Statistical Physics Exam

23rd April 2014

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/sPlanck’s constant h 6.63·10−34J · s

1

Problem 1 (25P)

1) (2P) Which thermodynamic potential becomes minimimal in equilibrium in each of the follow-ing situations:

• at constant number of particles, contant temperature, and constant volume. Helmholtzfree energy F• at constant number of particles, contant temperature, and constant pressure. Gibb’s freeenergy G

2) (3P) Consider two gas chambers A and B which are in contact. In equilibrium, which thermo-dynamic quantities are the same in A and B if

a) the systems can exchange heat. Temperatureb) the systems can exchange particles. Chemical potentialc) the wall between the two systems can move. Pressure

3) (4P) Write down the four steps of a Carnot cycle. How do the entropy and temperature changein each step?

a) ...Isothermic...expansion ∆T = 0 ; ∆S > 0

b) ...Adiabatic... expansion ∆T < 0 ; ∆S = 0

c) ...Isothermic...compression ∆T = 0 ; ∆S < 0

d) ...Adiabatic... compression ∆T > 0 ; ∆S = 0

Figure 1: Carnot cycle

4) (3P) How do vapour pressure, boiling temperature, and melting temperature of a mixturechange compared to the pure liquid?Vapour pressure decreases, boling temperature increases, melting emperature decreases.

5) (2P) Which corrections does the van-der-Waals gas introduce compared to the ideal gas?Particles can interact: a represents the attraction between the particles, b the excluded volume:(p = a

v2

)(v − b) = kT

6) (2P) How do energy and heat capacity of an ideal monoatomic gas depend on temperature?Energy E = 3

2NkBT ; Heat Capacity Cv = 3

2NkB; Cp = 5

2NkB (N : number of atoms; kB:

Boltzmann constant)

7) (4P) A system has three energy levels, the first of which with E1 = ε1, is non-degenerate,the second with E2 = 2ε1 is twofold-degenerate, and the third one with E3 = 3ε1 is three-fold degenerate. Write down the microcanonical partition function of a classical three-particlesystem for Etot = 4ε1. Ω =

∑Etot

= 4ε1

2

Figure 2: Possible states with Etot = 4ε1. Top: Particles non-disinguishable Ω = 2 Bottom: Particles distinguishableΩ = 6

8) (1P) How does the entropy change in an irreversible process?Entroopy increases.

9) (4P) Sketch the Maxwell-Boltzmann distribution of velocities p (|v|) for an ideal gas. What islarger: the average velocity or the most probable velocity?The Maxwell-Boltzmann distribution for the speed is depicted as below,

Figure 3: The Maxwell-Boltzmann distribution for the speed

The average speed is larger than the most probable speed.

3

Problem 2 (25P)

Consider a system of N spins subject to a magnetic field B. The spins are non-interactingwith the spin number s = 1, distinguishable, and have the non-degenerated energy eigenvaluesεm = −µBm per one spin, where µ is the magnetic moment per one spin, and m = −1, 0, 1.

a) (5 points): Calculate the partition function ZN .The single-spin partition function Z1:

Z1 =1∑

m=−1

eβµBm

= 1 + eβµB + e−βµB︸ ︷︷ ︸2 cosh(βµB)

= 1 + 2 cosh(βµB). (1)

The N -spin partition function ZN simply leads to

ZN = Z1N

= [1 + 2 cosh(βµB)]N . (2)

or =[1 + eβµB + e−βµB

]N. (3)

b) (5 points): Calculate the average energy E (2 points). For this, you may calculate E usingZN , or you may calculate E from the average of the total energy eigenvalues. The averageenergy is a monotonically increasing function of kBT : Find two asymptotic values of E in thelimit of T → 0 and T →∞ (2 points), and sketch E(kBT ) (1 point).The average energy E:

E = −[∂ ln(ZN)

∂β

]N,V

= −2NµB sinh(βµB)

1 + 2 cosh(βµB). (4)

or = NµBe−βµB − eβµB

1 + eβµB + e−βµB. (5)

or, using the average of the total energy eigenvalues,

E = N〈−µBm〉

= N

∑1m=−1−µBmeβµBm∑1

m=−1 eβµBm

= −2NµB sinh(βµB)

1 + 2 cosh(βµB).

or = NµBe−βµB − eβµB

1 + eβµB + e−βµB.

4

For T → 0:

E → −NµB. (6)

For T →∞:

E → 0. (7)

1 point : The average energy is plotted as below:

Figure 4: The average energy E as a function of kBT

c) (6 points): Sketch the heat capacity CB(kBT ). For this, you may calculate CB using E, oryou may calculate CB using the energy fluctuations, or you may use the sketch of E(kBT ).The heat capacity CB:

CB =

[∂E

∂T

]B

=2B2µ2N cosh

(BµkBT

)kBT 2

(2 cosh

(BµkBT

)+ 1) − 4B2µ2N sinh2

(BµkBT

)kBT 2

(2 cosh

(BµkBT

)+ 1)2

=2Nµ2B2

(2 + cosh

(µBkBT

))kBT 2

(1 + 2 cosh

(µBkBT

))2 . (8)

or, using the energy fluctuations,

CB =〈∆E2〉kBT 2

= N2 〈(−µBm)2〉 − 〈−µBm〉2

kBT 2

=2B2µ2N cosh

(BµkBT

)kBT 2

(2 cosh

(BµkBT

)+ 1) − 4B2µ2N sinh2

(BµkBT

)kBT 2

(2 cosh

(BµkBT

)+ 1)2

=2Nµ2B2

(2 + cosh

(µBkBT

))kBT 2

(1 + 2 cosh

(µBkBT

))2 .

