Energy and Entropy Physics 423

Home , Enthalpy of fusion, Exact differential, Inexact differential

David Roundy

Spring 2012 Draft 1/30/2012

Contents

Contents i

1 Monday: What kind of beast is it?1 1.1 Thermodynamic variables (Chapter 1 )...... 1 Activity: Thermodynamic variable cards (4 min) ...... 1 Required homework 1.1: steam-tables...... 2 Activity: Dimensions (5 min?) ...... 2 Activity: Conjugate pairs (10 min?) ...... 2 Activity: Intensive vs. extensive (10-15 min) ...... 2 Required homework 1.2: extensive-internal-energy...... 2 Activity: State variables and functions (5 min?) ...... 2 Activity: Conserved quantities (??? min) ...... 3 Thermodynamic equilibrium ...... 3 Lecture: (??? min) ...... 3 Equations of state ...... 3 Lecture: (3-4 min) ...... 3

Physics 423 i 2 Tuesday: 4 Activity: Math pretest (15 min) ...... 4 2.1 Partial derivatives as change of variables ...... 4 Lecture: (???) ...... 4 Activity: Paramagnetism (??? min) ...... 4 Activity: Ideal gas bulk modulus (??? min) ...... 4 2.2 Total differential ...... 5 Lecture: (11 min) ...... 5 BWBQ: (5 min) ...... 6 Lecture: (5 min) ...... 6 BWBQ: (15 min) ...... 6 Exact vs. inexact differentials ...... 7 BWBQ: (10 min) ...... 7 Lecture: (5 min?) ...... 7 Required homework 1.3: total-differentials...... 7 2.3 Chain rules ...... 7 Lecture: (10 min) ...... 7 Exam: Math pretest ...... 8 Question 1 ...... 9 Question 2 ...... 9 Question 3 ...... 10 Question 4 ...... 11

3 Wednesday: 12 Handout: Changes of variables...... 12 Homework 1 due ...... 14 Problem 1.1 Steam tables ...... 14 Problem 1.2 Extensive internal energy ...... 14 Problem 1.3 Total diﬀerentials ...... 14

4 Thursday: 15 4.1 Mixed partial derivatives ...... 15 Activity: BWBQ: Mixed partials (30-40 min?) ...... 15 Activity: Mixed partials vanishing (??? min) ...... 15 Handout: Mixed partial derivatives...... 15 Required homework 2.1: a-maxwell-relation ...... 19 Required homework 2.2: adiabatic-susceptibility...... 19 Required homework 2.3: summation-notation ...... 19 Required homework 2.4: homogeneous-function-theorem ...... 19 Monotonicity and invertibility ...... 19 Activity: Monotonicity lecture/discussion (10 min) ...... 19 Legendre transform ...... 20 Lecture: (??? min) ...... 20

Physics 423 ii 5 Friday: Lagrange multipliers for minimization 23 Homework 2 due ...... 23 Problem 2.1 A Maxwell relation ...... 23 Problem 2.2 Adiabatic susceptibility ...... 23 Problem 2.3 Summation notation...... 24 Problem 2.4 Euler’s homogeneous function theorem...... 24

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

7 Tuesday: First and Second Laws 28 Activity: Name-the-experiment pretest (10 min) ...... 28 System and surroundings ...... 28 Lecture: (2-3 min) ...... 28 BWBQ: (10 min) ...... 28 7.1 First Law ...... 28 Lecture: (10 min) ...... 28 SWBQ: (5 min) ...... 29 7.2 Second Law and Entropy ...... 29 SWBQ: (1 min) ...... 29 Lecture: (5 min) ...... 29 Fast and slow ...... 30 Lecture: (7-12 min) ...... 30 7.3 The thermodynamic identity (5.1.2.2 )...... 30 Lecture: (8-15 min) ...... 30 Activity: Name the experiment (20 min) ...... 30 7.4 Heat capacity ...... 31 SWBQ: (5 min (skipped in 2011)) ...... 31 Lecture: (10 min) ...... 31 BWBQ: (10 min (skipped in 2011)) ...... 32 Lecture: (5 min (skipped in 2011)) ...... 32 Activity: Entropy change of cooling coﬀee (20-30 min (skipped 2010)) . . . . 32 Required homework 3.1: adiabatic-ideal-gas ...... 33 Required homework 3.2: bottle-in-bottle...... 33 Exam: Pretest...... 33

Physics 423 iii Question 1 ...... 34 Question 2 ...... 34

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

9 Thursday: Heat and work 38 Activity: Expanding Gas Quiz (20-45 min) ...... 38 Handout: Quiz ...... 38 Question 1 Free expansion...... 39 Activity: Name the experiment with changing entropy (20 min) ...... 40 9.1 Work ...... 40 Lecture: (5 min) ...... 40 BWBQ: (10 min) ...... 40 Activity: Using pV plots (20-30 min) ...... 40 Activity: Using TS plots (15-20 min) ...... 41 Activity: Name the experiment for a rubber band (15 min) ...... 41

10 Friday: 42 10.1 Engines and Fridges ...... 42 Other views of the Second Law ...... 42 Lecture: (??? min) ...... 42 Carnot eﬃciency ...... 42 Lecture: (??? min) ...... 42 Activity: Carnot eﬃciency (??? min) ...... 44 Activity: Big money (15 min?) ...... 44 Lab 2 due...... 45 Question 2.1 Mass of ice remaining ...... 46 Question 2.2 Final temperature of water ...... 46 Question 2.3 Change in entropy of water...... 46 Question 2.4 Change in entropy of ice...... 47 Question 2.5 Net change...... 47 Question 2.6 Mass of ice remaining ...... 47 Question 2.7 Final temperature ...... 47 Question 2.8 Errors? ...... 47 Homework 3 due ...... 48 Problem 3.1 Adiabatic compression...... 48

Physics 423 iv Problem 3.2 A bottle in a bottle ...... 48 Practice homework 4.1: power-from-ocean ...... 49 Required homework 4.2: power-plant-river...... 49 Required homework 4.3: heat-pump...... 49

11 Monday: 50 Monotonicity and invertibility ...... 50 Activity: Monotonicity lecture/discussion (10 min) ...... 50 11.1 Thermodynamic potentials (Chapter 7 )...... 51 Legendre transform ...... 51 Activity: Legendre transform revisited (15 min?) ...... 51 Activity: Understanding the potentials (20 min) ...... 52 Lecture: (5 min?) ...... 52 Required homework 4.4: gibbs-free-energy ...... 53 11.2 Maxwell relations ...... 53 Lecture: (5 min) ...... 53 BWBQ: (15-25 min) ...... 53 Activity: Name the experiment with Maxwell relations (10 min) ...... 54

12 Tuesday: Lab 2: rubber band 55 Activity: Rubber band lab...... 55 Prelab 3 due...... 55 Question 3.1 Tension vs. temperature...... 56 Question 3.2 Isothermal stretch ...... 56 Lecture: (10 min) ...... 57 Required homework 4.5: free-expansion...... 58

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

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

15 Friday: Thermodynamics practice 66

Physics 423 v Activity: Black body thermodynamics (30 min) ...... 66 Activity: Applying the second law...... 66 Required homework 5.1: maine-entropy-2nd-law-spontaneous-metal . 67 Required homework 5.2: isothermal-adiabatic-compressibility . . . . . 67 Lab 3 due...... 67 Question 3.3 Tension vs. temperature...... 69 Question 3.4 Tension vs. length...... 70 ∂S Question 3.5 ∂L T vs. length...... 70 Question 3.6 Isothermal stretch ...... 70 Question 3.7 Adiabatic stretch...... 70

16 Monday: Statistical approach 71 Lecture: (6 min) ...... 71 16.1 Fairness function (Chapter 6 ) ...... 72 Lecture: (13 min) ...... 72 Activity: Combining probabilities (??? min) ...... 72 Lecture: (??? min) ...... 73 Activity: Demonstrating extensivity (??? min) ...... 73

17 Tuesday: Optimizing the fairness 74 Activity: Students as molecules (20 min) ...... 74 17.1 Least bias lagrangian ...... 74 Lecture: (10 min?) ...... 75 17.2 Weighted averages ...... 75 Lecture: (20 min?) ...... 75 17.3 Probabilities of microstates (Chapter 11 ) ...... 75 Lecture: (10 min) ...... 75 SWBQ: (10 min) ...... 77 SWBQ: (3 min) ...... 77 Lecture: (7 min) ...... 77 Required homework 5.3: boltzmann-ratio...... 77 Challenge homework 5.4: plastic-rod ...... 77

18 Wednesday: From statistics to thermodynamics 78 18.1 Thermodynamic properties from the Boltzmann factor . . . . . 78 Lecture: (5 min) ...... 78 Activity: Solving for maximum fairness (10 min) ...... 78 Lecture: (15 min) ...... 79 BWBQ: (skipped this) ...... 80 BWBQ: (skip this?) ...... 80 Activity: Entropy of microcanonical ensemble (20 min) ...... 81 Homework 5 due ...... 81 Problem 5.1 Hot metal...... 81

Physics 423 vi Problem 5.2 Isothermal and adiabatic compressibility...... 81 Problem 5.3 Boltzmann ratio ...... 82 Problem 5.4 A plastic rod (challenge)...... 82 Required homework 6.1: rubber-band-model...... 83 Required homework 6.2: rubber-meets-road ...... 83

19 Thursday: Statistical mechanics of air 84 19.1 Quantum spectra ...... 84 Lecture: (10 min) ...... 84 19.2 Diatomic gas ...... 85 Lecture: (20 min) ...... 85 Activity: Diatomic molecule from quantum up (1 hour 30 min without wrap-up) 87

20 Friday: 88 20.1 Diatomic gas wrapup ...... 88 Lecture: (20 min) ...... 88 Homework 6 due ...... 92 Problem 6.1 A rubber band model ...... 93 Problem 6.2 The rubber meets the road ...... 93 20.2 Third law ...... 94 Ice rules ...... 95 Handout: Measurement of entropy of water ...... 95 Handout: Pauling ice rules...... 98 Handout: Entropy of water revisited ...... 103 Activity: Concept diagram (10 min) ...... 109 Exam: Survey and post-test...... 109 Question 1 ...... 110 Question 2 ...... 110 Question 3 ...... 111 Question 4 Lab 1...... 112 Question 5 Lab 2...... 112 Question 6 Lab 3...... 112 Question 7 Class as a whole ...... 113

21 Monday: Final exam 114 Exam: Final exam ...... 114 Problem 1 Masses on a piston...... 115 Problem 2 Gibbs free energy ...... 116 Problem 3 Two processes ...... 117 Problem 4 Hanging Chain ...... 118 Problem 5 Insulated room ...... 119 Problem 6 Soap bubble ...... 120

Physics 423 vii Class schedule 121

Index 123

Physics 423 viii 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 × 5 cards (or larger?), and ask them to write the name of each variable at the top of the lined side. 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 particularly 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. In each of the following activities, we’ll ask students to divide up their cards, so each one must go into either one pile or the other. When we’ve established the correct answers, students will 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, we want heat and work to be here, as well as the standard p, V , T , S, U, N, M, and density. I might also like to add number density n, speciﬁc heat or heat capacity, and maybe coeﬃcient of thermal expansion or isothermal compressibility. “Are there any additional properties you might add if we were talking about a rubber band instead, 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 2 d) Would pV = (NkBT ) 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 thermodynamic 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 ﬁrst object? It will get shorter. Why? Is this always true, or will it in some cases get taller? Height and force in the previous example are a conjugate pair. In your groups, try to divide our stack of thermodynamic variables up into analogous conjugate 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 we might 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 up here. Q and W won’t, although they relate to some of these conjugate pairs. Similarly, those state variables that are simply products or ratios of other state variables don’t partake in these pairs.

Activity: Intensive vs. extensive (10-15 min) Ideally here I’d like to have two cups of water (the same amount), and ask what would happen to each of those thermodynamic quantities, if instead of asking about one cup of water, I put the two together and asked about a 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 have twice 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 things that I would need to tell you in order for you to reproduce my experiment. There may be redundant state variables. A state function is something that could be measured (or possibly computed) if you know the state of a system (i.e. a suﬃcient set of state variables.

Physics 423 2 Monday 4/18/2012 This may be more clear if we use an example from classical mechanics. We can consider a single point particle, which is moving in some external gravitational ﬁeld. We can split up the following variables: ~r, ~v, ~a, ~p, F~ , m, KE, PE, W . “Right down on your small whiteboard which of these variables are state variables, state functions (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 quantities are conserved?

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

Equations of state Lecture: (3-4 min) Many of our state variables are redundant: a few are needed to deﬁne the state, and the rest can be computed or measured based on those. An equation of state is how we express this computation. An equation of state is an equation that relates a set of mutually-dependent thermodynamic state variables, such as pressure-volume or tension-length. The most famous is the ideal 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 the total magnetization M, and the entropy S:

µB − µB e kB T − e kB T M = Nµ (2.1) µB − µB e kB T + e kB T ( µB − µB ) µB µB µB e kB T − e kB T k T − k T S = NkB ln 2 + ln e B + e B + (2.2) k T µB − µB B e kB T + e kB T Solve for the magnetic susceptibility, which is deﬁned as: ∂M χB = (2.3) ∂B T Also solve for the same derivative, taken with the entropy S held constant: ∂M (2.4) ∂B S Why does this come out as zero?

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

pV = NkBT (2.5)

Physics 423 4 Tuesday 4/19/2012 3 U = Nk T (2.6) 2 B ( " 3 # ) V mU 2 5 S = Nk ln + (2.7) B N 3πNh¯2 2

From these, solve for the following partial derivatives: ∂p B = −V (2.8) ∂V T ∂p BS = −V (2.9) ∂V S The former is the isothermal bulk modulus...

2.2 Total diﬀerential

Lecture: (11 min) Introduce total differentials here. f(x, y) (2.10) ∂f ∂f df = dx + dy (2.11) ∂x y ∂y x This may look pretty weird at first, but if you think of the differential dx as a small change in x, then this is just a nice notation for the first terms in a Tailor expansion of f(x, y), which is 2 0 0 ∂f 0 ∂f 0 1 ∂ f 0 2 f(x , y ) = f(x, y) + (x − x) + (y − y) + 2 (x − x) + ··· ∂x y ∂y x 2! ∂x y (2.12) 2 0 0 ∂f 0 ∂f 0 1 ∂ f 0 2 f(x , y ) − f(x, y) = (x − x) + (y − y) + 2 (x − x) + ··· (2.13) ∂x y ∂y x 2! ∂x y ∂f ∂f ∆f = ∆x + ∆y + ··· (2.14) ∂x y ∂y x Now if we take the limit that ∆x is small and ∆y is small, the higher-order terms vanish, and we are left with ∂f ∂f df = dx + dy (2.15) ∂x y ∂y x Note also, that we can study the total differential of functions with any number of arguments, so if we had f(a, b, c, e), we could write ∂f ∂f ∂f ∂f df = da + db + dc + de (2.16) ∂a b,c,e ∂b a,c,e ∂c a,b,e ∂e a,b,c

Physics 423 5 Tuesday 4/19/2012 One exciting thing about total derivatives in thermodynamics is that the derivatives ∂f ∂f ∂x and ∂y are often observable state variables themselves! Thus we often interpret a y x 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 = T dS + 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) dg df = dh (2.22) dh

BWBQ: (15 min) Given f(x, y) = ln (x2 + y2), ﬁnd df. Find dF when: X F = − Pi ln Pi (2.23) i Find dF if

F = U − TS (2.24) dU = T dS − pdV (2.25)

Physics 423 6 Tuesday 4/19/2012 Exact vs. inexact diﬀerentials BWBQ: (10 min) Given an expression for a total diﬀerential

df = ydx + xdy (2.26)

∂f ∂f what is ∂x and ∂y ? What is f(x, y)? y x “When you are given a total differential, you should be able to interpret it as two or more equations involving partial derivatives. This usually often the first step in making use of a total differential. ” Can you solve the same problem for

x x dg = ln + 1 dx − dy? (2.27) y y

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 dg are a exact differentials. I’ll try to always write inexact differentials with ad ¯, to make it clear that it is not the total differential of a function! Inexact differentials are useful, but not 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 1 = ∂b ∂b c ∂a c

Ordinary chain rule for total derivatives

dx dy dx = (2.29) dy dz dz

Physics 423 7 Tuesday 4/19/2012 Ordinary chain rule for partial derivatives Write on board:

∂a ∂b ∂a = ∂b d ∂c d ∂c d

∂a ∂a = ∂c d (2.30) ∂b ∂b d ∂c d

Cyclic chain rule for partial derivatives Write on board:

∂c ∂a = − ∂b a ∂b ∂c c ∂a b

∂a ∂c ∂c = −1 (2.31) ∂b c ∂b a ∂a b This is particularly handy when you’ve got something that is being held constant that you happen to know a derivative of, or if you’re taking a derivative of something you’d rather hold constant. Note: in most problems, it only helps to use the cyclic chain rule once. If you use it a second time, you’re starting to go in circles, and may wish to go back to square one.

Physics 423 8 Tuesday 4/19/2012 Math pretest

Question 1 The Gibbs Free Energy, G, is a function of the independent variables temperature T and pressure p, i.e., it can be written as G(T, p). The total diﬀerential of G can be 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 is equal to ∂S

∂p T

Question 2 Z is a function of the two variables x and y, i.e. Z = Z(x, y). Consider the following two expressions: 1 ∂Z α = − (2.32) Z ∂x 1 ∂Z β = (2.33) Z ∂y

∂α ∂β Show that in general ∂y + ∂x = 0. Explain your reasoning.

Physics 423 9 Tuesday 4/19/2012 Question 3 A curve has been traced out on the z-y graph below, and has been labeled Path A. Consider the integral I I1 ≡ zdy (2.34) A,clockwise where the integral is taken around Path A starting at point P , proceeding clockwise until reaching point P again.

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

z Path A

0 y

Refer again to the graph in part (a). We deﬁne H as a function of the independent variables z and y; i.e. H = H(z, y). Consider the integral I ~ I2 ≡ ∇H · d~r (2.35) A,clockwise where the integral is taken around Path A starting at point P , proceeding clockwise until reaching point P again. (Note: d~r = dyyˆ + dzzˆ)

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

Physics 423 10 Tuesday 4/19/2012 Question 4 This Pressure-Volume (p-V ) diagram represents a system consisting a ﬁxed amount of ideal gas that undergoes two diﬀerent quasistatic processes in going from state A to state B:

Process #1 State B Pressure

Process #2 State A

0 Volume

[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? Explain your answer.

