Solution to Statistical Physics Exam

29th June 2015

Name Student Number

Problem 1 Problem 2 Problem 3 Problem 4 Total Percentage Mark

Useful constants Gas constant R 8.31J/ (K · mol) −23 Boltzmann constant kB 1.38·10 J/K 23 −1 Avogadro number NA 6.02·10 mol Speed of light c 300·106m/s

1 (25P) Problem 1

1. (2P) Find the entropy of rolling an octahedral and a dodecahedral “die”, respectively.The entropy S reads n X 1 S = k p ln , B i p i=1 i

where pi is the probability that the system is in the i-th microstate, and n is the total number of the microstates. For the octahdral “die” one ﬁnds pi = 1/8 and n = 8, yielding S = kB ln 8 .

For the dodecahedral “die” one ﬁnds pi = 1/12 and n = 12, yielding S = kB ln 12 .

2. (1P) How does the phase space density of a microcanonical ensemble in equilibrium change with time? The phase space density of the microcanonical ensemble is constant in equilibrium. 3. (2P) Which are the units of the micro-canonical and the grand-canonical partition function, respectively? The partition functions are dimensionless, they are the number of states, in the microcanonical case, those which are compliant with a certain energy, in the grand-canonical case, weighted by the dimensionless Boltzmann factor.

4. (3P) Which quantities are constant in a

(a) micro-canonical ensemble ? N,V,E number of particles, volume, energy (b) canonical ensemble ?N,V,T number of particles, volume, temperature (c) grand-canonical ensemble?µ, V, T chemical potential, volume, temperature

5. (1P) How does the heat capacity of a classical ideal monoatomic gas depend on temperature? 3 Cv = 2 NkB no temperature dependence. 6. (1P) Which thermodynamic potential is constant during a Joule-Thomson process? Enthalpy H

7. (2P) Express the heat capacity at constant volume CV and the heat capacity at constant pressure CP as a second derivative of an appropriate thermodynamic potential (N=const) ∂Q dS ∂2F ∂F heat capacity at constant volume CV = ∂T = T dT = −T ∂T 2 since ∂T = V,N V V,N V,N −S (N, p, T )

∂Q dS ∂2G ∂G heat capacity at constant pressure Cp = ∂T = T dT = −T ∂T 2 since ∂T = p,N p p,N p,N −S (N, p, T ) 1. (2P) What is the qualitative diﬀerence between the phase diagram of water and that of a “normal substance”, e.g. carbon dioxide? The volume of solid water is larger than that of liquid water at most temperatures. Hence, the melting curve has a negative slope, whereas it has a positive slope for CO2. 2. (3P) Write down the total diﬀerential for the Helmholtz Free energy ? The Helmholtz free energy A dependes on number of particles N, volume V , and temperature T , hence ∂A ∂A ∂A dA = dN + dV + dT ∂N ∂V ∂T = µdN − P dV − SdT

2 3. (4P) Which quantities are intensive: volume, temperature, particle number, pressure, entropy? Temperature, pressure. 4. (3P) Sketch the isotherms for the van der Waals equation in a P,V (pressure,Volume) diagram for temperature T > Tc, T = Tc, and T < Tc where T C is the critical temperature.

5. (1P) What is the maximal occupation number of an energy level in an ideal Fermi gas? One for each spin level, or two if spin is not considered.

3 (25P) Problem 2

Consider a system of N ideal gas particles with mass m at a temperature T . The mean pressure is P . 3 Ng gas particles move freely in a volume V = Lg . Another Ns ideal gas particles are absorbed on a 2 surface with area Ls, forming a two-dimensional gas and N = Ng + Ns. The energy of an absorbed |p|2 particle is ε = 2m − ε0 where p is the two-dimensional momentum and ε0 is the binding energy per particle.

