The specific heats, $C_p$ and $C_v$ of a gas of diatomic molecules, A, are given (in units of $\text{J mol}^{-1}\text{K}^{-1}$) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then
Answer: (B) A has a vibrational mode but B has none
Compare $\gamma = C_p/C_v$ with the ideal values for a diatomic gas: $\frac75 = 1.40$ (rigid) and $\frac97 \approx 1.29$ (with one vibrational mode).
- A: $\gamma = \dfrac{29}{22} \approx 1.32$, below $1.40$, so its molecules have extra (vibrational) degrees of freedom.
- B: $\gamma = \dfrac{30}{21} \approx 1.43 \approx 1.40$, a rigid diatomic gas.
So A has a vibrational mode and B has none.
Solution by Sreeraj P, M.Sc Physics