Thermodynamics & Statistical Mechanics (Advanced)

Question 1 (12)

(a) From $U(S,V)$ with $dU = T\,dS - P\,dV$. First transform to enthalpy (swap $V\to P$): $H = U + PV$, so $dH = T\,dS + V\,dP$. (1) Then to Gibbs (swap $S\to T$): $G = H - TS = U + PV - TS$. (1) $dG = dH - T\,dS - S\,dT = T\,dS + V\,dP - T\,dS - S\,dT = -S\,dT + V\,dP.$ (1) Hence $\left(\frac{\partial G}{\partial T}\right)_P = -S$ (1) and $\left(\frac{\partial G}{\partial P}\right)_T = V$. (1)

(b) Since $dG$ is an exact differential, mixed second partials are equal: (1) $\frac{\partial}{\partial P}\left(\frac{\partial G}{\partial T}\right) = \frac{\partial}{\partial T}\left(\frac{\partial G}{\partial P}\right).$ (1) Substituting the first derivatives: $-\left(\frac{\partial S}{\partial P}\right)_T = \left(\frac{\partial V}{\partial T}\right)_P.$ (2) (Exactness holds because $G$ is a state function — its differential is path-independent.)

(c) At fixed $T,P$ the combined system+reservoir entropy change is $dS_{tot} = -dG/T \ge 0$, so spontaneous processes lower $G$; equilibrium is the minimum of $G$. Full marks for connecting minimisation of $G$ to second-law maximisation of total entropy. (3)

Question 2 (12)

(a) Maximise $S = -k_B\sum p_i\ln p_i$ with constraints $\sum p_i = 1$ and $\sum p_i E_i = \langle E\rangle$. (1) Lagrangian: $\mathcal{L} = -k_B\sum p_i\ln p_i - \alpha(\sum p_i -1) - \beta k_B(\sum p_i E_i - \langle E\rangle)$. $\frac{\partial \mathcal{L}}{\partial p_i} = -k_B(\ln p_i + 1) - \alpha - \beta k_B E_i = 0$. (2) Solving: $\ln p_i = -1 - \alpha/k_B - \beta E_i$, so $p_i \propto e^{-\beta E_i}$. (2) Normalisation gives $p_i = \frac{e^{-\beta E_i}}{Z}$, $Z=\sum e^{-\beta E_i}$; identifying with thermodynamics $\beta = 1/k_BT$. (1)

(b) $\langle E\rangle = \sum p_i E_i = \frac{1}{Z}\sum E_i e^{-\beta E_i} = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} = -\frac{\partial \ln Z}{\partial\beta}.$ (2) From $S = -k_B\sum p_i\ln p_i$ with $\ln p_i = -\beta E_i - \ln Z$: $S = k_B(\beta\langle E\rangle + \ln Z)$, hence $F = \langle E\rangle - TS = -k_BT\ln Z$. (2)

(c) A microstate is a complete specification of every particle's state (positions/momenta or quantum numbers); a macrostate is the set of microstates sharing the same macroscopic variables ($E,V,N$). $Z$ sums Boltzmann weights over all microstates, encoding how they aggregate into thermodynamics. (2)

Question 3 (10)

(a) Equipartition: each quadratic degree of freedom contributes $\tfrac12 k_BT$ to $\langle E\rangle$. (1) Diatomic (3 translational + 2 rotational) = 5 dof: $\langle E\rangle = \tfrac52 N_A k_BT = \tfrac52 RT$. (2) $C_V = \tfrac52 R \approx 20.8\ \text{J mol}^{-1}\text{K}^{-1}$. (1)

(b) $Z = 1 + e^{-\beta\varepsilon}$. (1) $\langle E\rangle = -\frac{\partial\ln Z}{\partial\beta} = \frac{\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}} = \frac{\varepsilon}{e^{\beta\varepsilon}+1}.$ (2) $C = \frac{d\langle E\rangle}{dT} = \frac{d\langle E\rangle}{d\beta}\frac{d\beta}{dT}$, with $\frac{d\beta}{dT}=-\frac{1}{k_BT^2}$. $\frac{d\langle E\rangle}{d\beta} = -\frac{\varepsilon^2 e^{\beta\varepsilon}}{(e^{\beta\varepsilon}+1)^2}$. (2) So $C = k_B\left(\frac{\varepsilon}{k_BT}\right)^2\frac{e^{\varepsilon/k_BT}}{(e^{\varepsilon/k_BT}+1)^2}$. (1)

