QJEEM25Ph16f6: JEE Main 2025 Physics - Class 11 JEEM25Ph16f6 Solution: For a diatomic gas, if for rigid molecules and for another…

Let's analyze the problem step by step:

For a rigid diatomic molecule (without vibrational modes):

Degrees of freedom:

Translation: 3

Rotation: 2

Total: 5

The molar specific heat at constant volume is given by:

$C_v = \frac{5}{2}R$

At constant pressure, it is:

$C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$

Therefore, the ratio is:

$\gamma_1 = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \approx 1.4$

For the diatomic molecule that also has vibrational modes (assuming the vibrational mode is fully excited):

Additional vibrational mode contributions:

Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential).

Therefore, the degrees of freedom become:

$3\ (\text{translation}) + 2\ (\text{rotation}) + 2\ (\text{vibration}) = 7$

The molar specific heat at constant volume now is:

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

And at constant pressure:

$C_p = C_v + R = \frac{7}{2}R + R = \frac{9}{2}R$

Thus, the ratio is:

$\gamma_2 = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} \approx 1.2857$

Comparison:

For the rigid molecule, $\gamma_1 \approx 1.4$ .

For the molecule with vibrational modes, $\gamma_2 \approx 1.2857$ .

Clearly, \gamma_2 < \gamma_1 .

Therefore, the correct option is:

Option A: \gamma_2 < \gamma_1