(9)

5

Figure 5: The heat capacity per the Boltzmann constant CB/kB as a function of kBT

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

F = − ln(ZN)

β

= −NkBT ln [1 + 2 cosh(βµB)] . (10)

e) (5 points): Sketch the magnetizationM(kBT ). For this, you may calculate the average of thetotal magnetic moment 〈−Nεm/B〉, or you may calculate M using the average energy E, oryou may calculateM using the Helmholtz free energy F .The magnetizationM:

M = −(∂F∂B

)N,T

=2Nµ sinh

(µBkBT

)1 + 2 cosh

(µBkBT

) . (11)

or, using the average energy E,

M = −E/B

=2Nµ sinh

(µBkBT

)1 + 2 cosh

(µBkBT

) .or, using the average of the total magnetic moment,

M = 〈−Nεm/B〉

=

∑1m=−1Nµme

βµBm∑1m=−1 e

βµBm

=2Nµ sinh

(µBkBT

)1 + 2 cosh

(µBkBT

) .6

Figure 6: The magnetization per the magnetic moment M/µ as a function of kBT

Problem 3 (25P)

Consider N spinless non-interacting free bosons of mass m in a volume V at temperature T .

a) (2 points) Write down the average occupation nε of single particle state with energy ε as afunction of the chemical potential µ and the temperature T .Bose-Einstein distribution

nε =1

e(ε−µ)/(kBT ) − 1(12)

b) (5 points) Find the density of states D(ε)dε.The density of state in phase space is

D(ε) =1

h3

∫ddqddpδ(ε−H(~q, ~p)) (13)

For a free particle the energy function is H(~q, ~p) = p2

2m. For a particle in 3D (d = 3) one gets:

D(ε) =4πV

h3

∫ ∞0

p2δ

(ε− p2

2m

)dp (14)

Now, one has to use the rule for composing a delta distribution with a function:

δ(g(p)) =∑

i,g(pi)=0

1

|g′(pi)|δ(p− pi) (15)

For g(p) = ε− p2

2m, g′(p) = − p

mand p0 =

√2mε, so:

δ

(ε− p2

2m

)=

m√2mε

δ(p−√

2mε)

(16)

7

The density of state becomes

D(ε) =2πV

h3(2m)3/2√ε (17)

c) (5 points) Sketch D(ε) and nε as a function of the single particle energy µ < ε (Rememberµ < 0). What happens to nε close to µ?

Close to ε = µ, nε diverges as the denominator in the distribution function approaches zero.

d) (3 points) Express the density of particles as an integral over energy.Intermediate result:

N

V=

2π

h3

(2m

β

)3/2 ∫ ∞0

x1/2

exp(x− βµ)− 1dx where β =

1

kBT(18)

N =

∫nεD(ε)dε

=

∫ ∞0

dε1

eβ(ε−µ) − 1

2πV

h3(2m)3/2√ε

(19)

Making the substitution x = βε the integral turns into:

N

V=

2π

h3

(2m

β

)3/2 ∫ ∞0

x1/2

ex−βµ − 1dx (20)

e) (5 points) Using the fact that the density does not change if the temperature is lowered,determine whether µ increases or decreases as T is lowered.

8

N

V=

2π

h3

(2m

β

)3/2

︸ ︷︷ ︸A

∫ ∞0

x1/2

ex−βµ − 1dx︸ ︷︷ ︸

B

(21)

As T is lowered, the density on the left hand side does not change, but the terms A and B,which do depend on the temperature change indivudially.

T → 0⇒ β =1

kBT→∞⇒ A→ 0 (22)

B has to change in such a way, that the product AB remains constant:

A→ 0⇒ B →∞⇒ e−βµ → 1 (23)

Since we know, that β →∞, µ has to go to 0 faster. Since initially µ < 0, this implies that µincreases towards 0 as T is lowered.

f) (5 points) At the critical temperature µ ≈ 0 and Bose-Einstein condensation sets in. Findthe critical temperature Tc as a function of the density. Leave your answer in terms of adimensionless integral.

g) At T = Tc, µ = 0.N

V=

2π

h3(2mkBTc)

3/2

∫ ∞0

x1/2

ex − 1dx︸ ︷︷ ︸

dimensionless integral I≈2.315

(24)

Solving for the critical temperature gives:

Tc =1

2mkB

(h3

2πI

N

V

)2/3

(25)

9

Figure 7: Qualitative results of the two series of experiments.

