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(B) The net work done in a cyclic process is equal to the area enclosed by the $P-V$ diagram.Since the cycle $P \rightarrow Q \rightarrow R \rightarro

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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 $P-V$ diagram.Since the cycle $P ightarrow Q ightarrow R ightarrow S ightarrow P$ is traversed in an anti-clockwise direction,the net work done is negative.Area of the rectangle $= 	ext{length} 	imes 	ext{width} = (300 - 100) 	ext{ cc} 	imes (200 - 100) 	ext{ kp}$.Converting units to $SI$: $\Delta V = 200 	ext{ cc} = 200 	imes 10^{-6} 	ext{ m}^3$ and $\Delta P = 100 	ext{ kp} = 100 	imes 10^3 	ext{ Pa}$.Work done $W = -(\Delta P 	imes \Delta V) = -(100 	imes 10^3 	ext{ Pa}) 	imes (200 	imes 10^{-6} 	ext{ m}^3)$.$W = -(10^5) 	imes (2 	imes 10^{-4}) 	ext{ J} = -20 	ext{ J}$.

$A$ thermodynamic system is taken through the cycle $PQRSP$ process as shown in the figure. The net work done by the system is .... $J$. (Note: $1 \text{ kp} = 10^3 \text{ Pa}$,$1 \text{ cc} = 10^{-6} \text{ m}^3$)

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