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(B) In a cyclic process,the net heat absorbed $\Delta Q$ is equal to the net work done $W$,which is equal to the area enclosed by the cyclic curve on

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$\frac{\pi}{2} 	ext{ J}$

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(B) In a cyclic process,the net heat absorbed $\Delta Q$ is equal to the net work done $W$,which is equal to the area enclosed by the cyclic curve on the $P-V$ diagram.The area of the circle is given by $A = \pi 	imes r_P 	imes r_V$,where $r_P$ is the radius along the $P$-axis and $r_V$ is the radius along the $V$-axis.From the graph:$P$ ranges from $50 	ext{ kPa}$ to $150 	ext{ kPa}$,so the diameter is $100 	ext{ kPa} = 10^5 	ext{ Pa}$. Thus,$r_P = 50 	ext{ kPa} = 5 	imes 10^4 	ext{ Pa}$.$V$ ranges from $20 	ext{ cc}$ to $40 	ext{ cc}$,so the diameter is $20 	ext{ cc} = 20 	imes 10^{-6} 	ext{ m}^3$. Thus,$r_V = 10 	ext{ cc} = 10^{-5} 	ext{ m}^3$.Area $= \pi 	imes (5 	imes 10^4 	ext{ Pa}) 	imes (10^{-5} 	ext{ m}^3) = \pi 	imes 0.5 	ext{ J} = \frac{\pi}{2} 	ext{ J}$.

$A$ system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is

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