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

Physics 423 11 Tuesday 4/19/2012 3 Wednesday:

Current board: ∂c ∂a 1 ∂a ∂b a = ∂a ∂b ∂a = − ∂b c ∂b = ∂b c ∂c ∂a c ∂b d ∂c d ∂c d ∂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 famous ideal gas law. The second is the internal energy of a monatomic ideal gas. The third is the Sackur-Tetrode equation for entropy, which is true for any ideal gas.

pV = NkBT 3 U = Nk T 2 B ( " 3 # ) V mU 2 5 S = Nk ln + B N 3πNh¯2 2

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

∂p bT = −V ∂V T Now ﬁnd the adiabatic bulk modulus, which involves holding the entropy constant:

∂p bS = −V ∂V S

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

µB − µB e kB T − e kB T M = Nµ µB − µB e kB T + e kB T ( µB − µB ) µB µB µB e kB T − e kB T k T − k T S = NkB ln 2 + ln e B + e B + k T µB − µB B e kB T + e kB T Solve for the magnetic susceptibility, which is deﬁned as:

∂M χB = ∂B T Also solve for the same derivative, taken with the entropy S held constant:

∂M

∂B S Why does this come out as zero?

Physics 423 13 Wednesday 4/20/2012 Energy and Entropy Homework 1 Due Wednesday 4/20

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

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

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

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

U = γV αSβ (3.1)

where α, β and γ are constants. The internal energy is an extensive quantity. What constraint 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 1 Due 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: ∂c ∂a 1 ∂a ∂b a = ∂a ∂b ∂a = − ∂b c ∂b = ∂b c ∂c ∂a c ∂b d ∂c d ∂c d ∂a b

4.1 Mixed partial derivatives

Activity: BWBQ: Mixed partials (30-40 min?) Compate ∂V !  ∂V  ∂ ∂ ∂p ∂T p and T (4.1) ∂p  ∂T  T p 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 !  ∂V  ∂ ∂ ∂p ∂T p = T (4.2) ∂p  ∂T  T p ∂2V = (4.3) ∂p∂T ∂2V = (4.4) ∂T ∂p 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 homework 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.

Physics 423 15 Thursday 4/21/2012 Mixed partial derivatives

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

V 3 c

a Volume V1

b p Pressure1 T3

T3 p2

T1 p3 Temperature Figure 4.1: The pV T surface for an ideal gas with ﬁxed N.

Physics 423 16 Thursday 4/21/2012 Partial Derivatives and Material Properties

P2 V2

V1 P3

T1 T2 T3

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

T3 V2

T2 V1

P1 P2 P3

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

©2006 University of Maine Physics Education Research Laboratory, Orono, ME A With two diﬀerent colors of ink, draw and identify the following features on the surface at point a. For each expression, explain in words what these features represent graphically.

∂V (4.5) ∂T p

∂V (4.6) ∂p T Do you expect the value of either of these two derivatives to change for diﬀerent values of pressure or temperature? Explain.

B Identify on the diagram the same two derivatives given in part A, only now at points b and c instead of at a. Do each of these two derivatives change as you move along the variable with respect to which they are diﬀerentiated? If so, how? Do each of these two derivatives change as you move along the variable that is held constant during diﬀerentiation? If so, how?

C Consider the graph of the projection of the ideal gas pV T surface onto the V -T plane ∂V on the next page. Identify ∂T p at temperature T2 for the three diﬀerent values of pressure depicted. ∂V Does ∂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 to you? Is this consistent with your response in part B?

∂V D Consider the projection onto the V -p plane. Identify ∂p at pressure p2 for the three T diﬀerent values of temperature depicted. ∂V Does ∂p at pressure P2 increase, decrease, or remain the same as you go from lower T to higher temperatures? Explain what this means physically. Does it make sense to you? Is this 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 the signs of the changes). Can you say anything about the magnitude of the change at this point?

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

Physics 423 18 Thursday 4/21/2012 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 how many variables are needed to define the state of a given system. It depends on what you are willing to allow to change. V and T would be sufficient, but you might also include N, and might want to include even more if you have more than one sort of molecule. Since 2 (or 3) variables fully define the state of a given system, we know that every other state variable could be written

as a function of those two variables. “But which two variables good(x) are those?” bad(x) We can reason about this pretty easily. “Which state variables can I directly control?” T , S, p, V . But which of these p could be the two variables that define the state of our system? Let us consider a simple plot of p versus V at constant temperature. “Which of these curves are physically reasonable?” V Talk about creating a box full of some material, say a gas or liquid. I could choose to fix the volume of the box, or alternatively I could give it a piston on top and exert a fixed force on it, thus determining the pressure. “Should it matter which approach I use?” This tells us that for any given pressure, there must be only one volume, and for any given volume, there can only be one pressure. Thus our curve must be monotonic. “How would our answer change if we were to plot p versus V at fixed pressure?” ∂p ≤ 0 (4.7) ∂V T ∂p ≤ 0 (4.8) ∂V S Now let’s think about temperature and entropy. The same argument applies, indicating that temperature 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 only need insulate the system and avoid doing anything irreversible. So the same argument shows that temperature and entropy are monotonically related. The sign we can work out from the property of temperature that hot things heat up cold things, and the thermodynamic definition of entropy change. ∂T ≥ 0 (4.9) ∂S p

Physics 423 19 Thursday 4/21/2012 ∂T ≥ 0 (4.10) ∂S V We will also see that the heat capacity is always positive, since it relates to precisely these derivatives. Note that monotonicity does not indicate proportionality!

Legendre transform Lecture: (??? min) Today we’ll be talking about Legendre transforms, but I’d like to begin 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 compress something: ∂p K = −V (4.11) ∂V A high bulk modulus means that a material is very stiff, and it requires a very high pressure to change the volume by a given fraction. The factor of volume in front is chosen to make the bulk modulus an intensive quantity, and the negative sign is chosen to make it a positive quantity. The dimensions of bulk modulus are those of pressure. But what should be held constant? We have two reasonable choices, we could either hold the temperature fixed, or we could hold then entropy fixed. ∂p KT = −V (4.12) ∂V T ∂p KS = −V (4.13) ∂V S 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 use the thermodynamic relation

dU = T dS − pdV (4.14) ∂U p = − (4.15) ∂V S Using this deﬁnition of p, we can quite easily ﬁnd the adiabatic compressibility, if we know U as a function of S and V . ∂p KS = −V (4.16) ∂V S ∂U ! ∂p ∂ = − ∂V S (4.17) ∂V ∂V S S

Physics 423 20 Thursday 4/21/2012 ∂2U = 2 (4.18) ∂V S So if we know U as a function of S and V , we’re golden. Let’s see how things work for the isothermal compressibility.

∂p KT = −V (4.19) ∂V T ∂U ! ∂p ∂ = − ∂V S (4.20) ∂V ∂V S T 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 naturally a function of temperature, but of entropy! So what we do is to look for something that is a function of temperature. The approach to ﬁnd somthing like this is called a Legendre transform. 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 − T dS − SdT (4.23) = T dS − pdV − T dS − 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 deﬁnition for pressure

∂F p = − (4.26) ∂V T ∂p KT = −V (4.27) ∂V T ∂F ! ∂p ∂ = − ∂V T (4.28) ∂V ∂V T T ∂2F = − 2 (4.29) ∂V T So the question naturally arises, “What is the Helmholtz free energy?” To ﬁnd one answer, we can look at the meaning of pdV which shows up in the total diﬀerential of both U and F . This term represents the tiny amount work done as the volume is changing by

Physics 423 21 Thursday 4/21/2012 a tiny amount. We can let the volume change by a larger amount by integrating. If we integrate while holding the entropy fixed (this is an adiabatic process), we find that the total work 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 equal to the change in free energy. That’s where the “free” part of its name comes in: it’s the amount of energy that is free for doing work. Back to the bulk moduli. We have two bulk moduli, adiabatic and isothermal. What do they mean, or how would we measure them? One way to measure a bulk modulus (somewhat indirectly) 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” that determines the speed of sound. But which bulk modulus determines this? In practice, it is the adiabatic bulk modulus that matters, because the thermal conductivity is sufficiently low 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 slowly would 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 T with p and V if you renamed F .

∂U = T (4.30) ∂S V ∂F = S (4.31) ∂T V ∂2U ∂T 2 = (4.32) ∂S V ∂S V ∂2F ∂S 2 = (4.33) ∂T V ∂T V ∂2U ∂2F 2 2 = 1 (4.34) ∂S V ∂T V

Physics 423 22 Friday 4/22/2012 5 Friday: Lagrange multipliers for minimization

Current board: ∂c ∂a 1 ∂a ∂b a = ∂a ∂b ∂a = − ∂b c ∂b = ∂b c ∂c ∂a c ∂b d ∂c d ∂c d ∂a b Energy and Entropy Homework 2 Due Friday 4/22

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

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

µB − µB e kB T − e kB T M = Nµ µB − µB e kB T + e kB T ( µB − µB ) µB µB µB e kB T − e kB T k T − k T S = NkB ln 2 + ln e B + e B + k T µB − µB B e kB T + e kB T Solve for the magnetic susceptibility, which is deﬁned as: ∂M χB = ∂B T Also solve for the same derivative, taken with the entropy S held constant: ∂M

∂B S

Physics 423 23 Friday 4/22/2012 Why does this come out as zero?

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

5 X x = 2n (5.1) n=0 b)

3 3 X X m x = (5.2) n n=1 m=1 c) ∞ X 1 x = (5.3) 2i i=0 d)

1 X A = bi (5.4) i=0 1 X ci x = (5.5) A i=0

Problem 2.4 Euler’s homogeneous function theorem a) Consider an extensive function X, which is a function of two extensive variables, B and D: X = X(B,D) What does this tell us about X(λB, λD)? b) If we deﬁne ∂X a ≡ ∂B D ∂X c ≡ ∂D B What is the total diﬀerential of X, expressed in terms of a, B, c and D?

Physics 423 24 Friday 4/22/2012 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 2 Due Friday 4/22

Physics 423 25 Monday 4/25/2012 6 Monday: Lab 1: Heat and Temperature

Current board: ∂c ∂a 1 ∂a ∂b a = ∂a ∂b ∂a = − ∂b c ∂b = ∂b c ∂c ∂a c ∂b d ∂c d ∂c d ∂a b

Lecture: (10 min) This class is called Energy and Entropy. You’ve already learned quite a bit about energy, but entropy is probably quite new to you. Thermodynamics is a ﬁeld that involves making experimental measurements of bulk substances, and using theory to connect those measurements with other measurements. We will begin this course by making measurements of how much energy is needed to heat up water (and ice) by a certain amount.

Activity: Energy/Heat equivalence (15 min now, 10 min later) We will use a resistive heating element to heat up some ice and water, measuring the current and the voltage to ﬁnd the power. We will measure its change in temperature to work out the heat capacity 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 about how you will be analyzing it. Many thermodynamic measurements are measurements of derivatives. The temperature you can measure directly, and the power of the heater (which

Physics 423 26 Monday 4/25/2012 is the energy dissipated into the water per unit time) we can work out directly. From the two of those, we will work out the heat capacity Write on board: dQ¯ “Cp = ” ∂T p

The heat (called Q) is the amount of energy which is thermally transfered, just like work is the amount of energy which is mechanically transfered. I write the derivative funny because Q is not a function of T . The p subscript just means we’re keeping the pressure constant.

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 course talking about what entropy “really is”, but for now, just know that you can measure entropy by measuring heat and integrating: Write on board: Z dQ¯ ∆S = reversible T

Dulong-Petit Law Lecture: (5 min) In 1819, shortly after Dalton had introduced the concept of atomic weight in 1808, Dulong and Petit observed that if they measured the speciﬁc heat per unit mass of a variety of solids, and divided by the atomic weights of those solids, the resulting per-atom speciﬁc heat was essentially constant. This is the Dulong-Petit law, although we have since given a name to that constant, which is 3R or 3kB, depending on whether the relative 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 write Dulong-Petit’s law as:

Cp = 3NkB (6.1)

where N is the total number of atoms.

Physics 423 27 Tuesday 4/26/2012 7 Tuesday: First and Second Laws

Current board: ∂a ∂b ∂a = dQ¯ ∂b d ∂c d ∂c d “Cp = ” ∂T p

∂c ∂a 1 ∂a ∂b a Z = = − dQ¯ reversible ∂b c ∂b ∂b c ∂c ∆S = ∂a c ∂a b T

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

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

BWBQ: (10 min) In yesterday’s lab, what composed the system, and what composed its surroundings? a) In the ﬁrst stage, when we were melting the ice, we could say that the ice was the system, and the water, heater, thermometer and the rest of the world were the surroundings. 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 thermometer (and the cup), since all these objects were being heated. Our measurement of Q really measured the heat transferred to the whole combined system. We just hoped that 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 conserved (or is a substance, to use Aristotle’s terminology). But it is useful to look at those

Physics 423 28 Tuesday 4/26/2012 two non-state variables work and heat. Both are changes in energy of a system, so we can write 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 about inﬁnitesimal 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 happen to its temperature? If students look puzzled, suggest they assume that as long as you do this 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 will energy be transfered by heating? What is the rule that governs this? Why can’t two objects that are at the same temperature spontaneously change temperature?

Lecture: (5 min) The second law of thermodynamics clariﬁes this rule, and extends it to 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:

Z dQ¯ ∆S = reversible (7.2) T

The Second Law of Thermodynamics simply states that for any possible process, the change 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.

Physics 423 29 Tuesday 4/26/2012 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 thisdQ ¯ ? As it turns out, we can deﬁne a state function S called entropy and so long as a process is done reversibly

dQ¯ = T dS only when change is quasistatic (7.6)

so ﬁnally we ﬁnd out that Write on board: dU = T dS − pdV The fact that the T in this equation is actually the physical temperature was originally an experimental observation. At this point, the entropy S is just some weird heat-related state function.

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

∂V S ∂p T ∂τ T ∂T V ∂T p ∂T τ ∂p S ∂τ T ∂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 20 minutes, but I think it was worthwhile. I might be tempted to put more groups on something ∂U like ∂p . S

Physics 423 30 Tuesday 4/26/2012 7.4 Heat capacity

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

dU =dQ ¯ −dW¯ (7.7) = T dS − pdV (7.8)

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

dQ¯ QS = T dS (7.9) Z dQ¯ ∆S = QS (7.10) T ∂U T = (7.11) ∂S V Any suggestions how we might go about measuring the entropy?

Lecture: (10 min) As we learned last week, heat capacity is amount of energy required to raise the temperature of an object by a small amount. dQ¯ C ∼ (7.12) ∂T 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, ∂U CV = (7.15) ∂T V

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

dQ¯ = T dS (7.16)

dQ¯ = CαdT +?dα (7.17)

T dS = CαdT +?dα (7.18) C ? dS = α dT + dα (7.19) T T Physics 423 31 Tuesday 4/26/2012 ∂S Cα = T (7.20) ∂T α This may look like an overly-tricky derivative, so let’s go through the ﬁrst law and check that we got it right in a few cases. I’ll do the CV case. We already know that dU =dQ ¯ − pdV (7.21) ∂U CV = (7.22) ∂T V ∂U ∂S = (7.23) ∂S V ∂T V ∂S = T (7.24) ∂T V where the second step just uses the ordinary chain rule.

BWBQ: (10 min (skipped in 2011)) Students work out the heat capacity Cp from the ﬁrst law. If it looks necessary, tell them to use the enthalpy, but hopefully they’ll remember this from Tuesday. dU =dQ ¯ − pdV (7.25) H = U + pV (7.26) dH =dQ ¯ + V dp (7.27) ∂H = Cp (7.28) ∂T p ∂H ∂S Cp = (7.29) ∂S p ∂T p ∂S = T (7.30) ∂T p

Lecture: (5 min (skipped in 2011)) We can ﬁnd a change in entropy from the heat capacity quite easily: Z 1 ∆S = dQ¯ (7.31) T QS Z C(T ) = dT (7.32) T

Activity: Entropy change of cooling coﬀee (20-30 min (skipped 2010)) Students work 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.

Physics 423 32 Tuesday 4/26/2012 Required homework 3.1: adiabatic-ideal-gas Required homework 3.2: bottle-in-bottle

Physics 423 33 Tuesday 4/26/2012 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 explicit about 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 explicit about how you would perform the measurement, and draw a sketch of your apparatus.

Physics 423 34 Tuesday 4/26/2012 8 Wednesday: Second Law lab

Current board:

∂c Z dQ¯ ∂a 1 ∂a ∂b a reversible ∆S +∆S ≥ 0 = = − ∆S = system surroundings ∂b ∂b c ∂c T ∂b c ∂a ∂a c b

∂a ∂b ∂a dQ¯ = “Cp = ” dU =dQ ¯ +dW ¯ dU = T dS − pdV ∂b d ∂c d ∂c d ∂T p

Activity: Melting ice (45 min) Let’s talk about entropy changes in irreversible processes. Let’s try doing this as a quickie lab! We’ll reuse the same materials as for Lab1. We’ll mass some water (possibly hot water) and measure its initial temperature. Then we add some 0◦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 of their ice water. They will also work out the change in entropy of the ice and the initial water for 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 in an insulated container (but at constant pressure)? What is the change in entropy? What is the final temperature? How much ice is left? What is the change in entropy of the ice? Of the water? The heat capacity of liquid water is Cp = 4.18 J/g/K and is roughly constant over this temperature range, and that the latent heat of fusion (a.k.a. the enthalpy of fusion) of ice is 333 J/g. In 2010, (when the students only computed the answers without a lab portion) this could have taken more than the 20 minutes it took, as students didn’t have time to even look at the entropy, 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).