Figure 8: Phase diagramme of a typical compound, e.g. carbon dioxide. Blue: boiling cure, red: sublimation curve,green: meling curve

Problem 4 (25P)

In two series of experiments the vapour pressure p of pure compound is measured at differenttemperatures T. In the first series, the gas is in equilibrium with its liquid, in the second series,the gas is in equilibrium with its solid phase. The results are qualitatively shown in Fig. 7.

a) (6P) Draw a phase-diagramm (p,T) of a typical compound (e.g. Carbon Dioxide) with gas,liquid and solid phase, and label all important points, lines, and areas.

b) (2P) Write down the differential equation which describes the coexistence curves.The gas-liquid and gas-solid equilibrium can be described by the Clausius Claperon equation

dp

dT=

∆H

T∆V(26)

where ∆H is the heat of vapourisation or heat of sublimation, respectively.

10

c) (3P) Which three approximations are introduced to solve the above equation for the gas-solidcoexistence curve?Neglecting the volume of the solid Vgas Vsolid; i.e. ∆V = Vgas − Vsolid ≈ Vgas,treating the vapour as an ideal gas Vgas = nRT

p,

and ngeglecting the temperature dependence of the heat of ∆H

we can integrate eq.26 (for n = 1)

1

pdp =

∆H

RT 2dT∫ p

p0

1

pdp =

∫ T

T0

∆H

RT 2dT

lnp

p0

= −∆H

R

(1

T− 1

T0

)(27)

ln p = −∆H

R

(1

T− 1

T0

)+ ln p0

Same approximaions can be made for the liquid-gas transition.

d) (4P) What do slope and intercept of the curves in Fig.7 tell you?In Fig.7 the linearised form of the integrated eq.26 is plotted:

ln p = −∆H

R· 1

T+

∆H

R· 1

T0

+ ln p0

the slope is −∆HR· and the intercept ∆H

R· 1T0

+ ln p0. Hence, from the slope one can determinethe heat of vapourisation. Using

∆G = −RT lnp

p0

and∆G = ∆H − T∆S

we get

lnp

p0

= −∆G

RT

lnp

p0

= −∆H

R· 1

T+

∆S

R

and hence the intercept is ∆SR

+ln p0 from which we get the change in entropy for the transitions.

e) (5P) Which of the two curves corresponds to the experiment of the gas-liquid equilibrium andwhich one to the gas-solid (explain briefly)?Blue: solid-gas; Red: liquid-gas. The enthalpy H is a state function, hence ∆Hs→g = ∆Hs→l +∆Hl→g Since H increases with the transition to the higher-temperature phase these ∆H are allpositive and herefore ∆Hs→g > ∆Hl→g. The slope of the curve corresponding to the solid-gasexperiment is thus steeper.

f) (5P) How can you determine at which pressure and temperature all three phases coexist basedon the results of the above experiments (explain)?

11

At the triple point, all three phases coexist. Hence all three coexistence lines intersect at thatpoint. From the plotted experimental data we get

lnp

p0

= −∆H

R

(1

T− 1

T0

)p = p0 exp

[−∆H

R

(1

T− 1

T0

)]for each of the two experiments. Setting the two curves equal gives the intersection:

ln p = −∆Hlg

R

(1

T− 1

T l0

)+ ln pl0

ln p = −∆Hsg

R

(1

T− 1

T s0

)+ ln ps0

−∆Hlg

R

(1

T− 1

T l0

)+ ln pl0 = −∆Hsg

R

(1

T− 1

T s0

)+ ln ps0

∆Hsg

R

(1

T− 1

T s0

)− ∆Hlg

R

(1

T− 1

T l0

)= ln ps0 − ln pl0

∆Hsg

RT− ∆Hsg

RT s0− ∆Hlg

RT− ∆Hlg

RT l0= ln ps0 − ln pl0

∆Hsg

RT− ∆Hlg

RT= ln ps0 − ln pl0 +

∆Hsg

RT s0+

∆Hlg

RT l0∆Hsg −∆Hlg

RT= ln ps0 − ln pl0 +

∆Hsg

RT s0+

∆Hlg

RT l0

1

T=

(ln ps0 − ln pl0 + ∆Hsg

RT s0

+∆Hlg

RT l0

)∆Hsg−∆Hlg

R

T =

∆Hsg−∆Hlg

R(ln ps0 − ln pl0 + ∆Hsg

RT s0

+∆Hlg

RT l0

)... and plug in T to get p

lnp

p0

= −∆H

R

(1

T− 1

T0

)

ln p = −∆Hlg

R

(

ln ps0 − ln pl0 + ∆Hsg

RT s0

+∆Hlg

RT l0

)∆Hsg−∆Hlg

R

− 1

T l0

+ ln pl0

p = pl0 exp

[−∆Hlg

R

1

T+

∆Hlg

R

1

T0

]

p = pl0 exp

−∆Hlg

R·

(ln

ps0pl0

+∆Hlg

RT s0

+ ∆Hsg

RT l0

)(

∆Hsg

R− ∆Hlg

R

) +∆Hlg

RT l0

12