1. (6P) Calculate the classical partition functions of the free gas Zg and of the adsorbed gas Zs. The particles are indistinguishable. In continuous space

X V ≈ d3p (1) h3 ˆ p

Such that V Z (1) ≈ d3p exp [−βε ] (2) h3 ˆ p

p2 Plugging in βεp = and integrating over the sphere 2mkB T

2 V 2 p Z (1) = 3 4π dp · p exp − (3) h ˆ 2mkBT

2 We now substitute x2 = p and obtain a standard integral 2mkB T

∞ V 3 2 2 2 Z (1) = 3 4π (2mkBT ) dx · x exp −x (4) h ˆ−∞ | {z } 1 √ 4 π such that

V 3 Z (1,V,T ) = (2πmk T ) 2 (5) g h3 B V Ng Zg (Ng,V,T ) = 3N (6) Ng!λ g

2 Ls Nsε0 Zs (1,V,T ) = 2 (2πmkBT ) · exp (7) h kBT 2N Ls Nsε0 Zs (Ns,V,T ) = 2N · exp (8) Ns!λ kBT

3Ng Ng 2 V 2πmkBT Zg = 2 Ng! h

2N 2Ng 2 Ls 2πmkBT Nsε0 Zs = 2 · exp Ns! h kBT

2. (4P) Find an expression for the Gibbs free energies Gg and Gs of the free and the absorbed gas, respectively, using the partition functions from above.

G = A + PV = −kBT ln Z + NkBT

4 where we have used the ideal gas law PV = NkBT. Hence

 3Ng  Ng 2 V 2πmkBT Gg = −kBT ln  2  + NgkBT Ng! h

Ng V 3Ng 2πmkBT = −kBT ln + ln 2 + NgkBT Ng! 2 h 3N 2πmk T = −k T ln V N − ln N ! + g ln B + N k T B g 2 h2 g B 3N 2πmk T ≈ −k T ln V N − N ln N − N + g ln B + N k T B g g g 2 h2 g B 3 2πmk T = −N k T ln V − ln N + ln B g B g 2 h2 N 3 2πmk T = N k T ln g − ln B g B V 2 h2

" # 2Ns N Ls 2πmkBT Nsε0 Gs = −kBT ln 2 · exp + NskBT Ns! h kBT

2Ns Ls 2πmkBT Nsε0 = −kBT ln + Ns ln 2 + + NskBT Ns! h kBT 2πmk T N ε 2Ns B s 0 = −kBT ln Ls − ln Ns! + Ns ln 2 + + NskBT h kBT 2N 2πmkBT Nsε0 ≈ −kBT ln Ls − Ns ln Ns − Ns + Ns ln 2 + + NskBT h kBT 2 2πmkBT ε0 = −NskBT ln Ls − ln Ns + ln 2 + h kBT Ns 2πmkBT ε0 = NskBT ln 2 − ln 2 − Ls h kBT

3. (4P) Find the chemical potentials µg and µs of the free and the absorbed gas, respectively. The G ∂G chemical potential is the Gibbs free energy per particle µ = N or also µ = ∂N p,T . Hence, V 3 2πmkBT µg = −kBT ln + ln 2 Ng 2 h and 2 Ls 2πmkBT ε0 µs = −kBT ln + ln 2 + Ns h kBT

4. (4P) At temperature T , the free gas particles and the absorbed gas particles are in equilibrium. Calculate the mean number of gas particles absorbed per unit area in terms of of the given conditions (temperature pressure, ...) In equilibrium, the two chemical potentials are equal

µg = µs 2 V 3 2πmkBT Ls 2πmkBT ε0 −kBT ln + ln 2 = −kBT ln + ln 2 + Ng 2 h Ns h kBT 2 V 3 2πmkBT Ls 2πmkBT ε0 ln + ln 2 = ln + ln 2 + (9) Ng 2 h Ns h kBT

5 N We are interested in the number of particles per area, i.e. solving for 2 : Ls 2 Ls 2πmkBT ε0 V 3 2πmkBT − ln = ln 2 + − ln − ln 2 Ns h kBT Ng 2 h 1 2πmkBT ε0 V = − ln 2 + − ln 2 h kBT Ng 1 2 2 Ns Ng h ε0 ln 2 = ln + ln + Ls V 2πmkBT kBT 1 2 2 Ns Ng h ε0 2 = · exp (10) Ls V 2πmkBT kBT