Physics 423 35 Wednesday 4/27/2012 Lab 1: Energy and temperature In this lab, we will be measuring how much energy it takes to melt ice and heat water. Materials: • 2 digital multimeters A • Styrofoam cup • Temperature guage V • Heating element • Scale • Ice and water

The setup

You will put some mass of ice (about 50g) and ice-cold water (about 150g) into your styrofoam cup. 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 heating element (which is just a resistor). Thus we can ﬁnd out how much energy was added to the water, and how this changes the temperature. In order to keep the temperature measurement reasonable, 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 the heater, you should make sure the computer is taking data. Turn on the heater, and write down the time you do so as well as the current and voltage, from which you can ﬁnd the power dissipated in the resistor. If the current or voltage changes during the course of the experiment, take note of the new values—and the time.

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

Question 1.2 Plot your data II Plot the heat capacity versus temperature. This will be a bit trickier. You can ﬁnd the heat capacity from the previous plot by looking at the slope. ∂Q Cp = (8.1) ∂T p This is what is called the heat capacity, which is the amount of energy needed to change the temperature by a given amount. The p subscript means that your measurement was made at constant pressure. This heat capacity is actually the total heat capacity of everything you put in the calorimeter, which includes the resistor and thermometer.

Physics 423 36 Wednesday 4/27/2012 Question 1.3 Speciﬁc heat From your plot of Cp(T ), work out the heat capacity per unit mass 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 required to 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 as the slow melting of ice), it is just

Q ∆S = (8.2) T where Q is the energy thermally added to the system and T is the temperature in Kelvin. What is was change in the entropy of the ice you melted? What was the change in entropy per molecule? What was the change in entropy per molecule divided by Boltzmann’s constant?

Question 1.5 Entropy for a temperature change Choose two temperatures that your water reached (after the ice melted), and ﬁnd the change in the entropy of your water. This change is given by Z dQ¯ ∆S = (8.3) T Z tf P (t) = dt (8.4) ti T (t) where P (t) is the heater power as a function of time and T (t) is the temperature, also as a function of time.

Physics 423 37 Wednesday 4/27/2012 9 Thursday: Heat and work

Current board:

∂c Z dQ¯ ∂a 1 ∂a ∂b a reversible ∆S +∆S ≥ 0 = = − ∆S = system surroundings ∂b ∂b c ∂c T ∂b c ∂a ∂a c b

∂a ∂b ∂a dQ¯ = “Cp = ” dU =dQ ¯ +dW ¯ dU = T dS − pdV ∂b d ∂c d ∂c d ∂T p

Activity: Expanding Gas Quiz (20-45 min) Let’s start the class with a little quiz about 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’t like the fact that entropy isn’t conserved.

Physics 423 38 Thursday 4/28/2012 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 with a movable piston, at a temperature of 300 K. Slowly the gas expands against the movable piston, while the cylinder is in contact with a thermal reservoir at 300 K. The temperature of 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. Five moles of the same ideal gas are confined to one half of the container, at a temperature of 300 K. The other half of the container is a vacuum. The plastic divider is suddenly removed and the gas expands to fill the container. Because it is a free expansion of an ideal gas (no work 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 allowed to 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 compare with ∆Sadiabatic? Please explain.

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

Physics 423 39 Thursday 4/28/2012 Activity: Name the experiment with changing entropy (20 min) “Name the experiment!” For each of the derivatives below, describe the experiment that you would perform in order to measure it.

∂S ∂S ∂S ∂S (9.1) ∂V T ∂T V ∂p T ∂T p In this activity, we’ll look again at what changing entropy means, and how we might measure that 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~ · dP~ , for a paramagnet −B~ · dM~ 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? What if 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 processes that happen at constant pressure, volume, temperature or entropy. This touches on the name-the-experiment discussion we had yesterday.

Physics 423 40 Thursday 4/28/2012 We talked about measurements such as ∂τ (9.2) ∂L S

∂L Pressure (9.3) ∂T τ (9.4)

In these measurements, we measure how a state variable changes as 0 Volume we vary another one. Another sort of measurement involves integrals rather than derivatives, and measures ﬁnite changes. To discuss this, it’s common to use what are called pV diagrams. For instance, consider the 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?

Activity: Using TS plots (15-20 min) Now let’s look at another cycle. Let’s consider The following ﬁgure, which looks similar, but is now a plot of T vs. S. What is this cycle? How would you go about running a cycle like

this? T What is the net heat transfer over one cycle of this process? How much was transfered on each step? What is the net work done after one cycle of this process? How much work was done at each step? 0 S 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 ∂τ ∂τ ∂τ ∂S ∂L ∂L (9.5) ∂L T ∂L S ∂T L ∂T S ∂L T ∂τ T ∂τ S ∂L ∂L ∂T ∂S ∂T ∂S ∂S (9.6) ∂T τ ∂T S ∂L S ∂τ T ∂τ S ∂T τ ∂T L In class 2010, we ended up spending most of our wrap-up talking about what it means to hold entropy ﬁxed. Students were also confused as to whether constant Q meands constant T .

Physics 423 41 Friday 4/29/2012 10 Friday:

Current board:

∂c Z dQ¯ ∂a 1 ∂a ∂b a reversible ∆S +∆S ≥ 0 = = − ∆S = system surroundings ∂b ∂b c ∂c T ∂b c ∂a ∂a c b

∂a ∂b ∂a dQ¯ = “Cp = ” dU =dQ ¯ +dW ¯ dU = T dS − pdV ∂b d ∂c d ∂c d ∂T p

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 Kelvin formulation states that No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. The Clausius formulation states that No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature. These formulations make it clear that if you could violate the Second Law, you could become ﬁlthy rich.

Carnot eﬃciency T Lecture: (??? min) A heat engine is a device that converts heat h Q into work. I will diagram heat engines as displayed in this picture. The h heat engine contains several parts.

• At the top and bottom are hot and cold heat sinks. The operator of W the engine has to keep these two heat sinks at ﬁxed temperature, which Q means burning fuel to warm up the hot sink, and using something like c a radiator to keep the cold sink cool. Tc

Physics 423 42 Friday 4/29/2012 • In the middle of the picture is the engine itself, which will contain some sort of a working substance that is (most likely) alternately heated and cooled.

• There is some amount of heat Qh transfered from the hot sink to the engine, and some other amount of heat Qc (both taken to be positive numbers) 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 world is generated by steam turbines1—which are simple heat engines. So it’s not really a 19th century application, although it was pretty well understood in the 19th century.

Eﬃciency The eﬃciency in general is what you get out divided by

what you put in. In this case, what we have to put in is the heat added Th to the hot resevoir Qh, and what we get out is the work W , so Qh W η = (10.2) Q h W Q − Q = h c (10.3) Qh Qc Q = 1 − c (10.4) Tc Qh So clearly, we’d like to minimize the amount of heat sent to the cold sink. This also has environmental advantages. If each step is done reversibly, then we could run a heat engine in reverse, and have a refridgerator. Thus we put do work on the fridge, and cool oﬀ the cold sink, while warming up the hot sink. The eﬃciency of any reversible heat engine must be the same as the

efficiency of any reversible refridgerator. We can see this by using a heat Th engine to drive a refridgerator. By choice the work done by the engine is Qh Qh the same as the work done on the fridge. If the efficiencies of the fridge and engine differ, then there will be a net transfer of heat either from hot sink to cold sink or from cold sink to hot sink. The former would W be reasonable and natural, but the latter would be crazy, which means

that the fridge cannot be more eﬃcient than the engine. However, if Qc Qc 1This consists of all coal-burning plants and nuclear power plants, and I’m not sure what else...T almostc certainly geothermal.

Physics 423 43 Friday 4/29/2012 both fridge and engine are reversible, then if the fridge is less eﬃcient than the engine, then we could run the thing in reverse, and get the crazy situation again happening, in which nothing changes except that heat is transferred from a cold place to a hot place... and that just isn’t natural! “So we can only conclude that every possible reversible heat engine must have the same eﬃciency!”

Activity: Carnot eﬃciency (??? min) For a reversible engine, the net entropy change of system plus surroundings must be zero over each cycle. Since the entropy change of the system is always zero, even for an irreversible engine, we only need consider the entropy change of the surroundings. “In your groups, on big white boards, use these properties to work out the eﬃciency of a reversible heat engine.” This could be done (more quickly) as lecture, in which case the following is what I’d do:

Qh ∆Sh = − (10.5) Th Qc ∆Sc = (10.6) Tc ∆Sh + ∆Sc = 0 (10.7) Q Q = c − h (10.8) Tc Th Q Q c = h (10.9) Tc Th Q T c = c (10.10) Qh Th Tc ηcarnot = 1 − (10.11) Th This is the Carnot efficiency, and is acheived by any reversible heat engine. This efficiency also is the upper bound on the efficiency of any heat engine. Note that this efficiency applies equally well to traditional cyclic heat engines, steam turbines, photovoltaics, and thermopower devices. There is also a corresponding limit on the efficiency of refridgeration devices, which has the same broad applicability.

Activity: Big money (15 min?) “Supposing you are able to violate the Second Law in the 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 in one direction, but not the other direction.

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

Physics 423 44 Friday 4/29/2012 c) You develop a nano-door that allows only high-energy molecules to pass through it in one direction, but allows only low-energy molecules to pass through in the other direction—and requires no energy input.

Physics 423 45 Friday 4/29/2012 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 you add 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 temperature of the water after it has been in the cup long enough to equilibrate with the cup. Then add around 100g of 0◦ ice to the cup and record its mass. Cover the cup and wait for the ice to melt. While you wait, work in your groups on the following questions using your big white boards. The heat capacity of liquid water is Cp = 4.18 J/g/K and is roughly constant over the temperature range we will be using, and that the enthalpy of fusion (a.k.a. latent heat) of ice is 334 J/g.

Group analysis

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

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

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

Physics 423 46 Friday 4/29/2012 Question 2.4 Change in entropy of ice Work out the change in entropy of the ice as it melted and (possibly) warmed up.

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

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 out and 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 the cup.

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

Physics 423 47 Friday 4/29/2012 Energy and Entropy Homework 3 Due Friday 4/29

Problem 3.1 Adiabatic compression Consider the isothermal expansion of a simple ideal 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 or nitrogen, 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 N to be constant. a) Use the first law to relate the inexact differential for work to the exact differential dT for 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 additional definition of workdW ¯ = −pdV to solve for the relationship between p, dp, V and dV for an adiabatic process. d) Integrate the above differential equation to find a relationship between the initial and final pressure and volume for an adiabatic process.

Problem 3.2 A bottle in a bottle The internal energy of helium gas at temperature T is to a very good approximation given by 3 U = Nk T (10.14) 2 B Consider a very irreversible process in which a small bottle of helium is placed inside a large bottle, which otherwise contains vacuum. The inner bottle contains a slow leak, so that the helium leaks into the outer bottle. The inner bottle contains one tenth the volume of the outer bottle, which is insulated. What is the change in temperature when this process is complete? How much of the helium will remain in the small bottle?

End of Energy and Entropy Homework 3 Due Friday 4/29

Physics 423 48 Monday 5/2/2012 Practice homework 4.1: power-from-ocean Required homework 4.2: power-plant-river Required homework 4.3: heat-pump

Physics 423 49 Monday 5/2/2012 11 Monday:

Current board:

∂c Z dQ¯ ∂a 1 ∂a ∂b a reversible ∆S +∆S ≥ 0 = = − ∆S = system surroundings ∂b ∂b c ∂c T ∂b c ∂a ∂a c b

∂a ∂b ∂a dQ¯ = “Cp = ” dU =dQ ¯ +dW ¯ dU = T dS − pdV ∂b d ∂c d ∂c d ∂T p

Monotonicity and invertibility Activity: Monotonicity lecture/discussion (10 min) Ask groups to discuss how many variables are needed to define the state of a given system. It depends on what you are willing to allow to change. V and T would be sufficient, but you might also include N, and might want to include even more if you have more than one sort of molecule. Since 2 (or 3) variables fully define the state of a given system, we know that every other state variable could be written

as a function of those two variables. “But which two variables good(x) are those?” bad(x) We can reason about this pretty easily. “Which state vari-

ables can I directly control?” T , S, p, V . But which of these p could 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?” V Talk about creating a box full of some material, say a gas or liquid. I could choose to fix the volume of the box, or alternatively I could give it a piston on top and exert a fixed force on it, thus determining the pressure. “Should it matter which approach I use?” This tells us that for any given pressure, there must be only one volume, and for any given volume, there can only be one pressure. Thus our curve must be monotonic. “How would our answer change if we were to plot p versus V at fixed pressure?”

∂p ≤ 0 (11.1) ∂V T

Physics 423 50 Monday 5/2/2012 ∂p ≤ 0 (11.2) ∂V S Now let’s think about temperature and entropy. The same argument applies, indicating that temperature 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 only need insulate the system and avoid doing anything irreversible. So the same argument shows that temperature and entropy are monotonically related. The sign we can work out from the property of temperature that hot things heat up cold things, and the thermodynamic definition of entropy change. ∂T ≥ 0 (11.3) ∂S p ∂T ≥ 0 (11.4) ∂S V We will also see that the heat capacity is always positive, since it relates to precisely these derivatives. Note that monotonicity does not indicate proportionality!

11.1 Thermodynamic potentials (Chapter 7 )

Legendre transform Activity: Legendre transform revisited (15 min?) Recall that: dU = T dS − pdV (11.5) As you may also recall, we worked out three other thermodynamic potentials using Legendre transforms: 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 = T dS + V dp

Write on board: dG = −SdT + V dp

Physics 423 51 Monday 5/2/2012 Activity: Understanding the potentials (20 min) Students work out in groups what the changes in H and F are, in processes that hold one or the other of their natural variables constant. Groups that are ahead can be asked to consider the Gibbs free energy. Z Z dH = T dS (11.9) constant p

∆H = Qconstant p (11.10)

What is Z Z dF = pdV (11.11) constant T ∆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 volume changes, 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 melt the 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 either case. Now suppose we have have some steam and we want to use it to do some work. How much work can we get out of it? Obviously, the internal energy is relevant here, since energy must be conserved. But suppose we want to achieve this work at constant temperature (so we won’t have to insulate things)? If we’re keeping things at constant temperature, then there is heating going on... so the amount of work can’t be the same as the change in internal energy. How much work can we do? The answer is given by the change in Helmholtz free energy. Finally we get to the Gibbs free energy. As you might gather, it’s helpful when you want to keep both temperature and pressure constant. You might wonder what you could possibly be changing, if you keep both pressure and temperature constant! One possibility is that you could be doing a phase transition, like melting ice at zero centigrade and atmospheric pressure. This is possible because the Gibbs free energies of ice and water are the same at that temperature and pressure. Another possibility is that you’re undergoing a chemical reaction... which is another sort of a phase transition. Reactions or phase transitions are reversible (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.

Physics 423 52 Monday 5/2/2012 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 !  ∂f  ∂ ∂ ∂y ∂x y = x (11.14) ∂y  ∂x  x y ∂2f ∂2f = (11.15) ∂x∂y ∂y∂x 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 = T dS − pdV (11.16) dF = −SdT − pdV (11.17) dH = T dS + 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, S is the entropy, F is the Helmholtz free energy, H is the enthalpy, and G is the Gibbs free energy.

BWBQ: (15-25 min) How are the following thermodynamic derivatives related? (Each group gets one or two) a) ∂T ∂S and ∂p V ∂V T b) ∂T ∂S and ∂p S ∂V p c) ∂T ∂p and ∂V p ∂S T In 2010 class, I turned some of these derivatives upside down, which made the problem considerably more challenging, which I think was useful. In 2011, I have the derivatives already pre-turned-upside-down.

Physics 423 53 Monday 5/2/2012 Activity: Name the experiment with Maxwell relations (10 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.

∂S ∂S (11.20) ∂V T ∂p T For extra fun... ∂τ ∂τ ∂L ∂L ∂S ∂S (11.21) ∂S L ∂S T ∂S T ∂S τ ∂τ L ∂L τ

Physics 423 54 Tuesday 5/3/2012 12 Tuesday: Lab 2: rubber band

Current board: ∂a 1 dQ¯ dH = T dS + V dp “Cp = ” ∆Ssystem+∆Ssurroundings ≥ 0 = ∂T ∂b c ∂b p ∂a c

∂a ∂b ∂a Z = dQ¯ reversible dG = −SdT + V dp ∂b ∂c ∂c ∆S = dU = T dS − pdV d d d T ∂c ∂a ∂b a = − ∂b c ∂c dU =dQ ¯ +dW ¯ dF = −SdT − pdV ∂a b

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

Physics 423 55 Tuesday 5/3/2012 Prelab 3: Thermodynamics of a rubber band Please answer the following questions before beginning the experiment. Think carefully before making your predictions, but don’t worry if you aren’t sure what will happen. In every case, give a reason for your prediction—guessing is perfectly acceptable, but if you are only guessing, just say so.

Question 3.1 Tension vs. temperature Will the tension increase or decrease with increasing temperature? Sketch a graph of what you expect to observe. Be sure to include zero 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 which you (expect to) have clean data.1

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

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

c) If the rubber band is reversibly isothermally stretched, what will the magnitude of W and Q be? Which is larger, and does this make sense to you? How do the magnitudes of 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, and see if you can come up for a justiﬁcation for reasonable values. The units of entropy are J/K, and all other quantities have dimensions of energy.

