JEE Physics 2025 Question with Solution

Given: $\gamma_1 = \frac{C_P}{C_V}$ for a rigid diatomic molecule and $\gamma_2 = \frac{C_P}{C_V}$ for a diatomic molecule with vibrational modes.

Find: The correct relation between $\gamma_1$ and $\gamma_2$.

For a gas,

$$\gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V} = 1 + \frac{R}{C_V}$$

So, as $C_V$ increases, the value of $\gamma$ decreases.

For a rigid diatomic molecule, the degrees of freedom are $f = 5$. Hence,

$$C_V = \frac{f}{2}R = \frac{5}{2}R$$

Therefore,

$$\gamma_1 = 1 + \frac{R}{(5/2)R} = 1 + \frac{2}{5} = \frac{7}{5}$$

When vibrational modes are active, additional degrees of freedom are included. Then,

$$f = 7$$

and so

$$C_V = \frac{7}{2}R$$

Therefore,

$$\gamma_2 = 1 + \frac{R}{(7/2)R} = 1 + \frac{2}{7} = \frac{9}{7}$$

Now compare:

$$\gamma_1 = \frac{7}{5} = 1.4, \qquad \gamma_2 = \frac{9}{7} \approx 1.29$$

Thus,

$$\gamma_2 < \gamma_1$$

Therefore, the correct option is D.

Given: A comparison of the specific heat ratios of two diatomic gases, one rigid and one with vibrational modes.

Find: Whether $\gamma_2$ is greater than, equal to, or less than $\gamma_1$.

Vibrational modes add extra degrees of freedom to the molecule. Because of this, more heat supplied goes into internal energy storage for the same rise in temperature, which increases $C_V$.

Since

$$\gamma = 1 + \frac{R}{C_V}$$

a larger value of $C_V$ makes $\gamma$ smaller.

So, for the diatomic gas having vibrational modes,

$$\gamma_2 < \gamma_1$$

Hence, the correct option is D.