JEE Physics 2025 Question with Solution

Given: List-I contains different ideal gases and List-II contains values of $\gamma=\frac{C_P}{C_V}$.

Find: The correct matching between the two lists.

Use

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

where $f$ is the degree of freedom.

For triatomic rigid gas:

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

So, A matches with III.

For diatomic non-rigid gas:

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

So, B matches with IV.

For monoatomic gas:

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

So, C matches with I.

For diatomic rigid gas:

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

So, D matches with II.

Therefore, the correct matching is A-III, B-IV, C-I, D-II. The correct option is A.

Given: $\gamma=\frac{C_P}{C_V}$ for ideal gases depends on the number of degrees of freedom.

Find: Match each gas in List-I with the correct value in List-II.

The relation is

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

Now evaluate each case:

1. Triatomic rigid gas has $f=6$.

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

Hence A \rightarrow III.

2. Diatomic non-rigid gas has $f=7$.

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

Hence B \rightarrow IV.

3. Monoatomic gas has $f=3$.

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

Hence C \rightarrow I.

4. Diatomic rigid gas has $f=5$.

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

Hence D \rightarrow II.

So the final matching is A-III, B-IV, C-I, D-II.

Note: One part of the solution text lists mismatched verbal statements, but the numerical working and the final answer consistently support option A.