Physics 423 56 Tuesday 5/3/2012 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 or infinite. It can be seen to be infinite, from the definition of Cp on the board, which means R that CpdT = Q, which is finite and positive. On the last problem, more of you had trouble. First, as in the previous problem, you needed to be sure to use T in Kelvins. If it ever makes a difference, you always need to use Kelvins, as the zero value of Celsius is arbitrary. Secondly, many of you struggled with actually doing the integral. Z t2 P (t) ∆S = dt (12.1) t1 T (t) Z t2 1 = P dt (12.2) t1 T (t) 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. Z t2 1 ∆S = dt (12.3) t1 T (t) X 1 = P ∆t (12.4) T i i X 1 = P (1s) (12.5) T i i This you can just plug into the spreadsheet and get the answer. This works especially well if your data doesn’t match a very nice analytic form. In most of your data, however, the temperature was quite linear, so you could write

T (t) = T0 + mt (12.6) Z t2 1 ∆S = P dt (12.7) t1 T (t) Z t2 1 = P dt (12.8) t1 T0 + mt (12.9) But this starts looking pretty nasty. A nicer approach would be to write dT = mdt (12.10) Z T2 1 dT ∆S = P (12.11) T1 T m P Z T2 1 = dT (12.12) m T1 T

Physics 423 57 Tuesday 5/3/2012 P = ln T |T2 (12.13) m T1 = C ln T |T2 (12.14) p T1 I should also note that this problem could be approached more directly, although my homework question didn’t lead you in this direction. Z dQ¯ ∆S = (12.15) T Z T2 C (T )dT = p (12.16) T1 T Z T2 dT = Cp (12.17) T1 T

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

Required homework 4.5: free-expansion

Physics 423 58 Wednesday 5/4/2012 13 Wednesday:

Current board: ∂a 1 dQ¯ dH = T dS + V dp “Cp = ” ∆Ssystem+∆Ssurroundings ≥ 0 = ∂T ∂b c ∂b p ∂a c

∂a ∂b ∂a Z = dQ¯ reversible dG = −SdT + V dp ∂b ∂c ∂c ∆S = dU = T dS − pdV d d d T ∂c ∂a ∂b a = − ∂b c ∂c dU =dQ ¯ +dW ¯ dF = −SdT − pdV ∂a b

Activity: Name another experiment with Maxwell relations (10-20 min) “Name another experiment!” For each of the following derivatives, we can already come up with a simple experiment. Use a Maxwell relation to come up with yet another way to measure this. ∂τ ∂τ ∂V ∂T ∂T ∂S (13.1) ∂L T ∂T L ∂T p ∂p V ∂V S ∂T L

Energy and Entropy Homework 4 Due Wednesday 5/4

Problem 4.1 Power from the ocean (practice) It has been proposed to use the thermal gradient of the ocean to drive a heat engine. Supoose that at a certain location the water temperature 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 two temperatures?

b) If the engine is to produce 1 GW of electrical power, what minimum volume of water 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 over this temperature range.

Physics 423 59 Wednesday 5/4/2012 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 waste energy 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 mountain river at 15oC. Roughly how much money can you make in a year by installing your imporved 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 the river increase?

e) To avoid this “thermal pollution” of the river the plant could instead be cooled by evaporation 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?

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

a) Define a coefficient of performance γ for a heat pump, which (like W the efficiency of a heat engine) is the ratio of “what you get out”

to “what you put in.” Qc

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

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

d) What is the ideal coeﬃcient of performance of a heat pump when the indoor temperature is 70◦F and the outdoor temperature is 50◦F? How does it change when the outdoor temperature drops to 30◦F?

Physics 423 60 Wednesday 5/4/2012 Problem 4.4 Using the Gibbs free energy You are given the following Gibbs free energy: aT 5/2 G = −kT N ln 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 temperature, 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 to be constant. Consider the free adiabatic expansion of an ideal gas to twice its volume. “Free expansion” means that no work is done, but also that the process is also neither quasistatic nor 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?

End of Energy and Entropy Homework 4 Due Wednesday 5/4

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

Physics 423 61 Thursday 5/5/2012 14 Thursday: Thermodynamics practice

Current board: ∂a 1 dQ¯ dH = T dS + V dp “Cp = ” ∆Ssystem+∆Ssurroundings ≥ 0 = ∂T ∂b c ∂b p ∂a c

∂a ∂b ∂a Z = dQ¯ reversible dG = −SdT + V dp ∂b ∂c ∂c ∆S = dU = T dS − pdV d d d T ∂c ∂a ∂b a = − ∂b c ∂c dU =dQ ¯ +dW ¯ dF = −SdT − pdV ∂a b

Activity: Simple cycle (40-110 min) Consider the following simple cycle for a monatomic ideal gas. 3 U = Nk T (14.1) 2 B pV = NkBT (14.2)

We start at V0, p0 and T0. We first allow the gas to expand to twice its original volume at fixed temperature. 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 very slowly felt that it was quite helpful.

Activity: Temperature change of dissolving salt (40 min?) The enthalpy of dis- 3 solution (dissolving) of NaCl in water is 4 × 10 J/mol. The speciﬁc heat cp of water is 4.2 J/g/K. What is the change in temperature of a cup holding 100 g of water (moderately well insulated), if you dissolve 30 g of salt in it?

Physics 423 62 Thursday 5/5/2012 What happened to the entropy of the isolated salt plus water system? Can we predict the value of the entropy change from the information given? Could we measure the change in 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.

4 × 103J/mol · 30g ∆H = (14.3) dis 23g/mol + 35.4g/mol = 2000J (14.4) Q ∆T ≈ (14.5) Cp 2000J = (14.6) 4.3J/g/K · 100g ≈ 5K (14.7)

Activity: Never, sometimes or always true (60 min?) “For each of the scenarios in the handout, work out in your groups whether it is never true, sometimes true or always true. But more importantly, provide reasoning for your answer, expressed in terms of the laws or principles of thermodynamics! ” Students will naturally expect that 1b is always true, neglecting the choice to keep pressure constant. I needed to talk with each group, and have them write down the First Law, in order to realize that they don’t know the sign or magnitude of the work, and thus don’t know whether the internal energy goes up or down. Students continue to be unclear as to which quantities are state functions, and also are vague on the process of using a reversible path with the same starting and ending points to determine the change in a state function for an irreversible process.

Physics 423 63 Thursday 5/5/2012 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 about what fundamental laws are responsible. If it is only sometimes true, try to give examples of each case.

Hot object in cool water If I put a hot object (of any material) into a tub of colder water 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 higher internal energy.

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

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

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

Physics 423 64 Thursday 5/5/2012 Bag of hot air? Consider an insulated spherical bag full containing a ﬂuid—which could be either liquid or solid. The bag is very strong, so it cannot be stretched, and also happens to 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 because I’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 will increase.

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 ﬂuid will increase.

Physics 423 65 Thursday 5/5/2012 15 Friday: Thermodynamics practice

Current board: ∂a 1 dQ¯ dH = T dS + V dp “Cp = ” ∆Ssystem+∆Ssurroundings ≥ 0 = ∂T ∂b c ∂b p ∂a c

∂a ∂b ∂a Z = dQ¯ reversible dG = −SdT + V dp ∂b ∂c ∂c ∆S = dU = T dS − pdV d d d T ∂c ∂a ∂b a = − ∂b c ∂c dU =dQ ¯ +dW ¯ dF = −SdT − pdV ∂a b

Activity: Black body thermodynamics (30 min) Let’s consider the following diagram. Here we’ve got two objects in an insulated en- insulation vironment, surrounded by vacuum, so the only way they can exchange Q energy is through electromagnetic radiation. Further, let’s assume that 1 each object is perfectly black, so that it absorbs any radiation incident on it. T1 T2 • Suppose T1 > T2, what can we say about Q1 and Q2 on the basis of the Second Law?

• Suppose T1 = T2, what can we say about Q1 and Q2? Q2 • Now let us suppose that we replace the right-hand object with one that only absorbs two thirds of incident radiation and reﬂects the other 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 one frequency in between the two objects, which means that at every frequency, the absorption must 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 mole of liquid water at several temperatures and of ice.

Physics 423 66 Friday 5/6/2012 • What happens if you throw an ice cube into water at room temperature? Is the reverse possible?

• What happens if you drop a room-temperature rubber band into ice-cold water? Is the 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

Physics 423 67 Friday 5/6/2012 Lab 3: Thermodynamics of a rubber band In this lab, we will measure the tension of a rubber band stretched to a ﬁxed length, as a function of temperature. By measuring this at several similar lengths, we will be able to determine the change in internal energy U and entropy S for a rubber band that is stretched at ﬁxed temperature.

Materials: • Vernier temperature guage

• One rubber band • Several clamps

• Tube • Boiling water

• Stopper with hook • Ice

• Vernier force guage • Pan

Background

The inexact diﬀerential of work for a system such as a rubber band that is stretched in just one direction is d−W = −τdL. (15.1) This follows naturally from the deﬁnition of work as force dotted with distance—provided one takes into account the sign convention for tension, which is opposite that of pressure. Thus the thermodynamic identity is:

dU = T dS + τdL (15.2)

We could use the thermodynamic identity directly, but since we are working at constant T , 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 ∂2F = (15.5) ∂L∂T ∂T ∂L ∂ ∂F ∂ ∂F = (15.6) ∂L ∂T L T ∂T ∂L T L ∂S ∂τ − = (15.7) ∂L T ∂T L

Physics 423 68 Friday 5/6/2012 By measuring the variation of tension with temperature at fixed length, we can find out how entropy 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 same isothermal stretch: ∂F τ = (15.8) ∂L T

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

The setup

You will stretch your rubber band between a force meter and a hook in the stopper in the bottom of a pipe. If you attach the rubber band to the force meter by means of a chain of paper clips, then you can ensure that when the pipe is ﬁlled with water, the rubber band is completely immersed. You will need to insert a thermometer probe into the top of the pipe in order to measure the temperature of your water—and thus the temperature of your rubber band.

Collect data

During the experiment you will pour water of various different temperatures into the pipe in order to heat up your rubber band, and record the tension as a function of length. When changing to a different temperature, you will need to empty your pipe and refill it. Each time, you should measure the temperature at the top and bottom to ensure that the water is well mixed.

Conclusions and Questions

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

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

Physics 423 69 Friday 5/6/2012 Question 3.4 Tension vs. length Plot the tension versus length for a few temperatures.

∂S ∂S Question 3.5 ∂L T vs. length Plot ∂L T versus length for the same set of temperatures you chose for Question 3.4.

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

a) What is the change in free energy for an isothermal stretch at this temperature from the 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 perform in order to answer Question 3.6 for a stretch that is adiabatic rather than isothermal?

Physics 423 70 Friday 5/6/2012 16 Monday: Statistical approach

Current board: ∂a 1 dQ¯ dH = T dS + V dp “Cp = ” ∆Ssystem+∆Ssurroundings ≥ 0 = ∂T ∂b c ∂b p ∂a c

∂a ∂b ∂a Z = dQ¯ reversible dG = −SdT + V dp ∂b ∂c ∂c ∆S = dU = T dS − pdV d d d T ∂c ∂a ∂b a = − ∂b c ∂c dU =dQ ¯ +dW ¯ dF = −SdT − pdV ∂a b

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 thermodynamic measurements to extract microscopic properties. From quantum mechanics, you know that given a Hamiltonian describing a system, you can solve for all the possible eigenstates and their energies. But how can you know which of those states a given system will be in? And given that state, how can you predict the result of interactions of the system with its surroundings, when you don’t know the hamiltonian or eigenstates of the surroundings? These are the questions that are answered by statistical mechanics.

Physics 423 71 Monday 5/9/2012 16.1 Fairness function (Chapter 6 )

Lecture: (13 min) The primary quantity in statistical mechanics is the probability Pi of ﬁnding the system in eigenstate i. The approach we are going to use is to state that the probabilities are those which maximize the fairness (or minimize the bias). So we need to deﬁne a fairness function F that we can maximize. First, let’s talk about some properties the 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 until they’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 to get students thinking about probabilities and how they combine. Consider two systems A and B, which you can think of as dice or quantum mechanical systems with several eigenstates. Each system has a set of possible states i. If the two are A B uncorrelated, then we can describe their separate probabilities as Pi and Pj . What is the probability Pij of finding A in state3 i and B in state j? A B Pij = Pi Pj (16.5) Suppose the two systems are correlated, and I know that the probability of finding the systems in states i and j respectively is Pij for any states i and j. How would I find the probability of finding A in state n, regardless of the state of B?

A X Pn = Pnj (16.6) j Write on board: all states X Fairness = −k Pi ln Pi i

Physics 423 72 Monday 5/9/2012 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 the fairness function is additive (and thus extensive), in the case of combining two uncorrelated systems with separate probabilities.

X FA = −k Pi ln Pi (16.7) i X FB = −k Pi ln Pi (16.8) i X FAB = −k PiPj ln (PiPj) (16.9) ij X = −k PiPj (ln Pi + ln Pj) (16.10) ij ! ! X X = −k PiPj ln Pi − k PiPj ln Pj (16.11) ij ij ! ! ! ! X X X X = −k Pi ln Pi Pj − k Pj ln Pj Pi (16.12) i j j i ! ! X X = −k Pi ln Pi − k Pj ln Pj (16.13) i j

= FA + FB (16.14)

Physics 423 73 Tuesday 5/10/2012 17 Tuesday: Optimizing the fairness

Current board:

∆Ssystem+∆Ssurroundings ≥ 0 ∂a 1 dQ¯ “Cp = ” = ∂T ∂b c ∂b p dG = −SdT + V dp ∂a c dU = T dS − pdV ∂a ∂b ∂a Z = dQ¯ reversible ∂b ∂c ∂c ∆S = d d d T dF = −SdT − pdV

∂c ∂a ∂b all states = − a X Fairness = −k Pi ln Pi ∂b c ∂c dU =dQ ¯ +dW ¯ dH = T dS + V dp ∂a b i

Activity: Students as molecules (20 min) Students, who represent molecules, will each get 1 ordinary die, and the class will be split up into either six or nine sections (perhaps tables?) Students will all start at a single table, and we’ll have the students at each table compute 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 go to the North, East, South or West respectively, otherwise they will remain stationary. We’ll recompute the fairness, and repeat. They can think of the diﬀerent tables as diﬀerent quantum eigenstates, and of themselves as distinguishable particles.

17.1 Least bias lagrangian

Lecture: I introduced the fairness function on Monday, and yesterday you made some measurements 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 go about maximizing it.

1The fairness of the room is deﬁned as the following, but let them work this out if possible... X F = N fi ln fi (17.1) i

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

Physics 423 74 Tuesday 5/10/2012 SWBQ: How would you ﬁnd the maximum of a function?

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

∂F = 0 (17.2) ∂Pi

“Using the formula for the fairness function, what can this tell us about Pi?” The answer is that the probabilities should all be zero._ ¨ There is a problem with this, which is that Pi can’t take just any values, because these are probabilities, and they have to add up to one. This is a constraint, and to solve a constrained maximization, we use the method of Lagrange multipliers. We ﬁrst deﬁne a Lagrangian2: ! X L = F + α 1 − Pi (17.3) i

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

17.2 Weighted averages

Lecture: (20 min?) Most thermodynamic quantities can be expressed as weighted averages over all possible eigenstates (or microstates). For instance, we the internal energy is given by: X U = PiEi (17.4) i

Note that this will probably not be an eigenvalue of the energy, but that’s okay. The energy eigenvalues are so close for the total energy of a macroscopic object that we couldn’t distinguish 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 normal Physics meaning for Lagrangian, but rather the definition from optimization theory, since we are optimizing.

Physics 423 75 Tuesday 5/10/2012 Lecture: (10 min) I’d like to now summarize what I hope you all ﬁgured out yesterday about the probabilities that come from maximizing the fairness. Except that this time I’ll use an internal energy constraint instead of a length constraint. ! ! X X X L = −kB Pi ln Pi + αkB 1 − Pi + βkB U − PiEi (17.5) i i i where α and β are the two Lagrange multipliers. We want to maximize this, so we set its derivatives to zero: ∂L = 0 (17.6) ∂Pi = −kB (ln Pi + 1) − kBα − βkBEi (17.7)

ln Pi = −1 − α − βEi (17.8) At this point, it is convenient to invoke the normalization constraint... X Pi = 1 (17.9) i X 1 = e−1−α−βEi (17.10) i X 1 = e−1−α e−βEi (17.11) i X e1+α = e−βEi (17.12) i (17.13) Write on board: all states X Z ≡ e−βi i Write on board: e−βi P = i Z

Boltzmann factor P = (17.14) i partition function At this point, we haven’t yet solved for β, and to do so, we’d need to invoke the internal energy constraint: X U = EiPi (17.15) i P E e−βEi U = i i (17.16) Z

Physics 423 76 Tuesday 5/10/2012 SWBQ: (10 min) Introduce two non-interacting identical systems with N eigenstates each, and talk about whether things will be extensive or intensive. “How many eigenstates does the combined system have?” Students answer N or 2N. Using spin 1 system, ask for what the eigenstates of two spin-1 particles are. Soon we realize that it’s N (2) = N 2, which means it’s neither extensive nor intensive.

SWBQ: (3 min) What are the energy eigenstates of the combined system, if the energy (1) eigenstates of the small systems are Ei ?

Lecture: (7 min) Talk about the partition function, and work out what the partition function 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 be a thermodynamic state variable, as we will see tomorrow.

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

Physics 423 77 Wednesday 5/11/2012 18 Wednesday: From statistics to thermodynamics

Current board:

dQ¯ “C = ” dU = T dS − pdV all states p X ∂a 1 ∂T p Fairness = −k P ln P = i i ∂b c ∂b i ∂a c Z dQ¯ ∆S = reversible dF = −SdT − pdV T ∂a ∂b ∂a all states = X Z ≡ e−βi ∂b d ∂c d ∂c d i dU =dQ ¯ +dW ¯ dH = T dS + V dp

∂c ∂a ∂b a = − −βi ∂b ∂c e c ∆Ssystem+∆Ssurroundings ≥ 0 dG = −SdT + V dp Pi = ∂a b 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 the probabilities of being in the various eigenstates, by assuming that the “fairest” distribution would 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 plus surroundings can increase or stay the same, but cannot decrease. The maximum value of the fairness 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 of the fairness (a.k.a. entropy), which is X F = −k Pi ln Pi (18.1) i X U = PiEi (18.2) i

Physics 423 78 Wednesday 5/11/2012 e−βEi P = (18.3) i Z “On your big white boards, solve for the fairness as a function of U, β and Z. i.e. eliminate the sum over i.” X Fmax = −kB Pi ln Pi (18.4) i X e−βEi e−βEi = −k ln (18.5) B Z Z i X e−βEi = −k (−βE − ln Z) (18.6) B Z i i −βEi −βEi X Eie X ln Ze = k β + k (18.7) B Z B Z i i

Fmax = kBβU + kB ln Z (18.8)

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

F 1 U = max − ln Z (18.9) kBβ β “We saw before that ln Z 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 also know that

dU = T dS − pdV (18.11) ∂U T = (18.12) ∂S V Since we have an equation for U in terms of S, we just need to ﬁgure out how to hold V constant, and we’ll know what T is! “What does it mean to hold V constant? It hasn’t shown up in any of our statistical equations?” If we change the volume, we change the energy eigenvalues, so if we hold V constant (and in general, do no work) then the energy eigenvalues are ﬁxed. So this derivative should be possible.

