David Ronis McGill University · 2019-10-24 · Josiah Willard Gibbs Ludwig Boltzmann Chemistry...

Josiah Willard Gibbs Ludwig Boltzmann

Chemistry 365:Statistical Thermodynamics

David Ronis

McGill University

Albert Einstein Peter Debye

Winter Term, 2019

Chemistry 365

Statistical Thermodynamics

David Ronis

McGill University

© 2019

All rights reserved. In accordance with Canadian Copyright Law,

reproduction of this material, in whole or in part, without the prior

written consent the author is strictly prohibitied.

Last modified on 24 October 2019

Winter Term, 2019

Table of Contents

1. General Information . . . . . . . . . . . . . . . . . . . . 3

1.1. Summary of Key Results . . . . . . . . . . . . . . . . . . 6

2. Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . 7

3. Stirling´s Formula . . . . . . . . . . . . . . . . . . . . 10

4. Some Notes on Assemblies and Distribution Functions . . . . . . . . . 13

4.3. Summary: Canonical Ensemble . . . . . . . . . . . . . . . 16

4.4. Summary: Grand Canonical Ensemble . . . . . . . . . . . . . 17

5. Ideal Bose & Fermi Gases . . . . . . . . . . . . . . . . . . 19

5.1. Introduction and General Considerations . . . . . . . . . . . . . 19

5.2. Results for Particles in a Cubic Box . . . . . . . . . . . . . . 22

5.2.1. The Case For Boltzmann Statistics . . . . . . . . . . . . . . 25

5.3. Results Specific to Fermions or Bosons . . . . . . . . . . . . . 29

6. Normal Mode Analysis . . . . . . . . . . . . . . . . . . . 37

6.1. Quantum Mechanical Treatment . . . . . . . . . . . . . . . 37

6.2. Normal Modes in Classical Mechanics . . . . . . . . . . . . . 39

6.3. Force Constant Calculations . . . . . . . . . . . . . . . . . 39

6.4. Normal Modes in Crystals . . . . . . . . . . . . . . . . . 41

6.5. Normal Modes in Crystals: An Example . . . . . . . . . . . . . 43

7. Gibbs-Duhem Relations . . . . . . . . . . . . . . . . . . . 48

8. Chemical Equilibrium . . . . . . . . . . . . . . . . . . . 50

9. Dense Gases: Virial Coefficients . . . . . . . . . . . . . . . . 54

10. Summary of Key Results . . . . . . . . . . . . . . . . . . 62

11. Problem Sets . . . . . . . . . . . . . . . . . . . . . . 63

11.1. Problem Set 1 . . . . . . . . . . . . . . . . . . . . . 63

11.2. Problem Set 2 . . . . . . . . . . . . . . . . . . . . . 65

11.3. Problem Set 3 . . . . . . . . . . . . . . . . . . . . . 66

11.4. Problem Set 4 . . . . . . . . . . . . . . . . . . . . . 68

11.5. Problem Set 5 . . . . . . . . . . . . . . . . . . . . . 70

11.6. Problem Set 6 . . . . . . . . . . . . . . . . . . . . . 72

12. Past Quizzes . . . . . . . . . . . . . . . . . . . . . . 75

12.1. Quiz I (2015) . . . . . . . . . . . . . . . . . . . . . 75

12.2. Quiz I (2016) . . . . . . . . . . . . . . . . . . . . . 78

12.3. Quiz I (2018) . . . . . . . . . . . . . . . . . . . . . 82

12.4. Quiz I (2019) . . . . . . . . . . . . . . . . . . . . . 84

12.5. Quiz II (2015) . . . . . . . . . . . . . . . . . . . . . 86

12.6. Quiz II (2016) . . . . . . . . . . . . . . . . . . . . . 90

12.7. Quiz II (2018) . . . . . . . . . . . . . . . . . . . . . 94

13. Past Final Exams . . . . . . . . . . . . . . . . . . . . 98

13.1. Final Exam (2015) . . . . . . . . . . . . . . . . . . . 99

13.2. Final Exam (2016) . . . . . . . . . . . . . . . . . . . 102

13.3. Final Exam (2018) . . . . . . . . . . . . . . . . . . . 105

13.4. Final Exam (2019) . . . . . . . . . . . . . . . . . . . 109

Winter Term, 2019

General Information -3- Chemistry 365

1. General Information

Chemistry 365: Statistical Thermodynamics

Molecular basis of thermodynamics with applications to ideal gases and simple solids. Topics

to be covered will include: calculation of thermodynamic functions, chemical equilibrium

constants, Einstein and Debye models of solids, absolute reaction rate theory, Debye-Huckel

theory of strong electrolytes.

Prerequisites: CHEM 345 or equivalent, or permission of instructor.

Professor: David Ronis

Office: Otto Maass Room 426

Office Hours: Whenever you want. Just drop by my office. If I’m not there, send me an e-mailand we’ll set up a time.

E-mail: [email protected](Help my mail program route your e-mail; Please put CHEM 365 somewhere in the subject.)

Grader: Christopher Hennecker

Lectures: Monday, Wednesday and Friday 11:35 A.M. - 12:25 P.M.,Location: Otto Maass 217

Course Web Site: https://ronispc.chem.mcgill.ca/ronis/chem365Case sensitive username and password are needed for full access.

This is 2 credit course. We will meet 3 times per week for 2/3 of the term. The last class will begiven on Wednesday, March 13, 2019 (for a total of 26 lectures). The first mid-term exam willtake place on

Tuesday, February 5, 2019, from 6-9 P.M., in Otto Maass 112.

The second mid-term exam will be given on

Wednesday, March 20, 2019, from 6-9 P.M., in Otto Maass 112.

Students may opt to sit a reguarly scheduled final exam, and if they do, the course grade will becomputed as detailed below in scheme II.

Winter Term, 2019

Chemistry 365 -4- General Information

GRADING SCHEME

The grade in this course will be computed in one of the following two ways:

CHEM 365 Grading Schemes

I II

Homework 10% 10%Midterm I 35% 25%Midterm II 55% 25%Final (optional) 0% 40%

Note that if you decide to take the final exam, I will use grading scheme II, no matter what theoutcome.

TEXT

T. L. Hill, An Introduction to Statistical Thermodynamics (Dover Publications)

SUPPLEMENTARY TEXTS

1. L.K. Nash, Elements of Statistical Thermodynamics

2. James R. Barrante, Applied Mathematics for Physical Chemistry, 3rd edition. This isn’t astatistical thermodynamics book, rather it reviews things taught in Cal. 1-3, using physicalchemistry examples. A good reference or refresher.

Notes:

• McGill University values academic integrity. Therefore, all students must understand themeaning and consequences of cheating, plagiarism and other academic offenses under theCode of Student Conduct and Disciplinary Procedures (see www.mcgill.ca/students/srr/honest/for more information). (Approved by Senate on 29 January 2003.)

• In accord with McGill University’s Charter of Students’ Rights, students in this course havethe right to submit in English or in French any written work that is to be graded. (approved bySenate on 21 January 2009)

• In the event of extraordinary circumstances beyond the University’s control, the content and/orevaluation scheme in this course is subject to change.

Winter Term, 2019

General Information -5- Chemistry 365

Tentative Outline: General Information, Winter Term, 2019

Lecture Topic Chapter(s) in Hill

Lecture 1 Statistical Mechanics: Why? and Basic Assumptions 1Lecture 2 The Canonical Ensemble: Partition Functions 2Lecture 3 " 2

Lecture 4 Microscopic Basis for Thermodynamics 2Lecture 5 " 2Lecture 6 Other Ensembles 2

Lecture 7 Bose-Einstein, Fermi-Dirac and Boltzmann Statistics 3, 22Lecture 8 " 3, 22Lecture 9 " 3, 22

Classical Statistical Mechanics:Maxwell-Boltzmann Velocity Distribution

Lecture 10 6

Lecture 11 Ideal Monatomic Gases 4Lecture 12 Vibration and Rotation in Diatomic Molecules 8

Lecture 13 " 8Lecture 14 Polyatomic Gases: Rotation 9

1st Midterm Exam: Tuesday, February 5, 2019, from 6-9 P.M., in Otto Maass 112.

Lecture 15 Polyatomic Gases: Vibration

Lecture 16 Polyatomic Gases: 9Chemical Equilibrium Revisited: 10

Equilibrium Constant Calculations 10Lecture 17

Lecture 18 Equilibrium Constant Calculations (continued) 10

Lecture 19 Ideal Crystals: Einstein & Debye Solids 5Lecture 20 " 5Lecture 21 Activated Rate Theory 11

Lecture 22 " 11Lecture 23 Real Gases: The Second Virial Coefficient 15Lecture 24 " 15

Lecture 25 Ionic Solutions: Debye-Huckel Theory 18Lecture 26 " 182nd Midterm Exam: Wednesday, March 20, 2019, from 6-9 P.M., in Otto Maass 112

Winter Term, 2019

Chemistry 365 -6- General Information

1.1. Summary of Key Results

Some Key Results for CHEM 365

Rotation

Diatomic/linear PolyatomicTranslation Vibration ElectronicType of Motion

h2(n2x + n2

y + n2z)

8mL2

h2

8π 2 IJ(J + 1) ? h− ω (n + 1/2) ε j

Energy levels

Degeneracy 1 (2J + 1) ? 1 ω j

0 Θrot ≡h2

8π 2 IkB

Θαrot ≡

h2

8π 2 Iα kB

Θvib ≡h− ωkB

∆ε /kB

Characteristic tem-

perature

2π mkBT

h2

3/2

V

N

N !

Molecular partition

function (except for

translation)

J=0Σ (2J + 1)e−Θrot J(J+1) /T

(heteronuclear)

? e−Θvib/2T

1 − e−Θvib/Tj

Σω je−β ε j ≈ ω0e−β ε0

"T

σ Θrot

π 1/2

σ ABCΠ

T

Θαrot

1/2T

Θvib

N/AHigh temperature

approximation to

partition function

Aα = −kBT ln(qα )Helmholtz free

energy per molecule

kBT ln(ρΛ3/e)

Sα =< Eα > −Aα

T

Entropy per mole-

cule

−kB ln(ρΛ3/e5/2)

3

2kBT kBT

3

2kBT kBT N/A

High temperature

energy per molecule

NkBT

V0 0 0 0

Contribution to pres-

sure

kBT ln[ρΛ3] −kBT ln[T /(σ Θrot)] −kBT ln

π 1/2

σ ABCΠ

T

Θαrot

1/2

h− ω2

+ kBT ln(1 − e−Θvib/T ) −kBT ln(qelectronic)Chemical potential

2π mkBT

h2

3/2T

σ Θrot

π 1/2

σ ABCΠ

T

Θαrot

1/2e−Θvib/2T

1 − e−Θvib/Tj

Σω je−β ε j

Equilibrium constant

contribution

Winter Term, 2019

Lagrange Multipliers -7- Chemistry 365

2. Lagrange Multipliers

Lagrange’s Method of Undetermined Multipliers

Here’s a heuristic proof of the validity of Lagrange’s method of undetermined multipliers(with apologies to the mathematicians for sloppiness about the requisite conditions for differen-tiability, etc.). First consider some function of n variables, F(x1, x2, . . . , xn), as you all know,extrema of F are found by solving the simultaneous set of equations:

Fi ≡

∂F

∂xi

x j≠i

= 0, i = 1, 2, . . . , n. (2.1)

In general, this will determine the function’s minima, maxima, and saddle (or inflection) points.Do you know why this works? Consider the Taylor series expansion for F for xi → x0

i + δ xi;i.e.,

δ F ≡ F − F (0) =n

i=1Σ Fiδ xi +

1

2

n

i, j=1Σ Fi, jδ xiδ x j+. . . , (2.2)

where Fi is given in Eq. (2.1) and

Fi, j ≡

∂2F

∂xi∂x j

. (2.3)

Since the linear term in Eq. (2.2) changes sign when δ xi → −δ xi , the only way we can have anextremum is to have the linear term vanish, no matter what we choose for the δ xi’s (other thanthem being small enough to be able to neglect the second order terms). Whether we have founda minimum, maximum or saddle point is determined by the sign of the second terms, but thiswon’t concern us here.

The preceding discussion is for the case where there are no constraints imposed on the varia-tions δ xi . Suppose there are; specifically, lets assume that only xi’s which satisfy

G( j)(x1, . . . , xn) = 0, for j = 1, . . . , m, (2.4)

where m is the number of constraints, are of interest. We can still use Eq. (2.2) to describe thechange in F near any point, although now, only those variations which are consistent with Eq.(2.4) must be considered. For small variations about a point which is consistent with the con-straints, we may approximately rewrite Eq. (2.4) as

n

i=1Σ G

( j)i δ xi = 0, for j = 1, . . . , m, (2.5)

where

Winter Term, 2019

Chemistry 365 -8- Lagrange Multipliers

G( j)i ≡

∂G( j)

∂xi

xk≠i

. (2.6)

Equation (2.5) has a nice geometric interpretation; namely, if we introduce n-dimensional vec-

tors, g( j) ≡ (G( j)1 , G

( j)2 , . . . , G( j)

n ), and δ x ≡ (δ x1, . . . , δ xn), then Eq. (2.5) becomes

δ x ⋅ g( j) = 0, (2.7)

and hence, the allowed variations in x are orthogonal to the vectors g( j), and are thus orthogonalto the subspace spanned by the g( j)′s. Similarly, we can represent the linear variation of F as

δ F = δ x ⋅ f, (2.8)

where f ≡ (F1, . . . , Fn). Unlike the unconstrained case, we cannot demand that δ F will vanishfor any δ x. Instead, it must vanish only for those δ x’s that are consistent with the constraints;i.e., those that satisfy Eq. (2.7). For this to happen, it is necessary and sufficient that f lie in thesubspace spanned by the g( j)′s; thus, we take the g( j)′s as a basis for this subspace and write

f = λ1g(1)+. . . +λ mg(m), (2.9)

where the λ j’s are called the undetermined multipliers, and are found by demanding that anysolutions to Eq. (2.9) also obey the m constraints. If you use the definitions given above, it’s rel-atively easy to show that Eq. (2.9) is equivalent to setting the partial derivatives of the function

F(x1, . . . , xn) −m

j=1Σ λ jG

( j)(x1, . . . , xn) (2.10)

to zero (where we treat the λ j’s as constants). This is known as Lagrange’s method of undeter-mined multipliers.

2.1. An Example

Let’s consider a 3 dimensional example. Let

F(x, y, z) = x2 + y2 + z2. (2.11)

With no constraints, Eq. (2.1) can be used to easily show that this function has a minimum atx = y = z = 0 (with value 0). First lets impose a single constraint, e.g.,

G(1)(x, y, z) = x − y − 1 = 0. (2.12)

According to Eq. (2.10), we should set partial derivatives of the function

x2 + y2 + z2 − λ1(x − y − 1) (2.13)

to zero. This gives

2x − λ1 = 0, 2y + λ1 = 0, and 2z = 0, (2.14)

Winter Term, 2019

Lagrange Multipliers -9- Chemistry 365

which has the solution

x =λ1

2, y = −

λ1

2, and z = 0. (2.15)

Returning to the constraint, Eq. (2.12), we see that only λ1 = 1 giv es a solution that satisfies theequation, and thus the constrained extremum is at

x =1

2, y = −

1

2, and z = 0, (2.15)

where F = 1/2. You should verify that first eliminating one of the variables using the constraintgives the same result.

Lets introduce a second constraint into our example, say

G(2)(x, y, z) = x − z = 0. (2.16)

Now we must consider partial derivatives of

x2 + y2 + z2 − λ1(x − y − 1) − λ2(x − z), (2.17)

or

2x − λ1 + λ2 = 0, 2y + λ1 = 0, and 2z + λ2 = 0, (2.18)

which gives

x =λ1 + λ2

2, y = −

λ1

2and z = −

λ2

2. (2.19)

The constraints must still be imposed, cf. Eqs. (2.12) and (16), and this gives

λ1 +λ2

2= 1 and

λ1

2+ λ2 = 0, (2.20)

which has the solution λ1 = 4/3 and λ2 = −2/3. Hence, using this in Eq. (2.19) shows that theconstrained extremum is now at x = z = 1/3, and y = −2/3, where F = 2/3. Again, check that thesame result is obtained when you use the constraints to eliminate two of the independent vari-ables.

Winter Term, 2019

Chemistry 365 -10- Stirling´s Formula

3. Stirling´s Formula

There is a simple way to obtain an asymptotic expansion for the factorial, n!, for largearguments. First define a new function, called the gamma-function, as

Γ(x + 1) ≡ x! ≡ ∫∞0

ds sxe−s = x x+1 ∫∞0

ds exp x[ln(s) − s], (3.1)

where the last equality follows by making the change of variables s → xs. The function Γ(x) iscalled the gamma function. It is easy to show, either by integrating by parts or by noting thatΓ(x + 1) = xΓ(x), that x! = n! when x = n, an integer.

The integrand appearing in Eq. (3.1) is shown in the following figure.

Fig. 3.1. The integrand in Eq. (3.1).

When x is large, the integrand is large and sharply peaked around the maximum. In this case, itis profitable to Taylor expand the integrand around the maximum of the argument to the expo-nential, i.e., about s = 1 in Eq. (3.1). Hence, we write

x[ln(s) − s] = −x − x∞

n=2Σ (−1)n(s − 1)n

n. (3.2)

If we only keep the n = 2 term in the sum in Eq. (3.2), it follows that the integrand in Eq. (3.1)decays like a Gaussian centered at s = 1, with width x−1/2. Remembering that x is large, we cannow extend the lower limit of integration in Eq. (3.1) to −∞ with exponentially small error.Finally, we make the change of variable s → 1 + ∆/x1/2 and rewrite Eq. (3.1) as

x! = x x+1/2e−x ∫∞−∞

d∆ exp−

1

2

∆2 −∞

n=3Σ (−1)n∆n

nx(n−2) / 2

. (3.3)

Winter Term, 2019

Stirling´s Formula -11- Chemistry 365

The terms in the sum are at least as small as x−1/2, and hence, to leading order can bedropped. The resulting integral is easily done, and hence,

x!∼(2π )1/2 x x+1/2e−x , (3.4a)

which implies that

ln(x!) ∼ x ln(x) − x + ln(2π x)1/2

. (3.4b)

Clearly, for x∼O(1023), the last logarithm is negligible, and the simple form of Stirling’s formulaln(x!)∼x[ln(x) − 1], is obtained. A comparison of the approximate and exact results is shown inFig. 3.2.

Fig. 3.2. Comparing Stirling’s approximations to ln(x!). The insetshows smaller x.

Finally, some of the numerical results used to make Fig. 3.2 are shown in the following table:

Table 3.1: Testing Stirling’s Approximation

ln(x!) x[ln(x) − 1] x[ln(x) − 1] + ln[(2π x)1/2](exact) Value ∆% Value ∆%

x

1 0 -1 ∞ -0.0810615 ∞5 4.78749 3.04719 36.4 4.77085 0.34810 15.1044 13.0259 13.8 15.0961 0.055250 148.478 145.601 1.94 148.476 1.12×10−3

100 363.739 360.517 0.886 363.739 2.29×10−4

500 2611.33 2607.3 0.154 2611.33 6.38×10−6

1000 5912.13 5907.76 0.074 5912.13 1.41×10−6

Winter Term, 2019

Chemistry 365 -12- Stirling´s Formula

With a little extra effort you can work out the next term in the expansion, by Taylorexpanding the exponential of the sum in Eq. (3.3) and keeping the next to leading order termswhich give nonvanishing integrals. If you were to carry out this procedure to arbitrarily largeorder, would you expect the resulting series to converge? Why?

Winter Term, 2019

Assemblies -13- Chemistry 365

4. Some Notes on Assemblies and Distribution Functions

4.1. Probabilities in Multiple Weakly Connected Canonical Assemblies

E

EE

EE

E

E

E

EEE

EE

EE

E E

E

E

E

E

EEE

E

EE E

E

E

E

A B

A A

A

A

A

B B

B B

B B

BBB

A B A

B

1 3 51

1

33

3

10

100 50

6

5

5644

22

10 66

12 75 6

Fig. 4.1. An assembly of A and B systems that can exchange energybetween themselves. The overall assembly is isolated from the sur-roundings. The numerical label denotes the quantum state assignedto the system. According to the figure, N A = 8 and NB = 12, whilenA

1 = 2, nA3 = nB

3 = 2, nB6 = 2. All the other states have ni = 1 or 0.

