Thermodynamics & Statistical Mechanics (Advanced)

Question 1 (10 marks)

(a) $dF = -S\,dT - P\,dV$. Since $F$ is a state function, mixed second derivatives are equal: $\frac{\partial}{\partial V}\left(\frac{\partial F}{\partial T}\right)_V = \frac{\partial}{\partial T}\left(\frac{\partial F}{\partial V}\right)_T \implies \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V.$ (2 marks: identify $dF$; 1 mark: equality of mixed partials; 1 mark: correct Maxwell relation)

From $P = \dfrac{Nk_BT}{V-Nb}$: $\left(\dfrac{\partial P}{\partial T}\right)_V = \dfrac{Nk_B}{V-Nb}$, so $\left(\frac{\partial S}{\partial V}\right)_T = \frac{Nk_B}{V-Nb}.$

(b) From $dU = TdS - PdV$ at fixed $T$: $\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T - P = T\cdot\frac{Nk_B}{V-Nb} - \frac{Nk_BT}{V-Nb} = 0.$ Hence $U$ is independent of $V$ at fixed $T$. (1 mark: $\partial U/\partial V$ expression; 1 mark: substitution; 1 mark: cancellation → 0)

(c) At fixed $T$: $\Delta F = -\int_{V_1}^{V_2} P\,dV = -\int_{V_1}^{V_2}\frac{Nk_BT}{V-Nb}dV$ $\Delta F = -Nk_BT\ln\!\left(\frac{V_2-Nb}{V_1-Nb}\right).$ (1 mark: $\Delta F = -\int P\,dV$; 1 mark: integrand; 1 mark: result)

Question 2 (11 marks)

(a) $Z_1 = 1 + e^{-\varepsilon/k_BT}$. Distinguishable: $Z = Z_1^N = (1+e^{-\varepsilon/k_BT})^N$. (1 + 1)

(b) $\langle U\rangle = -\dfrac{\partial \ln Z}{\partial \beta} = N\dfrac{\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}} = \dfrac{N\varepsilon}{e^{\beta\varepsilon}+1}$, with $\beta = 1/k_BT$. (1 mark: formula; 1 mark: differentiation; 1 mark: simplified result)

(c) Let $x = \varepsilon/k_BT$. $C_V = \frac{d\langle U\rangle}{dT} = Nk_B x^2\frac{e^{x}}{(e^{x}+1)^2}.$ (2 marks derivation; 1 mark result) Sketch: rises from 0, peaks around $k_BT \sim 0.42\,\varepsilon$ (i.e. $x\approx 2.4$), falls off as $T^{-2}$ at high $T$ — the Schottky anomaly. (1 mark)

(d) $T\to0$: all particles in ground state, $\Omega=1$, $S\to0$ (third law). $T\to\infty$: both levels equally populated, $\Omega = 2^N$, $S \to Nk_B\ln 2$. (1 + 1)

Question 3 (10 marks)

(a) Clapeyron: $\dfrac{dP}{dT} = \dfrac{L}{T\Delta V}$. With ideal vapour $\Delta V \approx V_{vap} = RT/P$ (per mole), negligible $V_{liq}$, constant $L$: $\frac{dP}{dT} = \frac{LP}{RT^2} \implies \frac{dP}{P} = \frac{L}{R}\frac{dT}{T^2}.$ Integrating: $\ln\dfrac{P_2}{P_1} = -\dfrac{L}{R}\left(\dfrac1{T_2}-\dfrac1{T_1}\right).$ (1 mark Clapeyron; 1 mark approximation; 1 mark separation; 1 mark integration)

(b) $\ln(3) = 1.0986$. $\dfrac1{T_2}-\dfrac1{T_1} = \dfrac1{330}-\dfrac1{300} = -3.0303\times10^{-4}$ K⁻¹. $L = -\frac{R\ln(P_2/P_1)}{(1/T_2 - 1/T_1)} = -\frac{8.314\times1.0986}{-3.0303\times10^{-4}} = 3.014\times10^4\ \text{J mol}^{-1}.$ $L \approx 30.1$ kJ/mol. (1 mark $\ln$; 1 mark reciprocal difference; 1 mark substitution; 1 mark answer)

