Statistical Physics Exam

17th April 2015

Name Student Number

Problem 1 Problem 2 Problem 3 Problem 4 Total Percentage Mark

Useful constantsgas constant R 8.31J/ (K ·mol)Boltzmann constant kB 1.38·10−23J/KAvogadro number NA 6.02·1023mol−1

speed of light c 300·106m/s

1

(25P) Problem 1

1. (1P) Write down the ideal gas equation of state. PV = NkBT

2. (2P) Find the entropy of one coin toss, and find the entropy of one dice rolling.The entropy Sreads

S = kB

n∑i=1

pi ln1

pi,

where pi is the probability that the system is in the i-th microstate, and n is the total numberof the microstates.

For the coin toss one finds pi = 1/2 and n = 2, yielding S = kB ln 2 . For the dice rolling one finds

pi = 1/6 and n = 6, yielding S = kB ln 6 .

1. (4P) Which quantities are intensive: volume, temperature, particle number, pressure, chemicalpotential, entropy, density and mass? temperature, pressure, chemical potential, and density

2. (1P) Write down the value of the thermal energy at 300 K in units of pN·nm. kB × T = 0.0138 pN · nm/K× 300K = 4.14 pN · nm

3. (3P) Which quantities are constant in a

(a) canonical ensemble ? N,V, T

(b) grand-canonical ensemble ?µ, V, T

(c) isobaric-isothermal ensemble?N,P, T

4. (1P) How are Gibbs free energy and chemical potential related?µ =(∂G∂N

)p,T

; µ: Gibbs freeenergy per particle

5. (2P) How does the temperature of a solvent in a mixture change compared to the pure solvent

(a) for boiling? higher in mixture

(b) for melting ? lower in mixture

6. (2P) In a phase diagram, what is meant by a

(a) triple point? Point (pressure and temperaure) where solid, liquid and gas phase coexist.

(b) critical point? Point (pressure and temperaure) where liquid and gas phase cannot bedistinduished (refelxion point in van der Waals isothermes).

7. (3P) Derive the Maxwell relation −(∂S∂P

)T

=(∂V∂T

)P

. You can assume the number of particlesN to be fixed. We use

dG =

(∂G

∂P

)dPT +

(∂G

∂T

)P

dT

= V dP − SdT (1)

The mixed derivatives, which are equal for state functions are(∂S

∂P

)T

=

(∂V

∂T

)P

(2)

2

8. (1P) How does the heat capacity of a classical ideal monoatomic gas depend on temperature?Cv = 3

2NkB no temperature dependence.

9. (3P) What corrections, compared to the ideal gas, does the van der Waals gas introduce ? Sketchthe van der Waals Potential (Energy vs. distance). Gas particles have finite volume such thatthe total volume is reduced. This is a repelling interaction. In addition, there is an attractiveinteraction that can be understood as increased pressure(

p+ aN2

V 2

)(V − bN) = NkBT (3)

10. (2P) Name an impotant assumption in the Einstein Theory for crystal phonons. All normalmodes (global motions of all atoms together) have the same frequency. This Einstein frequencyis different for different materials.

3

(25P) Problem 2Consider a system of N spins subject to a magnetic field B. The spins are non-interacting, distin-guishable, and have the non-degenerated energy eigenvalues εm = −µBm + ε0 per one spin, where µis the magnetic moment per one spin, and m = −1 or 1.

1. (5P) Calculate the partition function ZN . The single-spin partition function Z1:

Z1 =∑

m=−1,−1

eβµBm−βε0

= e−βε0

eβµB + e−βµB︸ ︷︷ ︸2 cosh(βµB)

= 2e−βε0 cosh(βµB). (4)

The N -spin partition function ZN simply leads to

ZN = Z1N

=[2e−βε0 cosh(βµB)

]N. (5)

or = e−Nβε0[eβµB + e−βµB

]N. (6)

2. (5P) Calculate the average energy E (2 points). For this, you may calculate E using ZN , oryou may calculate E from the average of the total energy eigenvalues. The average energy is amonotonically increasing function of kBT : Find two asymptotic values of E in the limit of T → 0and T →∞ (2 points), and sketch E(kBT ) (1 point). The average energy E:

E = −[∂ ln(ZN )

∂β

]N,V

= N [ε0 −Bµ tanh(βµB)] . (7)

or = −N((−Bµ− ε0) e−βBµ−βε0 + (Bµ− ε0) eβBµ−βε0

)e−βBµ−βε0 + eβBµ−βε0

. (8)

or, using the average of the total energy eigenvalues,

E = N〈−µBm+ ε0〉

= N

∑m=−1,1(−µBm+ ε0)eβµBm−βε0∑

m=−1,1 eβµBm−βε0

= −N((−Bµ− ε0) e−βBµ−βε0 + (Bµ− ε0) eβBµ−βε0

)e−βBµ−βε0 + eβBµ−βε0

.

or = N [ε0 −Bµ tanh(βµB)] .