Thus, ΩA = 8! / (2! × 2!) = 1008 and ΩB = 12! / (2! × 2!) = 119750400.

Consider an assembly of systems of types A, B, ... as cartooned in Fig. 4.1. nαi is the num-

ber of systems of type α in state i and Nα is the number of systems of type α . For the canonicalassembly, the systems can exchange energy with each other, but otherwise are closed (specifi-cally, mass or number and volume aren’t exchanged). In addition, we do not allow A and B sys-tems to exchange with each other, e.g., referring to Fig. 4.1, the upper left system is always A,the upper right is always B, etc.. From the point of view of any of the systems, the others in theassembly mimic a heat bath. Note that the A and B systems can be chosen arbitrarily and canhave wildly different quantum states and quantum numbers.

In general, the nαi are nonnegative integers that satisfy the following constraints:

iΣ nα

i = Nα , for α = A, B, . . . , (4.1a)

Winter Term, 2019

Chemistry 365 -14- Assemblies

where Nα is the number of systems of type α in the assembly, and

αΣ

iΣ nα

i Eαi = ε , (4.1b)

where Eαi is the energy of a system of type α in state i and ε is the total energy of the assembly.

There is only one energy constraint because we’re allowing energy to be exchanged between thesystems in the assembly.

Since the entire assembly is isolated from the rest of the universe, the equal a priori

hypythosis can be applied; i.e., all states with a given energy (ε ) are equally probable. Degener-ate energy states arise in two ways: 1) energy degeneracy in the individual systems (e.g., like wesee in the electronic states of hydrogen) and 2) the different ways in which we can assign the nα

i

states to the different members in the assembly. The former is in the i’s and Eαi ’s and are

accounted for explicitly when sums over states is done for a single system. The combinatorialdegeneracy results in a multinomial coefficient, namely,

Ω(nαi ) ≡

Nα !

iΠnα

i !. (4.2)

Nα ! is the number of ways the states can be assigned to the systems of type α . The factors ofnα

i ! correct for the number of ways the nαi can be assigned to the same systems. For example,

suppose systems 1, 4, and 10, only, are assigned to state 5. There are 3! ways to do this, but thedifferent order that they are assigned doesn’t result in a new state of the assembly; in the end sys-tems 1,4, and 10 will be in state 5. The nα

i ! factors correct for this over-counting in Nα !.Finally, note that the total degeneracy for the entire assembly is just Ω(nA

i )Ω(nBi ). . . .

We’re interested in the average number of systems of type γ in state j, which, with theequal a priori hypothesis, can be written as

⟨nγj ⟩ =

nAi

ΣnB

i Σ . . .′ n

γj αΠΩ(nα

i )

nAi

ΣnB

i Σ . . .′

αΠΩ(nα

i )∼ n

γj *, (4.3)

where the prime on the sums indicates that the n’s are chosen to obey the constraints given inEqs. (4.1a) and (4.1b), and where the * denotes the value of n

γj in the choice of n’s that maximize

the Ω’s subject to the constraints. The second relation is obtained when the maximum termmethod is used.

To find the maximum term we consider the logs of the Ω’s and use Lagrange’s method ofundetermined multipliers. Specifically, we consider the maximum of the auxiliary function

H ≡αΣ

ln(Nα !) −

iΣ

ln(nα

i !) + λα nαi + β nα

i Eαi

Winter Term, 2019

Assemblies -15- Chemistry 365

∼αΣ

ln(Nα !) −

iΣ

nα

i [ln(nαi ) − 1] + λα nα

i + β nαi Eα

i, (4.4)

where λα and β are the Lagrange multipliers and where the second relation is obtained by usingStirling’s formula. Finally, by setting the derivative of H with respect to one of the nα

i to zero, itfollows that the canonical distribution function becomes

Pαi =

⟨nαi ⟩

Nα∼ nα

i *

Nα=

e−λα

Nαe−β Eα

i =e−β Eα

i

Qα, (4.5)

where Qα ≡iΣ e−β Eα

i is the canonical partition function for a system of type α , and results when

Eq. (4.1a) is used to eliminate λα . The so-called joint probability, i.e., the probability that the Asystem is in state i and the B system is in state j, P

A,Bi, j , is obtained by looking at the average of

nAi nB

j . It is easy to show that PA,Bi, j = P A

i PBj with the same β , when the maximum term method is

used.

4.2. The Connection to Thermodynamics

Consider changes of the average energy of the A systems

d⟨E A⟩ =iΣ E A

i dP Ai − P A

i pAi dV = −

1

β iΣ[ln(P A

i ) + ln(Q)]dP Ai − ⟨ pA⟩dV =

dΦA

β− ⟨ pA⟩dV .

(4.6)

where pAi ≡ −(∂E A

i /∂V )N is the mechanical pressure in state i. The second equality is obtainedby reexpressing the energy in terms of the log of the probability, cf. Eq. (4.5), and the third bynoting that

iΣ dPα

i = 0; finally,

Φα ≡ −iΣ Pα

i ln(Pαi ). (4.7)

As we showed at the end of the last section, the probabilities are multiplicative for the assemblyunder consideration. That being the case, it is easy to show that ΦA,B,... = Σα Φα .

The First Law of Thermodynamics states, assuming that all changes are reversible, thatdE = dQHeat − pdV , where dQHeat is the heat absorbed by the system. Hence, by comparingwith Eq. (4.6) we see that dQHeat = dΦA/β . Going a step further, the Second Law tells us thatdS = dQHeat /T and therefore

dS A =1

β TdΦA. (4.8)

Since the entropy is a state function, Eq. (4.8) must correspond to an exact differential; i.e.,

dS A = F ′A(ΦA)dΦA, (4.9)

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Chemistry 365 -16- Assemblies

where (β T )−1 = F ′A(ΦA). By integrating Eq. (4.9) we see that S A = F A(ΦA), up to an unimpor-

tant additive constant.

Another property the entropy must have is that it is additive for weakly interacting (strictlyspeaking noninteracting) systems. In particular, for the assembly of systems considered hereStotal = Σα Sα or

Ftotal(Σα Φα ) =αΣ Fα (Φα ). (4.10)

By taking derivatives of Eq. (4.10) with respect to Φα ’s it follows that

F ′total(Σα Φα ) = F ′

1 (Φ1) = F ′2 (Φ2) = . . . = constant ≡ kB, (4.11)

where the primes now denote derivatives and where kB is a universal constant known as Boltz-mann’s constant (experimentally determined to be 1. 380622 × 10−23J /K ). Hence,Fα (Φα ) = kBΦα . Finally, by returning to Eq. (4.9) we see that β = 1/(kBT ).

4.3. Summary: Canonical Ensemble

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Summary: Canonical Ensemble -17- Chemistry 365

Probability Pi = e−β Ei /Q

Partition Function Q ≡iΣ e−β Ei =

levelsΣ ω levele

−β Elevel

Av erage Energy ⟨E⟩ =iΣ Pi Ei =

∂ ln Q

∂(−β )

N ,V

Pressure ⟨ p⟩ = −iΣ Pi

∂Ei

dV

N

= kBT

∂ ln Q

∂V

β ,N

Chemical Potential µ = −kBT

∂ ln Q

∂N

T ,V

Entropy S = −kBiΣ Pi ln Pi =

⟨E⟩ − A

T

Helmholtz free Energy A = −kBT ln Q(N , V , T ) = ⟨E⟩ − TS

Thermodynamics dA = −SdT − pdV + µdN

dE = TdS − pdV + µdN

Fluctuations σ 2E ≡ ⟨(E − ⟨E⟩)2⟩ = ⟨E2⟩ − ⟨E⟩2

=

∂⟨E⟩∂(−β )

N ,V

= kBT 2CV

4.4. Summary: Grand Canonical Ensemble

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Chemistry 365 -18- Summary: Grand Canonical Ensemble

Probability Pi(N ) =eβ µN−β Ei

Ξ

Partition Function Ξ ≡NΣ

iΣ eβ µN−β Ei(N ,V ) =

NΣ eβ µN Q(N , V , T )

Av erage Energy ⟨E⟩ =NΣ

iΣ Pi(N )Ei(N , V ) =

∂ ln Ξ∂(−β )

β µ,V

Pressure ⟨ p⟩ = −NΣ

iΣ Pi(N )

∂Ei

dV

N

= kBT

∂ ln Ξ∂V

β ,N

=ln ΞβV

Number of Molecules ⟨N ⟩ =

∂ ln Ξ∂β µ

V ,β

Fluctuations1 σ 2N ≡ ⟨(N − ⟨N ⟩)2⟩ = ⟨N 2⟩ − ⟨N ⟩2 =

∂⟨N ⟩∂β µ

T ,V

= ρkBTκ ⟨N ⟩

1 The isothermal compressiblity is defined as κ ≡ −1

V

∂V

∂p

N ,T

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Ideal Bose & Fermi Gases -19- Chemistry 365

5. Ideal Bose & Fermi Gases

5.1. Introduction and General Considerations

When we consider quantum statistical mechanics of identical particles, we abandon theconventional, classical, notion of states (i.e., as lists of quantum numbers assigned to the parti-cles) and instead focus on "occupation numbers", ni , the number of particles in 1-particle state i

(e.g., a single hydrogenic electron or a single particle in a box). Indistinguishably implies thatthe quantum probability density, |ψ (1, 2, 3, 4, . . .)|2, be symmetric under exchange of any particlelabels, and, in turn, this means that ψ (1, 2, . . .) = ± ψ (2, 1, . . .). Quantum field theory and specialrelativity show that the + sign is used for integer spin particles, generically called Bosons, whilethe − sign is used for half-odd integer spins, referred to as Fermions.

Tw o see how this works, consider the following 2-particle example: each particle can be inone of two states, |α , i > or |β , i >, i = 1, 2, with energies εα and ε β , respectively. If we ignorethe consequences of indistinguishability we can construct 4 possible states for this system asshown in the following table:

Table 5.1: Possible States and Energies for a Two Lev el System with Two Particles

State Wav efunction Energy nα nβ Particle

1 |α , 1 > |α , 2 > 2εα 2 0 Bosons2 2−1/2(|α , 1 > |β , 2 > + |β , 1 > |α , 2 >) εα + ε β 1 1 Bosons3 2−1/2(|α , 1 > |β , 2 > − |β , 1 > |α , 2 >) εα + ε β 1 1 Fermions4 |β , 1 > |β , 2 > 2ε β 0 2 Bosons

Q(N = 2, T , V ) =

e−εα /kBT + e−ε β /kBT

2

, for distinguishable particles (4 states)

e−(εα +ε β )/kBT , for Fermions (1 state)

e−2εα /kBT + e−(εα +ε β )/kBT + e−2ε β /kBT , for Bosons (3 states).

(5.1)

For systems with more than two particles we simply symmetrize or anti-symmetrize the wav e-function (i.e, add all possible permutations of the particle labels together) for Bosons or use theso-called Slater determinant for Fermions.1 This shows that ni = 0, 1, 2, . . . for Bosons, while

1For Bosons this means

Ψ(1, 2, . . . , N ) =Permutations

Σ P[ψ a(1) . . .ψ z(N )],

where, P permutes the particle labels and where ψ a(1) is an orbital for one particle in 1-particle state a

(e.g., for atoms, an atomic orbital). For Fermions we construct the so-called Slater determinant, i.e.,

Ψ(1, 2, . . . , N ) =

ψ a(1)

ψ b(1).

.

.ψ z(1)

ψ a(2)

ψ b(2).

.

.ψ z(2)

. . .

. . .

. . .

. . .

. . .

ψ a(N )

ψ b(N ).

.

.ψ z(N )

.

Chemistry 365 -20- Ideal Bose & Fermi Gases

ni = 0, 1 for Fermions, as required by the Pauli Exclusion Principle.1 In either case, the wav e-function is determined, up to an overall phase factor, by specifying the ni’s.

Since E = Σi niε i , it follows that the canonical partition function is

Q(N , V , T ) =

iΣ ni = N

1 or ∞

nα = 0Σ

1 or ∞

nβ = 0Σ . . . exp

−

iΣ ε i ni /kBT

, (5.2)

where the upper limits of the sums depend on whether Fermions or Bosons are under considera-tion. The sums aren’t easy to do when constrained to have Σi ni = N . Since N fluctuates in thegrand canonical ensemble, we expect that the constraint on N to disappear. Moreover, we’veshown that the choice of ensemble does not matter when calculating thermodynamic quantities.Hence, we switch to the grand canonical ensemble, with partition function

Ξ =∞

N = 0Σ eβ µN

iΣ ni = N

1 or ∞

n1 = 0Σ

1 or ∞

n2 = 0Σ . . . e

−βiΣ ε i ni

=1 or ∞

n1 = 0Σ

1 or ∞

n2 = 0Σ . . . e

−βiΣ(ε i−µ)ni

=i

Π

1 or ∞

ni = 0Σ eβ (µ−ε i)ni

=i

Π1 ± eβ (µ−ε i)

±1

, (5.3)

where we’ve gone back to our earlier notation of naming states by a roman index, β and µ havetheir usual meanings, and where the last equality follows by explicitly carrying out the sums forFermions (upper signs) or Bosons (lower signs).

Once we have the grand canonical partition function, thermodynamic functions areobtained in the usual manner, e.g.,

β pV = ln(Ξ) = ±iΣ ln

1 ± eβ (µ−ε i)

, (5.4)

⟨N ⟩ =

∂ ln(Ξ)

∂β µ

T ,V

=iΣ ⟨ni⟩ =

iΣ 1

eβ (ε i−µ) ± 1, (5.5)

and

⟨E⟩ =

∂ ln(Ξ)

∂ − ββ µ,V

=iΣ ⟨ni⟩ε i , (5.6)

Note that neither expression is normalized. The Slater determinant is odd under exchange of any two rowsand vanishes if two or more rows are the same, thereby having the required symmetry and Pauli exclusionprinciple.

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where the average occupation numbers are

⟨ni⟩ =1

eβ (ε i−µ) ± 1, (5.7)

and are known as the Fermi-Dirac (+ sign) or Bose-Einstein (- sign) distributions.

Finally consider the heat capacity. From Eqs. (5.6) and (5.7) it follows that

CV =kB

2

i, jΣ ⟨n j⟩

2eβ (ε j−µ)⟨ni⟩

2eβ (ε i−µ)[β (ε i − ε j)]

2

jΣ ⟨n j⟩

2eβ (ε j−µ)

, (5.8)

where the details of calculation are given in Appendix A. Note that CV is positive, as expected.

The quantity ⟨ni⟩2 exp[β (ε i − µ)] plays a key role, and is shown in Figs. 5.1 and 5.2, for Fermions

and Bosons, respectively. An approximate expression for Fermions for 4⟨ni⟩2ex can be obtained

by noting that the Taylor expansion of ln(4⟨ni⟩2ex)∼ − x2/4 + O(x4), where x ≡ β (ε i − µ); hence,

4⟨ni⟩2ex ≈ e−x2/4

Fig. 5.1. Mean occupation numbers, ⟨ni⟩, and 4⟨ni⟩2eβ (ε −µ) for

Fermions. The spin degeneracy is not included. Notice howthe orbitals change from filled to empty over a range ofβ (ε − µ) ∼ ± 5 or, for T = 300K , over a range∆(ε − µ) ∼ ± 2. 5kBT = ±0. 065 ev. The plot of ⟨ni⟩

2eβ (ε −µ)

shows that the main contributions to the heat capacity comefrom a narrow band of 1-particle states near the Fermi energy.Finally, the dotted brown line is the approximation to4⟨ni⟩

2eβ (ε −µ) discussed in the text.

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Fig. 5.2. Mean occupation numbers, ⟨ni⟩, and ⟨ni⟩2eβ (ε −µ) for

Bosons. The spin degeneracy is not included.

Equation (5.8) leads to an approximate expression for CV . Appendix A shows a more rig-orous approach2

5.2. Results for Particles in a Cubic Box

In order to proceed we need a concrete model for the 1-particle states and energies. Thesimplest is probably the particle with mass m in a cubical box of side length L. This has ener-gies:

ε nx ,ny,nz=

h2

8mL2(n2

x + n2y + n2

z), where ni = 1, 2, 3, . . . . (5.9)

Note that the ground state energy, ε1,1,1 = 3h2/(8mL2) → 0 as L → ∞. When the sum over sin-gle particle states (i.e., over positive ni , i = x, y, z,) is approximated as an integral and the resulttransformed to polar coordinates, we find that all the examples at the end of the previous sectioncan be written as

(2S + 1)π2 ∫

∞0

dn n2 f (λ , β ε n), (5.10)

where the factor of (2S + 1) accounts for the spin degeneracy of a particle with spin S, the factorof π /2 = 4π /8 is the area of the unit sphere in the octant where ni > 0, i = x, y, z, and λ ≡ eβ µ is

the proper activity. Next, change variables by letting n = √ 8m/h2 Lε 1/2 and rewrite Eq. (5.10) as

2See J.E. Mayer and M.G Mayer, Statistical Mechanics, (John Wiley & Sons, Inc., 1940), Sec. 16g.

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∫∞0

dε g(ε ) f (λ , β ε ), (5.11)

where

g(ε ) ≡ 2π (2S + 1)

2m

h2

3/2

V ε 1/2 (5.12)

is known as the density of states and can be interpreted as the number of states per unit energywith energies between ε and ε + dε . With this, the general results given at the end of the pre-ceding section become

β pV = ± ∫∞0

dε g(ε ) ln1 ± λe−β ε

, (5.13a)

⟨N ⟩ = ∫∞0

dε g(ε )λe−β ε

1 ± λe−β ε , (5.13b)

and

⟨E⟩ = ∫∞0

dε g(ε )ελe−β ε

1 ± λe−β ε . (5.13c)

Note that Eqs. (5.13b) and (5.13c) can be obtained by taking the usual derivatives of β pV , i.e.,ln(Ξ), cf. Eq. (5.13a).

Next we expand the integrands into Taylor series in λ , specifically,

± ln1 ± λe−β ε

= −+

∞

j = 1Σ (−+ λ) j

je− jβ ε (5.14a)

and

λe−β ε

1 ± λe−β ε = ±

∂ ln(1 ± λe−β ε )

∂β µ

β ,V ,N

= −+∞

j = 1Σ (−+ λ) je− jβ ε , (5.14b)

and change variables yet again by letting ε ≡ (kBT / j)x . This gives:

β pV = −+2

π 1/2

V

Λ3(2S + 1)

∞

j = 1Σ (−+ λ) j

j5/2Γ(3 / 2), (5.15a)

⟨N ⟩ = −+2

π 1/2

V

Λ3(2S + 1)

∞

j = 1Σ (−+ λ) j

j3/2Γ(3 / 2), (5.15b)

and

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⟨E⟩ = −+2

π 1/2

V

Λ3kBT (2S + 1)

∞

j = 1Σ (−+ λ) j

j5/2Γ(5 / 3), (5.15c)

where Λ ≡ h/√ 2π mkBT is the thermal de Broglie wav elength, and have expressed the remainingintegrals as Γ-functions,3 specifically, Γ(3

2) = 1

2π 1/2, for β pV and ⟨N ⟩, and Γ(5

2) = 3

4π 1/2 for ⟨E⟩.

When these are used in Eqs. (5.15a)−(5.15c) and the results rearranged, we find that

ρΛ3 = (2S + 1)G3/2(λ) (5.16a)

and

β pΛ3 =2β ⟨E⟩Λ3

3V= (2S + 1)G5/2(λ), (5.16b)

where

Gz(λ) ≡1

Γ(z) ∫∞0

dx x z−1 λe−x

1 ± λe−x= λΦ(−+ λ , z, 1) = −+

∞

j=1Σ j−z(−+ λ) j , (5.16c)

generalizes Hill’s Fz(α ),4 the function Φ(a, b, c) is discussed in Gradshetyn and Ryzhik,5 andρ = ⟨N ⟩/V is the number density. The first equality in Eq. (5.16b) holds for the individual statesfor the particle in a box, i.e., pn = 2ε n/3V .