(c) Gibbs phase rule: $F = C - P + 2$. Here $C=1$, $P=2$ (two coexisting phases): $F = 1 - 2 + 2 = 1$. One degree of freedom (the coexistence line). (1 + 1)

Question 4 (11 marks)

(a) $f(E) = \dfrac{1}{e^{(E-\mu)/k_BT}+1}$. At $T=0$: step function — $f=1$ for $E<\mu=E_F$, $f=0$ for $E>E_F$; all states below $E_F$ filled. (1 + 1)

(b) $N = V\int_0^{E_F} g(E)\,dE = V\cdot\dfrac{1}{2\pi^2}\left(\dfrac{2m}{\hbar^2}\right)^{3/2}\int_0^{E_F}E^{1/2}dE$ $= \frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\frac23 E_F^{3/2} = \frac{V}{3\pi^2}\left(\frac{2mE_F}{\hbar^2}\right)^{3/2}.$ Solve for $E_F$: $\dfrac{2mE_F}{\hbar^2} = (3\pi^2 n)^{2/3}$, so $E_F = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}.$ (1 mark integral setup; 1 mark integration; 1 mark rearrangement; 2 marks final result)

(c) $U = V\int_0^{E_F} E\,g(E)\,dE = \dfrac{V}{2\pi^2}\left(\dfrac{2m}{\hbar^2}\right)^{3/2}\int_0^{E_F}E^{3/2}dE$ $= \frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\frac25 E_F^{5/2}.$ Divide by $N$ expression: $\dfrac{U}{N} = \dfrac{(2/5)E_F^{5/2}}{(2/3)E_F^{3/2}} = \dfrac35 E_F$. Hence $U = \tfrac35 N E_F$. (1 mark integral; 1 mark integration; 1 mark ratio; 1 mark result)

Question 5 (8 marks)

(a) $Z = \sum_{n=0}^{\infty} e^{-n\beta h\nu} = \dfrac{1}{1-e^{-\beta h\nu}}$ (geometric series). $\langle E\rangle = -\frac{\partial \ln Z}{\partial\beta} = -\frac{\partial}{\partial\beta}\left[-\ln(1-e^{-\beta h\nu})\right] = \frac{h\nu\,e^{-\beta h\nu}}{1-e^{-\beta h\nu}} = \frac{h\nu}{e^{\beta h\nu}-1}.$ (1 mark sum; 1 mark closed form; 1 mark $\langle E\rangle$ formula; 1 mark differentiation; 1 mark result)

(b) High $T$: $\beta h\nu \ll 1$, $e^{\beta h\nu}-1 \approx \beta h\nu$, so $\langle E\rangle \approx \dfrac{h\nu}{\beta h\nu} = \dfrac1\beta = k_BT$. (2 marks) This matches equipartition: an oscillator has two quadratic degrees of freedom (kinetic + potential), each $\tfrac12k_BT$, giving $k_BT$. (1 mark)

[
  {"claim":"Q3: molar latent heat L ≈ 3.01e4 J/mol","code":"R=8.314; T1=300; T2=330; P1=1e4; P2=3e4; L=-R*log(P2/P1)/(1/T2-1/T1); result = abs(L-3.014e4) < 200"},
  {"claim":"Q3: Gibbs phase rule F=1 for 1 component 2 phases","code":"C=1; P=2; F=C-P+2; result = (F==1)"},
  {"claim":"Q2: high-T entropy limit is N kB ln2 (Omega=2**N)","code":"N=symbols('N',positive=True); kB=symbols('k_B',positive=True); S=kB*log(2**N); result = simplify(S - N*kB*log(2))==0"},
  {"claim":"Q4: U = (3/5) N E_F from ratio of integrals","code":"E,EF=symbols('E E_F',positive=True); num=integrate(E*sqrt(E),(E,0,EF)); den=integrate(sqrt(E),(E,0,EF)); result = simplify(num/den - Rational(3,5)*EF)==0"},
  {"claim":"Q5: high-T limit of Planck energy is kBT","code":"h,nu,kB,T=symbols('h nu k_B T',positive=True); E=h*nu/(exp(h*nu/(kB*T))-1); lim=limit(E,T,oo); result = simplify(lim-kB*T)==0"}
]