For T → 0:

E → −NµB +Nε0. (9)

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For T →∞:

E → Nε0. (10)

The average energy is plotted as below.

Figure 1: The average energy E as a function of kBT

1. (6P) Sketch the heat capacity CB(kBT ). For this, you may calculate CB using E, or you maycalculate CB using the energy fluctuations, or you may use the sketch of E(kBT ). The heatcapacity CB :

CB =

[∂E

∂T

]B

=N(

(−Bµ− ε0) e−BµkT −

ε0kT + (Bµ− ε0) e

BµkT −

ε0kT

)(e−

BµkT −

ε0kT

(BµkT 2 + ε0

kT 2

)+ e

BµkT −

ε0kT

(ε0kT 2 − Bµ

kT 2

))(e−

BµkT −

ε0kT + e

BµkT −

ε0kT

)2

−N(

(−Bµ− ε0) e−BµkT −

ε0kT

(BµkT 2 + ε0

kT 2

)+ (Bµ− ε0) e

BµkT −

ε0kT

(ε0kT 2 − Bµ

kT 2

))e−

BµkT −

ε0kT + e

BµkT −

ε0kT

=4B2µ2Ne

2BµkT

kT 2(e

2BµkT + 1

)2

=B2µ2Nsech2

(BµkT

)kT 2

. (11)

or, using the energy fluctuations,

CB =〈∆E2〉kBT 2

= N2 〈(−µBm+ ε0)2〉 − 〈−µBm+ ε0〉2

kBT 2

=4B2µ2Ne

2BµkT

kT 2(e

2BµkT + 1

)2

=B2µ2Nsech2

(BµkT

)kT 2

. (12)

5

0 1 2 3 4 5

0.0

0.1

0.2

0.3

0.4

0.5

kBTêBm

CB

êk

B

Figure 2: The heat capacity per the Boltzmann constant CB/kB as a function of kBT/Bµ

1. (4P) Write down the Helmholtz free energy A in terms of ZN and kBT (1 point). Calculate A(3 points). The Helmholtz free energy A:

A = − ln(ZN )

β

= −NkBT ln[2e−βε0 cosh(βµB)

]. (13)

2. (5P) Sketch the magnetization M(kBT ). For this, you may calculate the average of the totalmagnetic moment 〈−N(εm − ε0)/B〉, or you may calculate M using the Helmholtz free energyA. The magnetizationM:

M = −(∂A∂B

)N,T

= Nµ tanh(βBµ). (14)

or, using the average energy E,

M = −(E −Nε0)/B

= Nµ tanh(βBµ). (15)

6

0 2 4 6 8 10

0.0

0.2

0.4

0.6

0.8

1.0

kBTêBm

MêN

m

Figure 3: The magnetization per the total magnetic momentM/Nµ as a function of kBT/Bµ

(25P) Problem 3Consider the grand-canonical partition function for non-relativistic, ideal Fermi gases and Bose gases,respectively, in logarithmic form

ln Ξ = ± (2s+ 1) Σp ln (1± exp [−β (εp − µ)]) (16)

where εp = p2

2m is the energy corresponding to momentum state p, m is the mass of a particle, µ is thechemical potential, and β = 1

kBTwith T temperature and kB Boltzmann’s constant

1. (4P) In the classical limit, the logarithm of the partition function can be expanded.

(a) What does “classical limit” mean? Give a description in words and as a relation. Theclassical limit is low densities (large volume and/or low number of particles) and hightemperatures, exp [−β (εp − µ)]� 1

(b) Write down the first two terms of the expansion. Expand the logarithm as ln (x± 1) ≈±x− x2

2 ±x3

3 · · · for small x� 1

ln Ξ = ± (2s+ 1)∑p

exp [−β (εp − µ)]∓ 1

2exp [−2β (εp − µ)] + · · · (17)

2. (6P) Show that the zeroth order term, assuming continuous space, for particles with s = 0, is

ln Ξ =V

λ3exp (βµ) (18)

with λ =(

h2

2πmkBT

) 12

where h is Planck’s number and V is the volume. [You need the integral´∞−∞ dxx2 exp

[−x2

]=√π

4 .]In continuous space ∑

p

≈ V

h3

ˆd3p (19)