These sums converge absolutely for |λ | < 1 (µ < 0) and can be summed numerically,although the effort to compute the integrals numerically is comparable and must be used forFermions for λ > 1. Some results are shown in Fig. 5.3. When λ = 1 (i.e., µ = 0), the sums inEq. (5.16c) reduce to the Riemann zeta functions as shown in the following table:

Table 5.2: Connections to the Riemann- Zeta FunctionsGz(1): Formulas and Numerical Values

z Fermions: (1 − 21−z)ζ (z) Bosons: ζ (z)

3/2 0.7651 2.6125/2 0.8671 1.341

3One of the ways of defining the Γ function is via its integral representation

Γ(z) ≡ ∫∞

0dx x z−1e−x ,

which includes all of the remaining integrals. For more properties of these functions, see, e.g., the Hand-

book of Mathematical Functions, tenth edition, M. Abramowitz and I.S. Stegun, eds., ch. 6.4T. L. Hill, An Introduction to Statistical Thermodynamics (Dover Publications), p. 418.5I.S. Gradshetyn and I.M. Ryzhik, Table of Integrals, Series and Products (Academic Press. New York,1980), Sec. 9.55.

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Fig. 5.3. Proper activities, λ , and equations of state for ideal Boltz-mann, Fermi-Dirac, and Bose-Einstein gasses. The spin degeneracyfactor has not been included. Note that the figure was actually obtainedby choosing a range of λ’s, calculating ρΛ3, and transposing the plot.

The first thing to notice is that the curves are universal, in that all ideal systems’ properties,once corrections for the spin degeneracy are made, will fall on the same curves. Also notice thepositive (Fermions) and negative deviations (Bosons) from ideal gas behavior. For Fermions,these positive deviations arise because the Pauli exclusion principle which disallows two non-interacting particles to occupy the same state or space, and thus has an effect similar to stericrepulsion). Bosons don’t hav e to obey the exclusion principle and can have any number of parti-cles in the same state, thereby acting like an attraction.

5.2.1. The Case For Boltzmann Statistics

It turns out that there is a simple fix that allows us to treat the particles as if they were dis-tinguishable. Consider a separable Hamiltonian which has 1-particle energy levels, ε i = i, fori = 1, 2, 3, . . .. Let’s examine the states for a system comprised of three identical particles, whereHΨ(a, b, c) = 9Ψ(a, b, c)

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Table 5.3: States where HΨ(a, b, c) = 9Ψ(a, b, c)

Degeneracy

ΩFD ΩDistinguishable ΩBE

1-Particle States Nonzero ni’s

(7, 1, 1) n7 = 1 and n1 = 2 0 3 1(6, 2, 1) n6 = 1, n2 = 1, and n1 = 1 1 3! 1(5, 3, 1) n5 = 1, n3 = 1 and n1 = 1 1 3! 1(5, 2, 2) n5 = 1 and n2 = 2 0 3 1(4, 4, 1) n4 = 2 and n1 = 1 0 3 1(4, 3, 2) n4 = 1, n3 = 1, and n2 = 1 1 3! 1(3, 3, 3) n3 = 3 0 3! 1

The degeneracy for Fermions, ΩFD, vanishes whenever the Pauli exclusion principle is violated,i.e., whenever ni > 1, as expected. Since the wav efunction for Bosons can always be sym-metrized, any choice of the ni’s giv es a single wav efunction, and ΩBE = 1. Finally, the degenera-cies for the distinguishable cases have 0 < ΩDistinguishable ≤ N !, the equality holding when all the1-particle states are different. This discussion can be summarized by noting that

ΩFD ≤ΩDistinguishable

N !≤ ΩBE ,

where equality holds when all the 1-particle states are different.

If the number of available states is much larger than the number of particles, then most ofthe states included in the partition function obey the Pauli exclusion principle, and the distin-guishable calculation over-counts these by the same factor of N !. Hence, a corrected partitionfunction becomes

QBoltzmann =QDistinguishable

N !.

This approach is known as Boltzmann statistics. Note that each component of a mixture contrib-utes a N ! factor.

Returning to the particle in a box example, we now hav e

QBoltzmann =(V Λ−3)N

N !

and thus

β A = − log(Q) = −N log(V Λ−3) + N [log(N ) − 1] = N [log(ρΛ3) − 1],

where we have used Stirling’s formula and ρ ≡ N /V is the number density.6 The Gibbs paradox

is resolved!

The ideal gas model makes it fairly straightforward to quantify the condition necessary forBoltzmann statistics to be used, namely, that the number of accessible states, NStates, greatly

6Hill writes this as a single logarithm as N log(ρΛ3/e).

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exceeds the number of particles, N . As we hav e just shown, when this holds, the resolution ofGibbs’ paradox is to simply divide the partition function, calculated as if the particles were dis-tinguishable, by N !. The argument is simple and is essentially that presented in Hill.7 To begin,note that the particle in a box quantum numbers, ni , i = x, y, z, which when plotted, form a threedimensional cubic lattice in the positive octant, each lattice point corresponding to a state. Interms of the lattice cells, each has unit volume and each corresponds to a single state (to be sure,there are eight lattice points per cell, but each is shared with 8 neighbors), Thus, if we ignoreedge effects, the number of states with energies less than εmax is just the volume of of an eighthof a sphere with radius, n, less than the n corresponding to the εmax. By using Eq. (5.9), the cut-off radius is n = (8mεmax/h2)1/2 L, and the corresponding number of states is simply

NStates ≡1

8

4π n3

3=

4π3

2mεmax

h2

3/2

V .

By taking εmax = 3kBT /2 and requiring that NStates >> N , we see that

ρΛ3 <<

6

π

1/2

(5.17)

in order to use Boltzmann statistics. Perhaps more physically, the number of particles per vol-ume corresponding a size comparable to the thermal de Broglie wav elength is small, and hence,the number of particles within a thermal de Broglie wav elength, the scale at which quantumeffects (e.g., wav e-particle duality) become important, is small. Note that raising temperatureand/or mass, and/or lowering the density favor the use of Boltzmann statistics.

7T.L. Hill, op. cit., Sec. 4.1.

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Fig. 5.4. The temperature dependence of the thermal deBroglie wav elength for various gases. Note that with theexception of the electron, Λde Broglie is comparable to orsmaller than the size of an atom at room temperature.

Some specific results are given in the following table:8

Table 5.4: Is Boltzmann Statistics Valid?

Element (State) T (K) (π /6)1/2 ρΛ3

He (l) 4 1.6He (g) 4 0.11He (g) 20 2×10−3

He (g) 100 3.5×10−5

Ne (l) 27 1.1×10−2

Ne (g) 27 8.2×10−5

Ne (g) 100 3.1×10−6

Ar (l) 86 5.1×10−4

Ar (g) 86 1.6×10−8

Kr (l) 127 5.4×10−5

Kr (g) 127 2.0×10−7

300 1465electrons inmetals (Na)

Thus, we see that except for ultra-low temperatures and the lightest elements (both probablyaren’t of much interest to chemists), Boltzmann Statistics should be an excellent approximation.

8From, D.A. McQuarrie, Statistical Mechanics, (Harper and Row Pub., Inc., New York, NY, 1973), Table4-1, p. 72.

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There is one major exception, namely electrons around room temperature, something that hasmajor ramifications for bonding, etc..

5.3. Results Specific to Fermions or Bosons

5.3.1. Fermions

As we saw in the preceding section, cf. Fig. 5.1, the Fermi-Dirac distribution is basicallyflat for ε < µF and falls off rapidly for ε > µF , where µF is the chemical potential, known as theFermi energy to honor Enrico Fermi; hence, a reasonable approximation is to cut off the integra-tions at ε = µF and let ⟨ni⟩∼2S + 1 for ε < µF . For example, by using Eqs. (5.13b) and (5.13c) itfollows that

⟨N ⟩ ≈ ∫µF

0dε g(ε ) =

4π (2S + 1)

3

2mµF

h2

3/2

V (5.18)

or

µF ≈h2

8m

6ρπ (2S + 1)

2/3

. (5.19)

Similarly,

⟨E⟩ ≈ ∫µF

0dε g(ε )ε =

4π (2S + 1)

5

2m

h2

3/2

V µ5/2F =

3

5⟨N ⟩µF . (5.20)

As the temperature is reduced, the approximations we’ve just made become more accurate;hence, the simple results just obtained become more valid and represent the ground state configu-ration of the Fermi (ideal) gas. Finally, the approximations leading these results don’t lend them-selves well to quantities like like the heat capacity. As we saw above, CV , arises from a smallband of energy levels where ε ≈ µF ; these make CV nonzero and its calculation more compli-cated.

By using Eq. (5.8) and the approximate expression for ⟨ni⟩eβ (ε i−µ) it follows that

CV

kB

∼ kBT

8

∫∞−∞

dx g(kBTx + µ)e−x2/4 ∫∞−∞

dy g(kBTy + µ)e−y2/4(x − y)2

∫∞−∞

dx g(kBTx + µ)e−x2/4

∼ kBTg(µ)

4 ∫∞−∞

dx e−x2/4 x2 = kBTg(µ)π 1/2 = 2(2S + 1)

2π mkBT

h2

3/2

V

µ

kBT

1/2

,

where the last expression was obtained by using Eq. (5.12). By using Eq. (5.18) to eliminate thevolume in favor of ⟨N ⟩ it follows that

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CV

⟨N ⟩kB

∼ π 1/23kBT

2µF

=3τ

2π 1/2,

where τ ≡ π kBT /µF is a reduced temperature. The more rigorous calculation given in the Ap-pendix, cf. Eq. (A13), gives π τ /2, i.e., a 46% error. In any event, both calculations show that thatCV ∝T as T → 0, which is consistent with the Third Law of Thermodynamics. Also note thatkBT /µF << 1; e.g., if µF = 1 eV, and T = 300K then CV /kB ≈ 0. 069.

5.3.2. Bosons

The Bose gas thermodynamic properties can be obtained from the results given in Eqs.(5.13a−c) or (5.4−7). In particular, from Eq. (5.13b), i.e.,

⟨N ⟩ = ∫∞0

dεg(ε )

eβ (ε −µ) − 1, (5.21)

noting that the particle in a box model has ε ≥ 0 as L → ∞, it follows that µ < 0 or λ < 1 so thatthe integrand in Eq. (5.21) be positive and integrable. However, from the data shown in Fig. 5.3,it follows that λ = 1 for ρΛ3 ≈ 2. 612. For 4 He, using the experimental density of liquid helium(0. 145g/cm3) this occurs at T = 3. 14K . Experiment shows that there is a transition at 2.18K.(What contributes to the discrepancy?)

In order to resolve this problem, note that the exact number of particles in the ground state(ε = 0) in the grand canonical ensemble is

(2S + 1)λ

(1 − λ). (5.22)

On the other hand, the factor of g(ε )∝ε 1/2 in Eq. (5.21) shows that the ground state doesn’t con-tribute at all. This makes sense if the states form a continuum; the probability of finding anyexact state (not a narrow band of them) vanishes, and we must introduce probability densities.On the other hand, the ground state contribution to the partition function is the most divergent!

The way out of this seeming paradox is to postulate what is known as the two fluid model.We write

⟨N ⟩2S + 1

=λ

1 − λ+ 2π

2m

h2

3/2

V ∫∞0

dεε 1/2

eβ ε − 1(5.23a)

=λ

1 − λ+

V

Λ3ζ (3 / 2), (5.23b)

where the first term is the number of particles in the ground state, the so called Bose condensate,while the second accounts for those in excited states. Finally, if we assume that µ → 0 −, andexpand λ in the denominator of the first term in Eq. (5.23b) we find that

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β µ ∼ −Λ3

V

ρΛ3 − ζ (3 / 2)

, (5.24)

where we have taken S = 0 with ρΛ3 ≥ ζ (3 / 2). Thus, the number of particles in the ground stateis

N0 ≡λ

1 − λ= ⟨N ⟩

1 −ζ (3 / 2)

ρΛ3

= ⟨N ⟩1 −

T

T0

3/2, (5.25)

where T0 is the transition temperature (i.e., where ρΛ3 = ζ (3 / 2) = 2. 61 . . .). Notice that whenT = T0 no particles are in the ground state, but this rises as temperature is lowered until 100% ofthem are in the ground state at absolute zero. Having macroscopic occupation of a single statedrastically changes the physical properties of the system. Have a look at the demonstrations byAlfred Leitner https://www.youtube.com/watch?v=sKOlfR5OcB4 for an experimental survey ofsome of the more striking changes that arise in superfluids or athttps://www.youtube.com/watch?v=BFdq6IecUJc for superconductors.

5.3.3. Photons and Black Body Radiation

Photons can be wav e-like or particle-like. The former means that each photon has a fre-quency, ω , a wav e-vector, k, and amplitudes for the components of the photon’s electric, E, andmagnetic, H, fields. The amplitudes are arbitrary, except that the radiation must be transverse,i.e., k ⋅ E = 0, while the wav e-vector and frequency are related by the well known dispersionrelation

ω = kc, (5.26)

where c is the speed of light in vacuum. The condition of transversality means that radiation ismade up of two independent polarizations, e.g., left and right circular polarization. From theselection rule for absorption and emission, we know that the photon behaves as a spin-1 particle,modulo transversality, since a molecule’s angular momentum changes by ∆l = ±h− on absorptionor emission in all linear spectroscopies. Hence, photons are Bosons. Finally, the energy of thephoton is just ε = h− ω .

Note that the number of photons of a given frequency isn’t conserved (unlike the atoms ina chemical reaction) and, in particular, processes like

M photons at frequency ω →← N photons at frequency ω (5.27)

are possible. At equilibrium, the free energy is a minimum, and the usual argument requires that(N − M)µPhoton = 0, or µPhoton = 0.

With this introduction, consider the radiation in a closed cubical container of side length L.The radiation in the box comes to equilibrium with the walls and forms standing wav es in differ-ent directions and having different frequencies. If we assume that the electric field vanishes atthe walls and consider the field as sin-like (just like the particle in a box) it follows thatk = π /L(nx,ny,nz)

T , where ni = 1, 2, 3, . . .. By repeating the steps used above, it follows that

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n(ω ) the density of radiation between frequencies ω and ω + dω is

n(ω ) =V

π 2c3

ω 2

eβ h−ω − 1, (5.28)

where remember that the photon energy states are doubly degenerate owing to the two polariza-tion possibilities. The main differences between this and our earlier result are caused by the factthat ε = h− ω = h− kc. By multiplying this result by the photon energy, h− ω , giv es

E(ω ) = Vh−

π 2c3

ω 3

eβ h−ω − 1, (5.29)

where E(ω )dω is the energy found in the frequency interval ω to ω + dω and is known asPlanck’s formula.

Limiting behaviors are easy to find; specifically,

E(ω ) ∼

VkBTω 2

π 2c3for h− ω << kBT

Vh−

π 2c3ω 3e−β h−ω when h− ω >> kBT .

(5.30)

The low frequency (or high temperature) behavior is known as the Raleigh-Jeans radiation law,and was known before the discovery of quantum mechanics. One of the problems with the clas-sical theory, according to the Raleigh-Jeans formula, is that the total radiation energy in the boxdiverges; this is known as the ultraviolet catastrophe, and posed a very difficult problem forclassical mechanics and electrodynamics.

By integrating the full result, we find that

Etotal

V=

h−π 2c3 ∫

∞0

dωω 3

eβ h−ω − 1=

(kBT )4

(ch−)3π 2 ∫∞0

dxx3

ex − 1=

6ζ (4)(kBT )4

(ch−)3π 2, (5.31)

where the last two results were obtained by letting ω = kBT /h− x. The T 4 dependence wasdeduced classically using thermodynamics by Stephan and Wein. Wein actually proved some-thing more general called Wein’s Law; namely, E(ω )/V = ω 3 f (ω /T ), where f is not known.Note that Wein’s Law is obeyed by the Planck formula.

5.4. Appendix A: CV in an Ideal Bose or Fermi Gas

The mathematical details leading to the constant volume heat capacity, CV , cf. Eq. (5.8), are nowpresented. First, from thermodynamics, recall that

CV ≡

∂⟨E⟩∂T

⟨N ⟩,V

=

∂⟨E⟩∂T

β µ,V

+

∂⟨E⟩∂β µ

β ,V

∂β µ

∂T

⟨N ⟩,V

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=

∂⟨E⟩∂T

β µ,V

−

∂⟨E⟩∂β µ

β ,V

∂⟨N ⟩∂T

β µ,V

∂⟨N ⟩∂β µ

β ,V

, (A1)

where the second equality follows when we consider ⟨E⟩ to be a function of T or β , β µ, and vol-ume, V , while the last equality is obtained when we use the cyclic rule (implicit function differ-entiation).

Next we use Eqs. (5.5)−(5.7) to evaluate the derivatives that appear in Eq. (A1); i.e.,

∂⟨E⟩∂T

β µ,V

=1

kBT 2iΣ ⟨ni⟩

2eβ (ε i−µ)ε 2

i ,

∂⟨E⟩∂β µ

β ,V

=iΣ ⟨ni⟩

2eβ (ε i−µ)ε i ,

∂⟨N ⟩∂T

β µ,V

=1

kBT 2iΣ ⟨ni⟩

2eβ (ε i−µ)ε i ,

∂⟨N ⟩∂β µ

β ,V

=iΣ ⟨ni⟩

2eβ (ε i−µ),

which when used in Eq. (A1) gives

CV =1

kBT 2iΣ ⟨ni⟩

2eβ (ε i−µ)

ε 2i − ε i

jΣ ⟨n j⟩

2eβ (ε j−µ)ε j

jΣ ⟨n j⟩

2eβ (ε j−µ)

= i, jΣ ⟨ni⟩

2eβ (ε i−µ)⟨n j⟩

2eβ (ε j−µ)ε i(ε i − ε j)

kBT 2

jΣ ⟨n j⟩

2eβ (ε j−µ)

.

(A2)

Finally, we take half the double sum in the numerator of Eq. (A2), exchange the dummy sumindices, i→← j and add it to the unmodified half, resulting in Eq. (5.8). Note that there are alter-nate ways of writing Eq. (A2). For example,

⟨ni⟩2eβ (ε i−µ) =

1

4

sech((β (ε i − µ) / 2)) for Fermions

csch((β (ε i − µ) / 2)) for Bosons.(A3)

An alternate route to CV , is to eliminate µ in favor of ⟨N ⟩ in the energy. From Eq. (5.18)we see that

ρ = D ∫∞0

dεε 1/2

eβ (ε −µ) + 1, (A4)

where ρ ≡ ⟨N ⟩/V and D ≡ 2π (2S + 1)(2m/h2)3/2. By integrating by parts, we can rewrite Eq.(A4) as

ρ =2D

3

ε 3/2

eβ (ε −µ) + 1)

∞

0

+2D

3β ∫

∞0

dεε 3/2eβ (ε −µ)

[(eβ (ε −µ) + 1]2, (A5a)

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=2Dµ3/2

3 ∫∞−β µ

dx1 +

kBTx

µ

3/2ex

(ex + 1)2, (A5b)

where the boundary term in Eq. (A5a) vanishes and where we have changed variables tox ≡ β (ε − µ) to get Eq. (A5b). By using the Taylor expansion

(1 + x)α = 1 +∞

n=1Σ α (α − 1) . . . (α − n + 1)

n!xn, (A6)

we see that

ρ ∼ 2Dµ3/2

3

1 +

∞

n=1Σ 3

2

1

2. . .

5

2− n

kBT

µ

n

Zn

, (A7)

where

Zn ≡1

n! ∫∞−∞

dxxnex

(ex + 1)2=

0 for n odd

2(1 − 2−n)ζ (n), for n even,

(A8)

where ζ (n) is the Riemann zeta function.9 Note that the odd n terms vanish because ex /(ex + 1)2

is even in x.

In obtaining Eq. (A7) we’ve extended the lower integration limit to −∞. On one hand, this

only introduces exponentially small errors, since ∫−β µ

−∞dx xne−x . . . is exponentially small when

β µ >> 1; Nonetheless, while this makes sense for each term in the Taylor expansion, the summust (and does) diverge since (1 + kBTx/µ)3/2 is imaginary when x < −β µ. Our approach gener-ates what is known as an asymptotic expansion. If you plot the individual terms, they willdecrease until some minimum value is obtained, and increase thereafter causing the series todiverge. The trick is to sum no further. Another example of this is the Stirling formula for n!.