5. (2P) Keeping temperature T and total number of particles N the same, now the volume in which the free gas particles can move is increased. What consequences does this have for the equilibrium between the free gas particles and the gas particles absorbed on the surface, i.e. are more or less particles absorbed compared to the case with smaller volume? Justify your answer. 2 According to eq. 10 the number of absorbed particles per area (Ls) decreases. One can think of there is more space for particles to move freely, and hence the probability to “ﬁnd” space in the larger volume is larger. 6. (3P) What do partition function, Gibbs free energy, and chemical potential for gas particles absorbed on a one-dimensional “surface” look like? 1N 2N 2 Ls 2πmkBT Nlε0 Zl = 2 · exp Nl! h kBT Nl 1 2πmkBT ε0 Gl = NlkBT ln − ln 2 − Ll 2 h kBT 1 Ll 1 2πmkBT ε0 µl = −kBT ln + ln 2 + Nl 2 h kBT

7. (2P) The Nl particles absorbed on the one-dimensional surface are now in equilibrium with the free gas particles. (There are no particles on a two-dimensional surface and N = Ng + Nl ). How does the number of particles per unit length in this equilibrium compare to the number of particles per unit area (from the equilibrium above) under the same (temperature, pressure, mass...) conditions? µg = µl

1 V 3 2πmkBT Ll 1 2πmkBT ε0 ln + ln 2 = ln + ln 2 + Ng 2 h Ng 2 h kBT Nl 1 2πmkBT ε0 V 3 2πmkBT ln 1 = ln 2 + − ln − ln 2 Ll 2 h kBT Ng 2 h Ng 2πmkBT ε0 = ln − ln 2 + V h kBT 2 Nl Ng h ε0 1 = · exp Ll V 2πmkBT kBT Hence Nl 1 2 2 L1 h √ l = = λ Ns 2 2πmkBT Ls

6 (25P) Problem 3

Consider an ideal Fermi gas of N particles in a volume V and in a magnetic ﬁeld H at temperature T = 0. The energy of a particle is p2 ε = ± µ H (11) 2m B where µB is Bohr’s mageton.

1. (6P) Give an expression for the chemical potential µ0 in terms of the number of particles N for vanishing magnetic ﬁeld. For zero temperature, the Fermi-Dirac distribution is a step function. To get the number of particles N, we integrate up to the Fermi momentum pF

pF V 3 N = 2 · 3 d p h ˆ0 pF V 2 = 2 · 3 4π p dp h ˆ0 4πV = 2 · p3 3h3 F At T = 0 and for vanishing magnetic ﬁeld, the chemical potential equals the Fermi energy 2 pF µ0 = εF = 2m . Hence, 1 N 3 p = 3h3 F 2 · 4πV and 2 N 3h3 3 µ = 8πV (12) 0 2m 2 3 2 N 3 ~ 3π = V (13) 2m

2. (4P) Write down the Fermi momenta of the spins oriented parallel and anti-parallel, respectively, to the external magnetic field H 6= 0 For finite magnetic field p2 ε = F ± µ H (14) F 2m B

and with µ0 = εF 1 2 pF + = [2m (µ0 − µBH)] 1 2 pF − = [2m (µ0 + µBH)]

3. (8P) Find the total energy of spins oriented parallel E− and anti-parallel E+, respectively. The total energy of the spins oriented anti-parallel to the external magnetic ﬁeld is V pF + p2 E+ = 3 + µBH dp h ˆ0 2m pF + 2 4πV p 2 = 3 + µBH p dp h ˆ0 2m pF + 4 4πV p 2 = 3 + µBHp dp h ˆ0 2m 5 3 4πV p µBHp = F + + F + h3 10m 3

7 and the total energy of the spins oriented parallel to the external magnetic ﬁeld is

5 3 4πV p µBHp E = F − − F − − h3 10m 3

4. Bonus Task (3P) Show that for weak external magnetic ﬁelds, µ0 µBH, the average energy per particle " 2# E 3 5 µBH ≈ µ0 1 − (15) N 5 2 µ0 The average energy per particle is E E + E = + − N N 5 3 5 3 4πV p p µBH 4πV p p µBH = 1 F + + F + + F − − F − N h3 10m 3 h3 10m 3 5 5 3 3 ! 4πV p + p µBH p − p = F + F − + F + F − Nh3 10m 3

Using eq.12 4πV 1 3 = 3 Nh 2 2 · (2mµ0) we have 5 5 3 3 ! E 3 pF + + pF − µBH pF + − pF − = 3 + N 2 10m 3 2 · (2mµ0) we can write

