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(B) The net work done in a cyclic process is equal to the area enclosed by the cycle on the $PV$ diagram.Area $= 	ext{length} 	imes 	ext{width} = (

Vedclass · Accepted Answer

$-20$

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(B) The net work done in a cyclic process is equal to the area enclosed by the cycle on the $PV$ diagram.Area $= 	ext{length} 	imes 	ext{width} = (300 - 100) 	ext{ cc} 	imes (200 - 100) 	ext{ kp}$.Area $= 200 	ext{ cc} 	imes 100 	ext{ kp}$.Converting units to $SI$: $200 	imes 10^{-6} 	ext{ m}^3 	imes 100 	imes 10^3 	ext{ Pa} = 20 	ext{ J}$.Since the cycle $P ightarrow Q ightarrow R ightarrow S ightarrow P$ is traversed in an anticlockwise direction,the work done by the system is negative.Therefore,the net work done is $-20 	ext{ J}$.

$A$ thermodynamic system is taken through the cycle $PQRSP$ as shown in the $PV$ diagram. 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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