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(D) The work done in a cyclic process on a $P-V$ diagram is equal to the net area enclosed by the cycle.In the given figure,the cycle consists of two

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(D) The work done in a cyclic process on a $P-V$ diagram is equal to the net area enclosed by the cycle.In the given figure,the cycle consists of two triangles: $	riangle AOD$ and $	riangle BOC$.For the process $A ightarrow O ightarrow D$,the volume increases,so the work done is positive. The area of $	riangle AOD = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2V_0 - V_0) 	imes (2P_0 - P_0) = \frac{1}{2} 	imes V_0 	imes P_0 = \frac{P_0V_0}{2}$.For the process $B ightarrow O ightarrow C$,the volume decreases,so the work done is negative. The area of $	riangle BOC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2V_0 - V_0) 	imes (3P_0 - 2P_0) = \frac{1}{2} 	imes V_0 	imes P_0 = \frac{P_0V_0}{2}$.The total work done $W = W_{AOD} + W_{BOC} = \frac{P_0V_0}{2} + (-\frac{P_0V_0}{2}) = 0$.

$A$ thermodynamic system undergoes a cyclic process $ABCDA$ as shown in the figure. The work done by the system in the cycle is

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