Question 4 (12)

(a) $\langle n_i\rangle = \dfrac{1}{e^{\beta(E_i-\mu)}\pm 1}$: $+$ for Fermi–Dirac, $-$ for Bose–Einstein. (2) The $+$ arises from the Pauli exclusion principle (antisymmetric wavefunction, max one fermion per state); the $-$ from symmetric bosonic wavefunctions allowing unlimited occupation. (2)

(b) At $T=0$ electrons fill all states up to $k_F$. Number of states in a sphere of radius $k_F$ with spin factor 2: $N = 2\cdot\frac{V}{(2\pi)^3}\cdot\frac{4}{3}\pi k_F^3 = \frac{V k_F^3}{3\pi^2}.$ (3) So $n = N/V = k_F^3/3\pi^2 \Rightarrow k_F = (3\pi^2 n)^{1/3}$. (1) Then $E_F = \frac{\hbar^2 k_F^2}{2m} = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}.$ (2)

(c) Below $T_c$ the excited states cannot accommodate all bosons (their maximum capacity $\propto T^{3/2}$ saturates), so the remainder collapse into the single-particle ground state, giving macroscopic occupation — a phase transition driven purely by quantum statistics, not interactions. (2)

Question 5 (8)

(a) On a coexistence curve the two phases have equal chemical potential (molar Gibbs energy): $\mu_1(T,P)=\mu_2(T,P)$. (1) Along the curve $d\mu_1 = d\mu_2$. With $d\mu = -s\,dT + v\,dP$ (per mole): $-s_1 dT + v_1 dP = -s_2 dT + v_2 dP$. (1) $\Rightarrow \frac{dP}{dT} = \frac{s_2-s_1}{v_2-v_1} = \frac{\Delta s}{\Delta v}.$ (1) Latent heat $L = T\Delta s$, so $\frac{dP}{dT} = \frac{L}{T\Delta v}.$ (1)

(b) $f = C - P + 2$, single component $C=1$. (1) (i) Triple point: $P=3$ phases $\Rightarrow f = 1-3+2 = 0$ (invariant, fixed T and P). (1.5) (ii) Coexistence line: $P=2 \Rightarrow f = 1-2+2 = 1$ (univariant, one free variable). (1.5)

Question 6 (6)

import numpy as np
def thermo(E, T, kB=1.380649e-23):
    beta = 1/(kB*T)                 # inverse temperature 1/kT
    w = np.exp(-beta*(E - E.min())) # Boltzmann weights (shifted for stability)
    Z = w.sum()                     # partition function = sum of weights
    p = w/Z                         # normalized occupation probabilities
    Emean = np.sum(p*E)             # <E> = sum p_i E_i
    F = E.min() - kB*T*np.log(Z)    # F = -kT ln Z (undo the shift)
    return Z, Emean, F

Marks: correct $\beta$ (1), weights/Z (2), $\langle E\rangle$ (1.5), $F=-k_BT\ln Z$ (1.5). Numerical shift for stability is a bonus but not required.

[
  {"claim": "Two-level average energy = eps/(exp(beta*eps)+1)", "code": "beta,eps=symbols('beta eps',positive=True); Z=1+exp(-beta*eps); Emean=-diff(log(Z),beta); result = simplify(Emean - eps/(exp(beta*eps)+1))==0"},
  {"claim": "Two-level heat capacity is Schottky form", "code": "kB,T,eps=symbols('k_B T eps',positive=True); b=1/(kB*T); Z=1+exp(-b*eps); Emean=eps*exp(-b*eps)/Z; C=diff(Emean,T); target=kB*(eps/(kB*T))**2*exp(eps/(kB*T))/(exp(eps/(kB*T))+1)**2; result = simplify(C-target)==0"},
  {"claim": "Fermi energy from k_F: E_F=hbar^2/(2m)(3 pi^2 n)^(2/3)", "code": "hbar,m,n=symbols('hbar m n',positive=True); kF=(3*pi**2*n)**Rational(1,3); EF=hbar**2*kF**2/(2*m); target=hbar**2/(2*m)*(3*pi**2*n)**Rational(2,3); result = simplify(EF-target)==0"},
  {"claim": "Diatomic C_V = 5/2 R", "code": "R=symbols('R',positive=True); dof=5; Cv=Rational(dof,2)*R; result = simplify(Cv-Rational(5,2)*R)==0"}
]