The goal of this calculation is to obtain a low temperature expansion for µ; accordingly, welet

µ ≡ µ0(1 + c1τ + c2τ 2 + . . .), (A9)

where µ0 ≡ h2(6ρ /π (2S + 1))2/3/8m is the chemical potential at absolute zero (cf. Eq. (A19)) andwhere τ ≡ π kBT /µ0 is a reduced temperature. When Eq. (A9) is substituted into Eq. (A7),theresult expanded into a series in τ , and the coefficients of τ n, n > 0, set to zero, it follows that10

9See I.S. Gradshetyn and I.M. Ryzhik, op. cit., Eq. (3.411.3), p. 325.10The algebra becomes horrendous if you want to go much beyond the first correction; nonetheless, it’seasy for a symbolic algebra program, e.g., Mathematica or Maxima (http://maxima.sourceforge.net), to doso. Maxima was used here.

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µ = µ01 −

1

12τ 2 −

1

80τ 4 −

247

25920τ 6 −

16291

777600τ 8 −

1487

15360τ 10 + . . .

. (A10)

By starting with the general expression for the energy, i.e.,

⟨E⟩ = DV ∫∞0

dεε 3/2

eβ (ε −µ) + 1,

and repeating the steps that led from Eq. (A4) to (A7), it follows that

⟨E⟩ =2DV µ5/2

5

1 +

∞

n=1Σ 5

2

3

2. . .

7

2− n

kBT

µ

n

Zn

. (A11)

Finally, by using Eq. (A10) to eliminate µ it follows that

⟨E⟩ =3

5⟨N ⟩µ0

1 +

5

12τ 2 −

1

16τ 4 −

1235

36288τ 6 −

10367

155520τ 8 −

1478

5120τ 10 + . . .

, (A12)

which implies that

CV

⟨N ⟩kB

=π τ2

1 −

3

10τ 2 −

247

1008τ 4 −

10367

16200τ 6 −

4461

1280τ 8 + . . .

. (A13)

Note that CV ∼T as T → 0, which is consistent with the 3rd Law of Thermodynamics. The firstcorrection is O(T 3), just like the Debye-T 3 law for vibrations. That being the case, how can wedistinguish between vibrations and electronic contributions?

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Fig. 5.5. The low temperature, constant volume, heat capacity for asystem of ideal Fermions in a box as a function of the reduced tempera-ture, τ , introduced in the text.. The different curves correspond to thenumber of terms kept in Eq. (A13). As expected, all the curves reduceto the linear one as τ → 0. This doesn’t happen at higher temperatures,and moreover, it is obvious that something is breaking down sinceCV > 0.

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Normal Mode Analysis -37- Chemistry 365

6. Normal Mode Analysis

6.1. Quantum Mechanical Treatment

Our starting point is the Schrodinger wav e equation:

−

N

i=1Σ h−2

2mi

∂2

∂→r

2i

+ U(→r1, . . . ,

→r N )

Ψ(

→r1, . . . ,

→r N ) = E Ψ(

→r1, . . . ,

→r N ), (6.1.1)

where N is the number of atoms in the molecule, mi is the mass of the i’th atom, andU(

→r1, . . . ,

→r N ) is the effective potential for the nuclear motion, e.g., as is obtained in the Born-

Oppenheimer approximation.

If the amplitude of the vibrational motion is small, then the vibrational part of the Hamil-tonian associated with Eq. (6.1.1) can be written as:

Hvib ≈ −N

i=1Σ h−2

2mi

∂2

∂→∆

2

i

+ U0 +1

2

N

i, j=1Σ

↔Ki, j :

→∆i

→∆ j , (6.1.2)

where U0 is the minimum value of the potential energy,→∆i ≡ →

r i −→Ri ,

→Ri is the equilibrium posi-

tion of the i’th atom, and

↔Ki, j ≡

∂2U

∂→r i∂

→r j

→

r k =→Rk

(6.1.3)

is the matrix of (harmonic) force constants. Henceforth, we will shift the zero of energy so as tomake U0 = 0. Note that in obtaining Eq. (6.1.2), we have neglected anharmonic (i.e., cubic andhigher order) corrections to the vibrational motion.

The next and most confusing step is to change to matrix notation. We introduce a columnvector containing the displacements as:

∆ ≡ [∆x1 , ∆y

1 , ∆z1, . . . , ∆x

N , ∆yN , ∆z

N ]T , (6.1.4)

where "T " denotes a matrix transpose. Similarly, we can encode the force constants or massesinto 3N × 3N matrices, and thereby rewrite Eq. (6.1.2) as

Hvib = −h−2

2

∂∂∆

†

M−1

∂∂∆

+1

2∆†K∆, (6.1.5)

where all matrix quantities are emboldened, the superscript "†" denotes the Hermitian conjugate(matrix transpose for real matrices), and where the mass matrix, M, is a diagonal matrix with themasses of the given atoms each appearing three times on the diagonal.

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Chemistry 365 -38- Normal Mode Analysis

The Hamiltonian given by Eq. (6.1.5) is the generalization of the usual harmonic oscillatorHamiltonian to include more particles and to allow for "springs" between arbitrary particles.Unless K is diagonal (and it usually isn’t) Eq. (6.1.5) would seem to suggest that the vibrationalproblem for a polyatomic is non-separable (why?); nonetheless, as we now show, it can be sepa-rated. This is first shown using using the quantum mechanical framework we’ve just set up.Later, we will discuss the separation using classical mechanics. This is valid since, for harmonicoscillators, you get the same result for the vibration frequencies.

We now make the transformation→∆i → m−1/2

i

→∆i , which allows Eq. (6.1.5) to be reexpressed

in the new coordinates as

Hvib = −h−2

2

∂∂∆

†

∂∂∆

+1

2∆†K∆, (6.1.6)

where

K ≡ M−1/2KM−1/2. (6.1.7)

Since the matrix K is Hermitian (or symmetric for real matrices), it is possible to find aunitary (also referred to as an orthogonal matrix for real matrices) matrix which diagonalizes it;i.e., you can find a matrix P which satisfies

P†KP = λ or K = PλP†, (6.1.8)

where λ is diagonal (with real eigenvalues) and where

P P† = 1 (6.1.9)

where 1 is the identity matrix. Note that P is the unitary matrix who’s columns are the normal-ized eigenvectors. (See a good quantum mechanics book or any linear algebra text for proofs ofthese results).

We now make the transformation

∆ → P∆ (6.1.10)

in Eq. (6.1.5). This gives

Hvib = −h−2

2

∂∂∆

†

P†P

∂∂∆

+1

2∆† λ∆, (6.1.11)

Finally, Eq. (6.1.8) allows us to cancel the factors of P in this last equation and write

Hvib =3N

i=1Σ

−

1

2h−2 ∂2

∂∆2i

+1

2λ i∆2

i. (6.1.12)

Thus the two transformations described above hav e separated the Hamiltonian into 3N uncoupled

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harmonic oscillator Hamiltonians, and are usually referred to as a normal mode transformation.Remember that in general the normal mode coordinates are not the original displacements fromthe equilibrium positions, but correspond to collective vibrations of the molecule.

6.2. Normal Modes in Classical Mechanics

The starting point for the normal mode analysis in classical mechanics are Newton’s equa-tions for a system of coupled harmonic oscillators. Still, using the notation that led to Eq.(6.1.12), we can write the classical equations of motion as:

M∆(t) = −K∆(t), (6.2.1)

where components of the left-hand-side of the equation are the rates of change of momentum ofthe nuclei, while the right-hand-side contains the harmonic forces.

Since we expect harmonic motion, we’ll look for a solution of the form:

∆(t) = cos(ω t + φ )Y, (6.2.2)

where φ is an arbitrary phase shift. If Eq. (6.2.2) is used in Eq. (6.2.1) it follows that:

(Mω 2 − K)Y = 0, (6.2.3)

where remember that Y is a column vector and M, and K are matrices. We want to solve thishomogeneous system of linear equations for Y. In general, the only way to get a nonzero solu-tion is to make the matrix [Mω 2 − K] singular; i.e., we must set

det(Mω 2 − K) = det(ω 2 − M−1/2KM−1/2) det(M) = 0. (6.2.4)

Since, det(M) = (M1 M2. . . MN )3 ≠ 0 we see that the second equality is simply the characteristicequation associated with the eigenvalue problem:

Ku = λu, (6.2.5)

with λ ≡ ω 2 and with K given by Eq. (6.2.7) above. The remaining steps are equivalent to whatwas done quantum mechanically.

6.3. Force Constant Calculations

Here is an example of a force constant matrix calculation. We will consider a diatomicmolecule, where the two atoms interact with a potential of the form:

U(r1, r2) ≡1

2K

|r1 − r2| − R0

2

; (6.3.1)

i.e., a simple Hookian spring. It is easy to take the various derivatives indicated in the precedingsections; here, however, we will explicitly expand the potential in terms of the atomic displace-ments, ∆i . By writing, ri = Ri + ∆i (where Ri is the equilibrium position of the i’th nucleus), Eq.,

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(6.3.1) can be rewritten as:

U(r1, r2) =1

2K

R2

12 + 2R12 ⋅ (∆1 − ∆2) + |∆1 − ∆2|2

1/2

− R0

2

, (6.3.2)

where R12 ≡ R1 − R2. Clearly, the equilibrium will have |R12| = R0. Moreover, we expect thatthe vibrational amplitude will be small, and thus, the terms in the ∆’s in the square root in Eq.(6.3.2) will be small compared with the first term. The Taylor expansion of the square rootimplies that

√ A + B ≈ √ A +B

2√ A+. . . , (6.3.3)

we can write

U(r1, r2) =1

2K

R12 +

2R12 ⋅ (∆1 − ∆2) + |∆1 − ∆2|2

2R12

− R0

2

, (6.3.4)

which can be rewritten as

U(r1, r2) =1

2K [R12 ⋅ (∆1 − ∆2)]2, (6.3.5)

where the ˆ denotes a unit vector and where all terms smaller than quadratic in the nuclear dis-placements have been dropped. If the square is expanded, notice the appearance of cross terms

in the displacements of 1 and 2.

It is actually quite simple to finish the normal mode calculation in this case. To do so,define the equilibrium bond to point in along the x axis. Equation (6.3.5) shows that only x-components of the displacements cost energy, and hence, there will no force in thy y or z direc-tions (thereby resulting in 4 zero eigen-frequencies). For the x components, Newton’s equationsbecome:

m1

0

0

m2

⋅¨

∆x

1

∆x2

= −

K

−K

−K

K

⋅ ∆x

1

∆x2

, (6.3.6)

where mi is the mass of the i’th nucleus. This in turn leads to the following characteristic equa-tion for the remaining frequencies:

0 = det

m1

0

0

m2

ω 2 −

K

−K

−K

K

= (m1ω 2 − K )(m2ω 2 − K ) − K 2

and

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= ω 2(µω 2 − K ),

where µ ≡ m1m2/(m1 + m2) is the reduced mass. Thus we pick up another zero frequency and a

nonzero one with ω = √ K /µ, which is the usual result. (Note that we don’t count ± rootstwice--WHY?).

6.4. Normal Modes in Crystals

The main result of the preceding sections is that the characteristic vibrational frequenciesare obtained by solving for eigenvalues, cf., Eq. (6.2.5). For small to mid-size molecules this canbe done numerically, on the other hand, this is not practical for crystalline solids where the matri-ces are huge (3N × 3N with N∼O(N A))!

Crystalline materials differ from small molecules in one important aspect; they are peri-odic structures made up of identical unit cells; specifically, each unit cell is placed at

→R = n1

→a1 + n2

→a2 + n3

→a3, (6.4.1)

where ni is an integer and→ai is a primitive lattice vector. The atoms are positioned within each

unit cell at positions labeled by an index, α . Thus, the position of any atom in entire crystal isdetermined by specifying R and α . Hence, Eq. (6.2.5) can be rewritten as

λuR,α =R′Σ

α ′Σ KR,α ;R′,α ′uR′,α ′. (6.4.2)

The periodicity of the lattice implies that only the distance between unit cells can matter; i.e.,KR,α ;R′,α ′ = KR−R′;α ,α ′, and this turns Eq. (6.4.2) into a discrete convolution that can be simplifiedby introducing a discrete Fourier transform, i.e., we let

uk,α ≡RΣ eik⋅RuR,α . (6.4.3)

When applied to both sides of Eq. (6.2.5), this gives*

λ uk,α ≡αΣ Kk;α ,α ′uk,α ′, (6.4.4)

where

Kk;α ,α ′ ≡RΣ eik⋅RKR;α ,α ′. (6.4.5)

The resulting eigenvalue problem has rank 3Ncell , where Ncell is the number of atoms in a unitcell, and is easily solved numerically.

We hav en’t specified the values for the k’s. It turns out that a very convenient choice is to

use→k’s expressed in terms of the so-called reciprocal lattice vectors,

→G, i.e.,

*Strictly speaking, in obtaining Eq. (6.4.5) we have assumed some special properties of the lattice. Wedon’t use a finite lattice, rather one that obeys periodic boundary conditions (see below).

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→G ≡ m1

→b1 + m2

→b2 + m3

→b3, (6.4.6)

where the mi’s are integers, and where the reciprocal lattice basis vectors are defined as†

→b1 ≡ 2π

→a2 × →

a3

|→a1 ⋅ (

→a2 × →

a3)|,

→b2 ≡ 2π

→a3 × →

a1

|→a2 ⋅ (

→a3 × →

a1)|, and

→b3 ≡ 2π

→a1 × →

a2

|→a3 ⋅ (

→a1 × →

a2)|. (6.4.7)

Note the following:

a) The denominators appearing in the definitions of the→bi are equal; i.e.,

|→a1 ⋅ (

→a2 × →

a3)| = |→a2 ⋅ (

→a3 × →

a1)| = |→a3 ⋅ (

→a1 × →

a2)| = ν cell , (6.4.8)

where ν cell is the volume of the unit cell. The volume of the reciprocal lattice unit cell is(2π )3/ν cell .

b) By construction,

→ai ⋅

→b j = 2π δ i, j , (6.4.9)

where δ i, j is a Kronecker-δ .

c) The real and reciprocal lattices need not be the same, even ignoring how the basis vectorsare normalized; e.g., they are for the SCC, but the reciprocal lattice for the BCC lattice is aFCC lattice.

d) For any lattice vector,→R, and reciprocal lattice vector,

→G,

ei→R⋅

→G = e2π i(m1n1+m2n2+n3m3) = 1, (6.4.10)

cf. Eqs. (6.4.1), (6.4.6), and (6.4.9). Hence, adding any reciprocal a lattice vector to the→k

in Eq. (6.4.4) changes nothing, and so, we restrict→k to what is known as the First Brillouin

Zone; specifically,

→k = k1

→b1 + k2

→b2 + k3

→b3, (6.4.11)

where ki ≡ mi /Ni with mi = 0, 1, 2, . . . , Ni − 1. Note that N1 N2 N3 = N , the total numberof cells in the crystal. The Ni’s are the number of cells in the

→ai direction. With this

choice*, consider

1

N →k

Σ e−i→k⋅

→Ru →

k ,α =1

N →k

Σ→R′Σ ei

→k⋅(

→R′−

→R)u →

R′,α =→R′Σ u →

R′,α

3

i=1Π

1

Ni

Ni−1

mi=0Σ e2π imi(ni ′−ni)/Ni

.(6.4.12)

†Here we’re using the definition of C. Kittel, Introduction to Solid State Physics (Wiley, 1966), p. 53.There are others, see, e.g., M. Born and K. Huang, Dyamical Theory of Crystal Lattices (Oxford, 1968), p.69.*The more common choice for the range of the mi’s is mi = −Ni /2, . . . , 0, . . . , Ni /2, which, cf. Eq.(6.4.11), means that −1/2 ≤ ki ≤ 1/2.

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The sums in parenthesis are geometric series and give

1

Ni

Ni−1

mi=0Σ e2π imi(ni ′−ni)/Ni =

1

Ni

e2π i(ni ′−ni) − 1

e2π i(ni ′−ni)/Ni − 1

= δ ni ′,ni, (6.4.13)

which when used in Eq. (6.4.12) shows that

1

N →k

Σ e−i→k⋅

→Ru →

k ,α = u →R,α . (6.4.14)

The sums over→k are really sums over the mi’s, and thus, ki hardly changes as we go from

mi → mi + 1 for large Ni , cf. Eq. (6.4.11). This allows the sums over mi to be replaced byintegrals; i.e.,

→k

Σ . . . →3

i=1Π ∫

Ni−1

0dmi . . . =

V

(2π )31st Brillouin

zone

∫ ∫ ∫ d→k . . . , (6.4.15)

where V = Nν cell is the volume of the system.

The main goal of this discussion is to obtain the vibrational density of states. By using Eq.(6.4.15), it is easy to show that

g(ω ) =V

(2π )31st Brillouin

zone

∫ ∫ ∫ d→k δ (ω − ω (k)), (6.4.16)

where δ (x) is the Dirac δ -function. Once the frequencies are known, it is relatively easy tonumerically bin them by frequency as a function of k.

6.5. Normal Modes in Crystals: An Example

Consider a crystal with one atom of mass m per unit cell and nearest neighbor interactionsof the type considered in Sec. 6.3. Newton’s equations of motion for this atom become

m→∆0,0,0 = K [ex(∆x

1,0,0 + ∆x−1,0,0 − 2∆x

0,0,0) + ey(∆y0,1,0 + ∆y

0,−1,0 − 2∆y0,0,0)

+ez(∆x0,0,1 + ∆x

0,0,1 − 2∆x0,0,0)], (6.5.1)

where ei is a unit vector in the i direction. Note that this model implies that vibrations in the x, y,z, directions are separable. By using this in Eq. (6.4.5) we see that

Kk = −4ω 20

sin2(k1π )

0

0

0

sin2(k2π )

0

0

0

sin2(k3π )

, (6.5.2)

Winter Term, 2019


where ω0 ≡ (K /m)1/2 and where the identity, 1 − cos(x) = 2 sin2(x/2) was used. Since Kk is

already diagonal, we see that ω i(→k) = 2ω0 sin(kiπ ), i = 1, 2, 3, which is also the result for a one

dimensional chain (as was expected, given the separability of the vibrations in the x, y, and z

directions). The normal mode is just ei , which shows that our model potential is too simple.Consider the first eigen-vector. It corresponds to an arbitrary displacement in the x direction,with an eigenvalue that is independent of k y and kz . When k x = 0 we don’t get a single zero fre-quency, as expected, but one for any of the N y Nz values of (k y, kz). Thus, with 3N 2/3 zero fre-quencies, not 6, the crystal is unstable!

The problem is easily fixed by modifying the interaction potential, i.e., we replace Eq.(6.3.5) by

U ≡K

2(

→∆1 −

→∆2)2, (6.5.3)

which is invariant under rigid translations and rotations, and allows us to rewrite Eq. (6.5.1) as

m→∆0,0,0 = K (

→∆1,0,0 +

→∆−1,0,0 − 2

→∆0,0,0 +

→∆0,1,0 +

→∆0,−1,0 − 2

→∆0,0,0 +

→∆0,0,1 +

→∆0,0,1 − 2

→∆0,0,0). (6.5.4)

With this, it follows that

Kk = −4ω 20[sin2(k1π ) + sin2(k2π ) + sin2(k3π )]

↔1, (6.5.5)

where↔1 is a 3 × 3 identity matrix and where we have expressed

→k in terms of the reciprocal lat-

tice basis vectors, cf. Eq (6.5.11). The triply degenerate vibrational frequencies are simply

ω i(→k) = 2ω0[sin2(k1π ) + sin2(k2π ) + sin2(k3π )]1/2. (6.5.6)

If all the ki → 0, Eq. (6.5.6) becomes ω (→k) ∼ 2ω0kπ , where k ≡ (k2

1 + k22 + k2

3)1/2 which is theexpected linear dispersion law for long wevelength, acoustic vibrations. Some constant fre-quency surfaces in the First Brillouin Zone are shown in Figs. 6.5.1-6.5.3.

Winter Term, 2019


Fig. 6.5.1. A constant normal mode frequency sur-face for ω /ω0 = 1. The reciprocal lattice basis wasused which need not be orthogonal (although it is forthe SCC lattice). Also note that we’ve switched tothe other definition of the first Brillouin zone, with−1/2 ≤ ki ≤ 1/2. The surface is roughly spherical, asexpected for long wav elength acoustic phonons.

Fig. 6.5.2. As in Fig. 6.5.1 but with ω /ω0 = 2.Notice the C4 axises through the centers of any of theunit cell faces.