S 1 U = − ln Z (18.13) kBβ β

Physics 423 79 Wednesday 5/11/2012 ∂U T = (18.14) ∂S V 1 S ∂β ln Z ∂β 1 ∂Z = − 2 + 2 − (18.15) kβ kβ ∂S V β ∂S V Zβ ∂S V 1 S ∂β ln Z ∂β 1 ∂Z ∂β = − 2 + 2 − (18.16) kβ kβ ∂S V β ∂S V Zβ ∂β V ∂S V 1 1 S ln Z 1 ∂Z ∂β = + − + − (18.17) kβ β kβ β Z ∂β V ∂S V 1 1 S ln Z ∂β = + − + + U (18.18) kβ β kβ β ∂S V 1 = (18.19) kβ 1 β = (18.20) kT Then we can easily see now from Equation 18.9 that

U = TS − kT ln Z (18.21) −kT ln Z = U − TS (18.22)

F = −kBT ln Z (18.23)

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

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

N N X X (1) (1) (1) (1) S2 = Pi Pj ln Pi Pj (18.24) i=1 j=1 N N X X (1) (1) (1) (1) = Pi Pj ln Pi + ln Pj (18.25) i=1 j=1 N ! N X (1) X (1) (1) = 2 Pi Pj ln Pj (18.26) i=1 j=1 N X (1) (1) = 2 Pj ln Pj (18.27) j=1

= 2S1 (18.28)

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

Physics 423 80 Wednesday 5/11/2012 Activity: Entropy of microcanonical ensemble (20 min) Consider a system with W eigenstates, all of which have the same energy. What is the entropy?

W X 1 1 S = −k ln (18.29) B W W i=1

= kB ln W (18.30)

“This is the result that is on Boltzmann’s tombstone, and is sometimes taken as a deﬁnition of the entropy, although it is less general than the one that we’re using. ” Energy and Entropy Homework 5 Due Wednesday 5/11

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

a) During this time does the entropy of the metal increase, decrease, remain the same, or is 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 determinable with the given information? Explain your reasoning.

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

Problem 5.2 Isothermal and adiabatic compressibility The isothermal compressibility is deﬁned as 1 ∂V KT = − (18.31) V ∂p T

KT is be found by measuring the fractional change in volume when the the pressure is slightly changed with the temperature held constant. In contrast, the adiabatic compressibility is deﬁned as 1 ∂V KS = − (18.32) V ∂p S 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. Show that K C T = p (18.33) KS CV Where the heat capacities at constant pressure and volume are given by

∂S Cp = T (18.34) ∂T p

Physics 423 81 Wednesday 5/11/2012 ∂S CV = T (18.35) ∂T V

Problem 5.3 Boltzmann ratio At low temperatures, diatomic molecule can be well described as a rigid rotor. The Hamiltonian of such a system is simply proportional to the square of the angular momentum 1 H = L2 (18.36) 2I and the energy eigenvalues are l(l + 1) E =h ¯2 (18.37) lm 2I a) What is the energy of the ground state and the first and second excited states of the H2 molecule? b) At room temperature, what is the relative probability of finding a hydrogen molecule in 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 any one 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 force τ in a plastic rod at temperature T is given by its Equation of State

2 τ = aT (L − Lo)

where a is a positive constant and Lo is the rod’s unstretched length. For an unstretched rod (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 ) at some other temperature Tf and length Lf . a) Write U = U(T,L) and take the total derivative dU.

2 b) Show that the partial derivative (∂U/∂L)T = −aT (L − Lo). c) To integrate the resulting diﬀerential equation Line Integrate dU very carefully in the T,L plane, keeping in mind that CL = bT holds only at L = Lo. End of Energy and Entropy Homework 5 Due Wednesday 5/11

Physics 423 82 Thursday 5/12/2012 Required homework 6.1: rubber-band-model Required homework 6.2: rubber-meets-road

Physics 423 83 Thursday 5/12/2012 19 Thursday: Statistical mechanics of air

Current board:

∂c ∂a ∂b a = − −βi ∂b ∂c e c ∆Ssystem+∆Ssurroundings ≥ 0 dG = −SdT + V dp Pi = ∂a b Z

19.1 Quantum spectra

Lecture: (10 min) I’m going to quickly review and introduce the energy eigenvalues for some simple quantum mechanical problems. For each of the following, I’d like to sketch out the potential, then sketch the wavefunctions and the spacing of the energy levels. The ﬁrst problem you handled was a particle in an inﬁnite square well potential:

−h¯2 ∂2 H = (19.1) 2m ∂x2 h¯2π2n2 E = (19.2) n 2mL2

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

−h¯2 H = ∇2 (19.3) 2m h¯2π2 n2 + n2 + n2 E = x y z (19.4) nxnynz L2

Physics 423 84 Thursday 5/12/2012 The next moderately simple problem is the rigid rotator. In this case, the only energy in the Hamiltonian is the angular kinetic energy: −h¯2 H = L2 (19.5) 2I h¯2l(l + 1) E = (19.6) lm 2I Finally, we have a simple problem that hasn’t yet come up in the paradigms, which is the simple harmonic oscillator. In this case we have both kinetic and potential energy: −h¯2 ∂2 mω2 H = + 0 x2 (19.7) 2m ∂x2 2 1 E = n + hω¯ (19.8) n 2 0 Of course, you also studied the hydrogen atom, but its solution is less general than those we’ve listed here. Any diatomic molecule behaves like a rigid rotator and a simple harmonic oscillator, 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:

2 2 2 2 2 2 (1) h¯ π nx + ny + nz h¯ l(l + 1) 1 E = + + n + hω¯ 0 (19.9) nxnynznvlm 2mL2 2I 2 That’s an awful lot of quantum numbers, and that’s just one molecule, and we’re neglecting any possible electronic excited states! How does this change when we’ve got N molecules all conﬁned in the same box? We’ve already talked about how energies relate when we combine systems:

N X (1) Etot = Ei (19.10) i 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.

all states X Z = e−βEthis state (19.11) X −β E(1) +E(2) +··· = e nx1 ny1 nz1 nv1 l1m1 nx2 ny2 nz2 nv2 l2m2 (19.12)

nx1 ny1 nz1 nv1 l1m1,nx2 ny2 nz2 nv2 l2m2,···

Physics 423 85 Thursday 5/12/2012 . . . 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 can ﬁx this double-counting by multiplying by an N!, which gives us: 1 X −β E(1) +E(2) +··· Z = e nx1 ny1 nz1 nv1 l1m1 nx2 ny2 nz2 nv2 l2m2 (19.13) N! nx1 ny1 nz1 nv1 l1m1,nx2 ny2 nz2 nv2 l2m2,··· 1 X −βE(1) −βE(2) = e nx1 ny1 nz1 nv1 l1m1 e nx2 ny2 nz2 nv2 l2m2 ··· (19.14) N! nx1 ny1 nz1 nv1 l1m1,nx2 ny2 nz2 nv2 l2m2,···     1 X −βE(1) X −βE(2) = e nx1 ny1 nz1 nv1 l1m1 e nx2 ny2 nz2 nv2 l2m2 ··· N!     nx1 ny1 nz1 nv1 l1m1 nx2 ny2 nz2 nv2 l2m2 (19.15)  N 1 X −βE(1) = e nxnynznvlm (19.16) N!   nxnynznvlm

2 2 2 2 2 ! N  h¯ π n +n +n 2  ( x y z) h¯ l(l+1) 1 −β + + n+ ¯hω0 1 X 2mL2 2I ( 2 ) = e (19.17) N!   nxnynznvlm  N h¯2π2 n2 +n2 +n2 1 ( x y z) h¯2l(l+1) 1 X −β −β −β(n+ )¯hω0 = e 2mL2 e 2I e 2 (19.18) N!   nxnynznvlm  N h¯2π2 n2 +n2 +n2 1 ( x y z) h¯2l(l+1) 1 X −β X −β X −β(n+ )¯hω0 = e 2mL2 e 2I e 2 (19.19) N!   nxnynz nv lm

 N N N h¯2π2 n2 +n2 +n2 ! ! 1 ( x y z) h¯2l(l+1) 1 X −β X −β X −β(n+ )¯hω0 = e 2mL2 e 2I e 2 (19.20) N!   nxnynz nv lm Now, if we were computing the internal energy U, we’d be able to do something very similar: X U = PiEi (19.21) i 1 X = P E (19.22) N! ... tot nx1 ny1 nz1 nv1 l1m1,nx2 ny2 nz2 nv2 l2m2,··· 1 X (1) (2) = P... En n n n l m + En n n n l m + ··· N! x1 y1 z1 v1 1 1 x2 y2 z2 v2 2 2 nx1 ny1 nz1 nv1 l1m1,nx2 ny2 nz2 nv2 l2m2,··· (19.23)

N X (1) = P...En n n n l m (19.24) N! x1 y1 z1 v1 1 1 nx1 ny1 nz1 nv1 l1m1,nx2 ny2 nz2 nv2 l2m2,···

Physics 423 86 Thursday 5/12/2012 X (1) = N P E (19.25) nxnynznvlm nxnynznvlm nxnynznvlm 2 2 2 2 2 2 ! X h¯ π nx + ny + nz h¯ l(l + 1) 1 = N P + + n + hω¯ (19.26) nxnynznvlm 2mL2 2I 2 0 nxnynznvlm   2 2 2 2 2 2 X h¯ π nx + ny + nz X h¯ l(l + 1) X 1 = N P + P + P n + hω¯  nxnynz 2mL2 lm 2I nv 2 0 nxnynz lm nv (19.27) So you can see, if we can work out the average translational kinetic energy, rotational energy and vibrational energy of a molecule in this gas, then we’ll easily have the total internal energy of this system just by adding everything up and multiplying by N.

Activity: Diatomic molecule from quantum up (1 hour 30 min without wrapup) I’m going to divide you into groups. Each group will have a separate task, so that hopefully when we’re done as a class we’ll have an answer for the total internal energy of a diatomic 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 h¯2π2 β 1 (19.28) 2mL2 or the reverse. For the rotational energy, it will be h¯2 β 1 (19.29) 2I and for the vibrational energy, it will be

βhω¯ 0 1 (19.30) These energy scales have large gaps, so the translational energy may be in the high-temperature limit 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, in either 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 particular limit is that we’ve ignored the possibility that two of our molecules may be in precisely the same 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) there is another problem, which I also won’t go into. As long as our temperature is high enough that the probability of any given state is pretty low, we won’t run into these troubles (which could give us things like a Bose-Einstein condensate).

Physics 423 87 Friday 5/13/2012 20 Friday:

Current board:

∂c ∂a ∂b a = − −βi ∂b ∂c e c ∆Ssystem+∆Ssurroundings ≥ 0 dG = −SdT + V dp Pi = ∂a b Z

20.1 Diatomic gas wrapup

Lecture: (20 min) Yesterday we worked out the internal energy per molecule of a diatomic gas associated with translational kinetic energy, rotational kinetic energy, and vibrational energy (which has both a kinetic and potential component). For each case (except translation), you considered both the low- and high- temperature limits. In each case you had sums that looked like

X somethinge−βEi (20.1) i

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 was okay. For the low-temperature limit, you had an easier scenario, as all the Boltzmann factors were all very small compared with the ground state. So you could just truncate the sum after a few terms.

Physics 423 88 Friday 5/13/2012 Translation at high T

2 2 2 2 2 2 2 2 2 2 h¯ π (nx+ny+nz) U X h¯ π nx + ny + nz 1 −β = e 2mL2 (20.2) N 2mL2 Z nxnynz 2 2 2 2 2 2 2 2 2 2 Z ∞ Z ∞ Z ∞ h¯ π (nx+ny+nz) h¯ π nx + ny + nz 1 −β 2mL2 ≈ 2 e dnxdnydnz (20.3) 0 0 0 2mL Z 2 2 2 2 2 2 2 Z ∞ Z ∞ Z ∞ h¯ π (nx+ny+nz) h¯ π 1 2 2 2 −β 2mL2 = 2 nx + ny + nz e dnxdnydnz (20.4) 2mL Z 0 0 0 p hπ¯ ux = β √ nx (20.5) 2mL √ √ 3 2 2 ! ∞ ∞ ∞ U h¯ π 1 2mL k T Z Z Z 2 2 2 B 2 2 2 −(ux+uy+uz) = 2 ux + uy + uz e duxduyduz N 2mL Z hπ¯ 0 0 0 (20.6) √ 3 ∞ ∞ ∞ 2mL(k T ) 2 1 Z Z Z 2 2 2 B 2 2 2 −(ux+uy+uz) = ux + uy + uz e duxduyduz (20.7) hπ¯ Z 0 0 0 At this point we’ve extracted the physics from the integral. It’s clearly not zero, and it also isn’t inﬁnite, so it’s just some number that we can work out later. But we still need Z...

2 2 2 2 2 h¯ π (nx+ny+nz) X −β Z = e 2mL2 (20.8)

nxnynz 2 2 2 2 2 Z ∞ Z ∞ Z ∞ h¯ π (nx+ny+nz) −β 2 ≈ e 2mL dnxdnydnz (20.9) 0 0 0 r β hπ¯ ux = nx (20.10) √2m L Z ∞ Z ∞ Z ∞ 2mkBTL 2 2 2 Z = e−(ux+uy+uz)du du du (20.11) hπ¯ x y z √ 0 0 0 ∞ 3 2mk TL Z 2 = B e−u du (20.12) hπ¯ 0 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 these together (with a minimum of simpliﬁcation, we get:

√ 3 2 2mL(kB T ) 2 R ∞ 2 −u2 R ∞ −u2 U ¯hπ 3 0 u e du 0 e du = √ (20.13) 3 N 2mkB TL R ∞ −u2 ¯hπ 0 e du Z ∞ Z ∞ 2 −u2 −u2 = kBT 3 u e du e du (20.14) 0 0

Physics 423 89 Friday 5/13/2012 3 = k T (20.15) 2 B

Rotation at low T I’ll skip over this, as it isn’t very exciting (and it’s taking a long time 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 energy difference between the ground state and the first excited state.

Rotation at high T

∞ l 2 X X −β h¯ l(l+1) Z = e 2I (20.16) l=0 m=−l ∞ 2 X −β h¯ l(l+1) = (2l + 1)e 2I (20.17) l=0 Z ∞ 2 −β h¯ l(l+1) ≈ (2l + 1)e 2I dl (20.18) 0 As it turns out, we can relatively easily do this integral. However, the “+1” terms are insigniﬁcant, since l is integrated up to inﬁnity, and the large l contribution dominates. So we can:

Z ∞ 2 2 −β h¯ l Z ≈ 2le 2I dl (20.19) 0 hl¯ u = √ (20.20) 2IkBT Z ∞ 8IkBT −u2 Z = 2 ue u (20.21) h¯ 0 (20.22)

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

Z ∞ 2 2 2 2 U 1 h¯ l −β h¯ l ≈ 2l e 2I dl (20.23) N Z 0 2I 2 2 Z ∞ 1 h¯ 4IkBT 3 −u2 = 2 2 u e du (20.24) Z 2I h¯ 0 2 2 ∞ 2 2 ¯h 4IkB T R u3e−u du = 2I ¯h2 0 (20.25) 8IkB T R ∞ −u2 ¯h2 0 ue u R ∞ u3e−u2 du = 4k T 0 (20.26) B R ∞ −u2 0 ue u = kT (20.27)

Physics 423 90 Friday 5/13/2012 Here we can see much of the essential physics by recognizing that the energy is proportional to 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 diﬀerent approach, since I think showing the same approach for the third time in a row is a bit boring. ∞ X −β n+ 1 ¯hω Z = e ( 2 ) 0 (20.28) n=0 ∞ −β 1 ¯hω X −βn¯hω = e 2 0 e 0 (20.29) n=0 ∞ 1 n −β ¯hω0 X −β¯hω0 = e 2 e (20.30) n=0 This is just a harmonic series, so we can solve it using the standard trick, where I’ll call the series s: ∞ X n s ≡ e−β¯hω0 (20.31) n=0 ∞ X n e−β¯hω0 s = e−β¯hω0 e−β¯hω0 (20.32) n=0 ∞ X n = e−β¯hω0 (20.33) n=1 ∞ ! X n = e−β¯hω0 − 1 (20.34) n=0 e−β¯hω0 s = s − 1 (20.35) 1 = 1 − e−β¯hω0 s (20.36) 1 s = (20.37) 1 − e−β¯hω0 −β 1 ¯hω e 2 0 Z = (20.38) 1 − e−β¯hω0 So that’s nice. Of course, we still want to ﬁnd the energy. To do this, we can employ yet another trick—although it’s not so hard to do in the high-temperature limit the same way we solved the previous problem. We can recognize that X U = PiEi (20.39) i X e−βEi = E (20.40) i Z i

Physics 423 91 Friday 5/13/2012 1 X = E e−βEi (20.41) Z i i X Z = e−βEi (20.42) i ∂Z X = −E e−βEi (20.43) ∂β i Ei i ∂Z ∂β U = − Ei (20.44) Z So once we have the partition function, we could just take a derivative to ﬁnd the internal energy. So for the simple harmonic oscillator, we have:

−β 1 ¯hω ∂Z 1 e 2 0 −β¯hω0 = − hω¯ 0Z − 2 e hω¯ 0 (20.45) ∂β 2 −β¯hω0 Ei (1 − e ) 1 1 −β¯hω0 = − hω¯ 0Z − e hω¯ 0Z (20.46) 2 1 − e−β¯hω0 1 hω¯ 0 = − hω¯ 0Z − Z (20.47) 2 eβ¯hω0 − 1 U 1 hω¯ 0 = hω¯ 0 + (20.48) N 2 eβ¯hω0 − 1 This gives us an exact solution for the internal energy of a simple harmonic oscillator, but we still haven’t found the high-temperature limit. To ﬁnd that, we have to take βhω¯ 0 1. In this case, we can just use a simple Taylor’s expansion approach:

U 1 hω¯ 0 = hω¯ 0 + (20.49) N 2 eβ¯hω0 − 1 1 hω¯ 0 ≈ hω¯ 0 + (20.50) 2 1 + βhω¯ 0 − 1 1 hω¯ 0 = hω¯ 0 + (20.51) 2 βhω¯ 0 1 = hω¯ + k T (20.52) 2 0 B ≈ kBT (20.53)

This tells us that like the rotational energy, the vibrational energy approaches kBT per molecule at high temperatures. There is a general rule which occurs in the classical limit (which is the high-temperature 1 limit), that any quadratic term in the energy ends up providing 2 kBT to the internal energy. This is called the equipartition theorem. Since there are three translational degrees of freedom 2 3 (v in each direction), the kinetic energy gives us 2 kBT . There are two ways to rotate a diatomic molecule, which gives us an additional kBT . And ﬁnally, the vibration has both kinetic and potential energy, which each provide half of the ﬁnal kBT .

