JEE Physics 2024 Question with Solution

Given: The gas undergoes a cyclic process. For path $A$ to $B$, $PV^3 = RT$ and $T = 27^\circ\text{C} = 300\,\text{K}$. Also, $R = 8\,\text{J/mol\cdot K}$.

Find: Net work done in the complete cycle.

For path $AB$:

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

Assuming temperature to be constant,

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

For path $BC$:

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

For path $CA$:

$$W_{CA} = 0$$

Therefore,

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

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

Given: The process is cyclic and along $A$ to $B$ the gas obeys $PV^3 = RT$.

Find: 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. So we add the work contributions along each path.

Along $AB$, use

$$P = \frac{RT}{V^3}$$

and integrate with respect to volume from $V=2$ to $V=4$ to obtain $225\,\text{J}$.

Along $BC$, pressure is constant at $P=10$, so

$$W_{BC} = P\Delta V = 10(2-4) = -20\,\text{J}$$

Along $CA$, volume is constant, hence no work is done:

$$W_{CA} = 0$$

Thus, the enclosed net area, and hence the net work done in the cycle, is

$$205\,\text{J}$$

So the correct option is B.