Fig. 6.5.3. As in Fig. 6.5.1 but withω /ω0 = 3.

Normal mode frequencies for this model were computed numerically for a uniform sample ofki’s in the First Brillouin Zone and binned in order to get the vibrational density of of states asshown in Fig. 6.5.4

Winter Term, 2019


Fig. 6.5.4. Vibrational density of states for the isotropic model defined byEq. (6.5.3). Any single atom per unit cell lattice will give the same result.The data was obtained by numerically binning the frequencies in the FirstBrillouin Zone. Note that the maximum frequency for this model is 2√3ω0.

By noting that

δ ( f (x)) =iΣ δ (x − xi)

df (xi)

dxi

, (6.5.7)

where f (x) has zeros at x = xi , i = 1, . . ., and that

δ (x) = ∫∞−∞

ds

2πeixs, (6.5.8)

we can rewrite Eq. (6.4.16) as

g(ω )

3N= 2ω ∫

1

0dk1 ∫

1

0dk2 ∫

1

0dk3 δ (ω 2 − ω 2(k))

=ωπ

∞

∫−∞ds ∫

1

0dk1 ∫

1

0dk2 ∫

1

0dk3 eis(ω 2−ω 2(k))

=ωπ

∞

∫−∞ds eisω 2

Φ3(s),

where the extra factor of 3 is due to the triple degeneracy of each mode and where

Winter Term, 2019


Φ(s) ≡ ∫1

0dk e−4iω 2

0 s sin2(π k)

When we use Eq. (6.5.7) for the frequencies, it follows that the wav evector integrations separateand

g(ω ) = 2Vω ∫∞−∞

ds

2πeisω 2

Φ3(s), (6.5.9)

where

Φ(s) ≡ ∫dk1

2πe−is4ω 2

0 sin2(π k1) =e−2ω 2

0is

2π 2 ∫π

0dz e2ω 2

0is cos(z)

=e−2ω 2

0is

2π 2 ∫1

−1dz

e2ω 20isz

(1 − z2)1/2=

e−2ω 20is

2πJ0(2ω 2

0 s), (6.5.10)

where J0(x) is a Bessel function of the first kind.* When this is used in Eq. (6.5.10), we see that

g(ω ) =2Vω(2π )4 ∫

∞−∞

ds eis(ω 2−6ω 20)J3

0 (2ω 20 s) 6.5.11)

*The integral leading to the last equality can be found in I.S. Gradsheyn and I.M. Ryzhik, Table of Inte-

grals, Series, and Products, A. Jeffrey editor, (Academic Press, 1980), Eq. (3.387.2) on p. 321.

Winter Term, 2019

Chemistry 365 -48- Gibbs-Duhem Relations

7. Gibbs-Duhem Relations

Here’s a quick review of how the so-called Gibbs-Duhem relation is obtained in thermody-namics. Consider an extensive function like the energy of the system, E. Extensive means that

E(λextensive variables, intensive variables)

= λ E(extensive variables, intensive variables), (7.1)

where λ is arbitrary. For example, consider the energy as a function of the entropy, S, volume,V , and number of moles or molecules, N (for a one component system); Since each of these isextensive, and can be used to specify state of a one-component system, Eq. (7.1) implies that

E(λ S, λV , λ N ) = λ E(S, V , N ). (7.2)

By differentiating both sides with respect to lambda, and evaluating the result at λ = 1 we get:

S

∂E

∂S

V ,N

+ V

∂E

∂V

S,N

+ N

∂E

∂N

S,V

= E(S, V , N ). (7.3)

If we combine the first and second laws of thermodynamics we can write

dE = TdS − pdV + µdN , (7.4)

where T , p, and µ are the absolute temperature, pressure, and chemical potential, respectively.When this is used to evaluate the derivatives appearing on the left hand side of Eq. (7.3) it fol-lows that

E(S, V , N ) = TS − pV + µN . (7.5)

which was used in class.

By taking the differential of Eq. (7.5) we see that

dE = TdS + SdT − pdV − Vdp + µdN + Ndµ, (7.6)

which when compared with Eq. (7.4) implies that

0 = SdT − Vdp + Ndµ. (7.7)

This is known as the Gibbs-Duhem relation, and implies that changes in the intensive quantitiesare not independent. In particular, it implies that

∂µ

∂ p

T

=V

N≡ V (7.8)

Winter Term, 2019

Gibbs-Duhem Relations -49- Chemistry 365

and

∂µ

∂T

P

= −S

N≡ S. (7.9)

How does this discussion generalize for mixtures? Show that for any extensive quantityA(T , p, N1, N2, . . . ) ,

A(T , p, N1, N2, . . . ) = N1 A1 + N2 A2+. . . , (7.10)

where

Ai ≡

∂A

∂Ni

T ,P,N j≠i

(7.11)

is a partial molar quantity.

Winter Term, 2019

Chemistry 365 -50- Chemical Equilibrium

8. Chemical Equilibrium

8.1. Thermodynamics and Chemical Equilibrium

Here we will review some of the general concepts governing chemical equilibrium in ther-modynamics and in statistical thermodynamics. First, some notation. Consider the followingchemical reaction

r1 R1 + r2 R2+. . . →← p1P1 + p2P2+. . . , (8.1a)

where r( p)i are the stochiometric coefficients for the reactant (product) compound R(P)i . It isuseful to rewrite Eq. (8.1a) as

ν1 A1 + ν2 A2+. . . = 0, (8.1b)

where the stochiometric coefficients, ν i are positive for product compounds (i.e., ν i = pi) andnegative for reactant ones (i.e., ν i = − ri).

Consider a state where there are Ni molecules of compound i. As the reaction proceeds tothe left or right, the changes in the number of atoms occur in definite proportion, i.e.,

dN1

ν1

=dN2

ν2

= . . . = dξ , (8.2)

where ξ is known as the extent of the reaction. The reaction will proceed until until the GibbsFree Energy (G) is a minimum (at constant temperature and pressure). Moving the chemicalequilibrium from the minimum incurs a free energy cost:

dG = − SdT + VdP +iΣ µi dNi = − SdT + VdP +

iΣν iµi

dξ , (8.3)

where the second equality follows from Eq. (8.2). Temperature and pressure are constant, andhence, the first two terms vanish. On the other hand, the last term doesn’t hav e a definite sign,since the reaction in principle can move either to the right (dξ > 0) or left (dξ < 0); hence, wecan’t be at a minimum G unless the coefficient of dξ vanishes, i.e.,

iΣν iµi = 0. (8.4)

Eq. (8.4) is completely general and applies to any reaction (even to those that don’t inv olvebreaking and making bonds, e.g., as in phase equilibrium). Consider a chemical reaction in anideal gas mixture. We know that

µi = µ(0)i (T ) + kBT ln(pi), (8.5)

where pi is the partial pressure of compound i (in atmospheres) and µ(0)i (T ) is the standard

Winter Term, 2019

Chemical Equilibrium -51- Chemistry 365

chemical potential (free energy of formation per molecule) at 1atm. By using this in Eq. (8.4) wefind that

iΣν i ln(Pi) = −

iΣ ν iµ

(0)i (T )

kBT≡ −

∆G(0)(T )

kBT. (8.6)

Finally by combining the logs on the left and exponentiating, we obtain

iΠp

ν i

i = e−∆G(0)(T )/kBT ≡ KP(T ). (8.7)

By recalling that the ν i’s are negative for reactants, it follows that the quantity on the left of Eq.(8.7) is the usual stochiometric quotient; K p is the constant pressure equilibrium constant.

8.2. Chemical Equilibrium in Statistical Mechanics

Our treatment here uses the same thermodynamic principles as above, in particular Eq.(8.4), but now we know the chemical potentials (at least of dilute gases) in terms of molecularproperties. Recall that

µi =

∂A

∂Ni

T ,V ,N j≠i

= − kBT

∂ ln Q

∂Ni

T ,V ,N j≠i

. (8.8)

By noting that in Boltzmann statistics (assuming separability)

Q =i

Π qNi

Ni!, (8.9)

where the molecular partition function for component i, qi , can be written as

qi = qtransi qel

i [qroti qvib

j ]. (8.10)

Remember that not all molecules have rotational and/or vibrational degrees of freedom, and ifthey do, then their contribution depends on whether the molecule is linear or not and how manyvibrational modes it has [3n − 5 (linear) or 3n − 6 (nonlinear)]. Only one contribution dependson the volume, namely,

qtransi =

V

Λ3i

, (8.11)

where mi is the molecule’s mass and Λi = h/(2π mi kBT )1/2 is its thermal deBroglie wav elength.With this, using Stirling’s approximation for ln(N !), Eq. (8.9) becomes

β µi = ln(ρ i) − ln

qi

V

, (8.12)

where ρ i ≡ Ni /V is the number density of component i. Finally, by using Eq. (8.12) in (8.4) andrearranging the logarithms (essentially as was done in obtaining Eq. (8.7)), we find that

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Chemistry 365 -52- Chemical Equilibrium

iΠρν i

i =i

Π

qi

V

ν i

≡ Kρ . (8.13)

Since the gas is ideal, ρ i = β pi , and we can rewrite Eq. (8.13) as

iΠp

ν i

i = KP = Kρ (kBT ) iΣν i

. (8.14)

(Be careful with units here).

8.3. An Example

Consider the following chemical reaction in the gas phase:

N2(g) + 3H2(g) →← 2NH3(g). (8.15)

Table 8.1 contains expressions for the various partition functions needed for the reaction 8.15.

Table 8.1: Contributions to Kρ for the reaction in Eq. (8.15)

Compound NH3 N2 H2

ν i 2 -1 -3

Geometry pyramidal linear linear

qtrans/V Λ−3NH3

Λ−3N2

Λ−3H2

3 2 2Symmetry Number

3 2 2# Rot. DoF

qroti π 1/2

3

T 3

ΘANH3

ΘBNH3

ΘCNH3

1/2 T

2ΘN2

T

2ΘH2

6 1 1# vib. DoF

qvibi 6

i=1Π

e−Θvib,iNH3

/2T

1 − e−Θvib,iNH3

/T

e−ΘvibN2

/2T

1 − e−ΘvibN2 /T

e−ΘvibH2

/2T

1 − e−ΘvibH2 /T

qeli e−β ε el

NH3 e−β ε elN2 e−β ε el

H2

Notes:1) DoF = degrees of freedom.2) Only the electronic ground state is included.3) H2, N2 and NH3 all have non-degenerate electronic ground states.

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Chemical Equilibrium -53- Chemistry 365

Next we write Kρ = q2NH3

/(qN2q3

H2) = Ktrans Kel Krot Kvib, where

Ktrans =Λ−6

NH3

Λ−9H2

Λ−3N2

=

m2NH3

mN2m3

H2

3/2

h2

2π kBT

3

, (8.16a)

Kel = e−β ∆ε rxn , (8.16b)

Krot =16π9T

ΘN2Θ3

H2

ΘANH3

ΘBNH3

ΘCNH3

, (8.16c)

and

Kvib =

1 − e−Θvib

N2/T

1 − e−Θvib

H2/T

3

6

i=1Π

1 − e−Θvib,i

NH3/T

2, (8.16d)

where

∆ε rxn ≡ 2ε el

NH3+

6

i=1Σ 1

2h− ω vib,i

NH3

− ε el

N2+ 1

2h− ω vib

N2

− 3ε el

H2+ 1

2h− ω vib

H2

, (8.17)

is the observable or true energy of reaction, in that the Born-Oppenheimer ground state energieshave been raised to include the zero-point vibrational energies of all the molecules’ vibrationalmodes. Note the following limiting behaviors for Kvib:

Kvib ∼

1

T 8Θvib

N2(Θvib

H2)3

6

i=1Π(Θvib,i

NH3)2

for T << all Θvib

for T >> all Θvib.

(8.18)

The requirement on the high temperature limit is nontrivial given that ΘvibH2

is 6210 K (the othercompounds are much less), and some of our approximations will likely break down.

Winter Term, 2019

Chemistry 365 -54- Dense Gases: Virial Coefficients

9. Dense Gases: Virial Coefficients

One way to correct the ideal gas equation of states is to use the virial expansion, namely,

β p = ρ + B2(T )ρ2 + B3(T )ρ3+. . . , (9.1)

where ρ ≡ N /V is the number density and where Bn(T ) is known as the n’th virial coefficient.Our goal is to obtain a molecular expression for the virial coefficients (at least for B2(T )).

To proceed, we need to identify a useful small parameter that can be used to expand theequation of state. To do this, note that in Boltzmann statistics and ignoring all interactionsbetween molecules, the canonical partition function for a system of N molecules,QN (V , T ) = [q1(T )V ]N /N !, where V q1(T ) is the molecular partition function. Hence, the grandpartition function becomes

Ξ =∞

N=0Σ eβ µN [V q1(T )]N

N != exp

λV q1(T )

, (9.2)

where λ ≡ eβ µ, is the activity. By taking the appropriate derivatives of log(Ξ), it follows that

ρ ≡⟨N ⟩V

=1

V

∂ log(Ξ)

∂β µ

β ,V

=λV

∂ log(Ξ)

∂λβ ,V

= λ q1(T ), (9.3)

where the last equality is obtained when Eq. (9.2) is used. The ideal gas equation of state arisessince β p = V −1 log(Ξ) = λ q(T ) = ρ . Thus, we see that λ∝ρ as ρ → 0, where ideal gas behavioris expected. Since, in general,

Ξ = 1 +∞

N=1Σ λ N QN (V , T ), (9.4)

we can view the grand partition function as a power series in the small parameter λ , at least forlow densities.

Unfortunately, λ is not a particularly convenient quantity to control experimentally (unlikethe density). To deal with this, assume that the activity can be written as a power series in den-sity; i.e.,

λ = ρc1 + ρ2c2 + ρ3c3+. . . . (9.5)

From Eqs. (9.3) and (9.4) it follows that

ρV =

∞

N=1Σ N λ N QN (T , V )

1 +∞

N=1Σ λ N QN (T , V )

, (9.6)

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Dense Gases: Virial Coefficients -55- Chemistry 365

or equivalently,

ρV1 +

∞

N=1Σ λ N QN (T , V )

=∞

N=1Σ N λ N QN (T , V ). (9.7)

Equation (9.5) is used to rewrite Eq. (9.7) as

ρV +∞

N=1Σ ρ N+1(c1 + ρc2 + ρ2c3 + . . .)NVQN (T , V ) =

∞

N=1Σ N ρ N (c1 + ρc2 + ρ2c3 + . . .)N QN (T , V ),

(9.8)

and similar to what is done in the Frobenius method for ODE’s, we equate the coefficients ofeach power of ρ; i.e.,

V = c1Q1 for O(ρ), (9.9a)

c1VQ1 = c2Q1 + 2c21Q2 for O(ρ2), (9.9b)

c2VQ1 + c21VQ2 = c3Q1 + 4c1c2Q2 + 3c3

1Q3 for O(ρ3), (9.9c)

etc.. These equations are easily solved, giving

c1 =V

Q1

, (9.10a)

c2 = −(2Q2 − Q2

1)V 2

Q31

, (9.10b)

c3 = −(3Q1Q3 − 8Q2

2 + 5Q21Q2 − Q4

1)V 3

Q51

, (9.10c)

etc.. By substituting Eqs. (9.10a-c) into Eq. (9.5) we see that

λ =ρV

Q1

1 −

(2Q2 − Q21)ρV

Q21

−(3Q1Q3 − 8Q2

2 + 5Q21Q2 − Q4

1)ρ2V 2

Q41

+ . . ., (9.11)

which gives

β µ = log(λ) = log

ρV

Q1

−

2Q2 − Q2

1V ρ

Q21

+

Q4

1 − 6Q2Q21 − 6Q3Q1 + 12Q2

2V 2 ρ2

2Q41

+ . . . , (9.12)

Winter Term, 2019


expressing the non-ideal corrections to the Gibbs free energy (recall that µ = G in a one compo-nent system) in terms of canonical partition functions for systems of N = 2, 3, . . . particles; inaddition, we’ve made no assumptions about using classical or quantum mechanics. Similarly,remembering that β pV = log Ξ, we see that

β p = ρ −(2Q2 − Q2

1)V ρ2

2Q21

−(6Q1Q3 − 12Q2

2 + 6Q21Q2 − Q4

1)V 2 ρ3

3Q41

+ . . . . (9.13)

Note that in obtaining Eqs. (9.12) and (9.13) we have had to re-expand the logs in order to beconsistent with the omitted terms.*

Our results can be simplified if we assume that the internal electronic, vibrational, androtational degrees of freedom are separable, and that the translational ones are classical. Thus

QN ≡qN

1 ZN

N !,

where, as above, q1 is the molecular partition function for all intra-molecular degrees of freedom,including translation, divided by volume, and

ZN ≡ ∫ dr1. . . ∫ drN e−βU (N )(r1,...,rN ), (9.14)

where U (N )(r1, . . . , rN ) is the interaction potential for a system of N molecules and Z1 = V . ZN

is known as the configurational partition function. Note that the separability assumption is rea-sonable for electronic (in the absence of chemical reactions) and for vibrational degrees of free-dom, but is problematic for rotations if the molecule is aspherical. Fortunately, the rotationaldegrees of freedom can be treated classically for most of the cases of interest, and can be incor-porated into the definition of ZN , leaving only the rotational kinetic energy contributions in q1

(this is analogous to what we do with the translational contributions in q1 or with the rotationalpartition function when T >> Θrot).

With this change of notation, it follows that

β µ = log

ρq1

−∞

n=1Σ β n ρ n, (9.15)

where

*Specifically, we use the Taylor expansion

log(1 + x) = x −1

2x2 +

1

3x3 + . . . =

∞

n=1Σ (−1)n+1

nxn,

where x ≡ λ /(ρV /Q1), cf. Eq. (11). Thus, e.g., if we only keep terms to O(ρ) in λ we should only keepterms through the same order in the series for the logarithm; i.e.,

β µ = log(ρV /Q1) − (2Q2 − Q21)V ρ /Q2

1 + O(ρ2).

Winter Term, 2019


β1 ≡Z2 − V 2

V, (9.16a)

β2 ≡Z3V + 3Z2V 2 − V 4 − 3Z2

2

2V 2, (9.16b)

etc.. The β n’s are referred to as n’th order irreducible cluster integrals. By expressing the densityexpansion of the pressure in terms of the configurational partition functions and the irreduciblecluster integrals, we see that

β p = ρ −β1

2ρ2 −

2β2

3ρ3+. . . . (9.17)

Hence, B2(T ) = −β1/2, B3(T ) = −2β2/3.

A general connection between the irreducible cluster integrals and the virial coefficientscan easily be obtained from the Gibbs-Duhem relation, SdT − Vdp + Ndµ = 0, specifically,

∂β p

∂ρ

T

= ρ

∂β µ

∂ρ

T

= 1 −∞

n=1Σ nβ n ρ n, (18)

where the second equality is obtained when (9.15) is used. By integrating, remembering thatp = 0 when ρ = 0, we see that

β p = ∫ρ

0d ρ

∂ p

∂ρ

T

= ρ −∞

n=1Σ n

n + 1β n ρ n+1, (19)

and thus,

Bn+1 = −n

n + 1β n. (9.20)

It turns out that even though our definitions of β n explicitly contain the volume V, the vol-ume dependence disappears. For example, consider the second virial coefficient, B2(T ).According to Eqs. (9.16a) and (9.20),

B2(T ) = −Z2 − V 2

2V=

1

2V ∫ dr1 ∫ dr21 − e−β u12(r12)

, (9.21)

where u12(r12) is the interaction potential of a pair of molecules separated by a distance r12. Forneutral molecules, the interaction rapidly decays to zero once r12 is much larger than a few tensof angstroms, as will the integrand in Eq. (9.21); hence, it is useful to change one of the integra-tion variables, say r1 to a basis centered on r2. This allows Eq. (9.21) to be rewritten as

B2(T ) =1

2V ∫ dr2 ∫ dr121 − e−β u12(r12)

=

1

2 ∫ dr121 − e−β u12(r12)

. (9.22a)

Winter Term, 2019


Similar manipulations can be used in Eq. (9.16b) to show that B3 only depends on temperature.More generally, there are some sophisticated mathematical tools that prove this in general, butthey are well beyond the level of this class.*

For spherically symmetric interactions we can switch to polar coordinates, giving

B2(T ) = 2π ∫∞0

dr12 r212

1 − e−β u12(r12)

. (9.22b)

If the pair interactions are predominantly repulsive, i.e., u12(r12) > 0, the quantity in the paren-thesis will be positive, as will B2(T ) and the deviation from ideal gas behavior will be positive.If the interactions are predominantly attractive, the reverse will happen. Consider a more realis-tic potential where steric repulsions dominate at short distances and attractions dominate a largerdistances, e.g., as for the so called square well potential

u12(r) =

∞ for r < σ−ε for σ < r < γ σ0 otherwise,

where ε ≥ 0 and γ ≥ 1. The potential is shown in Fig. 9.1.