Physics 423 92 Friday 5/13/2012 Energy and Entropy Homework 6 Due Friday 5/13

Problem 6.1 A rubber band model Consider a model in which rubber is composed of segments of ﬁxed length a, each of which can be oriented in any of the six cardinal directions (±~xˆ, ±~yˆ or ±~ˆz). We will assume that the rubber band is aligned in the ~xˆ direction, and there is an energetic penalty ε for segments to be in any of the ~yˆ or ~ˆz directions, but that there is no interaction involving the orientations of neighboring segments.1

E~xˆ = 0 (20.54)

E−~xˆ = 0 (20.55)

Eanything else = ε (20.56)

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

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

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

d) What is the mean displacement in the ~xˆ direction L of a molecule composed of N segments under tension τ?

e) What is the entropy S of this system, as a function of β and τ? Keep in mind that the entropy is given by all states X S = −kB Pi ln Pi i 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 the Gibbs free energy of a model polymer. In this problem, you will use “reasonable” values for the 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, the length of each link in the polymer is a = 1 nm, and the energy of vertically-oriented links is 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 DNA in solution.

Physics 423 93 Friday 5/13/2012 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 value for a single molecule.

a) Find a value for Nwide that gives a reasonable (within an order of magnitude) prediction for 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—which is not when the rubber band itself has zero length. b) Plot the predicted length L at room temperature versus the tension τ, over the range of tensions you measured experimentally. Also include on your plot (possibly with an offset) your experimental data for this temperature. Comment on your agreement or disagreement. c) For a specific tension, plot the predicted length versus temperature, along with experimental data points. d) For a specific length, plot your predicted tension versus temperature. This will probably 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. You computed from your measurement the work done, the heat transfered, and the change in internal energy and entropy. Using the model for the same change in tension and at the same temperature, what are Q, W , ∆U and ∆S? f) Discuss the agreements and disagreements between the theory and the model. Suggest specific problems and possible improvements in the model. End of Energy and Entropy Homework 6 Due Friday 5/13

20.2 Third law

Lecture:

The third law of thermodynamics The entropy of any perfect crystal at zero temperature is zero. X S = −kB Pi ln Pi (20.57) i “When is this zero?” It’s equivalent to saying that any perfect crystal at zero temperature is in a non-degenerate ground state. It doesn’t sound so deep when looked at from a stat-mech viewpoint, 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.

Physics 423 94 Friday 5/13/2012 • 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 right temperature, so that the process is reversible.

• Slowly cool down the resulting CO2 to zero temperature. If we integrate Z dQ ∆S = reversible (20.58) T we ﬁnd no change in the entropy! This means that we really can tabulate the absolute entropy of any material we like. All we have to do to ﬁgure out the zero of entropy is cool it down to zero Kelvin.

Ice rules

Physics 423 95 Friday 5/13/2012 Physics 423 Friday 5/13/2012 Physics 423 Friday 5/13/2012 2680 LINUSPAULING Vol. 57

summary 4. On the addition of 18.5% of methyl alcohol 1. The velocity constant for the &acetone al- Eact increases by over 1700 cal. but the general cohol decomposition in the presence of dilute so- character of the EaCt-T curve remains little dium hydroxide has been measured at 5' inter- changed to 30". vals from 0 to 50". 5. The corresponding B values from the inte- 2. The constancy of ratio of velocity constant grated Arrhenius equation, In k = 2.3 B - (Eact/ to sodium hydroxide concentration has been RT), both in water and the methyl alcohol solu- confirmed over a limited concentration range tion parallel these Ea,, values and furnish experi- at 25'. mental evidence for abandoning the unfortunate 3. The energy of activation, calculated from term "temperature independent constant" for the Arrhenius equation for a series of tempera- this quantity. ture intervals, has been shown to be a function of 6. The data show that the collision theory is temperaturl: well outside the limits of error. inadequate and that the entropy of activation is E,,, increases consistently from a value of 15,850 an important quantity in considering solution re- cal. at 5" to 17,250 at 32.5 and then decreases by actions. about 400 cal. at 45'. NEWYORK, N. Y. RECEIVEDAUGUST 15, 1935

[CONTRIBUTION FROM THE GATESCEEMICAL LABORATORY,CALIFORNIA INSTITUTE OF TECHNOLOGY,NO. 5061 The Structure and Entropy of Ice and of Other Crystals with Some Randomness of Atomic Arrangement BY LINUSPAULING

Investigations of the entropy of substances at covery of the hydrogen bond4 that the unusual low temperatures have produced very important properties of water and ice (high melting and information regarding the structure of crystals, boiling points, low density, association, high di- the work of Giauque and his collaborators being electric constants, etc.) owe their existence to particularly noteworthy. For example, the ob- hydrogen bonds between water molecules. The served entropy of crystalline hydrogen shows that arrangement of oxygen atoms (but not of hydro- even at very low temperatures the molecules of gen atoms) in crystals of ice is known from x-ray orthohydrogen in the crystal are rotating about studies;6 it is not a close-packed arrangement (as as freely as in the gas;' subsequent to this dis- of sulfur atoms in the high-temperature form of covery the phenomenon of rotation of molecules in hydrogen sulfide), but a very open one, like that crystals was found to be not uncommon. Also of the silicon atoms in high-tridymite. Each the entropy values of carbon monoxide2 and oxygen atom in ice is tetrahedrally surrounded by nitrous oxide3 show that in crystals of these sub- four other oxygen atoms at the distance 2.76 A., stances the molecules are not uniquely oriented, and it has been assumed that it is bonded to these but have instead a choice between two orienta- atoms by hydrogen bonds, the number of hydrogen tions, presumably the opposed orientations CO atoms being just that required to place one hydro- and OC or NNO and ONN along fixed axes. It is gen atom between each pair of oxygen atoms. pointed out in this note that the observed entropy (Similarly in high-tridymite there is an oxygen of ice at low temperatures provides strong support atom between each pair of silicon atoms; we for a particular structure of ice, and thus gives an might say that each silicon atom is attached to answer to a question which has been extensively discussed during the past few years. four others by oxygen bonds.) It has been generally recognized since the dis- The question now arises as to whether a given hydrogen atom is midway between the two oxygen (1) W. F. Oiauque and H. L. Johnston, TIIISJOURNAL, 60, 3221 (1928); L. Pauling, Phys. Rev., 86, 430 (1930). (4) W. M. Latimer and W. H. Rodebusb, THIS JOURNAL,42, (2) J. 0. Clayton and W. F. Giauque, Tms JOURNAL,64, 2610 1419 (1920). (1932). (5) D. M. Dennison, Phys. Reo., 17, 20 (1921); W. H. Bragg, (3) R. W. Blue and W. F. Giauque, ibid., 67, 991 (1936); K. Puoc. Phys. SOC.(London), 34, 98 (1922); U'. H. Barnes, Proc. Clusius, Z. Elcklrochem., 40, 99 (1934). Roy. SOC.(London), Al26, 670 (1929). Physics 423 Friday 5/13/2012 Dec., 1935 THESTRUCTURE AND ENTROPYOF ICE 2681

atoms it connects or closer to one than to the one oxygen atom to another one 0.95 A. from an other. The answer to this is that it is closer to adjacent oxygen atom. It is probable that both one than to the other. In the gas molecule the processes occur. As an example of a change from 0-H distance is 0.95 A., and the magnitudes of one to another configuration, we may consider one the changes in properties from steam to ice are not of the puckered rings of six oxygen atoms occur- sufficiently great. to permit us to assume that this ring in ice distance is increased to 1.38 8. In ice each 0-H 0 /O H-0 \H H hydrogen atom is about 0.95 A. from one oxygen H atom and 1.81 4.. from another. O/H 0 \O do \H We now ask whether each hydrogen atom (or H rather hydrogen nucleus) has a choice of two \O H-0 0-H O/H positions along its oxygen-oxygen axis, inde- Change from one of the two cyclic arrangements of pendent of the positions of the other hydrogen hydrogen nuclei to the other is permitted by our nuclei. The answer is in the negative; for we postulates. The fact that at temperatures above know that the concentration of (OH)-and (H30)+ about 200A. the dielectric constant of ice is of ions in water is very small, and we expect the the order of magnitude of that of water shows that situation to be essentially unchanged in ice. the molecules can orient themselves with consider- Hence the hydrogen nuclei will assume positions able freedom, the crystal changing under the such that each oxygen atom (with at most very influence of the electric field from unpolarized to few exceptions) will have two hydrogen atoms polarized configurations satisfying the above attached to it. conditions. On cooling the crystal to low tem- Let us now make the following assumptions (to peratures it freezes into some one of the possible be supported later by a discussion of the entropy) configurations; but it does not go over (in a regarding the structure of ice. reasonable period of time) to a perfect crystal (1) In ice each oxygen atom has two hydrogen with no randomness of molecular orientation. atoms attached to it at distances of about 0.95 8., It will have at very low temperatures the entropy forming a water molecule, the HOH angle being L In W,in which W is the number of configurations about 105' as in the gas molecule. accessible to the crystal. (2) Each water molecule is oriented so that its Let us now calculate W,using two different two hydrogen at oms are directed approximately methods. toward two of the four oxygen atoms which sur- There are N molecules in a mole of ice. A given round it tetrahedrally, forming hydrogen bonds. molecule can orient itself in six ways satisfying (3) The orientations of adjacent water mole- condition 2. However, the chance that the ad- cules are such that only one hydrogen atom lies jacen t molecules will permit a given orientation is approximately along each oxygen-oxygen axis. inasmuch as each adjacent molecule has two (4) Under ordinary conditions the interaction hydrogen-occupied and two unoccupied tetra- of non-adjacent molecules is not such as to appre- hedral directions, making the chance that a given ciably stabilize any one of the many configura- direction is available for each hydrogen of the tions satisfying the preceding conditions with original molecule I/t1 and the chance that both can reference to the others. be located in accordance with the given orienta- Thus we assurne that an ice crystal can exist tion 1/4. The total number of configurations for N in any one of a large number of configurations,B molecules is thus W = (6/4)'" = (3/2)N. each corresponding to certain orientations of the The same result is given by the following equiva- water molecules. The crystal can change from lent argument. Ignoring condition 1, there are one configuration to another by rotation of some 22N configurations with hydrogen bonds between of the molecules or by the motion of some of the adjacent oxygen atoms, each hydrogen nucleus hydrogen nuclei, each moving a distance of about having the choice of two positions, one near one 0.86 8. from a potential minimum 0.95 8.from oxygen atom and the other near the other. Some (6) One of the large number of configurations is represented by the of these are ruled out by condition 1. Let us now 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 consider a given oxygen atom and the four sur- temperatures just below the melting point (but not at lower tem- rounding hydrogen atoms. There are sixteen peratures) the molecular arrangement might be partially or largely irregular. arrangements of the four hydrogen nuclei. Of Physics 423 Friday 5/13/2012 2682 LINUSPAULING VOl. 57

these ten are ruled out by condition l-one gives crystals will be found to have residual entropy at (H40)++, four give (HsO)+, four (OH)- and one very low temperatures as a result of some random- 0--. This condition for each oxygen atom thus ness of atomic arrangement. It is probable, permits only 6/la = 3//8 of the configurations, the however, that experimental verification of the total number of permitted configurations being residual entropy would be difficult for most of the accordingly W = 22"(3/8)N= (3/.JN. cases mentioned below. The contribution of this lack of regularity to Diaspore, "02, contains groups OH0 with the entropy of ice is thus R in 3//z = 0.805 E. U. hydrogen bonds.1° As in water, we expect the The observed entropy discrepancy of ice at low hydrogen nucleus to be closer to one oxygen atom temperatures is 0.87 E. U., obtained by subtracting than to the other. Inasmuch as there is no re- the entropy difference of ice at very low tempera- striction on the choice of the two positions, each tures and water vapor at standard conditions, for oxygen atom forming only one hydrogen bond, this which the value 44.23 E. U. has been calculated would lead to a residual entropy of R In 2 per from thermal data by Giauque and Ashley,? from mole of AlH02. The same residual entropy is the spectroscopic value 45.101 E. U. for the expected per HOz group for other crystals con- entropy of water vapor given by Gordon.8 The taining this group (staurolite). agreement in the experimental and theoretical Hydrogen bonds may not always lead to residual entropy vaSues provides strong support of the entropy. For example, the crystal lepidocro- postulated structure of icess cite,l' FeOOH, contains infinite strings of oxygen The structure of ice is seen to be of a type inter- atoms joined by hydrogen bonds, oHoHoHoHoH. mediate between that of carbon monoxide and If we assume that the choice of two positions by a nitrous oxide, in which each molecule can assume hydrogen nucleus is restricted by the requirement either one of two orientations essentially inde- that each oxygen have one and only one attached pendently of the orientations of the other mole- hydrogen, then there are two accessible configura- cules in the crystal, and that of a perfect molecu- tions per string. This lack of definiteness does lar crystal, in which the position and orientation not lead to an appreciable residual entropy for a of each molecule are uniquely determined by the large crystal. other molecules. In ice the orientation of a given In the double molecules of formic acid, with the molecule is dependent on the orientations of its O-H- four immediate neighbors, but not directly on the structure12 HC< O-H- >CH and in other mono- orientations of the more distant molecules. carboxylic acids forming double molecules, we ex- It should be pointed out that a finite residual pect that the interaction within a carboxyl group entropy calculated for a substance from experi- is strong enough to permit only the two configura- mental data obtained at temperatures extending tions in which one hydrogen nucleus is attached down to a certain temperature, with extrapolation to each carboxyl group to be stable at low tem- below that point, may arise either from failure of peratures. This would lead to a residual entropy the experimenter to obtain thermodynamic equi- of '/z R In 2 per RCOOH. librium in his calorimetric measurements or from By a similar argument we calculate for 0-oxalic error in the extrapolation. Measurements made acid (anhydrous), which contain finite strings13of under ideal conditions and extended to sufficiently molecules low temperatures would .c-c<-H->c-c< O-H- >C-.e,, presumably lead to zero entropy for any system. O-H- O-H- O-H- the residual entropy In 2 per mole of (COOH)z, Other Crystals with Residual Entropy R whereas a-oxalic acid, containing layers with the Our knowledge Of the structure of Crystals per- dealized structure18 in which there are infinite mits the prediction to be made that Other sequences of carboxyl groups attached by hydro- (7) W. F. Giauque and M. F. Ashley, PAYS. Rev., 4% 81 (1933). gen bonds, would have no appreciable residual (8) A. R. Gordon, J. Chem. Phys., 2, 65 (1934). (9) The existence of this residual entropy of ice at very low tem- (10) F. J. Ewing, J. Chem. Phys., 3, 203 (1935). peratures was discovered by Giauque and Ashley (ref. 7), who pre- (11) F. J. Ewing, {bid., 3,420 (1935). liminarily ascribed it to the pefsistance of rotation of ortho-water (12) L. Pauling and L. 0. Brockway, Proc. Naf. Acad. Sci., 20, molecules (comprising of the total) about their electric-moment 336 (1934). axes, giving an entropy of %/a R In 2 = 1.03 E. U. (13) S. B. Hendricks, Z. Krist., 91, 48 (1935). Physics 423 Friday 5/13/2012 Dec., 1935 ”HE STRUCTUREAND ENTROPYOF ICE 2683 ted apparently at random among O\C/O O\C/O I the positions provided, of which ‘c they occupy the fraction 1 - 6. H/O” “\‘H dH/o’yo \H In the cubic tungsten bronzes,16 Na,WOa, the sodium ions oc- I I cupy the fraction x of the avail- CI able positions. These crystals o/c\o db o/yo 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 it 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 hgstrijm 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 HF2group; 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- H~~~H F F, the two-minimum function and the num and three silicon atoms per formula are dis- H~~~H 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 bromide22Prepxed 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 In = ‘L.558E. u. per mole of RzO3. In (16) G. Hsgg, ibid., Bag, 192 (1935). R (17) L. Helmholz, J. Chem. Phys., 3, 740 (1935). Fe1-6S~ the atoms have a hex- (18) S. B. Hendricks and L. Pauling, THISJOURNAL, 47, 2904 agonal arrangemsent,the iron atoms being distribu- (1925). (19) T. F. W. Barth and E. Posnjak, J. Washington Acad. Sci., (14) G. Hagg and G.SBderhohn, 2.physik. Chem., 819, 88 (1935); 21, 255 (1931); F. Machatschki, Z. Kuist., 80, 416 (1031). G. Hagg, ibid., B29, 95 (1935); E. J. W. Verwey, Z. Krisf., 91, 65 (20) L. Pauling, Proc. Naf. Acad. Sci., 16, 123 (1930); W. W. (1935); E. Kordes, ibzd., 91, 193 (1935). Jackson and J. West, Z. Kvist., 76, 211 (1930). (15) G HAyg and I. Sucksdorff, 2. physik. Chem., B22, 444 (21) J. M. Bijvoet and W, Nieuwenkamp, ibid., 86, 466 (1933). (1933) (22) J. A. A. Ketelaar, ibid., 88, 26 (1934). Physics 423 Friday 5/13/2012 2684 LINUSPAULING AND L. 0.BROCKWAY Vol. 57

inappreciable, inasmuch as the number of layers in arrangement. The x-ray data show that the a crystal (other than an extremely small crystal) is sequence of layers is not completely random, the very small compared with the number of atoms. structure being essentially ABABAB-; it is In this connection it might be mentioned that possible, however, that a change in the sequence, there exists the possibility that ice may crystallize corresponding to twinning on the basal plane, with such an alternating layer structure. The occurs occasionally. oxygen-atom arrangement assigned to ice corre- I am indebted to Professor W. F. Giauque for sponds to superimposing double oxygen layers in discussing the question of the structure and the sequence ABAB-(A at 00, B at '/s 2/~, C at entropy of ice, as well as related questions, with z//3 of a hexagonal net). The sequence ABC- me. ABCABC-- would also lead to an arrangement Summary (diamond) such that each oxygen atom is surrounded by four others arranged tetrahedrally, It is suggested that ice consists of water mole- which is indeed, so far as I can see, just as satis- cules arranged so that each is surrounded by four factory as the reported arrangement. There is no others, each molecule being oriented in such a good evidence that such a cubic modification of way as to direct its two hydrogen atoms toward ice has been observed. However, the arbitrari- two of the four neighbors, forming hydrogen ness of orientation which we have found to exist bonds. The orientations are further restricted for the water molecules in ice suggests that there by the requirement that only one hydrogen atom may also be an arbitrariness in the sequence of lie near each 0-0 axis. There are (3/2)N such double oxygen layers, with configurations such as configurations for N molecules, leading to a resid- ABABABCBCB- occurring. Such an alter- ual entropy of R In 3/2 = O.SO5 E. U., in good nating layer structure would have hexagonal agreement with the experimental value 0.S7 E. U. symmetry, might develop faces at angles corre- The structure and entropy of other crystals sponding to the axial ratio c/a = 1.63, and would showing randomness of atom arrangement are not be distinguishable so far as residual entropy is discussed. concerned from a crystal with fixed oxygen atom PASADENA, CALIF. RECEIVEDSEPTEMBER 24,1935