5 5 p5 + p5 [2m (µ − µ H)] 2 + [2m (µ + µ H)] 2 F + F − = 0 B 0 B 10m 10m 3 5 3 5 (2m) 2 (µ − µ H) 2 + (2m) 2 (µ + µ H) 2 = 2m 0 B 0 B 10m 3 h 5 5 i 2 2 2 (2m) (µ0 − µBH) + (µ0 + µBH) = 5 and

µBH µBH 3 3 p3 − p3 = [2m (µ − µ H)] 2 − [2m (µ + µ H)] 2 F + F − 3 3 0 B 0 B

For µ0 µBH we can expand

h 5 5 i 5 5 3 5 5 3 (µ − µ H) 2 + (µ + µ H) 2 ≈ µ 2 − µ 2 µ H... + µ 2 + µ 2 µ H... 0 B 0 B 0 2 0 B 0 2 0 B 5 2 ≈ 2µ0 (16) and

h 3 3 i 3 3 1 3 3 1 (µ − µ H) 2 − (µ + µ H) 2 = µ 2 − µ 2 µ H... − µ 2 + µ 2 µ H... 0 B 0 B 0 2 0 B 0 2 0 B 1 2 ≈ −3µ0 µBH (17)

8 Such that plugging this in

3 5 ! E 3 (2m) 2 2µ 2 µ H 3 1 0 B 2 2 ≈ 3 + (2m) −3µ0 µBH N 2 5 3 2 · (2mµ0) 5 ! 2 1 3 2µ0 µBH 2 = 3 + −3µ0 µBH 2 5 3 2 · (µ0) 3 2µ0 µBH = − (µBH) 2 5 µ0 2 ! 3 5 (µBH) = µ0 − 5 2 (µ0) 2 ! 3 5 (µBH) = µ0 1 − 2 (18) 5 2 (µ0)

∂M 5. Bonus Task (4P) Calculate the susceptibility χ = ∂H for weak external magnetic ﬁelds, where M is the average magnetisation per volume. µ (N − N ) M = B − + V 4πV 3 4πV 3 µB 3 p − 3 p = 3h F − 3h F + V 3 3 4πV 2 2 µB 3h3 [2m (µ0 + µBH)] − [2m (µ0 − µBH)] = V and again with eq. 12 4πV 1 3 = 3 3h N 2 2 · (2mµ0) thus 3 3 N 2 2 µB 3 [2m (µ0 + µBH)] − [2m (µ0 − µBH)] M = 2·(2mµ0) 2 V N 3 3 2 2 = µB 3 [(µ0 + µBH)] − [(µ0 − µBH)] 2 2V (µ0)

In the limit of low magnetic ﬁeld and µBH µ ≈ εF we expand again

h 3 3 i 3 3 1 3 3 1 (µ + µ H) 2 − (µ − µ H) 2 = µ 2 + µ 2 µ H + ... − µ 2 − µ 2 µ H... 0 B 0 B 0 2 0 B 0 2 0 B 1 2 ≈ 3µ0 µBH

1 N 2 M = µB 3 3µ0 µBH 2 2V (µ0) 3N = µB (µBH) (19) 2V (µ0)

3N 2 = µBH 2V µ0

9 Figure 1: Pressure-Volume (left) and temperature-entropy (right) diagram of a three-step cyclic process.

(25P) Problem 4

A.) Figure 1 shows a three-steps cyclic process as Pressures-volume (P,V ) and temperature-entropy (T,S) diagram, respectively. Consider 1 mol of an ideal gas which has initially volume V1, pressure P1, and temperature T1, undergoing such a cyclic process. 1. (3P) Describe the steps 1 → 2; 2 → 3; and 3 → 1 sketched in the two diagrams: Which thermodynamic quantity changes, which is kept constant ? 1 → 2 isochoric heating to T2;2 → 3 isentropic (adiabatic) expansion to V3; and 3 → 1 isobaric cooling to T1 2. (12P) For each step calculate the performed work W and the heat transfer Q in terms of pressure P1 and volumes V1,V3. 1 → 2: V2 W12 = P1dV = 0 ˆV1

since V2 = V1

Q12 = CV (T2 − T1) V = C 1 (P − P ) V R 2 1 V1P1 P2 = CV − 1 R P1 V1P1 P2 = CV − 1 R P3 γ V1P1 V3 = CV − 1 R V2 γ V1P1 V3 = CV − 1 R V1

