JEE Physics 2025 Question with Solution

Given: The path $ABCD$ is shown on a $P$-$V$ diagram.

Find: The work done by the ideal gas along the path $ABCD$.

For a process on a $P$-$V$ diagram, work done by the gas is

$$W = \int P \, dV$$

That is, the work equals the signed area under the path. The sign depends on the direction of traversal.

The solution states that the path is traced counterclockwise, so the net work done by the gas is negative.

Using the segment-wise contribution given in the solution:

$$W = P_0 \times 2V_0 + 2P_0 \times (-V_0) + 2P_0 \times (-V_0) + P_0 \times 0$$

So,

$$W = 2P_0V_0 - 2P_0V_0 - 2P_0V_0$$

Hence,

$$W = -3P_0V_0$$

Therefore, the correct option is D.

Given: A path $ABCD$ on the $P$-$V$ plane with characteristic values $P_0$ and $V_0$.

Find: The net work done by the gas.

On a $P$-$V$ diagram, only horizontal segments contribute directly to $\int P \, dV$ because vertical segments have $dV = 0$.

• Along a rightward horizontal path, $dV > 0$, so work is positive.
• Along a leftward horizontal path, $dV < 0$, so work is negative.
• Along a vertical path, $dV = 0$, so work is zero.

From the provided working, the path is counterclockwise overall, which corresponds to net compression and therefore negative work done by the gas.

Adding the contributions exactly as given:

$$W = P_0 \times 2V_0 + 2P_0 \times (-V_0) + 2P_0 \times (-V_0) + P_0 \times 0 = -3P_0V_0$$

Therefore, the work done by the gas is $-3P_0V_0$, so the correct option is D.