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

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$-20$

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(B) The net work done by the system in a cyclic process is equal to the area enclosed by the cycle on the $P-V$ diagram.Since the cycle $PQRS P$ is traversed in an anti-clockwise direction,the work done is negative.Area of the rectangle $= 	ext{length} 	imes 	ext{breadth} = (V_Q - V_P) 	imes (P_S - P_P)$$V_P = 100 	ext{ cc} = 100 	imes 10^{-6} 	ext{ m}^3$,$V_Q = 300 	ext{ cc} = 300 	imes 10^{-6} 	ext{ m}^3$$P_P = 100 	ext{ kPa} = 100 	imes 10^3 	ext{ Pa}$,$P_S = 200 	ext{ kPa} = 200 	imes 10^3 	ext{ Pa}$$\Delta V = (300 - 100) 	imes 10^{-6} 	ext{ m}^3 = 200 	imes 10^{-6} 	ext{ m}^3$$\Delta P = (200 - 100) 	imes 10^3 	ext{ Pa} = 100 	imes 10^3 	ext{ Pa}$$	ext{Work} = -(\Delta P 	imes \Delta V) = -(100 	imes 10^3 	imes 200 	imes 10^{-6}) = -(20000 	imes 10^{-3}) = -20 	ext{ J}$.

$A$ thermodynamic system is taken through the cycle $PQRS P$ process as shown in the figure. The net work done by the system is ......... $J$.

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