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:
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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