Physics · Kinetic Theory of Gases

JEE Main 2025 — 22 January, Evening Shift — Question 54

For a diatomic gas, if $\gamma_{1}=\left(\frac{\mathrm{Cp}}{\mathrm{Cv}}\right)$ for rigid molecules and $\gamma_{2}=\left(\frac{\mathrm{Cp}}{\mathrm{Cv}}\right)$ for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct ? ( Cp and Cv are specific heats of the gas at constant pressure and volume)

Option A:
$\gamma_{2}>\gamma_{1}$
Option B:
$\gamma_{2}=\gamma_{1}$
Option C:
$2 \gamma_{2}=\gamma_{1}$
Option D:
$\gamma_{2}<\gamma_{1}$
Correct

Answer: D

Step-by-step solution

The ratio of specific heat capacities ( $\gamma = \frac{C_p}{C_v}$ ) for a gas depends on its total degrees of freedom ( $f$ ) through the formula:

\gamma = 1 + \frac{2}{f}

Rigid Diatomic Gas ( $\gamma_1$ )

A rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom (vibrational modes are inactive).

Total degrees of freedom: $f_1 = 3 + 2 = 5$

Adiabatic index:

\gamma_1 = 1 + \frac{2}{5} = \frac{7}{5} = 1.4

Non-Rigid Diatomic Gas with Vibrational Modes ( $\gamma_2$ )

Each vibrational mode contributes 2 additional degrees of freedom (1 for kinetic energy, 1 for potential energy).

Total degrees of freedom: $f_2 = 5 + 2n = 7$ (assuming $n = 1$ vibrational mode).

Adiabatic index:

\gamma_2 = 1 + \frac{2}{7} = \frac{9}{7} \approx 1.286

Conclusion

Since introducing vibrational modes increases the total degrees of freedom ( $f_2 > f_1$ ), the fraction $\frac{2}{f}$ becomes smaller, resulting in a lower ratio of specific heat capacities:

\gamma_2 < \gamma_1

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Equipartition Law of Energy and Degrees of Freedom