VKC Notes · Statistical thermodynamics · IFoS P-I

6.4 Statistical Thermodynamics: Boltzmann Distribution, Partition Functions, Sackur–Tetrode Equation and Heat Capacity

★★★ 2021–2025: 5 Mains parts, 55 marks, all IFoS Paper I, one every year and always 10–15 marks: a derivation (rotational q, C_V from Q), a calculation (Sackur–Tetrode entropy of N₂) or a model system (Einstein crystal, three-level spin). Civil Services and Geo-Scientist do not have this syllabus line.

Syllabus line. IFoS: “Micro and macro states; canonical ensemble and canonical partition function; electronic, rotational and vibrational partition functions and thermodynamic quantities; chemical equilibrium in ideal gas reactions” (Thermodynamics and statistical thermodynamics).

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6.4.1 Microstates, weight and the Boltzmann distribution ★★★

(6.4.1)

Macrostate: what thermodynamics sees (N, V, U or T). Microstate: one particular way of giving the molecules their energies. A configuration {n₀, n₁, …} (how many molecules in each level) can be reached by W microstates; W is its weight.SN
For large N one configuration has overwhelmingly the largest W. Maximising ln W (Stirling: ln N! ≈ N ln N − N) with fixed N and U (Lagrange multipliers α, β) gives the Boltzmann distribution (Eq. 6.4.1); β turns out to be 1/kT.SN
Features: (i) the population falls exponentially with energy; (ii) a degenerate level holds g_i times more; (iii) at T → 0 only the ground level is occupied; (iv) at T → ∞ all states become equally occupied, but never inverted (Fig. 6.4.1).SN
Worked example: HCl (θ_R = 15.2 K) at 298 K. n_J=3/n_J=0 = 7 e^−12θ_R/T = 7 × e^−0.613 = 3.79: the degeneracy (7) outweighs the Boltzmann factor (0.54), so J = 3 is more populated than J = 0 (Fig. 6.4.2). For a two-level gap of 500 K at 300 K, n₁/n₀ = e^−500/300 = 0.19.SN

Fig. 6.4.1 A two-level system (gap ε/k = 500 K): populations against T.

Draw it in 60 s: two curves from the left: one starting at 1 and falling, one starting at 0 and rising; both flatten towards a dashed line at ½ and never cross it.

6.4.2 The molecular partition function and its meaning ★★★

(6.4.2)

Meaning: q is the effective number of thermally accessible states at T. q → g₀ as T → 0 and grows without limit for an unbounded ladder; a closely spaced ladder (translation) gives a huge q, a widely spaced one (vibration) gives q ≈ 1.SN
If the energy is a sum of independent parts, ε = ε^T + ε^R + ε^V + ε^E, the exponential factorises and q is a product (Eq. 6.4.2). ln q is then a sum, so every thermodynamic function is a sum of contributions.SN
Canonical ensemble: a very large number of imaginary copies of the system at the same N, V, T, in thermal contact. The canonical partition function Q = Σ_i e^−βE_i runs over the energies E_i of the whole system. For independent molecules Q = q^N; for a gas, swapping identical molecules gives no new state, so divide by N!.SN

Fig. 6.4.2 Populations of the rotational levels of HCl (θ_R = 15.2 K) at 298 K and 1000 K, computed from Eq. 6.4.1 with g_J = 2J + 1.

Draw it in 60 s: bars rising and falling with J; the 298 K set peaks at J = 3; the hotter set is lower, wider and peaks further right; label J_max ≈ (T/2θ_R)^1/2 − ½.

6.4.3 Translational partition function and the Sackur–Tetrode equation ★★★

(6.4.3)

Particle in a 3D box, levels so close that the sum becomes an integral: each direction gives X(2πmkT)^1/2/h. Λ is the thermal wavelength: N₂ at 298 K, Λ = 19.1 pm, so q^T_m = 3.6 × 10³⁰ for 1 mol at 1 bar.SN
With Q = q^N/N! and Stirling, S = (U − U(0))/T + k ln Q gives the Sackur–Tetrode equation (Eq. 6.4.3) for the translational (whole, for a monatomic gas) entropy. The 5/2 = 3/2 (from U = 3/2 RT) + 1 (from the N! term). Heavier molecules and larger volume give more states, hence more entropy. Ar at 298 K: 155 J K⁻¹ mol⁻¹.SN

Solved PYQ IFoS 2023 Q4(a) 10m

Translational contribution to S°_m of N₂ (ideal) at 25 °C and 1 bar.

m = 0.028/(6.023 × 10²³) = 4.649 × 10⁻²⁶ kg; Λ = h/(2πmkT)^1/2 = 1.910 × 10⁻¹¹ m.
kT/p = (1.381 × 10⁻²³ × 298.15)/10⁵ = 4.116 × 10⁻²⁶ m³; kT/(pΛ³) = 5.90 × 10⁶.
S_m = R[ln(5.90 × 10⁶) + 2.5] = 8.314 × (15.59 + 2.50) = 150.4 J K⁻¹ mol⁻¹. The table value of S°(N₂) is larger because rotation adds about 41 J K⁻¹ mol⁻¹.SN

6.4.4 Rotational partition function ★★★

(6.4.4)

Solved PYQ IFoS 2022 Q4(b) 15m

Deduce the rotational partition function of a rigid heteronuclear diatomic rotor.

Rigid rotor: I = μr_e², ε_J = h²J(J + 1)/8π²I = kθ_RJ(J + 1), each level (2J + 1)-fold degenerate (M_J = −J…+J).
q^R = Σ(2J + 1)e^{−θ_RJ(J+1)/T}. When T ≫ θ_R the terms change slowly: replace the sum by an integral, put y = J(J + 1), dy = (2J + 1)dJ: q^R = ∫₀^∞e^−θ_Ry/Tdy = T/θ_R = 8π²IkT/h².SN
Heteronuclear: σ = 1. Check: U^R = NkT²(∂ln q/∂T) = RT, C_V,m^R = R (two rotational degrees of freedom). HCl, θ_R = 15.2 K: q^R = 19.6 at 298 K (exact sum 19.9; the next correction is +1/3).

Symmetry number σ: the number of indistinguishable orientations; 2 for a homonuclear diatomic (H₂, N₂) or symmetric linear molecule (CO₂), 1 for HCl, CO. A homonuclear molecule can use only odd or only even J for a given nuclear-spin state, which halves the sum.SN
H₂ (θ_R = 88 K) is too light for the high-T formula at room temperature; its ortho (odd J) and para (even J) forms give an anomalous heat capacity below 300 K. Rotational spectra and θ_R values: 13.6 Rotational spectra.

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