Thermodynamics Part 5

Concept:Work done in a cyclic process equals the area enclosed by the loop on the $p$-$V$ diagram.For an anticlockwise cycle, the net work is negative.Explanation:The cycle goes $A \rightarrow B \rightarrow C \rightarrow A$ anticlockwise.Equation of the curved path: $(V-2)^2 = 4ap$.At $V = 2$, $p = 0$ (point $B$).At $V = 1$ and $V = 3$, $p = \frac{1}{4a}$ (points $A$ and $C$).Work done = area under straight line $CA$ minus area under curved path $A \to B \to C$.Area under line $CA$ = $p \times \Delta V = \frac{1}{4a} \times (3-1) = \frac{1}{2a}$.Work along curve: $\int_1^3 p\,dV = \int_1^3 \frac{(V-2)^2}{4a} dV = \frac{1}{4a} \left[ \frac{(V-2)^3}{3} \right]_1^3 = \frac{1}{4a} \left( \frac{1}{3} - \frac{(-1)^3}{3} \right) = \frac{1}{4a} \times \frac{2}{3} = \frac{1}{6a}$.Net work $W = \frac{1}{6a} - \frac{1}{2a} = -\frac{1}{3a}$.Answer:Option D: $-\frac{1}{3a}$.