Q 11-12-035JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
For a diatomic gas, if $\gamma_1 = \left(\dfrac{C_p}{C_v}\right)$ for rigid molecules and $\gamma_2 = \left(\dfrac{C_p}{C_v}\right)$ for another diatomic molecule, but also having vibrational modes, then which one of the following options is correct? ($C_p$ and $C_v$ are specific heats of the gas at constant pressure and volume)
Answer: (C) $\gamma_2 < \gamma_1$
$\gamma = 1 + \dfrac{2}{f}$, where $f$ is the number of degrees of freedom.
Rigid diatomic: $f = 5$, $\gamma_1 = \dfrac{7}{5} = 1.4$.
With vibration, $f = 7$: $\gamma_2 = \dfrac{9}{7} \approx 1.29$.
More degrees of freedom give a smaller $\gamma$, so $\gamma_2 < \gamma_1$.
Solution by Sreeraj P, M.Sc Physics