JEE Physics 2025 Question with Solution

Given: A monoatomic gas has $\gamma = \frac{5}{3}$ and is compressed adiabatically to $V_2 = \frac{V_1}{8}$.

Find: The ratio $\frac{P_2}{P_1}$.

For a thermally insulated process, the adiabatic relation is

$$P_1 V_1^\gamma = P_2 V_2^\gamma$$

Substitute $\gamma = \frac{5}{3}$ and $V_2 = \frac{V_1}{8}$:

$$P_1 V_1^{\frac{5}{3}} = P_2 \left(\frac{V_1}{8}\right)^{\frac{5}{3}}$$

Now simplify:

$$P_1 V_1^{\frac{5}{3}} = P_2 \cdot \frac{V_1^{\frac{5}{3}}}{8^{\frac{5}{3}}}$$

Cancel $V_1^{\frac{5}{3}}$ from both sides:

$$P_1 = P_2 \cdot \frac{1}{8^{\frac{5}{3}}}$$

Therefore,

$$\frac{P_2}{P_1} = 8^{\frac{5}{3}}$$

Since $8 = 2^3$,

$$8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = 2^5 = 32$$

Therefore, the ratio of final pressure to initial pressure is $32$. The correct option is B.

Given: $\gamma = \frac{5}{3}$ and the volume becomes $\frac{1}{8}$ of the initial value.

Find: $\frac{P_2}{P_1}$.

In an adiabatic process, pressure varies inversely as $V^\gamma$, so

$$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma$$

Here,

$$\frac{V_1}{V_2} = 8$$

So,

$$\frac{P_2}{P_1} = 8^{\frac{5}{3}}$$

Write $8$ as $2^3$:

$$8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = 2^5 = 32$$

Therefore, the correct option is B.