Fig. 9.1. The square well (solid) and more realistic Lennard-Jones (L-J) 6-12 (dashed) potentials, respectively. The L-J potential,u(r) = 4ε [(σ /r)12 − (σ /r)6], is more realistic, in that it includes theeffects of London dispersion forces at larger distances.

The integral in Eq. (9.22b) is easily done and gives

B2(T ) =2π σ 3

3

1 − (γ 3 − 1)(eβ ε − 1)

. (9.23)

*See, e.g., J.P. Hansen and I.R. MacDonald, Theory of Simple Liquids, (Academic Press, London, 1976),ch. 4.

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From this it is easy to show that the Boyle temperature, where B2(TB) = 0, is

TB = −ε

kB log(1 − γ −3).

The van der Waals model gives B2(T ) = b − β a, where a and b are the van der Waals a and bcoefficients. Both models show that the effects of steric repulsion and attraction are additive.Other than that, the temperature dependence of the attractive terms differ slightly, although theyapproach each other when β ε → 0 where eβ ε − 1∼β ε .

The integration in Eq. (9.22b) for the more realistic Lennard-Jones 6-12 potential is morecomplicated and is given in the Appendix.

Fig. 9.2. The reduced second virial coefficient for the Lennard-Jones poten-tial. B*

2 ≡ B2/B2, hard sphere, where B2, hard sphere ≡ 2π σ 3/3, is the second virialcoefficient for the hard-sphere potential, and where the reduced temperatureis T * ≡ kBT /ε . As expected B*

2 > (<) 0 for high (low) reduced temperatures,where repulsions (attractions) dominate. The reduced Boyle temperature isapproximately 3.43. Note that B*

2∝(T *)−1/4 as T * → ∞ and is not constant(why?).

The integral in our expression for B2(T ), cf. Eq. (9.22b), imposes some important limita-tions on the interaction potential, u12(r); specifically, for large separations u12(r) → 0, and thusr21 − exp[−β u12(r)]∼r2 β u12(r) showing that u12(r) must decay faster than 1/r3 in order thatthe integral converges. Charge-charge, charge-dipole, and dipole-dipole interactions decay liker−1, r−2, and r−3, respectively! Charge-dipole and dipole-dipole interactions depend on the orien-tation of the dipole moments (i.e., angular degrees of freedom), and when these are includedproperly the integral converges. Ionic interactions are more problematic; in short, the systemmust be electrically neutral, and even then, the basic assumption about the nature of the virialexpansion (a power series in density) breaks down, giving rise to ρ1/2 terms.

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9.1. Appendix: B2(T ) for the Lennard-Jones potential

This section is based on the extensive discussion of the viral expansion by Hirshfelder,Curtiss, and Bird* First, express B2 in a dimensionless form by writing

B2(T ) =2π σ 3

3B*

2(T *), (9.24)

where we have let r → σ r, T * ≡ kBT /ε is the reduced temperature and where the reduced 2ndvirial coefficient is

B*2(T *) ≡ 3 ∫

∞0

dr r21 − exp

−

4

T *(r−12 − r−6)

, (9.25a)

=24

T * ∫∞0

dr r2(2r−12 − r−6) exp−

4

T *(r−12 − r−6)

, (9.25b)

=24

T * ∫∞0

dr r2(2r−12 − r−6) exp−

4

T *r−12

∞

n=0Σ 1

n!

4r−6

T *

n

,(9.25c)

where the second equality is obtained by integration by parts and where the last expression isobtained by expanding exp(4r−6/T *) in a Taylor series (i.e., we treat the attractions in the Boltz-mann factor as a perturbation).

Next we change variables by letting

z ≡4

T *r−12 or r =

T *

4z

−1/12

, (9.26)

thereby implying that

dr = −

T *

4

−1/12z−13 / 12

12dz. (9.27)

When these are used in Eq. (9.25c), the latter becomes

B*2(T *) =

∞

n=0Σ 2n+1/2(T *)−(2n+3) / 4

n! ∫∞0

dz z(2n−3) / 4(√z√ T * − 1)e−z (9.28a)

=∞

n=0Σ 2n+1/2(T *)−(2n+3) / 4

n!

√ T *Γ

2n + 3

4

− Γ

2n + 1

4

(9.28b)

*J.O. Hirshfelder, C.F. Curtiss, and R.P. Bird, Molecular Theory of Gases and Liquids, 2nd ed. (John Wileyand Sons Inc, N.Y., 1964), ch. 3.

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= −∞

n=0Σ

2n−3/2Γ

2n − 1

4

n!

1

T *

(2n+1) / 4

, (9.28c)

where

Γ(z) = ∫∞0

dt t z−1e−t , (9.29)

is known as the Γ-function. The last equality was obtained by noting that Γ(x + 1) = xΓ(x) andredefining the sum index.

Winter Term, 2019

Chemistry 365 -62- Summary of Key Results

10. Summary of Key Results

Some Key Results for CHEM 365

Rotation

Diatomic/linear PolyatomicTranslation Vibration ElectronicType of Motion

h2(n2x + n2

y + n2z)

8mL2

h2

8π 2 IJ(J + 1) ? h− ω (n + 1/2) ε j

Energy levels

Degeneracy 1 (2J + 1) ? 1 ω j

0 Θrot ≡h2

8π 2 IkB

Θαrot ≡

h2

8π 2 Iα kB

Θvib ≡h− ωkB

∆ε /kB

Characteristic tem-

perature

2π mkBT

h2

3/2

V

N

N !

Molecular partition

function (except for

translation)

J=0Σ (2J + 1)e−Θrot J(J+1) /T

(heteronuclear)

? e−Θvib/2T

1 − e−Θvib/Tj

Σω je−β ε j ≈ ω0e−β ε0

"T

σ Θrot

π 1/2

σ ABCΠ

T

Θαrot

1/2T

Θvib

N/AHigh temperature

approximation to

partition function

Aα = −kBT ln(qα )Helmholtz free

energy per molecule

kBT ln(ρΛ3/e)

Sα =< Eα > −Aα

T

Entropy per mole-

cule

−kB ln(ρΛ3/e5/2)

3

2kBT kBT

3

2kBT kBT N/A

High temperature

energy per molecule

NkBT

V0 0 0 0

Contribution to pres-

sure

kBT ln[ρΛ3] −kBT ln[T /(σ Θrot)] −kBT ln

π 1/2

σ ABCΠ

T

Θαrot

1/2

h− ω2

+ kBT ln(1 − e−Θvib/T ) −kBT ln(qelectronic)Chemical potential

2π mkBT

h2

3/2T

σ Θrot

π 1/2

σ ABCΠ

T

Θαrot

1/2e−Θvib/2T

1 − e−Θvib/Tj

Σω je−β ε j

Equilibrium constant

contribution

Winter Term, 2019

Summary of Key Results -63- Chemistry 365

11. Problem Sets

Note that the due dates are last year’s. This year’s will be announced in class and on the web

site.

11.1. Problem Set 1

DUE: Friday, January 21, 2019

1. Consider the binomial coefficient

N

m

≡N !

m!(N − m)!.

For large values of N and m, show that it has its largest value when m = N /2 (HINT: useStirling’s formula). What is the maximum value? For values of m near N/2, show that thebinomial coefficient is approximately Gaussian in form (HINT: Expand the log of the bino-mial coefficient in a Taylor series in m around the maximum value). What is the magnitudeof the width of the Gaussian? Given these results, would using the maximum term methodfor series involving the binomial coefficients be valid? Why?

2. a) Hill, Introduction to Statistical Thermodynamics, problem 1-5 (p.32).

b) Rather than using the equal-apriori probability postulate, Gibbs assumed that theentropy is always given by

S = −kB Σ P ln(P),

where the sum is over all the things which can fluctuate in the ensemble. In a systemwith constant T, N, and V, the thermodynamic equilibrium state is that which mini-mizes the Helmholtz free-energy, A ≡ E − TS; minimize the free-energy with respect toP j and compare your result with the canonical distribution. (Remember that the prob-abilities must sum to unity.)

3. Use the expression for the energy levels of a particle in a cubical box to show that

p j =2E j

3V

where p j is the pressure associated with the j’th state. Av erage both sides of the equationand use the result (to be shown later) that < E >= 3NkBT /2 to derive the ideal gas equationof state.

4. A model for a spin in a magnetic field can be obtained from statistical mechanics. Theenergy of a spin-I nucleus in a magnetic field is:

Hmag = − γ h− HIz

Winter Term, 2019

Chemistry 365 -64- Problem Sets

where h− is Planck’s constant, γ is a constant called the "gyromagnetic ratio", H is the exter-nal field, and Iz is the projection of the total spin along the direction of the external field(i.e., it is the magnetic quantum number and can take on values −I , −I + 1, . . . , I − 1, I ).

a) Work out an expression for the canonical partition function. HINT: Recall that for geo-metric series:

B

j=AΣ x j =

xB+1 − x A

x − 1

and that

sinh(x) =1

2

(ex − e−x).

b) By using a canonical ensemble, derive expressions for the average magnetization (i.e.,γ h− < Iz >), energy, and entropy per spin in the presence of the external field. Ignore allinteractions between different spins.

c) What is the expression for the constant external field heat capacity? How does itbehave at high and low temperatures? HINT: note that

csch2(x) ≡1

sinh2(x)≈

1

x2−

1

3+ O(x2), as x → 0,

4e−2x + O(e−4x), as x → ∞.

d) Sketch CH as a function of temperature.

5. An approximate canonical partition function for a dense gas is:

Q(N , V , T ) =1

N !

2π mkBT

h2

3N /2

(V − Nb)N exp[aN 2/VkBT ],

where m is the mass of the particles and a and b are molecular parameters (which are inde-pendent of temperature). Calculate the energy, entropy, pressure, and chemical potential forthis system. To what well known thermodynamic approximate equation of state does youranswer correspond?

Winter Term, 2019

Problem Set 2 -65- Chemistry 365

11.2. Problem Set 2

DUE: Monday, January 28, 2019

1. Hill, Introduction to Statistical Thermodynamics, problem 2-5. What general thermody-namic principle could be used to justify the second equality without using statistical thermo-dynamics?

2. For the grand canonical ensemble we’ve obtained two expressions for the pressure:

P =kBT

Vln Ξ or P = kBT

∂ ln Ξ∂V

β µ,β

,

where Ξ is the grand canonical partition function. Check that the derivative does not givethe first expression exactly. Nonetheless, the pressure calculated in the two ways agree--at

least to any reasonable accuracy. Why? (HINT: Suppose the two expressions are equiva-lent, and derive and solve a differential equation for ln Ξ. What does your result say aboutthe dependence of the pressure on V , β µ and β ?).

3. Hill, Introduction to Statistical Thermodynamics, problem 3-1.



Winter Term, 2019

Chemistry 365 -66- Problem Set 3

11.3. Problem Set 3

DUE: Monday, February 4, 2019

1. A system consists of two particles, each of which has two possible quantum states withenergies ε0 and ε1. Write the complete expressions for the partition function and for theheat capacity if:

a) The particles are distinguishable.

b) They are Fermions.

c) They are Bosons.

d) They obey Boltzmann statistics.

2. Calculate the entropy and heat capacity per mole of a 1atm gas of Argon at 300K usingBoltzmann statistics and 3000K and compare your answer with experiment. (You can findthe experimental numbers in the American Institute of Physics Handbook). What are themain sources of disagreement? Repeat the calculation for O vapor (HINT: Be careful here,atomic oxygen has some low energy excited electronic states and these will be populated atmoderately high temperature. You can find the spectroscopic data for these in the NISTdatabase at: http://physics.nist.gov/PhysRefData/ASD/index.html).

3. Calculate the exact molecular partition function for the hydrogen atom given that the energylevels in hydrogen are

En = −2π 2me4

n2h2, n = 1, 2, . . .

and that each level has degeneracy 2n2. How is this seemingly paradoxical resultexplained?

4. Consider a gas in equilibrium with a surface. The surface can adsorb the gas moleculesonto any of M independent, distinguishable sites (distinguishable because the binding sitesare localized to specific parts of the surface). The molecular partition function for anadsorbed molecule is q(T ) ≡ exp[−β Asurface].

a) Assume that the adsorbed molecules are independent and use the canonical ensemblefor N adsorbed molecules to compute the chemical potential for the molecules on thesurface. Remember to include the number of ways to adsorb the molecules onto thesurface. Also remember that the molecules themselves are indistinguishable.

b) Use our results for an ideal gas to compute the chemical potential in the gas.

c) If the gas and surface phases are in equilibrium, how does the fractional coverage,N /M , depend on the pressure in the gas. Your expression is known as the Langmuiradsorption isotherm.

5. (HARD) Consider a system of M independent and distinguishable macromolecules onwhich any number from 0 to m small molecules may bind. Let q j be the partition function

Winter Term, 2019


of the macromolecule when j molecules are bound. If there are N small molecules (e.g.,ions) and M macromolecules (e.g., proteins), compute the canonical partition function forthe system. Express your result as a sum over the number of macromolecules which have0, 1, . . . , m small molecules bound (assuming that binding to different macromolecules stillresults in the same qi’s). Show that the semi-grand partition function (i.e., M is fixed, but Nis not) is:

Ξ(M , T , µ) = ξ (µ, T )M

where

ξ (µ, T ) ≡m

j=0Σ q j λ

j ,

where λ ≡ eβ µ. Interpret this result. What happens if the macromolecules are indistinguish-able?

HINT: Consider the number of ways N small molecules can be attached to M polymerssuch that there are ni polymers with i small molecules attached. You’ll also need the gener-alization of the binomial formula, the so-called multinomial formula, i.e.,

m

i=0Σ ni=M

nimi=0

Σ M!m

i=0Π

xni

i

ni!=

m

i=0Σ xi

M

.

Winter Term, 2019


11.4. Problem Set 4

DUE: Friday, February 15, 2019

1. Hill, Statistical Thermodynamics, Problem 22-6.

2. Prove the equipartition theorem; i.e., consider a classical system with a Hamiltonian, H , thatis a quadratic function of the momenta and coordinates (collectively referred to asX ≡ x1, . . . , xN ):

H(X) ≡1

2

N

i, j=1Σ xi x j Hi, j , (1)

where Hi, j = H j,i are real constants. Ignore the factor of N ! (why might there not be one?) andshow that

q =

(2π kBT )N

hα det(Hi, j)

1/2

, (2)

where det(Hi, j) is the determinant of the matrix whose elements are the Hi, j parameters, andwhere α is the number of degrees of freedom described by the Hamiltonian in Eq. (1). (usuallyα = N /2). In order to show (2), you will need the following integral:

∫∞−∞

dx1 ∫∞−∞

dx2 . . . ∫∞−∞

dxN exp−

N

i, j=1Σ xi x j Ai, j

=

π N

det(Ai, j)

1/2

, (3)

where A is any symmetric, positive definite matrix.

Finally, use (2) to show that < E >= NkBT /2 and that CV = NkB/2. Hence, for every momentumor coordinate that is described by a quadratic Hamiltonian one gets a contribution of kB/2 to theheat capacity. How would this result change if only some of the coordinates or momenta weregoverned by a quadratic Hamiltonian?

3. Hill, Statistical Thermodynamics, Problem 6-7. How does your answer for the average energycompare with the equipartition theorem you proved above?

4. If a moving atom radiates light of wav elength λ0, the wav elength, λ , of the light detected by astationary detector

λ = λ o

1 −v

c

,

where v is the component of the velocity in the direction of the observer and where, c is thespeed of the light. This effect is known as the Doppler shift. (You have encountered it in thechanges in the tone of a jet engine as it passes overhead or in the sound of a train whistle). Useclassical statistical mechanics to show that I (λ)dλ , the intensity of radiation observed between

Winter Term, 2019


wavelengths λ and dλ , is

I (λ) ∝ exp−

mc2(λ − λ0)2

2λ20kBT

.

The spreading of the about the transition at λ0 is known as Doppler broadening. What will thewidth of a microwave, Doppler broadened line be for HCl at room temperature?

5. Consider a classical ideal gas of point dipoles in an electric field, E. In polar coordinates, theorientation dependent part of the Hamiltonian is

H =1

2I

p2θ +

p2φ

sin2(θ )

− µE cosθ ,

where I is the moment of inertia, θ and φ are the usual polar coordinates, pθ and pφ are theangular mementa for rotation in the θ and φ directions, respectively, and µ is the dipole moment.The external field is taken to be the z direction, i.e., θ is the angle between the dipole and theexternal field. Use classical statistical mechanics to compute the dipolar contributions to theenergy and heat capacity. Ignore electrostatic interactions between the dipoles. (HINT: do the θintegration last.)




9. (This was a final exam question worth 25%)

a) Derive the expressions for the contributions to the entropy per molecule due to electronic,translational, rotational and vibrational degrees of freedom in monatomic and diatomicgases.

b) Use your answers in a) to explain the order of the experimentally observed entropies of thefollowing molecules in the gas phase at 298K and 1 atm:

Entropy ΘR Θv

(J/K/mole) (K) (K)gas

Ar 154.84 - -Hg 174.96 - -N2 191.61 2.88 3374Br2 245.46 0.116 463

Be brief in your answers. You may evaluate the expressions if you like, but this is not neces-sary.

Winter Term, 2019


11.5. Problem Set 5

DUE: Friday, February 22, 2019

1. As was mentioned in class, for H2, we hav e to be careful with the overall symmetry of thenuclear molecular wav e-function. Since the proton has spin 1/2, the two nuclei are Fermions,and we have to write the molecular partition function as:

qnuc/rot =J evenΣ (2J + 1) exp

−

Θrot

TJ(J + 1)

+ 3J oddΣ (2J + 1) exp

−

Θrot

TJ(J + 1)

,

where the factor of 3 accounts for the 3 symmetric nuclear spin wav e-functions (there is only oneantisymmetric one for the H2 nuclei) and where Θrot = 85. 4 oK . The hydrogen molecules in thestates included in the first sum (even rotational, odd spin states) are called para-hydrogen, whilethose in the second (odd rotational, even spin states) are called ortho-hydrogen.

a) Derive an expression for the ratio of hydrogen molecules in the two states, fractions of themolecules, Northo and N para as a function of temperature. What does the ratio Northo andN para do for high or low temperatures.

b) Next show that the rotational contribution to the energy can be written as

< Erot >

N=

Northo(T )

NEortho(T ) +

N para(T )

NE para(T ), (1)

where

Eortho/para(T ) ≡ kBΘrotJ odd/even

Σ J(J + 1)(2J + 1) exp−

Θrot

TJ(J + 1)

J odd/evenΣ (2J + 1) exp

−

Θrot

TJ(J + 1)

. (2)

c) We next define rotational heat capacities for p- and o-hydrogen as:

Croto,p(T ) ≡

∂Eortho/para(T )

∂T. (3)

Derive molecular expressions for these. Compare this with the exact expression for therotational heat capacity. (HINT: first note that kBT 2CV =< (E− < E >)2 > also applies tothe various energies and heat capacities defined here.)

d) Write a program to plot the low temperature dependence of Croto,p, the full heat capacity and

the apparent heat capacity obtained by using the high temperature ratio of ortho to para

hydrogen. Plot your heat capacities in units of NkB for the range 0.01 to 5 T /Θrot (it’soverkill but, keep several hundred terms in the sums). You should reproduce the figure onthe cover of the book.

Winter Term, 2019


2. The Hamiltonian for the small amplitude vibrations of a one dimensional linear triatomic mol-ecule can be written as:

H =3

i=1Σ p2

i

2mi

+1

2k1,2|δ x1 − δ x2|

2 +1

2k2,3|δ x2 − δ x3|

2,

where δ xi is the displacement of the i’th atom from its equilibrium position and pi is its momen-tum. What are the normal modes and normal mode frequencies for this molecule? To what typeof classical motion do they correspond (the algebra is pretty messy here, and so, you may assumethat m1 = m2 = m3 = m and that k1,2 = k2,3 = k)? What is the vibrational partition function forthis hypothetical molecule?

