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(B) The net work done in a cyclic process is equal to the area enclosed by the cycle on the $P-V$ diagram.Since the cycle $PQRSP$ is traversed in an a

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$-20$

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(B) The net work done in a cyclic process is equal to the area enclosed by the cycle on the $P-V$ diagram.Since the cycle $PQRSP$ is traversed in an anti-clockwise direction, the net work done by the system is negative.Area of the rectangle $PQRS = 	ext{length} 	imes 	ext{width} = (V_Q - V_P) 	imes (P_R - P_Q)$.Given: $V_P = 100 \, cc = 100 	imes 10^{-6} \, m^3$, $V_Q = 300 \, cc = 300 	imes 10^{-6} \, m^3$.$P_Q = 100 \, kPa = 100 	imes 10^3 \, Pa$, $P_R = 200 \, kPa = 200 	imes 10^3 \, Pa$.Work done $W = - (\Delta V 	imes \Delta P) = - (300 - 100) 	imes 10^{-6} \, m^3 	imes (200 - 100) 	imes 10^3 \, Pa$.$W = - (200 	imes 10^{-6}) 	imes (100 	imes 10^3) = - 20 \, J$.Thus, the correct option is $B$.

$A$ thermodynamic system is taken through the cycle $PQRSP$ as shown in the $P-V$ diagram. The net work done by the system is ....... $J$.

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