Q160 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
If a non-linear polyatomic gas has $f$ vibrational nodes, the value of $\gamma = C_p/C_v$ is ($C_p$ and $C_v$ are specific heats at constant pressure and volume respectively)
- A$\dfrac{4+f}{2+f}$
- B$\dfrac{3+f}{4-f}$
- C$\dfrac{f-4}{3+f}$
- Correct answerD$\dfrac{4+f}{3+f}$
Solution
For a non-linear polyatomic gas, the number of translational degrees of freedom is $3$ and the number of rotational degrees of freedom is $3$.
Each vibrational mode contributes $2$ degrees of freedom, one associated with kinetic energy and the other with potential energy. Therefore, $f$ vibrational modes contribute $2f$ degrees of freedom.
The total number of degrees of freedom $N$ is:
$N = 3 + 3 + 2f = 6 + 2f$
The molar specific heat at constant volume $C_v$ is given by $\dfrac{N}{2}R$.
$C_v = \dfrac{6 + 2f}{2}R = (3 + f)R$
Using Mayer's relation, the molar specific heat at constant pressure $C_p$ is:
$C_p = C_v + R$
$C_p = (3 + f)R + R = (4 + f)R$
The ratio of specific heats $\gamma$ is given by $\dfrac{C_p}{C_v}$.
$\gamma = \dfrac{(4 + f)R}{(3 + f)R}$
$\gamma = \dfrac{4 + f}{3 + f}$
Each vibrational mode contributes $2$ degrees of freedom, one associated with kinetic energy and the other with potential energy. Therefore, $f$ vibrational modes contribute $2f$ degrees of freedom.
The total number of degrees of freedom $N$ is:
$N = 3 + 3 + 2f = 6 + 2f$
The molar specific heat at constant volume $C_v$ is given by $\dfrac{N}{2}R$.
$C_v = \dfrac{6 + 2f}{2}R = (3 + f)R$
Using Mayer's relation, the molar specific heat at constant pressure $C_p$ is:
$C_p = C_v + R$
$C_p = (3 + f)R + R = (4 + f)R$
The ratio of specific heats $\gamma$ is given by $\dfrac{C_p}{C_v}$.
$\gamma = \dfrac{(4 + f)R}{(3 + f)R}$
$\gamma = \dfrac{4 + f}{3 + f}$
Hence, the correct answer is option: 'D'.