Questions: Statistical Thermodynamics: Properties from Partition Functions

Question 1

A diatomic ideal gas is at a temperature where kT is much larger than the rotational energy spacing but much smaller than the vibrational energy spacing. What is the molar heat capacity C_V?

• A. 3/2 R (translation only)
• B. 5/2 R (translation + rotation)
• C. 7/2 R (translation + rotation + vibration)
• D. R (rotation only)

Answer: (B)

When kT >> rotational level spacing, rotational modes are fully thermally populated and each contributes R/2 per degree of freedom (2 rotational DOF for a linear molecule = R total). When kT << vibrational level spacing, the vibrational mode is frozen out — it cannot be thermally excited and contributes ~0 to C_V. The translational contribution is always 3/2 R. Total: 3/2 R + R = 5/2 R. At very high temperatures where kT >> hν_vib, vibration contributes an additional R (one for KE + one for PE), giving 7/2 R.

Question 2

Even well below the characteristic vibrational temperature θ_vib, vibrational modes still contribute approximately R to the molar heat capacity.

• T. True
• F. False

Answer: False

Vibrational modes are quantum mechanically 'frozen out' when kT << hν_vib (equivalently, when T << θ_vib = hν/k). In this regime, nearly all molecules remain in the vibrational ground state and thermal fluctuations lack enough energy to excite the first vibrational level. The contribution to C_V drops exponentially toward zero, not to R. This freezing out is a purely quantum effect — classical equipartition, which predicts R per vibrational mode at all temperatures, fails to describe this behavior.

Question 3

How does statistical thermodynamics provide a molecular interpretation of the third law of thermodynamics (S → 0 as T → 0)?

Model answer: At absolute zero, all molecules occupy the unique ground state, so the number of accessible microstates W = 1. Since S = k ln W, this gives S = 0.

The Boltzmann entropy formula S = k ln W connects macroscopic entropy to the number of ways W of arranging a system at a given energy. As temperature approaches zero, thermal energy becomes insufficient to populate any state above the non-degenerate ground state, so W → 1. The logarithm of 1 is 0, giving S = 0. This is a natural result of quantum mechanics (discrete energy levels) that cannot be derived from classical thermodynamics alone. If the ground state is degenerate (W > 1), a residual entropy k ln W persists at 0 K.