Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid, and have a mass m. Molecules of the gas B have an additional vibrational mode, and have a mass m/4 . The ratio of the specific heats (C_V^A and C_V^B ) of gas A and B, respectively is :

Solution

Physics sir · 4 steps

Count the energy terms per molecule, and ignore the mass: molar C_V does not care how heavy a molecule is.

1. Rigid diatomic A. A rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom. By equipartition each one carries 1/2RT per mole, so U = 5/2RT and C_V^A = 5/2R.
2. Add vibration for B. One vibrational mode adds two energy terms, kinetic and potential, each worth 1/2RT per mole. So the vibration adds R in total, giving C_V^B = 5/2R + R = 7/2R.

   C_V^A/C_V^B = (5/2R)/(7/2R) = 5/7

3. Mass is a distractor. Equipartition gives the same energy per degree of freedom whatever the molecule's mass. So m and m/4 do not change the molar specific heats.
4. So. C_V^A : C_V^B = 5 : 7, which is option (b).

The idea to keep. Each translational or rotational degree of freedom adds 1/2R to molar C_V. Each vibrational mode adds a full R, because it stores both kinetic and potential energy.

Answer from the NTA final answer key. Solution by PrepWiser.