Questions: Degrees of Freedom in Polyatomic Molecules

Question 1

A diatomic gas like N₂ has a measured molar heat capacity at constant volume of approximately (5/2)R at room temperature. If this same gas is heated to 5000 K, what would you expect C_v to approach?

• A. (5/2)R — the value stays constant because the molecular structure doesn't change
• B. (7/2)R — vibrational modes become thermally accessible at very high temperatures
• C. (3/2)R — only translational modes matter at high temperatures
• D. (9/2)R — all degrees of freedom double their contribution at high temperature

Answer: (B)

At room temperature, N₂'s vibrational mode is frozen out (kT ≪ ℏω), so only 3 translational + 2 rotational degrees of freedom contribute: C_v = (5/2)R. At 5000 K, kT becomes comparable to the vibrational quantum, and the one vibrational mode (contributing R, not (1/2)R, because it has both KE and PE) begins to fully activate, pushing C_v toward (5/2 + 1)R = (7/2)R. Option A is the classic misconception: molecular structure is unchanged, but which degrees of freedom are thermally accessible changes with temperature.

Question 2

A nonlinear triatomic molecule (3 atoms) like H₂O has how many vibrational modes, and how much does each mode contribute to C_v per mole at high temperature?

• A. 3 modes, each contributing (1/2)R — same as a rotational degree of freedom
• B. 3 modes, each contributing R — because each vibrational mode has both kinetic and potential energy
• C. 2 modes, each contributing (1/2)R — linear and nonlinear molecules have the same vibrational count
• D. 4 modes, each contributing R — nonlinear molecules gain an extra mode compared to linear

Answer: (B)

A nonlinear molecule with N atoms has 3N − 6 vibrational modes: 3(3) − 6 = 3 modes for H₂O. Each vibrational mode is a harmonic oscillator with both a kinetic energy term and a potential energy term, each averaging (1/2)kT per molecule by equipartition. Together that's kT per molecule, or R per mole — double what a rotational degree of freedom (only kinetic) contributes. This is the crucial asymmetry: vibrations count double.

Question 3

A linear triatomic molecule (like CO₂) has more vibrational modes than a nonlinear triatomic molecule (like H₂O) with the same number of atoms.

• T. True
• F. False

Answer: True

Linear molecules have 3N − 5 vibrational modes; nonlinear molecules have 3N − 6. For N = 3: a linear molecule has 4 vibrational modes, a nonlinear molecule has 3. Linear molecules lose one rotational degree of freedom (rotation about the bond axis contributes negligible energy), so that degree of freedom 'becomes' an additional vibrational mode instead.

Question 4

At room temperature, a diatomic ideal gas has the same molar heat capacity at constant volume as it would at 5000 K, because its molecular structure is unchanged.

• T. True
• F. False

Answer: False

Molecular structure is unchanged, but which degrees of freedom are thermally accessible depends on temperature. Vibrational modes freeze out when kT ≪ ℏω (the vibrational quantum). At room temperature for most diatomic gases, this condition holds and C_v ≈ (5/2)R. At 5000 K, vibrational modes become active and C_v approaches (7/2)R. This temperature dependence of heat capacity is a quantum mechanical effect that classical equipartition alone cannot explain.

Question 5

Why does each vibrational mode contribute twice as much energy per mole to a gas's heat capacity as each rotational degree of freedom?

Model answer: Each rotational degree of freedom has only kinetic energy, which by equipartition averages (1/2)kT per molecule, contributing (1/2)R per mole. A vibrational mode is a harmonic oscillator with both a kinetic energy term and a potential energy term — each averaging (1/2)kT — for a total of kT per molecule and R per mole. The asymmetry arises because vibration stores energy in two quadratic terms, not one.

The equipartition theorem assigns (1/2)kT to each independent quadratic term in the energy. Rotation contributes one quadratic term (kinetic). Vibration contributes two: (1/2)mv² for kinetic and (1/2)kx² for potential. This is why the fully classical heat capacity of a diatomic gas — if all modes were active — would be (3/2 + 1 + 1)R = (7/2)R, not (5/2)R.