γ γ where we have used the ideal gas law PV = RT and the adiabatic relation V1 P1 = V2 P2 and V1 = V2 together with P1 = P3. 2 → 3: Q23 = 0

10 since S3 = S2 (adiabatic process). Hence ∆U = W and

W23 = U3 − U2

From the deﬁnition of the heat capacity at constant volume we rewrite

U3 − U2 = CV (T3 − T2) C = V (P V − P V ) R 1 3 2 1

where we have used the ideal gas law again as P 3V 3=RT 3=P1V 3 and P2V2 = RT2=P2V1. Using the adiabatic relations we then arrive at CV P1 P2 W23 = V3 − V1 R P1 CV P1 P2 = V3 − V1 R P3 γ CV P1 V3 = V3 − V1 R V2 γ CV P1 V3 = V3 − V1 R V1

3 → 1: V1 W31 = P1dV = P1 (V1 − V3) ˆV3 From the ﬁrst law we know Q31 = W31 + U1 − U3 plugging in the expression for work and rearranging

(U1 + P1V1) − (U3 + P1V3) = (H1 − H3)

= CP (T1 − T3)

= γCV (T1 − T3)

where we have used the deﬁnition if the enthalpy, and the heat capacity at constant pressure such that we ﬁnally obtain γC P Q = − V 1 (V − V ) < 0 31 R 3 1

B.) Bonus Task For a low-density gas the virial expansion can be terminated at ﬁrst order in the density and the equation of state is: Nk T N P = B [1 + B (T )] (20) V V 2 where P :pressure, V : volume, N: number of particles, T :temperature and kB: Boltzmann’s constant. B2(T ) is the second virial coeﬃcient. The heat capacity will have corrections to its ideal gas value. We can write it in the form 3 N 2k C = Nk − B f(T ) (21) V,N 2 B V

11 1. (3P) Find the entropy and internal energy as partial derivatives of the Helmholtz free energy, A (Hint: use the relation between pressure P and Helmholtz free energy at constant temperature). Regarding to Helmholtz free energy for canonical ensemble we have ∂A P = −( ) (22) ∂V T Nk T N 2k T P = B + B B (T ) = P + δP (23) V V 2 2 ideal Using equ.22 the free energy is

N 2 A = A + k TB (T ) (24) ideal B 2 V Then ∂A N 2 S = −( ) = S + δS, δS = −k [B (T ) + TB0 (T )] (25) ∂T V ideal B V 2 2 Since we know both entropy and Helmholtz free energy, we ﬁnd the internal energy as

N 2 N 2 U = A + TS = U + k TB (T ) − k T [B (T ) + TB0 (T )] = U + δU (26) ideal B 2 V B V 2 2 ideal N 2 δU = −k T 2B0 (T ) (27) B 2 V

2. (4P) Find the form that f(T ) must have in order for the two equations to be thermodynamically consistent. As a result of subtask 1,

∂U N 2 C = ( ) = Cideal + δC =⇒ δC = −k T [2B0 (T ) + TB00(T )] (28) V ∂T V V V V B V 2 2 We ﬁnd in this way, 0 2 00 f(T ) = 2TB2(T ) + T B2 (T ) (29)

3. (3P) Find the heat capacity at constant pressure and constant number of partivles CP,N . Let us express the equation of state as V (P,T ). Since the density is assumed to be small in the correction one can use equation for the ideal gas to ﬁnd the the volume. We have, Nk T V = B + NB (T ) (30) P 2 Consequently, the entropy can be expressed as

S = Sideal + δS1 (31)

with replacing equ.30 in δS in equ.25 we have

B + TB0 δS = −k N 2 2 (32) 1 B k T B + B P 2 now ∂δS C = Cideal + T ( 1 ) (33) P P ∂T P

12 0 0 kB T kB 0 0 (B2 + TB2) ( P + B2) − ( P + B2)(B2 + TB2) = −kBN (34) kB T 2 ( P + B2) N 2 = Cideal − k [T (2B0 + TB00) − (B + TB0 )] (35) P B V 2 2 2 2 N 2 = Cideal + δC + k (B + TB0 ) (36) P V B V 2 2 As a result, N 2 C − C = (C − C )ideal + k (B + TB0 ) (37) P V P V B V 2 2