The following was from a past final exam:

3. (25%) What are the symmetry numbers and the number of vibrational degrees of freedom forthe following molecules:

NH3, C6 H6, C6 H5 D, CH3 D, and HCN .

What are the high-temperature (with respect to all vibrational temperature) values for CV and CP

per molecule?

Winter Term, 2019


11.6. Problem Set 6

DUE: Wednesday, February 28, 2018

1. Consider the chemical reaction:

N2O4 → 2NO2

a) Calculate the moments of inertia and rotational temperatures for N2O4 using the followingstructural information:

NN1.75 A

O

O O

O

1.18

A

134 o

Fig. 1. The geometry of N2O4.

b) Use the data in the following table to compute the electronic contribution to ∆Hrxn at298.15K. What happens to your result if you include zero-point vibrational energy?Finally, are your results enthalpies or entropies?

Ground Point Rotational Vibrational ∆H f

State Group Temperatures Frequencies (298.15K)(K) (cm−1) (kJ/mole)

Molecule

NO2 X2 A1 C2v 33.1811.5, 0.624,0.590

648, 1320, 1621

N2O4 ? Vh 9.08from part a) 82, 261.3,265.0, 425.0,487.7, 657.0,751.61, 811.7,1261.08, 1383,1717, 1756.76

c) Use your result in parts a) and b) and the data in the table to compute the equilibrium con-stant for this reaction in the 100 - 500 K range. Plot ln(KP) vs. 1/T . Would any deviationsfrom a straight line be observable in the experiment done in the physical chemistry lab?How do your results compare?

Winter Term, 2019


2. (Past exam question worth 25%) What is the expression for the equilibrium constant for thegas phase reaction:

H2 + Br → HBr + H

Be sure to define any symbols you use and explain any approximations you make.

3. (Past exam question worth 25%) Consider the gas-phase radical reaction:

CH2 + H2 → CH4.

a) Draw structures for the reactants and products. Beside each structure, give the symmetrynumber and number of vibrational degree’s of freedom.

b) Write out the molecular expression for the equilibrium constant, Kρ , for the reaction. Besure to define any symbols you use, and to state what approximations you are making.

c) How does your expression in part b) simplify if the temperature is much lower than all thevibrational temperatures for reactants and products? What happens when the temperature ismuch higher than all the vibrational temperatures?

d) Finally, briefly describe how your answers change the reaction is modified by using D2;i.e.,

CH2 + D2 → CH2 D2.

4. Hill, Problem 10-1



7. Consider a one dimensional lattice of lattice spacing "a" composed of N atoms each of massM. Each atom interacts with its nearest neighbors through a harmonic force with force constantK. Finally, the ends of the chain are assumed to obey "periodic boundary conditions;" i.e.,

xN+1(t) = x1(t),

where xi(t) is the displacement of the i’th atom from its equilibrium position.

a) By assuming that the normal modes correspond to one dimensional wav es of the form:

xn(t)∝ei(ω t−kan)

find the form of the dispersion law; i.e., the relationship between the frequency of the modeand its wav e vector.

b) How does your answer in a) change if the mass of every second atom is changed to M ′?

c) Since we know how many vibrational degrees of freedom there are in our system, what isthe maximum wav e vector of the highest frequency mode?

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d) Derive an expression for the vibrational partition function in the limit N → ∞.

8. Hill, Problem 5-11.

9. (From a past final, worth 25%) The Debye theory treats a crystal as a continuum body, andhence the relation between the frequency and wav e-vector of the vibrations (i.e., the dispersionrelation) is ω = kc, where c is the speed of sound in the body, and k is the wav e-vector. In ferro-magnetic materials at low temperature, there are quantized, vibration-like, wav es of magnetiza-tion called spin wav es; the dispersion relation for these types of wav es is ω ∝k2, where the k’s arequantized as for sound. By assuming that the underlying Hamiltonian is separable, find the formof spin-wav e contribution to the heat capacity at low temperatures.

All other things being equal, will the spin wav e contribution, or the Debye vibrational contribu-tion be more important at low temperatures?

Can a low temperature heat capacity measurement distinguish between changes in the forms ofthe dispersion relation and in the density of states (e.g., as in last year’s exam question)?

Winter Term, 2019

Past Quizzes -75- Chemistry 365

12. Past Quizzes

12.1. Quiz I (2015)

Chemistry 365: Quiz I (2015)

Useful Information

kB = 1. 380662 × 10−23J /K h = 6. 6256 × 10−34J ⋅ sec

Av ogadro´s Number = 6. 022169 × 1023 1atm = 1. 01325 × 105J /m3

Geometric Series:M

i=NΣ xi =

(xN − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for N = 0 and |x| < 1

Exponential Series: ex = 1 + x +x2

2+. . . =

∞

i=0Σ xi

i!

Stirling´s Approximation: ln(N !) ≈ N ln(N ) − N ∫∞−∞

dx e−ax2

= √ π /a

This is a closed book exam. Calculators are unnecessary. Answer all parts of all four (4) ques-tions, each is of equal value. The exam has 3 pages, including this one, and are printed on bothsides of the paper.

NOTE: If you want me to mark anything on the left-hand side of the exam book, you must

indicate this clearly. The default is to ignore them.

Remember to log out when you leave and in particular, be sure to record the number of

exam books you are turning in.

You may keep the exam.

Winter Term, 2019

Chemistry 365 -76- Quiz I (2015)

1. (25%)

a) Show that

⟨A⟩G.C =∞

N=0Σ P(N )⟨A⟩C,N , (1.1)

where ⟨. . .⟩GC denotes an average in a grand canonical ensemble, P(N ) is the probability offinding a system with N particles in the grand canonical ensemble, and ⟨. . .⟩C,N is an averagein a canonical ensemble with N particles. What is P(N )?

b) Show that

⟨(N − ⟨N ⟩)2⟩⟨N ⟩2

=ρkBTκ

⟨N ⟩, (1.2)

where all averages are grand canonical,

κ ≡ −1

V

∂V

∂P

T ,N

, (1.3)

is the isothermal compressibility, and ρ is the number density, ⟨N ⟩/V .

c) How do your answers in parts a) and b) show that the canonical and grand canonical ensem-bles are equivalent?

2. (25%)

a) What are the equal a priori and ergodic hypotheses? What does this imply for the form ofthe microcanonical distribution function?

b) What is the maximum term method and when is it valid?

c) Show how the you can use Lagrange’s method of undetermined multipliers to derive the formof the canonical distribution function.

d) Show how thermodynamics can be used to identify the entropy and the undetermined multi-pliers for your answer in part c).

e) Given the canonical partition function, Q(N , V , T ), in a one component system, how wouldyou compute the Helmholtz free energy, pressure, energy and chemical potential? Deriva-

tions are not necessary.

3. (25%, From the homework) A model for a spin in a magnetic field can be obtained from sta-tistical mechanics. The energy of a spin-I nucleus in a magnetic field is:

Hmag = − γ h− HIz (3.1)

where h− is Planck’s constant, γ is a constant called the "gyromagnetic ratio", H is the externalfield, and Iz is the projection of the total spin along the direction of the external field (i.e., it isthe magnetic quantum number and can take on values −I , −I + 1, . . . , I − 1, I ).

Winter Term, 2019

Quiz I (2015) -77- Chemistry 365

a) Work out an expression for the canonical partition function. HINT: Recall that for geometricseries:

B

j=AΣ x j =

xB+1 − x A

x − 1(3.2)

and that

sinh(x) ≡1

2

(ex − e−x). (3.3)

b) By using a canonical ensemble, derive expressions for the average magnetization (i.e.,γ h− < Iz >), energy, and entropy per spin in the presence of the external field. Ignore all inter-actions between different spins.

c) What is the expression for the constant external field heat capacity? How does it behave athigh and low temperatures? HINT: note that

csch2(x) ≡1

sinh2(x)≈

1

x2−

1

3+ O(x2), as x → 0,

4e−2x + O(e−4x), as x → ∞.(3.4)

d) How does your result generalize for a system of N spins that interact with the external fieldonly. Assume that the spins are attached to identical atoms in a crystal.

4. (25%)

a) Consider a system of identical non-interacting particles. Show how this leads to a Hamilton-ian that is separable. What does this imply for the wav e-function, energies of the states,canonical partition function and Helmholtz free energy?

b) What determines whether the particles are Bosons or Fermions and how does this affect youranswer in part a)?

c) Show how the correct treatment of the identity of the particles leads to states that are fullycharacterized by the occupation numbers ni , the number of particles in 1-particle state/orbitali. What values can the ni’s take for Bosons or Fermions?

d) Calculate the grand partition function for the system described in part c) and use your resultto derive general expressions for β pV , < N >, < E > and < ni >. Note that by general Imean without specifying the 1-particle energy levels to a specific mode (e.g., particle in abox).

Winter Term, 2019


12.2. Quiz I (2016)


Useful Information

kB = 1. 380662 × 10−23J /K h = 6. 6256 × 10−34J ⋅ sec


Geometric Series:M

i=NΣ xi =

(xN − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for N = 0 and |x| < 1


2+. . . =

∞

i=0Σ xi

i!


dx e−ax2

= √ π /a

This is a closed book exam. Calculators are unnecessary. Answer all parts of all four (4) ques-tions, each question is of equal value. The exam has 3 pages, including this one, and are printedon both sides of the paper.





Keep the Exam.

Winter Term, 2019


1. (25%)

a) What is meant by the term separable Hamiltonian and what does it imply for the wav e func-tion, energy levels, and canonical partition function?

b) Compute the vibrational molecular partition function for a diatomic molecule and use yourresult to calculate the vibrational contributions to the Helmholtz free energy, energy, entropy,pressure, and chemical potential.

c) Calculate the grand partition function, ⟨N ⟩, ⟨E⟩ and the pressure for an ideal structureless gasin a cubic box of side length L. Note that the energy levels of a particle of mass m in a cubicbox are

ε nx ,ny,nz=

h2

8mL2(n2

x + n2y + n2

z), where ni = 1, 2, 3, . . . . (1.1)

Be sure to define your symbols and explain any approximations or assumptions you make.

2. (25%)

a) To what thermodynamic functions do the logarithms of the microcanonical, canonical, andgrand canonical partition functions correspond?

b) All other things being equal, order the degeneracies for a given energy level for the threekinds of particles (i.e., Fermions, Bosons, or distinguishable) we have considered.

c) Briefly discuss the conditions necessary for the validity of using Boltzmann statistics in anideal gas of point particles

d) What is ⟨ni⟩, the average number of particles in single particle state i in a grand canonicalensemble, for an ideal gas of fermions? Use your result to obtain an expression for ⟨N ⟩.

3. (25%)

a) In the canonical ensemble, show that

σ 2E ≡ ⟨(E − ⟨E⟩)2⟩ = kBT 2CV , (3.1)

where CV is the system’s constant volume heat capacity.

b) How does your answer in part a) imply the equivalence of thermodynamic quantities calcu-lated in the canonical and microcanonical ensembles?

c) In the grand canonical ensemble, show that

⟨(N − ⟨N ⟩)(E − ⟨E⟩)⟩ = kBT

∂⟨E⟩∂µ

β ,V

=

∂⟨N ⟩∂(−β )

β µ,V

(3.2)

Winter Term, 2019


and that

⟨(E − ⟨E⟩)2⟩ = kBT 2CV +

∂E

∂N

T ,V

⟨(N − ⟨N ⟩)(E − ⟨E⟩)⟩. (3.3)

Why do Eqs. (3.1) and (3.3) differ?

Winter Term, 2019


4. (25%, From the homework) Consider a gas in equilibrium with a surface. The surface canadsorb the gas molecules onto any of M independent, distinguishable sites (distinguishablebecause the binding sites are localized to specific parts of the surface). The molecular partitionfunction for an adsorbed molecule is q(T ) ≡ exp[−β Asurface].

a) Assume that the adsorbed molecules are independent and use the canonical ensemble for N

adsorbed molecules to compute the chemical potential for the molecules on the surface.Remember to include the number of ways to adsorb the molecules onto the surface. Alsoremember that the molecules themselves are indistinguishable.


c) If the gas and surface phases are in equilibrium, how does the fractional coverage, N /M ,depend on the pressure in the gas. Your expression is known as the Langmuir adsorptionisotherm.

Winter Term, 2019


12.3. Quiz I (2018)


Useful Information

kB = 1. 380662 × 10−23J /K h = 6. 6256 × 10−34J ⋅ sec


Geometric Series:M

i=NΣ xi =

(xN − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for N = 0 and |x| < 1


2+. . . =

∞

i=0Σ xi

i!


dx e−ax2

= √ π /a

This is a closed book exam. Calculators are unnecessary. Answer all parts of all five (5) ques-tions, each question is of equal value. The exam has 2 pages, including this one, and is printed onboth sides of the paper.





Keep the Exam.

Winter Term, 2019


12.3.1.(20%, From the Homework) Consider the binomial coefficient

N

m

≡N !

m!(N − m)!. (12.3.1.1)

For large values of N and m, show that it has its largest value when m = N /2 (HINT: use Stir-ling’s formula). What is the maximum value? For values of m near N/2, show that the binomialcoefficient is approximately Gaussian in form (HINT: Expand the log of the binomial coefficientin a Taylor series in m around the maximum value). What is the magnitude of the width of theGaussian? Given these results, would using the maximum term method for series involving thebinomial coefficients be valid? Why?

12.3.2.(20%)

a) What is meant by the term Separable Hamiltonian?

b) What are the implications of having a separable Hamiltonian for the wav e function andenergy eigenvalues?

c) What are the implications of having a separable Hamiltonian for the canonical partition func-tion and thermodynamic functions?

12.3.3.(20%)

a) What is meant by the term Boltzmann Statistics?

b) Calculate the canonical partition function for a gas of N structureless particles in a cubic box,with energy levels

ε nx ,ny,nz=

h2

8mL2

(n2x + n2

y + n2z), ni = 1, 2, . . . , i = x, y, z, (12.3.3.1)

if the particles are distinguishable and by using Boltzmann Statistics.

c) Briefly explain why one of your results in part b) is obviously wrong.

d) When is Boltzmann Statistics valid?

e) What does the criterion in part d) become for the particle in a box model (show the details ofcalculation)?

12.3.4.(20%)

a) What is the method of Lagrange multipliers?

b) Derive the form of the canonical distribution function using the method of Lagrange multipli-ers.

c) Use thermodynamics to determine the identity of the Lagrange multipliers. In particular,show how the entropy arises?

12.3.5.(20%) Show how the form of the canonical partition function leads to the second and third lawsof thermodynamics. What assumptions are you making?

Winter Term, 2019


12.4. Quiz I (2019)


Useful Information

kB = 1. 380662 × 10−23J /K h− = 1. 054571628 × 10−34J sec

Av ogadro´s number = 6. 022169 × 1023 1atm = 1. 01325 × 105J /m3

Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1


2+. . . =

∞

i=0Σ xi

i!


dx e−ax2

= √ π /a

1. This is a closed book exam; no books or fact sheets are allowed.

2. Calculators are unnecessary.

3. There are five questions and each is worth 20%. Answer all parts of all questions.

4. The exam has 2 pages (including this one) and is printed on both sides.

5. If you want me to grade anything on the left-hand sides of the exam book, you must


6. Remember to sign-out when you leave; in particular, be sure to record the number of


7. KEEP THE EXAM.

8. Good Luck

Tuesday, February 5, 2019


1. (20%)

a) Use the equal apriori probability postulate to derive the mathematical form of the grand

canonical ensemble probability function, P j(N ). Be sure to clearly indicate what the vari-ous symbols mean and why they appear.

b) Show how the unknown parameters that appear in your answer in a) connect with thermo-dynamics.

2. (20%) In the canonical ensemble, show that

⟨(E − ⟨E⟩)2⟩ = kBT 2CV ,

where CV is the system’s constant volume heat capacity. How does this result imply the equiva-lence of the canonical and microcanonical ensembles?

3. (20%)

a) What is meant by Boltzmann statistics? When is it valid?

b) By using Boltzmann statistics, calculate the partition function for a non-interacting gas ofparticles of mass m in a cubic box of side length L. The energies are

ε nx ,ny,nz=

h2

8mL2(n2

x + n2x + n2

x) ni = 1, 2, . . . .

c) Use your result in part b) to calculate the pressure, ⟨E⟩, CV , and the chemical potential, µ.

4. (20%, (From the homework) A system consists of two particles, each of which has two pos-sible quantum states with energies ε0 and ε1. Write the complete expressions for the partitionfunction and for the constant volume heat capacity if:

a) The particles are distinguishable.

b) They are Fermions.

c) They are Bosons.

What are the average energy and heat capacity for your answers in parts a)−c)?

5. (20%)

a) What makes a particle a Boson or a Fermion? What implications does this have for thequantum mechanics of the system, in particular, on the total wav e function?

b) For each of the two possibilities, derive an expression for the grand canonical partitionfunction. Be sure to define any symbols.

c) Use your result in part b) to obtain the average energy and average number of particles.

Winter Term, 2019

Chemistry 365 -86- Quiz II (2015)

12.5. Quiz II (2015)

Chemistry 365: Quiz II (2015)

Useful Information

kB = 1. 380662 × 10−16erg/K h = 6. 6256 × 10−27erg ⋅ sec


Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1


2+. . . =

∞

i=0Σ xi

i!

Stirling´s Approximation: ln(N !) ≈ N ln(N ) − N

∞

−∞∫ dx e−ax2

= √ π /a



3. There are four questions and one bonus question. Each is worth 25%. Answer all parts ofquestions 1-4, and any or all of the bonus question, number 5.


5. If you want me to mark anything on the left-hand side of the exam book, you must indi-

cate this clearly. The default is to ignore them.



7. KEEP THE EXAM.

8. Good Luck

Tuesday, March 17, 2015

Quiz II (2015) -87- Chemistry 365

1. (25%) Define or give an equation defining (non-standard quantities appearing in your defini-tion must also be defined) as indicated for the following terms:

a) b)Translational energy density of states foran ideal gas. (Give an equation)

The rotational partition function for anonlinear molecule at high temperature.(Give an equation). Be sure to definethe symbols!

c) d)The equipartition theorem. The vibrational density of states in theDebye model. (Give an equation)

e) Ortho vs Para Hydrogen

2. (25%) (BASED ON HOMEWORK) Here we’ll extend the discussion of Langmuir adsorption(problem set 3) by allowing for multilayer adsorption. Consider a one component gas that is inequilibrium with a surface containing M distinguishable binding sites. Unlike the homeworkproblem, we now assume that multilayer adsorption is possible, and thus each adsorption site canhave a stack of n molecules n = 0, 1, 2, . . .. Let q0 be the molecular partition function of the baresurface, q1 be the molecular partition function of a molecule in direct contact with the surface,while q2 is the partition function of a molecule in the 2nd and subsequent layers. This last condi-tion is clearly an approximation. This model is known as the B.E.T. model after Brunnauer,Emmett, and Teller.

a) What is the partition function of a stack containing 0 or 1, or n > 1 adsorbate molecules?Assume separability.

b) Suppose that the system contains n0 empty sites and n j occupied sites with a stack ofj = 1, 2, 3, . . . molecules. Show that

M =∞

j=0Σ n j and N =

∞

j=1Σ j n j , (2.1)

where N is the total number of adsorbed molecules.

c) What is the degeneracy of the adsorbed surface state? (HINT: assume that the adsorbates aredistinguishable because of the surface lattice and small differences in the actual q j’s. Howmany ways can you choose the binding sites and layers and assign the N adsorbates? Whathappens to the N dependent factors when you use Boltzmann statistics to calculate the canon-ical partition function?

d) Compute the semi-grand partition function for the system (i.e., where the system is open tothe adsorbates but closed with respect to the surface sites). Use Boltzmann statistics. Indoing this, you will need the generalization of the binomial theorem, i.e.,



(a0 + a1 + a2+. . . )M = M!

∞

j = 0Σ n j = M

ni∞i = 0

Σ∞

j = 0Π a

n j

j

n j!(2.2)

e) Use your result in part d) to show that in the B.E.T. model, the semi-grand partition functionis

Ξ = qM0

1 + λ(q1 − q2)

1 − λq2

M

, (2.3)

where λ ≡ eβ µ.