Such that in zeroth orderln Ξ ≈ V

h3

ˆd3p exp [−β (εp − µ)] (20)

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Plugging in βεp = p2

2mkBTand integrating over the sphere

ln Ξ ≈ V

h34π

ˆdp · p2 exp

[− p2

2mkBT

]exp [βµ] (21)

We now substitute x2 = p2

2mkBTand obtain a standard integral

ln Ξ =V

h34π (2mkBT )

32

ˆ ∞−∞

dx · x2 exp[−x2

]︸ ︷︷ ︸

14

√π

exp [βµ] (22)

such that

ln Ξ =V

h3(2πmkBT )

32 exp [βµ] (23)

=V

λ3exp [βµ] (24)

3. (2P) Use eq. 18 to express the chemical potential as a function of number of particles. For thegrand-canonical partition function we have

PV = kBT ln Ξ

and for an ideal gas PVkBT

= N , hence ln Ξ = N in the classical limit. Therefore

N =V

λ3exp [βµ]

exp [βµ] =V

Nλ3

βµ = lnV

Nλ3

µ = kBT lnV

Nλ3

4. Use the first-order corrected logarithmic partition function

ln Ξ = ± (2s+ 1)

[V

λ3exp (βµ)∓ 1

252

V exp (2βµ)

λ3

](25)

(a) (4P) Calculate a first-order corrected number of particles . We need

N =1

β

∂ ln Ξ

∂µ(26)

and get

N =1

β

∂

∂µ±(

(2s+ 1)

[V

λ3exp [βµ]∓ 1

2

V

λ3

1

232

exp [2βµ]

])= ± (2s+ 1)

[V

λ3exp [βµ]∓ 1

232

V

λ3exp [2βµ]

](27)

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(b) (4P) Obtain a first-order corrected expression for the pressure P in the form PVNkBT

forFermi particles with s = 1

2 and for Bose gas particles with s = 0. We again use PVkBT

= ln Ξand the N obtained above. to obtain the correction we determine the difference in thefirst-order corrections ln Ξ1and N1(in zeroth order ln Ξ = N)

ln Ξ1 −N1 = ± (2s+ 1)

[∓ 1

252

V

λ3exp [2βµ]± 1

232

V

λ3exp [2βµ]

]= ± (2s+ 1)

[(∓ 1

252

± 1

232

)V

λ3exp [2βµ]

](28)

Hence,

ln Ξ = N ± 2s+ 1

252

V

λ3exp [2βµ] (29)

and thereforepV

NkBT= 1± 2s+ 1

252

V

Nλ3exp [2βµ] (30)

Fermi:pV

NkBT= 1 +

1

232

V

Nλ3exp [2βµ] (31)

Bose:pV

NkBT= 1− 1

252

V

Nλ3exp [2βµ] (32)

(c) (3P) Write down a first-order energy for an ideal Bose gas. For ideal gases we have E =32NkBT . Hence, E = 3

2kBT[Vλ3 exp [βµ] + 1

232

Vλ3 exp [2βµ]

]. Also E = 3

2PV such that

E =3

2NkBT

[1− 1

252

V

Nλ3exp [2βµ]

]=

3

2kBT

[N − 1

252

V

λ3exp [2βµ]

]=

3

2kBT

[V

λ3exp [βµ] +

1

232

V

λ3exp [2βµ]− 1

252

V

λ3exp [2βµ]

]=

3

2kBT

[V

λ3exp [βµ] +

2

252

V

λ3exp [2βµ]− 1

252

V

λ3exp [2βµ]

]=

3

2kBT

[V

λ3exp [βµ] +

1

252

V

λ3exp [2βµ]

](33)

5. (2P) How does the first-order corrected pressure of a Fermi gas/Bose gas compare to a classicalgas (lower, higher, equal)? The corrected pressure, compared to the classical ideal gas, is largerfor a Fermi gas (plus sign in partition function) and smaller for a Bose gas (minus sign in partitionfunction).

