The magnitude of heat exchanged by a system f | vedclass

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(B) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$Q = \Delta U + W$,so $Q = W$. The

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$5 \pi$

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(B) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$Q = \Delta U + W$,so $Q = W$. The work done $W$ is equal to the area enclosed by the cycle $ABCA$ in the $PV$ diagram.The process $ABCA$ consists of a semicircle of radius $r$ and a straight line $CA$.The diameter of the semicircle along the $P$-axis is $400 	ext{ kPa} - 200 	ext{ kPa} = 200 	ext{ kPa}$. Thus,the radius $r_P = 100 	ext{ kPa} = 10^5 	ext{ Pa}$.The diameter of the semicircle along the $V$-axis is $400 	ext{ cc} - 200 	ext{ cc} = 200 	ext{ cc} = 200 	imes 10^{-6} 	ext{ m}^3$. Thus,the radius $r_V = 100 	ext{ cc} = 10^{-4} 	ext{ m}^3$.The area of the semicircle is $A = \frac{1}{2} \pi r_P r_V = \frac{1}{2} 	imes \pi 	imes (10^5 	ext{ Pa}) 	imes (10^{-4} 	ext{ m}^3) = \frac{1}{2} 	imes \pi 	imes 10 = 5 \pi 	ext{ J}$.Since the cycle is traversed in the counter-clockwise direction,the work done is negative,but the magnitude of heat exchanged is $5 \pi 	ext{ J}$.

The magnitude of heat exchanged by a system for the given cyclic process $ABCA$ (as shown in the figure) is (in $SI$ units).

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