Use your result to show that the fractional coverage,

Θ ≡< N >

M=

λ(q1 − q2)

1 + λ(q1 − q2)+

λq2

1 − λq2

=λq1

[1 + λ(q1 − q2)](1 − λq2). (2.4)

How is λ related to the pressure in the gas next to the surface? What is happening asλq2 → 1 and what does this imply (if anything) for the pressure in the gas phase next to thesurface? Finally, note that the Langmuir result is obtained when q2 = 0.

3. (25%, FROM THE HOMEWORK)

a) Derive the expressions for the contributions to the entropy per molecule due to electronic,translational, rotational and vibrational degrees of freedom in monatomic and diatomicgases.

b) Use your answers in a) to explain the order of the experimentally observed entropies of thefollowing molecules in the gas phase at 298K and 1 atm:

Entropy ΘR Θv

(J/K/mole) (K) (K)gas

Ar 154.84 - -Hg 174.96 - -N2 191.61 2.88 3374Br2 245.46 0.116 463

Be brief in your answers.



4. (25%) Consider the gas phase reaction

C6 H6 + Br2→← C6 H5 Br + HBr. (4.1)

a) Fill in the following table (copied to your answer book):

Compound C6 H6 Br2 C6 H5 Br HBr

Structure:

Proper (i.e., noreflections or inver-sions) point-groupsymmetry opera-tions:

Symmetry number:

# vibrational modes:

b) Write down expressions for Kel , Kvib, Krot , Ktrans, and Kρ , defining any symbols youintroduce, and explaining any assumptions you make.

c) What does the equilibrium constant become for T << all Θvib,i or T >> all Θvib,i?

5. (25%, Bonus Question)

a) Write down the general expressions for the vibrational free-energy, energy, and heat capac-ity for a monatomic crystal containing N atoms. Express your answer as an integral overvibrational frequencies.

b) What does your answer in part a) for the heat capacity do in the high temperature limit.What do you mean by high temperature?

c) What are the Debye and Einstein approximations for the vibrations of a harmonic crystal?

d) Show what each model predicts for the asymptotic behavior of Cvib(T ) as T → 0.

e) Which model is correct? Why?

Winter Term, 2019


12.6. Quiz II (2016)


Useful Information

kB = 1. 380662 × 10−23J /K h− = 1. 054571628 × 10−34J sec


Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1


2+. . . =

∞

i=0Σ xi

i!


dx e−ax2

= √ π /a



3. There are four questions and one bonus question. Each is worth 25%. Answer all parts ofquestions 1-4, and any or all of the bonus question, number 5.


5. If you want me to mark anything on the left-hand sides of the exam book, you must




7. KEEP THE EXAM.

8. Good Luck



1. (25%) Define or give an equation defining (non-standard quantities appearing in your defini-tion must also be defined) as indicated for the following:

a) b)The second virial coefficient for a classi-cal gas of point particles. (Give an

equation)

The Fermi-Dirac distribution functionfor a gas of non-interacting fermions.(Give an equation). Be sure to definethe symbols!

c) d)The rate constant for the reactionA + B→←(AB)‡ → products in transitionstate theory (State the assumptions beingmade).

The number of vibrational modes in anonlinear polyatomic molecule contain-ing N atoms.

e) The symmetry number. (In particular,when and why do need it?).

2. (25%)

a) Derive an expression for the heat capacity of the quantum mechanical harmonic oscillator.

b) What does your answer in a) become at high or low temperatures? (What is meant here by"high" or "low"?)

c) What is meant by the term "normal mode transformation" and what does it imply for theheat capacity of a crystalline solid?

d) By using parts a) and c) write down the expression for the heat capacity of a Debye solid.

e) What does your answer in part d) give for high or low temperature (again explain what ismeant by high or low).

3. (25%)

a) Derive an expression for the rotational/nuclear molecular partition function for a homo-nuclear diatomic molecule in which the nuclei have spin I .

b) Use your expression in a) to derive high and low temperature expressions for the rota-tional/nuclear contribution to the Helmholtz free energy and energy per molecule at highand low temperatures.




C2 H2 + 2H2→← C2 H6. (4.1)


Compound: C2 H2 H2 C2 H6

Structure:

Symmetry operationsthat contribute to thesymmetry number:

Symmetry number:



c) What does the equilibrium constant become for T << all Θvib,i or T >> all Θvib,i? Whathappens at intermediate temperatures, where T >> Θvib,i for some i’s and T << Θvib, j forthe other modes?

5. (25%, Bonus Question, FROM THE HOMEWORK) Consider a one dimensional lattice oflattice spacing "a" composed of N atoms each of mass M. Each atom interacts with its nearestneighbors through a harmonic force with force constant K. Finally, the ends of the chain areassumed to obey "periodic boundary conditions;" i.e.,

xN+1(t) = x1(t), (5.1)

where xi(t) is the displacement of the i’th atom from its equilibrium position.

a) By assuming that the normal modes correspond to one dimensional wav es of the form:

xn(t)∝ei(ω t−kan) (5.2)

find the form of the dispersion law; i.e., the relationship between the frequency of the modeand its wav e vector. What values can the wav e numbers take on?

b) How many vibrational degrees of freedom are there in our system and what is the maximumwave vector of the highest frequency mode?



c) Derive an expression for the vibrational partition function in the limit N → ∞.

Winter Term, 2019


12.7. Quiz II (2018)


Useful Information

kB = 1. 380662 × 10−23J /K h− = 1. 054571628 × 10−34J sec


Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1


2+. . . =

∞

i=0Σ xi

i!


dx e−ax2

= √ π /a



3. There are four questions and each is worth 25%. Answer all parts of all questions.


5. If you want me to mark anything on the left-hand sides of the exam book, you must




7. KEEP THE EXAM.

8. Good Luck



1. (25%) Draw structures, state the number of vibrational modes, the rotational symmetry num-bers, and the high and low temperature (with respect to vibrational temperatures) heat capacities(CV ) for each of the following molecules:

a) C2 H4 b) C2 H6

c) B2 H6 d) NH3

e) PF5


CH4 + D2→← CH3 D + HD. (2.1)


Compound: CH4 D2 CH3 D HD

Stochiometriccoefficient:

Structure:

Symmetry opera-tions that contributeto the symmetrynumber:

Symmetry number:




3. (25%, From the homework) The Hamiltonian for the small amplitude vibrations of a onedimensional linear triatomic molecule can be written as: A Hamiltonian for the small amplitudevibrations of a one dimensional linear triatomic molecule can be written as:

H =3

i=1Σ p2

i

2m+

1

2k |δ x1 − δ x2|

2 +1

2k |δ x2 − δ x3|

2



where δ xi is the displacement of the i’th atom from its equilibrium position and pi is its momen-tum. (Here, unlike in the homework, we’ve taken all the masses and force constants to be thesame). Derive expressions for the normal mode frequencies. What is the vibrational molecularpartition function for this hypothetical molecule?



4. (25%)

a) Derive the forms of the Fermi-Dirac and Bose-Einstein distributions for an ideal gas with1-particle states, i, and 1-particle energies, ε i .

b) Sketch your result as a function of energy.

c) Use your result in part a) to calculate the density, ρ ≡ ⟨N ⟩/V for an ideal gas of noninter-acting electrons. HINT: the energy levels of a particle in a one dimensional box are givenby

ε nx=

h2n2x

8mL2, nx = 1, 2, 3, . . . .

d) Derive the expression for the pressure in the system considered in part c).

Winter Term, 2019

Chemistry 365 -98- Past Final Exams

13. Past Final Exams

Winter Term, 2019

Final Exam (2015) -99- Chemistry 365

13.1. Final Exam (2015)

Faculty of Science

McGill University


FINAL EXAMINATION

Examiner: Professor David Ronis Friday, April 24, 2015Associate Examiner: Professor P. Kambhampati 9:00 A.M. − Noon

1. No books or notes are permitted. Translation dictionaries and molecular models are permit-ted. Calculators aren’t needed.

2. Answer all questions in the exam book and show all work clearly.

3. There are 3 pages (including this one) and 4 questions, each of equal value.




exam books you are turning in on the cover of book 1.

6. Keep the exam (please).

7. Useful Information:

kB = 1. 380662 × 10−23J /K h = 6. 6256 × 10−34J ⋅ sec


Gas Constant, R=8.314 J/K mol

Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1

Exponential Series:∞

n=0Σ xn

n!= ex


dx e−ax2

= √ π /a

Friday, April 24, 2015

Chemistry 365 -100- Final Exam (2015)

8. Good Luck.

1. (25%)

a) Derive the expression for the vibrational contribution to the molecular partition function for adiatomic molecule. The atoms have masses m1 and m2 and the bond has a force constant K .

b) Use your result in a) to calculate the expression for the vibrational contribution to the heatcapacity and chemical potential.

c) Show how your expressions in b) behaves at high and low temperatures? What do you meanby high and low temperature?

d) Derive the expression for the vibrational contribution to the molecular partition function inclassical statistical mechanics. How does it compare with what you obtained in part a)?

2. (25%, Fr om the Homework) The Hamiltonian for the small amplitude vibrations of a onedimensional linear triatomic molecule can be written as:

H =3

i=1Σ p2

i

2m+

k

2|δ x1 − δ x2|

2 +k

2|δ x2 − δ x3|

2

where δ xi is the displacement of the i’th atom from its equilibrium position and pi is its momen-tum. (Notice that unlike the homework problem, the masses and force constants are all the samefor each atom and bond, respectively.) What are the normal modes and normal mode frequenciesfor this molecule? To what type of classical motion do they correspond? What is the vibrationalpartition function for this hypothetical molecule?

3. (25%) Consider the gas phase reduction of 1,4-benzoquinone with hydrogen to get 1,4-hydro-quinone, i.e.,

+ H

OH

OH

O

O

2

(Actually this is not done in the gas phase.)

a) For each compound, give the proper symmetry operations, the symmetry number, and thenumber of vibrational modes.

b) Write down the expression for the equilibrium constant. State any assumptions that youare making and why they are appropriate.

c) What does your expression become if the temperature is much higher or lower than all thevibrational temperatures?

Friday, April 24, 2015


d) What changes if you replace H2 by HD?

4. (25%, Fr om the Homework) Consider a gas in equilibrium with a surface. The surface canadsorb the gas molecules onto any of M independent, distinguishable sites (distinguishablebecause the binding sites are localized to specific parts of the surface). The molecular partitionfunction for an adsorbed molecule is q(T ) ≡ exp[−β Asurface].

a) Assume that the adsorbed molecules are independent and use the canonical ensemble forN adsorbed molecules to compute the chemical potential for the molecules on the surface.Remember to include the number of ways to adsorb the molecules onto the surface. Alsoremember that the molecules themselves are indistinguishable.


c) If the gas and surface phases are in equilibrium, how does the fractional coverage,θ ≡ N /M , depend on the pressure in the gas. Sketch your result.

Winter Term, 2019


13.2. Final Exam (2016)

Faculty of Science

McGill University


FINAL EXAMINATION

Examiner: Professor David Ronis Wednesday, April 20, 2016Associate Examiner: Professor P. Kambhampati 6:00 P.M. − 9:00 P.M.

1. No calculators, books or notes are permitted. Translation dictionaries and molecular modelsare permitted.









kB = 1. 380662 × 10−23J /K h = 6. 6256 × 10−34J ⋅ sec



Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1


n=0Σ xn

n!= ex


dx e−ax2

= √ π /a

Wednesday, April 20, 2016


8. Good Luck.

1. (25%)

a) Draw the structure of ethane.

b) What are the rotational symmetry number, the number of vibrational modes, and high-tem-perature limit of the constant volume heat capacity per molecule for ethane?

c) One of the vibrational modes corresponds to a twisting or torsional motion around theC − C bond; as a function of the dihedral angle, φ , the full potential energy for this motionhas the form:

U(φ ) = U0 +1

2V0[1 − cos(3φ )]. (1)

i) Sketch the potential energy.

ii) Discuss how the motion and the contribution to the heat capacity associated with thedihedral angle changes for T << Θ and for T >> Θ, where Θ ≡ V0/kB.

iii) Use classical statistical mechanics to derive an expression for the molecular partitionfunction and heat capacity contributions associated with the torsional motion. Leave

your answers in terms of integrals!

iv) What do you expect your classical expression to get right and wrong? Why?

2. (25%) Consider the chemical reaction:

C6 H6(g) + Br2(g) → C6 H5 Br(g) + HBr(g)

Give a molecular expression for the equilibrium constant. Be sure to define any symbols youuse.

3. (25%) The low temperature behavior of amorphous solids (e.g., glasses) is considerably differ-ent from that of crystalline solids. Low temperature experimental results may be obtained bypostulating the following density of states

g(ν ) =

α ν 0.3, for ν ≤ νmax

0, for ν > νmax

.

Wednesday, April 20, 2016


a) Determine the constant α for a system of N atoms.

b) Calculate the vibrational partition function, the average energy and the heat capacity at con-stant volume. Determine the low and high temperature limits of CV . What are the corre-sponding limits for a Debye crystal? Does this treatment describe a universal temperaturedependence of CV for amorphous solids? You might find the following integral useful,

∫∞0

dx x2.3 e−x

(1 − e−x)2= Γ(3. 3)ζ (2. 3) = 3. 84380. . . ,

where Γ(z) is the gamma function and ζ (z) is the Riemann zeta function.

4. (25%, From the Homework) A model for a spin in a magnetic field can be obtained from sta-tistical mechanics. The energy of a spin-I nucleus in a magnetic field is:

Hmag = − γ h− HIz

where h− is Planck’s constant, γ is a constant called the "gyromagnetic ratio", H is the externalfield, and Iz is the projection of the total spin along the direction of the external field (i.e., it isthe magnetic quantum number and can take on values −I , −I + 1, . . . , I − 1, I ).

a) Work out an expression for the canonical partition function. HINT: Recall that for geomet-ric series:

N

j=0Σ x j =

xN+1 − 1

x − 1

and that

sinh(x) = (ex − e−x)/2

b) By using a canonical ensemble, derive an expressions for the average magnetizationenergy and entropy per spin in the presence of the external field. Ignore all interactionsbetween different spins.

c) What is the expression for the constant external field heat capacity? How does it behave athigh and low temperatures?

d) Sketch CH as a function of temperature.

Winter Term, 2019


13.3. Final Exam (2018)

Faculty of Science

McGill University


FINAL EXAMINATION

Examiner: Professor David Ronis Tuesday, April 24, 2018Associate Examiner: Professor P. Kambhampati 9:00 A.M. − 12:00 P.M.

1. No calculators, books or notes are permitted. Translation dictionaries and molecular modelsare permitted.









kB = 1. 380662 × 10−23J /K h = 6. 6256 × 10−34J ⋅ sec



Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1


n=0Σ xn

n!= ex


dx e−ax2

= √ π /a

Tuesday, April 24, 2018


8. Good Luck.

1. (25%, Fr om the Homework) Consider a gas in equilibrium with a surface. The surface canadsorb the gas molecules onto any of M independent, distinguishable sites (distinguishablebecause the binding sites are localized to specific parts of the surface). The molecular partitionfunction for an adsorbed molecule is q(T ) ≡ exp[−β Asurface].

a) Assume that the adsorbed molecules are independent and use the canonical ensemble for N

adsorbed molecules to compute the chemical potential for the molecules on the surface.Remember to include the number of ways to adsorb the molecules onto the surface. Alsoremember that the molecules themselves are indistinguishable.


c) If the gas and surface phases are in equilibrium, how does the fractional coverage, N /M ,depend on the pressure in the gas. Your expression is known as the Langmuir adsorptionisotherm.

2. (25%) Consider the so-called water-gas reaction:

CO2(g) + H2(g) →← CO(g) + H2O(g). (1)

a) Fill in the following table (copied to your exam booklet).

Table 2.1: Some molecular properties

Compound: CO2 H2 CO H2O


Structure:


Symmetry number:





3. (25%)

a) Derive expressions for the molecular vibrational partition function assuming that the vibra-tions are harmonic and using Boltzmann statistics.

b) Derive the molecular contribution to the heat capacity for the calculation in part a). Whathappens at high and low temperatures?

c) In the spectroscopic studies of vibrational transitions (i.e., infrared spectroscopy), theselection rule is ∆n = ±1, and the observed intensities are proportional to the populationdifferences between the n’th and and n + 1’st vibrational levels.

i) At a giv en temperature, what fraction of the molecules are in the n’th vibrationalstate?

ii) How does the the intensity of the n → n + 1 transition depend on temperature?

iii) Finally, for the ideal harmonic potential note that all the transitions will occur at thesame frequency. Hence, how does the total intensity depend on temperature?Assume that the proportionality constants relating intensity to population differencesare independent of n. (Strictly speaking, the intensities are proportional to dipole-moment matrix elements which weakly depend on n for harmonic oscillators).

4. (25%)

a) What is the exact form for the rotational partition function for a hetero-nuclear diatomicmolecule? What is the rotational contribution to the heat capacity?

b) How does the rotational contribution to the heat capacity a) behave at low temperatures?

c) Derive the approximate high-temperature form for the rotational partition function, startingFrom the exact expression in part a). Clearly state any assumptions or approximations youare making.

d) Qualitatively discuss how the heat capacity of a poly-atomic molecule changes with tem-perature starting from absolute zero. Clearly state any assumptions you are make about therelative magnitudes of the molecular parameters.

Winter Term, 2019


13.4. Final Exam (2019)

Faculty of Science

McGill University


FINAL EXAMINATION

Examiner: Professor David Ronis Monday, April 15, 2019Associate Examiner: Professor P. Kambhampati 18:30 − 21:30 P.M.

1. No calculators, cell phones, books, or notes are permitted. Translation dictionaries andmolecular models are permitted.









kB = 1. 380662 × 10−23J /K h = 6. 6256 × 10−34J ⋅ sec



Geometric Series:M

i=0Σ xi =

(1 − xM+1)

(1 − x)→ (1 − x)−1 as M → ∞ for |x| < 1


n=0Σ xn

n!= ex


dx e−ax2

= √ π /a

Monday, April 15, 2019


8. Good Luck.

1. (25%) Consider the gas phase isomerization of cyclohexane:

C6 H12(g)

chair

→← C6 H12(g)

boat. (1.1)

a) Fill in the following table (copied to your exam booklet).

Table 1.1: Some molecular properties

Compound:C6 H12(g)

chair

C6 H12(g)

boat


Structure:


Symmetry number:




2. (25%)

a) Derive the Debye model for the heat capacity of a harmonic crystal. Be sure to define yourvariables and explain the physical assumptions being made.

b) Show how the Debye model behaves at low temperatures; in particular, what does heatcapacity become? What is meant by low temperatures?

c) What does the model predict for the heat capacity at high temperature? Is this surprising,why or why not?



3. (25%)

a) Derive exact expressions for the average energy and average number of particles, in a gasof non-interacting fermions (bosons).

b) Show that the average occupation of a single-molecule state, i, is

ni =1

eβ (ε i−µ) ± 1, (3.1)

where Fermions (Bosons) use the + (−) sign.

c) Photons are bosons and have µ = 0. In addition, they are transverse (i.e., their electric andmagnetic fields are perpendicular to each other and to the direction of propagation) but canexist in one of two polarization states (e.g., left or right hand polarization). They hav eenergy ε = h− ω , dispersion relation ω = kc, where c is the speed of light in vacuum, andcan be assumed to have vanishing electric and magnetic fields at the boundaries of the sys-tem.

Show that

⟨N ⟩ =V

π 2c3 ∫∞0

dωω 2

eβ h−ω − 1and that ⟨E⟩ =

h− V

π 2c3 ∫∞0

dωω 3

eβ h−ω − 1. (3.2)

To what famous expression does the last expression in (3.2) correspond?

d) At any temperature, show that

⟨N ⟩∝T 3 and that ⟨E⟩∝T 4. (3.4)

The latter is known as the Stephan-Boltzmann law.

4. (25%) (From the Homework, Hill Problem 5-11)

a) What is the chemical potential of an atom in an Einstein crystal? Assume that the energyof the vibrational well minimum is Nφ0, which, in particular, includes the minimum elec-tronic and lattice energies. Ignore the N dependence of φ0 and of the Einstein tempera-ture. (NOTE that φ0 is not the φ (0) discussed by Hill.)

b) Use your answer in part a) to derive an equation for the vapor pressure of an Einstein crys-tal, assuming the vapor is an ideal gas.


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