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(25P) Problem 41. (4P) Consider a system, divided into two sub-systems 1 and 2, each characterised by their

entropy, volume, and number of particles (S1, V1, N1) and (S2, V2, N2), respectively. The totalsystem is isolated and is at equilibrium. the two subsystems are also in equilibrium with eachother and can exchange energy and particles, nut the total energy and total number of particlesremains constant. The subsystems can also change their volume prvided the total volume isfixed. Show that at equilibrium the two subsystems have the same pressure P1 = P2, the sametemperature T1 = T2, and the same chemical potential µ1 = µ2.For the total system we always have

S = S1 + S2

V = V1 + V2 (34)N = N1 +N2

E = E1 + E2

since the total system is isolated. This means that also

dS = dS1 + dS2 = 0

dV = dV1 + dV2 = 0

dN = dN1 + dN2 = 0

dE = dE1 + dE2 = 0 (35)

Starting from the first law of thermodynamics we write

dE1 = T1dS1 − P1dV1 + µ1dN1 (36)

and

−dE2 = −T2dS2 + P2dV2 − µ2dN2 (37)

such that withdE1 = −dE2

T1dS1 − P1dV1 + µ1dN1 = −T2dS2 + P2dV2 − µ2dN2 (38)= −T2 (−dS1)− P2 (−dV1) + µ2 (−dN1)

= T2dS1 + P2dV1 − µ2dN1

which holds only for

T1 = T2

P1 = P2

µ1 = µ2 (39)

2. (10P) A cylinder contains an ideal gas in thermodnamic equilibrium at pressure P , Volume V ,temperature T , internal energy E, and entropy S. The cylinder is surrounded by a very largeheat reservoir at the same temperature T. By moving a piston a small change in volume ±∆V

can be made. The cylinder walls and the piston can be switched to

10

be either perfect thermal conductors or perfect thermal insulators. For each of the five processesbelow, state whether the changes in the quantities P, V, T,E, S (after re-establishment of theequilibrium) have been positive, negative or zero. “Slow” (“fast”) means during the volumechange, the speed of the piston is much less (greater) than the speed of the gas particles attemperature T .

(a) slow volume increase, thermal conductor isothermal expansion∆T = 0; ∆E = 0; ∆S =∆VV > 0;∆P < 0

(b) slow volume increase, thermal insulator reversible, adiabatic expansion∆Q = 0, hence ∆S =0 ; ∆E < 0 because of external work; ∆P < 0; ∆T < 0

(c) fast volume increase, thermal insulator adiabatic free expansion, irreversible∆S > 0;∆P <0;∆T = 0; ∆E = 0

(d) fast volume increase thermal conductor isothermal free expansion, irreversible∆S > 0;∆P <0;∆T = 0; ∆E = 0;

(e) fast volume decrease, thermal conductor isothermal (free) compression, ∆P > 0;∆T = 0;∆E = 0; ∆S < 0

3. (7P) Draw (and label) a thermodynamic cycle consisting of three processes involving an ideal

gas in a pressure-volume diagram.

(a) Leg 1-2: quasi-static adiabatic compressionLeg 2-3: isobaric coolingLeg 3-1: quasi-static isothermal expansion

(b) Determine the sign of the heat and work transfers and change in internal energy for eachleg and for the cycle as a whole. Leg 1-2: quasi-static adiabatic compression

Q12 = 0, W12 =3

2(P2V2 − P1V1), because P2 < P1 and V2 < V1 then W12 < 0.

On the other hand, we know: E12 = Q12 −W12 =⇒ E12 > 0

(c) Leg 2-3: isobaric cooling

W23 = P (V3−V2) and E23 =3

2(P3V3−P2V2), because P3 = P2 and V3 < V2 then W32 < 0

and E32 < 0.Then regarding to E32 = Q32 −W32 =⇒ Q32 < 0

(d) Leg 3-1: quasi-static isothermal expansion

E31 = 0, Q31 = W31.

For work done by system we have: W31 =V3´V1

PdV and for ideal gas we could rewrite:

W31 =V1´V3

NkBT

VdV = NkBT ln(

V 1

V 3), because V3 < V1 then W31 > 0 and then Q31 > 0.

11

4. (4P) Consider an ideal monoatomic gas with a mass of m=28g/mol and a heat capacity ofcV = 4 J

mol·K . For simplicity, in this task, use a gas constant of R = 3 Jmol·K .

(a) How much heat is required to heat 1kg of that gas from 210K to 420K at constant pressure?For constant pressure we need cp = cV +R

Q = n (cv +R) ∆T

with n = 1000g28g/mol , hence

Q =1000g

28g/mol(4 + 3)

J

mol ·K· 210K

=1000

4mol

J

mol ·K· 210K

= 52500J

(b) How much has the internal energy of that gas increased?

∆E = ncv ·∆T

∆E =1000g

28g/mol· 4J

mol ·K· 210K

=1000

7mol

J

mol ·K· 210K

= 30000J

(c) How much external work was done?

∆E = Q−WW = Q−∆E

= 52500J − 30000J

= 22500J

(d) The gas is heated by a heat pump, what is the efficiency of that heat pump? Effficiency ofa heat pump ηH

ηH =Thigh

Thigh − Tlow

=420K

420K − 210K= 2

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