JEE MAIN - Physics (2025 - 7th April Morning Shift - No. 20)

Match the List I with List II

List - I List - II
(A) Triatomic rigid gas (I) $\frac{C_p}{C_v}=\frac{5}{3}$
(B) Diatomic non-rigid gas (II) $\frac{C_p}{C_v}=\frac{7}{5}$
(C) Monoatomic gas (III) $\frac{C_p}{C_v}=\frac{4}{3}$
(D) Diatomic rigid gas (IV) $\frac{C_p}{C_v}=\frac{9}{7}$

Choose the correct answer from the options given below:

A-III, B-IV, C-I, D-II
A-II, B-IV, C-I, D-III
A-IV, B-II, C-III, D-I
A-III, B-II, C-IV, D-I

Explanation

The explanation uses the equation for the heat capacity ratio, $ \gamma = 1 + \frac{2}{\text{f}} $, where $ \text{f} $ is the degrees of freedom of the gas. Here's how it applies to different types of gases:

Triatomic Rigid Gas:

Degrees of freedom ($ \text{f} $) = 6

$ \gamma = 1 + \frac{2}{6} = \frac{4}{3} $

Diatomic Non-Rigid Gas:

Degrees of freedom ($ \text{f} $) = 7

$ \gamma = 1 + \frac{2}{7} = \frac{9}{7} $

Diatomic Rigid Gas:

Degrees of freedom ($ \text{f} $) = 5

$ \gamma = 1 + \frac{2}{5} = \frac{7}{5} $

Monoatomic Gas:

Degrees of freedom ($ \text{f} $) = 3

$ \gamma = 1 + \frac{2}{3} = \frac{5}{3} $

Therefore, when matching List I with List II, the correct pairing is:

A (Triatomic rigid gas) - III $(\frac{4}{3})$

B (Diatomic non-rigid gas) - IV $(\frac{9}{7})$

C (Monoatomic gas) - I $(\frac{5}{3})$

D (Diatomic rigid gas) - II $(\frac{7}{5})$

Thus, the correct answer from the options given is:

A-III, B-IV, C-I, D-II

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