the Boltzmann factor

Homework 5

19Thursday

5/12

Statistical mechanics of airQuantum spectraDiatomic gas

20Friday5/13

Diatomic gas wrapupThird law

Ice rulesHomework 6

21Monday

5/16Final exam


Index

adiabatic, 30

Boltzmann factor, 76

carnot efficiency, 44

efficiency, 43eigenstates, 71enthalpy, 52, 53entropy, 30, 31, 53equation of state, 3exact differential, 7

fairness function, 72

Gibbs free energy, 52, 53

heat, 27heat capacity, 31heat engine, 42heat sinks, 42Helmholtz free energy, 52, 53helmholtz free energy, 21

inexact differential, 7internal energy, 30, 53irreversible, 30

Lagrange multiplier, 75Lagrangian, 75Law of thermodynamics

First, 29Second, 29, 42

Clausius formulation, 42Kelvin formulation, 42

Third, 94

partition function, 76

quasistatic, 30

reversible, 30

sackur-tetrode equation, 4spontaneous, 30state function, 2state variable, 2surroundings, 28system, 28

total differential, 5

work, 40working substance, 43

Phys 423 123 Name

energy and entropy - sites.science.oregonstate.edu

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