Q 11-12-119JEE MainJEE Main 2020 (9 Jan, Shift 1)Medium
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 $\dfrac{m}{4}$. The ratio of the specific heats $(C_V)_A$ and $(C_V)_B$ of gas $A$ and $B$, respectively is:
Answer: (D) $5 : 7$
Here $C_V$ is the molar specific heat, $C_V = \dfrac{f}{2}R$.
Gas $A$ (rigid diatomic): $3$ translational $+ 2$ rotational $= 5$ degrees of freedom, $C_V = \dfrac{5}{2}R$.
Gas $B$: the vibrational mode adds kinetic and potential energy terms, so $f = 7$ and $C_V = \dfrac{7}{2}R$.
$$\frac{(C_V)_A}{(C_V)_B} = \frac{5}{7}$$
The molecular masses do not affect molar heat capacities.
Solution by Sreeraj P, M.Sc Physics