JEE Physics 2024 Question with Solution

Given: The gas undergoes a cyclic process. Along path $A$ to $B$, it obeys $PV^3 = RT$. Temperature is taken as $T = 300 \, \text{K}$ since $27^\circ \text{C} = 300 \, \text{K}$, and $R = 8 \, \text{J/mol\cdot K}$.

Find: Net work done in the complete cycle.

Work done from $A$ to $B$ is

$$W_{AB} = \int P \, dV$$

Assuming $T$ to be constant,

$$W_{AB} = \int \frac{RT}{V^3} \, dV$$

Using the given limits from $V=2$ to $V=4$,

$$W_{AB} = RT \int_{2}^{4} V^{-3} \, dV$$ $$W_{AB} = 8 \times 300 \times \left( -\frac{1}{2} \left[ \frac{1}{4^2} - \frac{1}{2^2} \right] \right)$$ $$W_{AB} = 225 \, \text{J}$$

For path $B$ to $C$,

$$W_{BC} = P \int_{4}^{2} dV = 10(2-4) = -20 \, \text{J}$$

For path $C$ to $A$,

$$W_{CA} = 0$$

Therefore, the net work done in the cycle is

$$W_{\text{cycle}} = W_{AB} + W_{BC} + W_{CA} = 225 - 20 + 0 = 205 \, \text{J}$$

Therefore, the net work done in the complete cycle is $205 \, \text{J}$. The correct option is B.

Given: The process is cyclic and the path $A$ to $B$ follows $PV^3 = RT$.

Find: The net work done over one complete cycle.

The work done in a cyclic process is the area enclosed by the loop on the $P-V$ diagram. That means we add the work contributions along each segment.

From the provided working, the curved path $A \to B$ contributes $225 \, \text{J}$, the horizontal path $B \to C$ contributes $-20 \, \text{J}$, and the vertical path $C \to A$ contributes $0$ because $dV=0$ on that path.

Hence,

$$W_{\text{total}} = 225 - 20 = 205 \, \text{J}$$

So, the net work done in the complete cycle is $205 \, \text{J}$.