Level 2 — RecallThermodynamics & Statistical Mechanics (Advanced)

Thermodynamics & Statistical Mechanics (Advanced)

30 minutes40 marksprintable — key stays hidden on paper

Level 2 Paper — Recall & Standard Problems

Time Limit: 30 minutes
Total Marks: 40


Instructions: Answer all questions. Use kBk_B for Boltzmann's constant, RR for the gas constant. Show working where required.


Q1. State the four thermodynamic potentials UU, HH, FF, GG and write each in terms of UU using the appropriate variables. (4 marks)

Q2. Starting from dF=SdTPdVdF = -S\,dT - P\,dV, derive the Maxwell relation associated with the Helmholtz free energy. (4 marks)

Q3. State the Gibbs phase rule and use it to determine the number of degrees of freedom for a single-component system at its triple point. (4 marks)

Q4. Write down Boltzmann's entropy formula. A system has Ω=4\Omega = 4 equally likely microstates in state A and Ω=8\Omega = 8 in state B. Compute the entropy difference SBSAS_B - S_A in units of kBk_B. (4 marks)

Q5. For a two-level system with energies 00 and ε\varepsilon, write down the canonical partition function ZZ for a single particle and derive an expression for the average energy E\langle E\rangle. (5 marks)

Q6. State the equipartition theorem. Hence give the total internal energy and the molar heat capacity CVC_V of one mole of an ideal diatomic gas (translational + rotational degrees of freedom only). (5 marks)

Q7. Write the Clausius–Clapeyron equation and state the physical meaning of each symbol. (4 marks)

Q8. Write down the Bose–Einstein and Fermi–Dirac occupation-number distributions. State one key physical difference between the two. (5 marks)

Q9. Write the Planck distribution for the spectral energy density u(ν)u(\nu) of blackbody radiation and state the limit it reduces to at low frequency. (5 marks)


End of Paper

Answer keyMark scheme & solutions

Q1. (4 marks)

  • UU = internal energy, natural variables (S,V)(S,V). (1)
  • H=U+PVH = U + PV, natural variables (S,P)(S,P). (1)
  • F=UTSF = U - TS, natural variables (T,V)(T,V). (1)
  • G=U+PVTS=HTSG = U + PV - TS = H - TS, natural variables (T,P)(T,P). (1)

Why: Each is a Legendre transform swapping a natural variable pair to make different variables independent.


Q2. (4 marks)

  • dF=SdTPdVdF = -S\,dT - P\,dV so S=(F/T)VS = -\left(\partial F/\partial T\right)_V, P=(F/V)TP = -\left(\partial F/\partial V\right)_T. (1)
  • Equality of mixed second partials: 2FVT=2FTV\frac{\partial^2 F}{\partial V\,\partial T} = \frac{\partial^2 F}{\partial T\,\partial V}. (1)
  • (SV)T=(PT)V-\left(\frac{\partial S}{\partial V}\right)_T = -\left(\frac{\partial P}{\partial T}\right)_V. (1)
  • Result: (SV)T=(PT)V\boxed{\left(\dfrac{\partial S}{\partial V}\right)_T = \left(\dfrac{\partial P}{\partial T}\right)_V}. (1)

Q3. (4 marks)

  • Gibbs phase rule: F=CP+2F = C - P + 2, where CC = components, PP = phases, FF = degrees of freedom. (2)
  • Triple point: C=1C=1, P=3P=3F=13+2=0F = 1 - 3 + 2 = 0. (1)
  • Interpretation: invariant point, no freedom to vary TT or pressure. (1)

Q4. (4 marks)

  • S=kBlnΩS = k_B \ln \Omega. (1)
  • SA=kBln4S_A = k_B\ln 4, SB=kBln8S_B = k_B\ln 8. (1)
  • SBSA=kBln(8/4)=kBln2S_B - S_A = k_B\ln(8/4) = k_B\ln 2. (1)
  • 0.693kB\approx 0.693\,k_B. (1)

Q5. (5 marks)

