In the following p-V diagram, the equation of | vedclass

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(D) The work done in a closed $p-V$ cycle is equal to the area enclosed by the path. Since the cycle is traversed in a counter-clockwise direction, th

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$-\frac{1}{3a}$

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(D) The work done in a closed $p-V$ cycle is equal to the area enclosed by the path. Since the cycle is traversed in a counter-clockwise direction, the work done by the gas is negative.The area under the curve $(V-2)^2 = 4aP$ from $V=1$ to $V=3$ is given by $\int_{1}^{3} P \, dV$.From the equation, $P = \frac{(V-2)^2}{4a}$.Area under the curve $= \int_{1}^{3} \frac{(V-2)^2}{4a} \, dV = \frac{1}{4a} \left[ \frac{(V-2)^3}{3} ight]_{1}^{3} = \frac{1}{12a} [(3-2)^3 - (1-2)^3] = \frac{1}{12a} [1 - (-1)] = \frac{2}{12a} = \frac{1}{6a}$.At $V=1$ or $V=3$, the pressure $P_0$ is given by $(1-2)^2 = 4aP_0$, so $P_0 = \frac{1}{4a}$.The area of the rectangle formed by the top horizontal line (at $P_0$) and the $V$-axis from $V=1$ to $V=3$ is $P_0 	imes (3-1) = \frac{1}{4a} 	imes 2 = \frac{1}{2a}$.The area enclosed by the cycle is the area of the rectangle minus the area under the curve:Area $= \frac{1}{2a} - \frac{1}{6a} = \frac{3-1}{6a} = \frac{2}{6a} = \frac{1}{3a}$.Since the cycle is counter-clockwise, the work done $W = -\frac{1}{3a}$.

In the following $p-V$ diagram, the equation of state along the curved path is given by $(V-2)^2 = 4aP$, where $a$ is a constant. The total work done in the closed path is:

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