Questions: Equipartition Theorem and Molecular Heat Capacities

Question 1

The measured molar heat capacity at constant volume (C_V) for N₂ at room temperature is approximately (5/2)R, not (7/2)R. What is the best explanation?

• A. N₂ only has 5 atoms, so it has fewer degrees of freedom than predicted
• B. The vibrational degree of freedom is frozen out because the quantum energy spacing is much larger than kT at room temperature
• C. Rotation about the bond axis is included in the (5/2)R value, accounting for the 'missing' (7/2)R
• D. N₂ is not an ideal gas, so the equipartition theorem does not apply

Answer: (B)

Equipartition classically predicts (7/2)R for a diatomic: (3/2)R translational + R rotational + R vibrational. But vibrational energy levels are quantized with spacing hν. For N₂ at room temperature, kT ≈ 2.5 kJ/mol while the vibrational quantum for N₂ is much larger — the vibrational mode is 'frozen out' because thermal energy cannot populate excited vibrational states. So only translation and rotation contribute, giving (5/2)R. The classical theorem gives the high-temperature ceiling; quantum mechanics determines when each mode reaches it.

Question 2

A monatomic noble gas like neon is heated. According to the equipartition theorem, what is its molar heat capacity C_V?

• A. (1/2)R — only one translational mode
• B. (3/2)R — three translational degrees of freedom, no rotational or vibrational modes
• C. (5/2)R — three translational plus two rotational modes
• D. R — one degree of freedom per atom

Answer: (B)

A monatomic atom has three translational degrees of freedom (motion along x, y, z), each contributing ½R, giving C_V = (3/2)R ≈ 12.5 J/(mol·K). It cannot rotate in the quantum-mechanically relevant sense (no internal axes with significant moment of inertia) and has no vibrational modes. This prediction matches experiment extremely well for noble gases — they are the cleanest test of equipartition because there are no complications from frozen modes or quantum corrections.

Question 3

According to the equipartition theorem, each vibrational mode contributes ½R to the molar heat capacity of a molecule at room temperature.

• T. True
• F. False

Answer: False

A fully active vibrational mode contributes R (not ½R) because each vibrational mode has two quadratic energy terms — one kinetic (½μv²) and one potential (½kx²), each contributing ½R. However, at room temperature most vibrational modes are frozen out by quantum effects (hν >> kT), so the actual contribution is essentially zero, not ½R. The ½R per quadratic term is the raw equipartition result, but vibration has two quadratic terms and is typically not thermally accessible anyway.

Question 4

At very high temperatures (thousands of kelvin), a diatomic gas like H₂ should approach a C_V of (7/2)R as vibrational modes become thermally accessible.

• T. True
• F. False

Answer: True

As temperature rises, kT eventually becomes comparable to the vibrational energy spacing hν, and the vibrational mode gradually 'thaws.' In the classical (high-T) limit, the vibrational mode contributes its full R (two quadratic terms), bringing C_V from (5/2)R (translation + rotation only) to (7/2)R (all modes active). For H₂, this transition begins noticeably above 1000 K. This is a striking experimental confirmation of quantum mechanics — the stepwise activation of modes as temperature increases is inexplicable in classical physics.

Question 5

Why do rotational degrees of freedom contribute to the heat capacity of N₂ at room temperature while vibrational degrees of freedom do not, even though both are present?

Model answer: Both are quantized, but their energy spacings are vastly different. Rotational energy levels are closely spaced (separation << kT at room temperature), so many excited rotational states are thermally populated and the mode behaves classically, contributing R (two modes). Vibrational energy levels are widely spaced (separation >> kT at room temperature) because bond stretching frequencies are much higher than rotational frequencies. The vibrational mode is frozen out — virtually all molecules are in the vibrational ground state — so it contributes essentially nothing to C_V.

The key is comparing the quantum energy spacing to the thermal energy kT. Rotational spacings scale as ℏ²/(2I); for heavy, long molecules like N₂, the moment of inertia I is large, giving small rotational spacings that are easily exceeded by kT at room temperature. Vibrational frequencies are much higher — stretching a chemical bond requires far more energy than rotating the molecule — so hν >> kT, and the Boltzmann factor suppresses excited-state population to negligible levels. This is a fundamental lesson: the equipartition theorem gives the classical (high-T) limit, but quantum mechanics controls which modes actually reach that limit.