  • Z=e0+eβε=1+eβεZ = e^{0} + e^{-\beta\varepsilon} = 1 + e^{-\beta\varepsilon}, with β=1/(kBT)\beta = 1/(k_BT). (2)
  • E=lnZβ\langle E\rangle = -\dfrac{\partial \ln Z}{\partial\beta}. (1)
  • lnZβ=εeβε1+eβε\dfrac{\partial \ln Z}{\partial\beta} = \dfrac{-\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}}. (1)
  • E=εeβε1+eβε=εeβε+1\langle E\rangle = \dfrac{\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}} = \dfrac{\varepsilon}{e^{\beta\varepsilon}+1}. (1)

Q6. (5 marks)

  • Equipartition: each quadratic degree of freedom contributes 12kBT\tfrac12 k_BT to average energy. (1)
  • Diatomic (trans + rot): f=3+2=5f = 3 + 2 = 5 degrees of freedom. (1)
  • Per mole: U=52RTU = \tfrac{5}{2}RT. (2)
  • CV=(U/T)V=52RC_V = \left(\partial U/\partial T\right)_V = \tfrac{5}{2}R. (1)

Q7. (4 marks)

  • dPdT=LTΔV\dfrac{dP}{dT} = \dfrac{L}{T\,\Delta V} (or ΔSΔV\dfrac{\Delta S}{\Delta V}). (2)
  • dPdT\dfrac{dP}{dT} = slope of coexistence curve; LL = latent heat of transition; TT = transition temperature; ΔV\Delta V = volume change between phases. (2)

Q8. (5 marks)

  • Bose–Einstein: nˉ=1e(εμ)/kBT1\bar{n} = \dfrac{1}{e^{(\varepsilon-\mu)/k_BT} - 1}. (2)
  • Fermi–Dirac: nˉ=1e(εμ)/kBT+1\bar{n} = \dfrac{1}{e^{(\varepsilon-\mu)/k_BT} + 1}. (2)
  • Difference: fermions obey Pauli exclusion (nˉ1\bar n \le 1, sign +1+1); bosons can multiply-occupy a state (sign 1-1). (1)

Q9. (5 marks)

  • Planck: u(ν)dν=8πhν3c31ehν/kBT1dνu(\nu)\,d\nu = \dfrac{8\pi h\nu^3}{c^3}\dfrac{1}{e^{h\nu/k_BT}-1}\,d\nu. (3)
  • Low-frequency (Rayleigh–Jeans) limit: ehν/kBT1hν/kBTe^{h\nu/k_BT}-1 \approx h\nu/k_BT, giving u(ν)8πν2c3kBTu(\nu) \to \dfrac{8\pi\nu^2}{c^3}k_BT. (2)

[
  {"claim":"S_B - S_A = k_B ln 2 (coefficient ln2)","code":"import sympy as sp\nkB=sp.Symbol('kB',positive=True)\nSA=kB*sp.log(4); SB=kB*sp.log(8)\nresult = sp.simplify(SB-SA - kB*sp.log(2))==0"},
  {"claim":"Two-level average energy equals eps/(e^{beta eps}+1)","code":"beta,eps=sp.symbols('beta eps',positive=True)\nZ=1+sp.exp(-beta*eps)\nE=-sp.diff(sp.log(Z),beta)\nresult = sp.simplify(E - eps/(sp.exp(beta*eps)+1))==0"},
  {"claim":"Diatomic (5 dof) molar CV = 5R/2","code":"R,T=sp.symbols('R T',positive=True)\nU=sp.Rational(5,2)*R*T\nCV=sp.diff(U,T)\nresult = sp.simplify(CV - sp.Rational(5,2)*R)==0"},
  {"claim":"Triple point degrees of freedom F=0 for C=1,P=3","code":"C,P=1,3\nF=C-P+2\nresult = (F==0)"},
  {"claim":"Planck low-freq limit gives Rayleigh-Jeans 8 pi nu^2 kB T / c^3","code":"h,nu,kB,T,c=sp.symbols('h nu kB T c',positive=True)\nu_planck=(8*sp.pi*h*nu**3/c**3)/(sp.exp(h*nu/(kB*T))-1)\nlim=sp.limit(u_planck,h,0)\nresult = sp.simplify(lim - 8*sp.pi*nu**2*kB*T/c**3)==0"}
]