In the P-V diagram shown in the figure,ABC is | vedclass

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(B) The work done in a $P-V$ diagram is equal to the area under the curve.For the given semicircle,the diameter is along the pressure axis from $P = 1

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$\frac{\pi}{2} 	ext{ atm-L}$

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(B) The work done in a $P-V$ diagram is equal to the area under the curve.For the given semicircle,the diameter is along the pressure axis from $P = 1 	ext{ atm}$ to $P = 3 	ext{ atm}$.Thus,the diameter $d = 3 - 1 = 2 	ext{ atm}$,so the radius $r = 1 	ext{ atm}$.The volume changes from $V = 2 	ext{ L}$ to $V = 1 	ext{ L}$ and back to $V = 2 	ext{ L}$.Since the process goes from $V = 2 	ext{ L}$ to $V = 1 	ext{ L}$ (compression) and then from $V = 1 	ext{ L}$ to $V = 2 	ext{ L}$ (expansion),the area under the expansion part is greater than the area under the compression part.The net work done is equal to the area of the semicircle.Area of semicircle $= \frac{1}{2} \pi r^2 = \frac{1}{2} 	imes \pi 	imes (1 	ext{ atm})^2 	imes (1 	ext{ L}) = \frac{\pi}{2} 	ext{ atm-L}$.Since the cycle is traversed in a clockwise direction,the work done is positive.

In the $P-V$ diagram shown in the figure,$ABC$ is a semicircle. The work done in the process $ABC$ is

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