[ CONTRIBUTJONFROM GATESCHEMICAL LABORATORY, CALIFORNIA INSTITUTE OF TECHNOLOGY,KO. 5071 The Radial Distribution Method of Interpretation of Electron Diffraction Photographs of Gas Molecules

BY LINUSPAULING AND L. 0. BROCKWAY

Introduction ward process of determining the structure of a The only method of interpretation of electron molecule from the analysis of experimental re- diffraction photographs of gas molecules which has sults, but consists instead in the testing (and been used to any great extent is the so-called rejection or acceptance) of any structures which visual method, involving the correlation of apparent may be formulated, a tedious calculation being maxima and minima on the photographs with required for each structure. maxima and minima on simplified theoretical We have developed a new method of interpreta- curves calculated for various models of the mole- tion of the photographs which does not suffer cule under consideration. This method of inter- from this disadvantage. This radial distribution pretation, originally developed by Wierl, ' has been method, which is closely related to the method of thoroughly tested by Pauling and Brockway2 who interpretation of x-ray diffraction data developed have shown it to yield values of interatomic dis- by Zernike and Prim3 for the study of the struc- tances accurate to within about 1yo (estimated ture of liquids and applied by Warren and Ging- probable error). The main disadvantage of the rich4 to crystals, consists in the calculation (from method is that it does not involve a straightfor- (3) F. Zernike and J. A. Pn'ns, 2. Phyvik, 41, 184 (1927); see also P. Debye and H. Menke, Ergab. Tech. Rontgenkunde, Akad. (1) R Wierl Ann Physik, 8, 521 (19311, 13, 453 (1932) Verlagsges., Leipzig, Vol. 11, 1931. (2) L Pauling and L 0 Brockway, J Chem Phis , 2, 867 (1934) Warren and S. Gingrich, Phys. REI., 368 Physics 423(4) B. E. N. Friday46, 5/13/2012 (1934). 1144 W. F. GIAUQUEAND J. W. STOUT VOl. 58

Jenkins4 and Bell and The latter Here the shift in the band decreases as the weight authors have shown in benzene solutions of con- of the halogen increases until in iodoacetic acid centrations from M = 0.01 to M = 1.5 trichloro- it occurs in the same position as acetic. This acetic acid is completely associated. We assume, suggests that there may be two opposing factors therefore, that under the conditions of our operative, one a weight factor and the other per- measurements all the acids were entirely present haps connected with the electron affinity or polari- as dimers. zability of the substituted group. It is noticeable The three infra-red absorption bands which that there is a correlation between the strength can be investigated in this solvent correspond to of the acid in aqueous solution and the position the frequencies vl, vz and vs of the previous paper. of this band. This is perhaps to be associated The first of these (VI) (which has been omitted in with changes in the contribution of ionic states Fig. 1 to conserve space) is mainly due to the C-H to the energy of the molecule. valence vibration, but is also in part made up of The remaining band, v3, is shifted in each case overtone and combination vibrations from other in the opposite direction, but the displacement is parts of the molecule. Thus it occurs even in smaller. This behavior was observed also in the trichloroacetic acid, but considerably displaced investigation on the effect of association.2 It and much weaker. The remaining two bands are does not seem possible in the light of our present to be attributed to vibrations within the ring knowledge to give an exact interpretation of these formed by the union of the two single molecules. band shifts although they seem too uniform to be Attempts to measure some of these bands in the fortuitous. single molecules at high temperatures were un- The author is grateful to Dr. Farrington Daniels successful on account of decomposition. for helpful suggestions and for continued interest It may be seen by reference to Table I that the in this problem. band vz is shifted toward higher frequencies by groups recognized by the organic chemist as Summary electronegative, while it is shifted to lower fre- 1. The infra-red absorption spectra of acetic quencies by groups electropositive in the same acid and eight substituted acetic acids have been sense. The uniform shift in the chloroacetic measured in carbon tetrachloride solution. Cer - acids discovered by Bennett and Daniels is contain uniform shifts in the absorption bands have firmed. The effect of the weight of the groups been observed and correlated with other measure- may be seen by considering the monohalogen acids. ments. (4) Bury and Jenkins, J. Chem. Soc., 688 (1934). (5) Bell and Arnold, ibid., 1432 (1935). MADISON,WISCONSIN RECENEDMAY 5, 1936

[CONTRIBUTION FROM THE CHEMICAL LABORATORYOF 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. GIAUQUEAND J. w.STOUT It has long been known to those interested in ton, Overstreet, and one of us1 on the entropies the accurate application of the third law of ther- of hydrogen chloride, hydrogen, oxygen and, chlo- modynamics that measured entropy changes in rine, combined with very accurate determinations reactions involving water did not agree kth of the heats of reactions those calculated from low temperature heat ca- Hz + '/SO*= HzO pacity data. In early comparisons the inaccuracy and 2HC1 4- 1/202= HzO + Cln of the available data seemed sufficient to explain (1) (a) Giauque and Wiebe, (HCI), THISJOURNAL, 60, 101 the disagreement, but even after more accurate (1928); (b) Giauque and Johnston, (Hz), ibid., 10, 3221 (1928); experiments were performed discrepancies still (c) Giauque and Johnston, (a),ibid., 61, 2300 (1929); (d) Giauque, Physics 423(Hs),ibid., 68, 4816 (1930); (e) Giauque Friday and Overstreet, 5/13/2012 (HC1, remained. The investigations of Wiebe, Johns- CIz), ibid., 64, 1731 (1932). July, 1936 THEHEAT CAPACITY OF ICE 1145 made by Rossini2 and various equilibrium data, the basis of these comparisons a small correction to the including those relating to the reactions3 original calibration was readily made. Helium gas was introduced into the space between the HgO + Hz = Hg + Hz0 two walls by means of a German silver tube, A. A similar Hg + '/Oz HgO German silver tube was soldered by means of Wood's metal into the cap, B. The sample, C, was transferred indicated that the C,d In T for water did not through this tube into the calorimeter, and helium gas at give the correct entropy.JT That this was so be- one atmosphere pressure admitted. The German silver came a certainty when Giauque and Ashley4calcu- tube was then heated and removed from the cap, leaving the hole sealed with Wood's metal. After the measure- lated the entropy of gaseous water from its band ments on the full calorimeter had been completed, the spectrum and showed that an entropy discrep- calorimeter was heated to the melting point of the Wood's ancy of about one calorie per degree per mole metal (72°C.) and the water completely pumped out with- existed. They presumed this to be due to false out dismantling the apparatus. The heat capacity of the equilibrium in ice at low temperatures. empty calorimeter was thenI measured. The remainder of the Water is a substance of such importance that heat capacity apparatus, we considered further experimental investigation the method of making the to be desirable not only to check the above dis- measurements and calcula- crepancy but especially to see whether slow cool- tions, and accuracy consid- ing or other conditions favorable to the attain- erations were similar to those previously de- ment of equilibrium could alter the experimental scribed.'".' result. Purification of Water,- Apparatus.-In order to prevent strains in the resistance Distilled water from the thermometer when the water was frozen a double-walled laboratory still was trans- calorimeter, Fig. 1, was constructed. The outside wall ferred into the vacuum- was of copper, 0.5 mm. thick, 4.4 cm. 0.d., and 9 cm. long. tight purification apparatus The inside copper wall, 0.5 mm. thick, was tapered, being constructed from Pyrex 3.8 cm. 0.d. at the bottom and 4.0 cm. 0. d. at the upper glass. The apparatus was end. The top of the inner container was made from a thin evacuated to remove dis- copper sheet, 0.2 mm. thick, which prevented the trans- solved gases, and flushed mission of strains to the resistance thermometer. The out several times with he- neck for filling the calorimeter was in the center of this lium gas. The water was sheet. A series of thin circular slotted vanes of copper distilled into a receiving were soldered to the inner container, and the assembly bulb, the first fraction being forced inside the outer tube. A heavy copper plate, 1 discarded The calorimeter mm. thick inside the inner wall, and 2 mm. thick between had previously been at- the walls, served as the bottom of both tubes. The ther- tached to the purification mocouple was soldered into tube D by means of Rose's system and evacuated. metal. When sufficient water had D A resistance thermometer-heater of No. 40 double silk collected in the receiving Fig. 1.-Calorimeter covered gold wire containing about 0.1% silver was wound bulb, it was transferred into on the outside of the calorimeter. The resistance was 310 the calorimeter. Helium gas at one atmosphere pressure ohms at 290°K. and dropped to about 17 ohms at G0K. was admitted to the calorimeter which was then sealed off The resistance thermometer was calibrated during the as described above. measurements by means of copper-constantan thermo- A series of short heat capacity measurements were made couple No. 16 which had been compared with a hydrogen in the temperature region immediately below the melting gas thermometer.6 However one of the five parallel con- point in order to determine the pre-melting effect due to stantan wires in the thermocouple had accidentally been liquid-soluble solid-insoluble impurity. From these meas- broken since the original calibration. This wire was dis- urements it was calculated that the mole fraction of im- carded and after the completion of the measurements the purity was three parts in a million, thermocouple was compared with the oxygen and hydro- The Heat Capacity of Ice.-The results of the gen vapor pressure thermometers.6b The thermocouple was also checked against the melting point, 54.39"K. heat capacity measurements are given in Table I. and higher transition point, 43.76'K., of oxygen.'' On The data are shown in Fig. 2.

(2) (a) (HzO), Bur. Standards J. Research, 6, 1 (1931); (b) (HCI), In the calculations one 15O calorie was taken ibid.,9, 883 (1932). as equal to 4.1832 international joules. The (3) See summary by Eastman, Circular 6125, U. S. Dept. of Comm., Bur. of Mines (1929). calorimeter contained 72.348 g. of ice. (4) Giauque and Ashley, Phys. Rev., 4S, 81 (1933). In order to allow time for the establishment of Physics(5) (a) Giauque, 423 Buffington and Schulze, THISJOURXAL, 49, 2343 Friday 5/13/2012 (1927): (h) Giauque, Johnston and Kelley, ibid., 49, 2867 (1927). an equilibrium state in the solid, the ice was cooled 1116 W. F. GIAUQUEAND J. W. STOUT Vol. 5s very slowly. The following are temperatures 43.96 4.469 1.641 i1 reached at various times after the ice was frozen : 48.52 4.571 1.837 i1 0 hours, 273.1'; 12 hours, 246'; 37 hours, 203'; 52.98 4.361 2.014 i1 57.66 5.041 2.203 11 GO hours, 180'; 84 hours, 168'; 92 hours, 156'; 62.63 5.228 2.418 11 108 hours, 105'; 120 hours, 91'; 156 hours, 90'. G7.83 4.910 2.612 11 The sample was then cooled from 90 to 68' in 70.61 5.403 2.723 1 about three hours and the heat capacity measure- 73.01 5.737 2.821 11 ments of series I taken. Next the calorimeter 75.60 4 I 638 2.922 1 78.51 4.991 3.016 11 was cooled to the temperatures of liquid hydrogen 79.98 4.133 3.070 1 at the following rate: 0 hours, 85'; 1.5 hours, 81.44 5.538 3.115 111 72'; 2 hours, 60'; 2.5 hours, 56'; 7 hours, 39'; 82.42 4.860 3.163 I v (liquid hydrogen evaporated) 17.5 hours, 49' ; 83.72 5.438 3.191 11 22 hours, 50' (more liquid hydrogen added) ; 23 83.94 3.765 3.199 1 86.66 4.893 3.286 111 hours, 41'; 27 hours, 13'. The measurements of 87.25 4.756 3.336 IV series I1 were then made. 89.20 5.557 3,389 11 91.32 4.394 3.488 111 I 91.93 4.651 3.532 IV I---- 94.93 5.233 3.649 11 95.85 4.649 3.660 111 97.37 6.234 3.724 IV 99.57 4.778 3.814 11 100.69 4.980 3.832 111 104.69 5.497 3.985 11 110.13 5.373 4.136 11 115.84 6.031 4.315 11 121.74 5.908 4.489 11 127.54 5.813 4.fi55 11 133.50 6.005 4.808 11 I 139.48 5.952 4,978 11 01 / I I I 145.43 5.928 5.135 11 0 80 160 240 151.43 6.240 5.306 11 Temperature, OK. 157.48 5.837 5.466 11 Fig. 2.-Heat capacity in calories per degree per mole 163.52 5.851 5.G63 11 of ice. 169.42 5.908 5.842 11 During this series of measurements which ex- 175.36 5.996 (i,007 11 181.25 5.678 6.185 11 tended from 15'K. to the melting point, and 187.20 5.983 6.359 11 covered a period of eighty hours, the calorimeter 192.96 5.658 A. 530 11 was under constant observation. To make cer- 199.11 6.133 6.710 11 tain that no unusual thermal situation was pres- 205.32 6.309 6.935 11 ent in the solid near the melting point, the heat 211.56 6.554 7.119 11 217.97 6.200 7.326 11 of fusion was determined at the end of the above 224.36 6.935 7.519 11 series of measurements. The value obtained, 230.08 6.068 7.711 11 1436 cal./mole, agrees well with that which has 236.19 6.101 7.887 11 been chosen for the entropy calculation. 242.40 6.795 8.048 11 249.31 6.903 8.295 11 TABLEI 256.17 6.591 8.526 11 HEATCAPACITY OF ICE 262.81 6.303 8.732 11 (Molecular weight, 18.0156) OOC. = 273.10"K. 267.77 4,465 8.909 11 T,OR. AT C, cal./deg./mole Series 16.43 1.403 0.303 I1 In the heat capacity measurements between 18.37 1.729 .410 I1 85 and lOO'K., the attainment of temperature 20.78 2.964 .528 I1 equilibrium in the solid was much less rapid than 24.20 3.815 .700 I1 at other temperatures. This observation is of 28.05 3.596 .883 I1 considerable interest and some of its implications 31.64 3.578 1.OG5 I1 Physics35.46 423 4 073 1.251 I1 will be discussed below. Friday 5/13/2012 39 62 4 242 1.440 IT 'fo study possible effects due to rapid cooling. July, 1936 THEHEAT CAPACITY OF ICE 1147 the sample was cooled from 273.1 to 90' in four Professor Sirnon12informs us that, of the values hours, The heat capacities in series I11 were listed in the "Handbuch der Physik," only the then measured. Slow equilibrium was again en- one at 10'R. is based on the results of his own countered] and the heat capacities were not sig- measurements. The heat capacities at higher nificantly different from those previously ob- temperatures were calculated from the work of tained. earlier observers. The experimental results of To determine whether the ice was being trans- Simon1$are in excellent agreement with the con- formed from an unstable to an equilibrium condi- tinuation of the curve drawn through our measure- tion at an appreciable rate, measurements ex- ments. We have included in Table I1 a value for tending over several hours each were made of the the heat capacity of ice at 10'K. picked from a rate of temperature drift at 60, 72, 78 and 85'K. smooth curve through Simon's data. After correcting for the known heat interchange TABLEI1 with the surroundings, the rate of evolution of HEATCAPACITY OF ICE heat due to internal changes was zero within (Molecular Weight, 18.0156) 0°C. = 273.10'K. the limits of experimental error. An amount Values taken from smooth curve through observations of 0.0008 calorie per mole per minute could CP, cal./ Deviations previous results-This research, % have been detected. deg./ Nernst Pollitzer D. and M. and B. and T,"K. mole 1910 1913 0. 1915 W. 1925 M. 1830 The calorimeter was allowed to stand for four 10'8 0,066 days at temperatures between 60 and 80' before 20 .490 -13.3 taking the measurements in series IV. Again 30 .984 - 0.8 the measurements were not significantly different 40 1.466 f13.1 from those previously obtained. 50 1.896 + 4.1 60 2.304 + 2.8 In Table I1 are listed the values of the heat ca- 70 2.701 + 8.5 pacity at even temperatures as read off a smooth 80 3.075 + 4.4 + 6.6 curve through the observations. These values 90 3.448 + 1.3 - 0.4 -29.6 are compared with those of previous investigators. 100 3.796 - 0.6 -22.4 A review of the measurements prior to 1913 has 110 4.130 - 1.9 -16.8 120 4.434 - 2.3 -11.8 been given by Dickinson and Osborne.'j Of these 130 4.728 - 2.4 - 7.6 the measurements of Nernst and co-workers are 140 4.993 - 1.8 - 3.8 most important. The values used for comparison 150 5.265 - 1.5 - 0.8 in Table I1 are calculated from an equation pro- 160 5.550 - 1.4 + 1.3 posed by Nernst, Koref and Lindemann? as repre- 170 5.845 - 1.5 + 2.5 180 6.142 - 1.5 + 3.4 senting these data. Pollitzers made measure- 190 6.438 - 1.5 + 3.9 ments between 20 and 90'K. His measurements 200 6.744 - 1.5 + 3.9 -2.1 were plotted and a smooth curve drawn through 210 7.073 - 1.7 + 3.3 -0.9 them to give the values compared in Table 11. 220 7.391 - 1.7 f 2.6 - .I The very accurate work of Dickinson and Os- 230 7.701 - 1.3 -0.5 + 1.9 + .3 240 8.013 - 0.4 - .2 + 0.9 + .2 borne6between and -4O'C. has been calculated 0 250 8.326 + 1.3 + .1 - 0.3 - .4 from an equation proposed by these authors. 260 8.642 + 6.1 + .3 - 1.6 -1.3 Maass and Waldbauerg and Barnes and Maasslo 270 8.960 f43.8 + .5 - 3.2 -2.6 have measured the total change in heat content The Entropy of Water.-Values of the heat from various low temperatures to 25'C., and of fusion, heat capacity of the liquid and heat of from the results have proposed equations for the (12) Simon, personal communication. Measurements of the heat specific heat of ice. Simon1' lists values for the capacity of ice between Q and 13OK. were made in 1923 by Simon, of from 1O0K. but the results have not as yet been published. Since these values heat capacity ice to the melting are of considerable interest in connection with the present investiga- point. tion, we are, with Professor Simon's kind permission, presenting them here. (6) Dickinson and Osborne, Bull. U.S. Bur. Standards, l2, 49 (1915). HEAT CAPACITYOF ICE BETWEEN9 AND 13'K. MEABUREMENTS (7) Nernst, Koref and tindemann. Silebcr. Berlin Akad. Wiss., OF F. SIMON 247 (1910). T,OK. 9.47 9.88- 10.46 11.35 ll.5j0 12 10 12.85 (8) Pollitzer, 2. Elcktrochcm., 19, 513 (1913). C, cal./deg. j mole 0.056 0.063 o 075 0.096 0.102 o 118 n 141 Physics(9) Maass 423and Waldbauer, THISJOURNAL, 47, 1 (1925). Friday 5/13/2012 (10) Barnes and Maass, Can. J. Research, 3, 205 (1930). a 0bt:iined in an independent second experiment. (1 I) Simon, "Hunilhlich der Physik," Vol. X, 1926, p, 363. (1.7) Ctrlcilluted fromdataof Simon, see Ref. 12. 114s W. F. GIAUQUEAND J. W. STOUT lTol.58

