Nitrogen gas (N₂) at room temperature has a measured molar heat capacity at constant volume of approximately (5/2)R, not the (7/2)R you might expect for a diatomic molecule. What is the best explanation?
AN₂ has fewer than 5 degrees of freedom at room temperature because it is a simple molecule
BThe vibrational modes are quantum mechanically frozen out at room temperature because kT is much smaller than the vibrational energy quantum
CRotational degrees of freedom do not contribute to heat capacity in diatomic gases
DThe equipartition theorem does not apply to diatomic gases
N₂ has 3 translational + 2 rotational + 2 vibrational (kinetic + potential) = 7 degrees of freedom in principle, predicting C_V = (7/2)R. But the vibrational mode's energy quantum ℏω_vib is roughly 0.1–0.5 eV, far exceeding kT ≈ 0.026 eV at 300 K. Quantum mechanics keeps the vibration locked in its ground state, contributing nothing to heat capacity. Rotational modes are active (ℏω_rot ≪ kT), so we get C_V ≈ (5/2)R. This is a quantum effect classical physics cannot explain.
Question 2 Multiple Choice
A monatomic ideal gas and a diatomic ideal gas (at a temperature where only translational and rotational modes are active) are each given the same amount of heat. Which gas experiences the larger temperature rise?
AThe monatomic gas, because it has fewer degrees of freedom to absorb the energy
BThe diatomic gas, because more degrees of freedom means each one absorbs more heat
CBoth experience the same temperature rise because they obey the ideal gas law
DThe diatomic gas, because its higher heat capacity means it stores more energy per degree
Heat capacity C_V = (f/2)R, where f is the number of active degrees of freedom. A monatomic gas has f = 3, so C_V = (3/2)R. A diatomic gas with active translation and rotation has f = 5, so C_V = (5/2)R. Since the diatomic gas requires more energy per mole per degree of temperature rise, the same amount of heat produces a smaller temperature increase in the diatomic gas. The monatomic gas, with fewer modes to distribute energy into, heats up faster.
Question 3 True / False
According to the equipartition theorem, a diatomic gas at very high temperature (where vibrational modes are fully active) should have C_V = (7/2)R at most temperatures.
TTrue
FFalse
Answer: False
C_V = (7/2)R only when all 7 degrees of freedom (3 translational, 2 rotational, 2 vibrational) are active. But degrees of freedom freeze out at low temperatures — a quantum effect. At room temperature, vibrational modes are frozen (kT ≪ ℏω_vib), giving C_V ≈ (5/2)R. Only at high temperatures where kT ≫ ℏω_vib does the full (7/2)R emerge. Classical physics predicts (7/2)R at all temperatures; the observed staircase in C_V vs. T requires quantum mechanics.
Question 4 True / False
Each translational degree of freedom of a molecule contributes (1/2)kT to its average thermal energy, according to the equipartition theorem.
TTrue
FFalse
Answer: True
This is precisely the equipartition theorem: every independent quadratic term in the energy — each degree of freedom — contributes exactly (1/2)kT to the average energy per molecule, where k is Boltzmann's constant and T is temperature. For three translational modes, the total translational kinetic energy is (3/2)kT per molecule, giving the average kinetic energy for a monatomic ideal gas.
Question 5 Short Answer
Why does the heat capacity of a diatomic gas increase as temperature rises, rather than remaining constant?
Think about your answer, then reveal below.
Model answer: At low temperatures, only translational degrees of freedom are active; rotational modes activate at intermediate temperatures; vibrational modes activate at high temperatures. Each activation adds to the heat capacity in steps.
Each mode has a characteristic quantum energy ℏω. A mode only contributes to heat capacity when kT is comparable to or exceeds ℏω. Rotational modes have low ℏω_rot (activate below ~100 K for most diatomics), so they are always on at practical temperatures. Vibrational modes have high ℏω_vib (~0.1–0.5 eV), so they only activate at hundreds to thousands of kelvin. The result is a staircase: C_V ≈ (3/2)R (translation only) → (5/2)R (+ rotation) → (7/2)R (+ vibration) as temperature rises. This step structure was inexplicable classically and was one of the early confirmations of quantum mechanics.