Questions: Partition Function Applications: From Molecular Properties to Thermodynamics

Question 1

Classical equipartition predicts C_v = (7/2)R for a diatomic gas at all temperatures. Experimentally, H2 at 100 K shows C_v ≈ (3/2)R. What explains this discrepancy?

• A. At 100 K, hydrogen molecules partially dissociate into atoms, reducing the effective degrees of freedom
• B. The equipartition theorem applies only to solids; gases require a different classical treatment at low temperature
• C. Rotational and vibrational energy levels are quantized; at 100 K, kBT is smaller than their level spacing, so these modes are frozen out and contribute nothing to C_v
• D. Experimental error; quantum corrections to heat capacity only matter below 10 K for any gas

Answer: (C)

This is the central triumph of quantum statistical mechanics over classical physics. Equipartition assigns (1/2)R per quadratic degree of freedom regardless of temperature — giving (7/2)R for a diatomic with 3 translational, 2 rotational, and 2 vibrational degrees of freedom. But when thermal energy kBT is much smaller than the quantum energy gap between ground and first excited state, exp(-hν/kBT) ≈ 0 and the mode stays in the ground state, contributing nothing to C_v. At 100 K, rotation for H2 and certainly vibration are largely frozen out, leaving only the (3/2)R contribution from translation.

Question 2

A chemist calculates the rotational partition function for O2 and gets a value twice as large as the experimentally-inferred value. What did the chemist most likely overlook?

• A. That O2 has two atoms, so the rotational partition function must be halved to account for reduced mass
• B. The symmetry number σ = 2 for homonuclear diatomics, which corrects for overcounting the indistinguishable orientations of the molecule
• C. That O2 is paramagnetic and has an electronic degeneracy that modifies the rotational partition function
• D. The zero-point rotational energy, which shifts the partition function by a factor of two at room temperature

Answer: (B)

For homonuclear diatomics like O2, N2, or H2, rotating the molecule by 180° produces a physically indistinguishable configuration. Counting both orientations as separate states overcounts the distinct quantum states by a factor of 2. The symmetry number σ = 2 divides the classical rotational partition function to correct for this indistinguishability. For heteronuclear diatomics like HCl or CO, the two orientations are distinguishable, so σ = 1. This is a quantum indistinguishability effect tied directly to molecular symmetry.

Question 3

The partition function formalism predicts that a diatomic molecule's heat capacity increases stepwise with temperature — adding a rotational contribution, then later a vibrational contribution — whereas classical equipartition predicts the same C_v at all temperatures.

• T. True
• F. False

Answer: True

Because different modes have different characteristic temperatures (θ_rot = ħ²/2IkB for rotation; θ_vib = hν/kB for vibration), they become thermally active at different thresholds. Rotation activates at tens to hundreds of K; vibration typically activates at thousands of K. Classical equipartition has no mechanism for temperature-dependent activation — it treats all modes as always fully active. The quantum partition function naturally produces the experimentally observed stepped C_v(T) curve, reproducing a prediction that classical physics fundamentally cannot make.

Question 4

Zero-point vibrational energy contributes to the temperature dependence of heat capacity because it represents an irreducible energy offset that shifts with temperature.

• T. True
• F. False

Answer: False

Zero-point energy (1/2)hν is a constant — it does not depend on temperature. Since C_v = dU/dT, a temperature-independent energy term has zero derivative and contributes nothing to heat capacity. Similarly, zero-point energies cancel in energy differences (reaction enthalpies, equilibrium constants) for properly referenced calculations. The misconception conflates 'contributing to absolute energy' with 'contributing to temperature-dependent properties.' Only the temperature-dependent part of U matters for C_v.

Question 5

Explain why the partition function formalism successfully predicts the temperature dependence of heat capacity in diatomic gases, while classical equipartition cannot.

Model answer: Classical equipartition assigns (1/2)kBT to every quadratic energy term with no temperature threshold — all modes are treated as always active, giving a temperature-independent C_v. The partition function formalism incorporates quantization: each mode has discrete energy levels separated by a gap hν (or ħ²/2I for rotation). When kBT << hν, the Boltzmann factor exp(-hν/kBT) is negligibly small — the mode stays in the ground state and contributes nothing to C_v. As temperature rises and kBT becomes comparable to the level spacing, the mode activates and its contribution asymptotically approaches the classical equipartition value. This quantization-gated activation is why different modes (translation, rotation, vibration) switch on at very different temperatures, producing the experimentally observed stepped C_v(T) curve.

The partition function formalism thus replaces classical equipartition's single-temperature prediction with a first-principles calculation that correctly captures the full temperature dependence of thermodynamic properties from molecular constants alone — mass, moments of inertia, and vibrational frequencies.