vaporization of water are so accurately known that IO’K. was calculated by means of the Debye further investigation was unnecessary. equation, using hv/k = 192. A summary of the The heat of fusion of ice has been accurately entropy calculation is given in Table 111. determined by a number of workers. The measurements prior to 1913 have been critically sum- TABLErrr marized by Dickinson, Harper and Osborne,14who CALCULATIONOF ENTROPYOF U~ATER also made a number of measurements of the heat O-1O0K., Debye function hv/k = 192 0.022 of fusion both by an electrical method and by the 10-273. 10°K.,graphical 9.081 method of mixtures. Dickinson and Osborne6 Fusion 1435.7/273.10 ,5,257 273.10-298.1OGK.,graphical 1 ,580 measured the heat of fusion in an aneroid calo- Vaporization 10499/298.10 35,220 rimeter, using electrical heating. The measure- Correction for gas imperfection O,OfL? ments in which energy was introduced electrically Compression R In 2.3756/760 -6.88Ci were recalculated by us on the basis of 1 int. Cal./dcg./~iiole44.28*0.05 joule = 4.1832 calories (15’). A weighted average of all the reported values yields 1435.7 cal./ The value of the entropy given in Table I11 mole with an estimated accuracy of *0.9 cal./ may be compared with that calculated from mole, for the heat of fusion. spectroscopic data. Giauque and Ashley4 util- Fiock16 has reviewed the measurements of the ized the preliminary molecular constants of water heat of vaporization of water and compared them as given by Mecke and Baumann20 to determine with the results of determinations at the Bureau the entropy of water. Later Gordon21 recaleu- of Standards16extending down to 50’. All meas- lated the thermodynamic quantities for water urements were converted into international joules. using the revised moments of inertia of Freuden- Of the data considered by Fiock, those of Grif- berg and Mecke.22 He obtained So~gB.l= 45.10 fiths, of Smith and of Henning contained meas- cal./deg./mole. The difference between the urements in the neighborhood of 25’C. Giving spectroscopic and calorimetric values is 0.82 equal weight to the result of each of the above cal./deg. /mole. three observers and to the value extrapolated The Problem of the False Equilibrium in from the Bureau of Standards measurements, Ice.-To account for the discrepancy between and taking 1 calorie (15’) = 4.1832 int. joules, the calorimetric and spectroscopic values for the we obtained an average value of 10,499 * 3 entropy of water Giauque and Ashley‘ offered an (av. dev.) calories/mole for the heat of vaporiza- explanation based on the assumption that the tion of water at 25’. ortho and para molecular states, which are known The “I.C.T.” values for the heat capacity of to exist in gaseous water, had persisted in the liquid waterI7 were plotted against the logarithm crystalline state at low temperatures. The situa- of the absolute temperature and integrated tion was assumed to be similar to that which ac- graphically to obtain the entropy between 0 and counts for the entropy discrepancy in the case of 25’. The value for the vapor pressure at 25’ was solid The ortho water was as- also obtained from the “I.C.T.”lS Using Berthe- sumed to have non-polar clockwise and counter lot’s equation of state and thermodynamics it clockwise rotations in ice, since the dielectric con- can be shown that the entropy correctionla to the stant of ice at low temperatures corresponds to ideal gas state is almost negligible in this case that of non-polar substances. This leads to a The critical constants18 used were Tc = 647.1°K. calculated discrepancy of 3/4RIn 2 = 1.03 cal./ and P, = 217.7 atm. deg. /mole. The entropy between 10 and 273.10’K. was We have had many interesting private discus- obtained by graphical integration of the measured sions with Professor Linus Pauling who has con- heat capacities. The entropy between 0 and sistently objected to the ortho-para explanation. (14) Dickinson, Harper and Osborne, Bull. U.S. Bur. Slandards, During the course of the present investigation 10, 235 (1014). Pauling2aoffered an alternative explanation based (15) Fiock, Bur. Standards J. &search, 6, 481 (1930). (16) (a) Osborne, Stimson and Fiock, ibid., 6, 411 (1930); (b) (20) Mecke and Baumann, (a) Nafuvwiss.. 20, 657 (1932); (b) Fiock and Ginnings, ibid., 8, 321 (1932). Phys. Z.,89, 833 (1932). (17) “International Critical Tables,” McGraw-Hill Book Co., (21) Gordon, J. Chem. Phys., 2,65 (1934). New York, Vol. V, 1926, p. 113. (22) Freudenberg and Mecke, Z.Phrsik, 81,465 (1933). Physics(18) “I. C. 423T.,” Vol. 111, p. 211. (23) Pauling, (a) personal communication; Friday (b) THIS 5/13/2012JOURNAL, SIB (19) “I. C. T.,”Vol. III,p. 248. 2680 (1935) July, 1936 THEHEAT CAPACITY OF ICE 1149

on the random orientation of hydrogen bonds in and found no appreciable heat capacity in this ice. Crystal structure investigations have shown region. that each oxygen atom is surrounded by four One of the purposes of the present work was to equivalent oxygen atoms. Although the posi- investigate the possibility of more complete at- tions of the hydrogen atoms have not been de- tainment of equilibrium. We have mentioned termined experimentally it seems reasonable to above that various experiments, in which ice was assume that they are located on lines joining ad- cooled slowly or rapidly to low temperatures, or jacent oxygens. Since the oxygen-dxygen dis- was allowed to stand for long periods of time at tance is considerably greater than twice the sepa- low temperatures, resulted in heat capacities ration which is characteristic of an oxygen-hydro- which were not appreciably different in the vari- gen linkage, it has been assumed by Bernal and ous series of measurements. At temperatures Fowlerz4that each oxygen is joined to two close between 85 and 100’K. the attainment of thermal and two distant hydrogen atoms. At higher tem- equilibrium in the solid was very much less rapid peratures the bond directions of the close or dis- than at other temperatures. For this reason the tant hydrogen atoms of a given oxygen are a mat- heat capacity measurements in this region are ter of chance. We quote from the paper of Ber- somewhat less accurate than the others. This nal and Fowler. “Therefore it is quite conceivable slow equilibrium presumably is due to the initial and even likely that at temperatures just below stages of excitation of some new degrees of free- the melting point the molecular arrangement is dom. From the value of the heat capacity this still partially or even largely irregular, though pre- evidently is connected with motion of the hydro- serving at every point tetrahedral coordination gen atoms. At temperatures below the region and balanced dipoles. In that case ice would be of slow equilibrium the dielectric constant2*is of crystalline only in the position of its molecules but the order of magnitude characteristic of a non- glass-like in their orientation. Such a hypothesis polar substance. At higher temperatures the may be still necessary to explain the dielectric dielectric constant rises rapidly and the orienta- constant and the absence of pyroelectricity.” tion time of the dipole decreases. From this it Pauling assumes further that when ice is cooled would appear that the new degrees of freedom re- to low temperatures, it fails to attain the ordered ferred to above are associated with the dipole ori- arrangement which would correspond to zero en- entation mechanism. tropy. He shows that the discrepancy corre- From the temperature drift experiments at sponding to the above lack of order would be R 60, 72, 78 and 85OK. it appears that no great In 6/4 = 0.806 cal./deg./mole. This is in very change in the heat content would be expected if close agreement with the experimentally deter- ice were kept at these temperatures for a con- mined discrepancy of 0.82 * 0.05 cal./deg./mole. siderable period of time. However, it should be pointed out that in two We may mention in passing that low tempera- previous cases, carbon monoxidez5 and NNOZ6 ture heat capacity measurements on some of the where random molecular orientations led to en- high pressure forms of ice would undoubtedly lead tropy discrepancies, the experimental discrep- to correct values for the entropy of water. ancy was a few tenths of a unit lower than the It is perhaps worth noting that while the theory calculated value due to partial attainment of an of random bond orientation would lead to the ordered state. If this were the case in ice, the same discrepancy, R In 3/2 = 0.806 cal./deg./ ortho-para explanation might be correct. How- mole in both hydrogen and deuterium oxides, the ever, we consider that the explanation advanced theory of molecular rotation in ortho states of by Pauling is the more plausible as well as being water would lead to different values for the cal- in better agreement with the experimental re- culated discrepancy. For hydrogen oxide the sults. value is 3/4 R In 2 = 1.033, for deuterium oxide, It is of interest to note that MacDougall and 1/3 R In 2 = 0.459 cal./deg./mole. Although Giauquez7 investigated ice from 0.2 to 4’K. we believe the latter theory to be less plausible it will be of interest to make the comparison when (24) Bernal and Fowler, J. Chcm. Phys., 1, 515 (1933). (25) Clayton and Giauque, Tars JOURNAL, 54,2610 (1932). measurements on deuterium oxide become avail- (26) (8) Clusius, Z. Elektrochcm., 40, 99 (1934); (b) Blue and able. PhysicsGiauque, THIS 423JOURNAL, 17,991 (1935). Friday 5/13/2012 (27) hbacDougak1 and Giauque, ibid., IS, 1032 (1936). (28) “I. C. T.,”Vol VI, page 78. Activity: Concept diagram (10 min) Students list concepts we’ve learned, and we form concept diagram on board.

Physics 423 109 Friday 5/13/2012 Survey and post-test

Question 1 How would you perform a measurement of the quantity

∂S (20.59) ∂τ T for a rubber band without making a measurement of heat transfer (e.g. if you don’t have any 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 = (20.61)

What is the internal energy U at temperature T ?

Physics 423 110 Friday 5/13/2012 Question 3 A curve has been traced out on the z-y graph below, and has been labeled Path A. Consider the integral I I1 ≡ zdy (20.62) A,clockwise where the integral is taken around Path A starting at point P , proceeding clockwise until reaching point P again.

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

z Path A

0 y

Refer again to the graph in part (a). We deﬁne H as a function of the independent variables z and y; i.e. H = H(z, y). Consider the integral I ~ I2 ≡ ∇H · d~r (20.63) A,clockwise where the integral is taken around Path A starting at point P , proceeding clockwise until reaching point P again. (Note: d~r = dyyˆ + dzzˆ)

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

Physics 423 111 Friday 5/13/2012 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 ice remaining and ﬁnal temperature of water?

Did you like this lab? Do you have any suggestions as to how we could improve it?

Question 6 Lab 3 What did you learn from Lab 3, in which we measured the tension of a rubber band?

Did you like this lab? Do you have any suggestions as to how we could improve it?

Physics 423 112 Friday 5/13/2012 Question 7 Class as a whole Do you like Thermal Physics? Why or why not?

Physics 423 113 Friday 5/13/2012 21 Monday: Final exam

Current board:

∂c ∂a ∂b a = − −βi ∂b ∂c e c ∆Ssystem+∆Ssurroundings ≥ 0 dG = −SdT + V dp Pi = ∂a b Z

Physics 423 114 Monday 5/16/2012 Final exam

Problem 1 Masses on a piston An ideal gas is contained in a cylinder with a tightly ﬁtting piston. Several small masses are on the piston. (Neglect friction between the piston and cylinder walls.) The cylinder is placed in an insulating jacket, and a large number of masses are then added to the piston. Tell whether the pressure, temperature, and volume of the gas will increase, decrease, or remain 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 independent variables T and p, i.e., it can be written as G(T, p). The total diﬀerential of G can be written 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 is equal to ∂S

∂p T

Physics 423 116 Name Problem 3 Two processes This Pressure-Volume (P -V ) diagram represents a system consisting a ﬁxed amount of ideal gas that undergoes two diﬀerent quasistatic processes in going from state A to state B:

Process #1 State B Pressure

Process #2 State A

0 Volume

[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? Explain your answer.

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

Phys 423 117 Name Problem 4 Hanging Chain The upper end of a hanging chain is fixed while the lower end is attached to a mass M. The (massless) links of the chain are ellipses with major axes l + a and minor axes l − a, and can place themselves only with either the major axis or the minor axis vertical. The figure below shows a four-link chain in which the major axes of the first and fourth links and the minor axes of the second and third links are vertical. Assume that 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?

Physics 423 118 Name Problem 5 Insulated room An object is placed in a thermally insulated room that contains air. The object and the air in the room are initially at diﬀerent temperatures. The object and the air in the room are allowed to exchange energy with each other, but the air in 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, remain the 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 your answer.

c) During this process, does the entropy of the object plus the entropy of the air in the room [Sobject + Sair] increase, decrease, remain the same, or is this not determinable with the given information? Explain your answer.

d) During this process, does the entropy of the universe [Suniverse] increase, decrease, remain the same, or is this not determinable with the given information? Explain your answer.

Phys 423 119 Name Problem 6 Soap bubble Consider the reversible expansion (stretching) of a film of soapy water. The film’s internal energy is proportional to its area—this is why bubbles are spherical, since this minimizes their surface area. The infinitesimal of work done by changing the area of a film is given by the surface tension σ: dW¯ = σdA (21.1)

a) Show how the surface tension is related to the internal energy, and to the Helmholtz free 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 free energy 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 given bubble, and ﬁnds that over a wide range of temperatures and areas,

∂σ = λ(T − T3) (21.2) ∂T A

−4 −2 −2 where λ is a constant with the value 2 × 10 Jm K , and T3 has the value of 600 K. What is the change in entropy when this bubble is isothermally expanded from an area of 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. If not, explain what additional information you would need.

Physics 423 120 Name Tentative schedule

Day Topic/subtopic Reading Homework due 1 What kind of beast is it? Monday Thermodynamic variables Chapter 1 Thermodynamic equilibrium 4/18 Equations of state

2 Partial derivatives as change of variables Tuesday Total diﬀerential 4/19 Exact vs. inexact diﬀerentials Chain rules 3 Wednesday Homework 1 4/20

4 Mixed partial derivatives Thursday Monotonicity and invertibility 4/21 Legendre transform 5 Friday Lagrange multipliers for minimization Homework 2 4/22 6 Monday Lab 1: Heat and Temperature 4/25 Dulong-Petit Law

First and Second Laws System and surroundings 7 First Law Tuesday Second Law and Entropy 4/26 Fast and slow The thermodynamic identity 5.1.2.2 Heat capacity 8 Wednesday Second Law lab Lab 1 4/27 9 Thursday Heat and work 4/28 Work

10 Engines and Fridges Lab 2 Friday Other views of the Second Law Homework 3 4/29 Carnot eﬃciency

Phys 423 121 Name 11 Monotonicity and invertibility Monday Thermodynamic potentials Chapter 7 5/2 Legendre transform Maxwell relations 12 Tuesday Lab 2: rubber band Prelab 3 5/3 13 Wednesday Homework 4 5/4 14 Thursday Thermodynamics practice 5/5 15 Friday Thermodynamics practice Lab 3 5/6 16 Monday Statistical approach 5/9 Fairness function Chapter 6

17 Optimizing the fairness Tuesday Least bias lagrangian 5/10 Weighted averages Probabilities of microstates Chapter 11 18 From statistics to thermodynamics Wednesday Homework 5 5/11 Thermodynamic properties from the Boltzmann factor

19 Statistical mechanics of air Thursday Quantum spectra 5/12 Diatomic gas

20 Diatomic gas wrapup Friday Third law Homework 6 5/13 Ice rules 21 Monday Final exam 5/16

Physics 423 122 Name Index

adiabatic, 30 partition function, 76

Boltzmann factor, 76 quasistatic, 30 carnot efficiency, 44 reversible, 30 efficiency, 43 sackur-tetrode equation,4 eigenstates, 71 spontaneous, 30 enthalpy, 52, 53 state function,2 entropy, 30, 31, 53 state variable,2 equation of state,3 surroundings, 28 exact differential,7 system, 28 fairness function, 72 total differential,5

Gibbs free energy, 52, 53 work, 40 working substance, 43 heat, 27 heat capacity, 31 heat engine, 42 heat sinks, 42 Helmholtz free energy, 52, 53 helmholtz free energy, 21 inexact diﬀerential,7 internal energy, 30, 53 irreversible, 30

Lagrange multiplier, 75 Lagrangian, 75 Law of thermodynamics First, 29 Second, 29, 42 Clausius formulation, 42 Kelvin formulation, 42 Third, 94

Phys 423 123 Name