Q 11-11-074JEE MainJEE Main 2024 (9 Apr, Shift 2)Hard
A real gas within a closed chamber at $27\,^\circ\text{C}$ undergoes the cyclic process as shown in the figure. The gas obeys the $PV^3 = RT$ equation for the path $A$ to $B$. The net work done in the complete cycle is (assuming $R = 8\ \text{J/mol K}$):
Answer: (B) $205\ \text{J}$
On path AB, $P = \dfrac{RT}{V^3}$ with $T = 300\ \text{K}$, so $RT = 8\times300 = 2400\ \text{J}$:
$$W_{AB} = \int_2^4\frac{RT}{V^3}dV = \frac{RT}{2}\left(\frac1{2^2} - \frac1{4^2}\right) = 1200\times\frac{3}{16} = 225\ \text{J}$$
$B \to C$ (isobaric at $10\ \text{N/m}^2$): $W_{BC} = 10\times(2 - 4) = -20\ \text{J}$. $C \to A$ is isochoric: $W_{CA} = 0$.
$$W_{net} = 225 - 20 = 205\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics