A gas is taken through a cyclic process ABCA | vedclass

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(A) For a cyclic process,the change in internal energy is zero,so the net heat supplied is equal to the net work done: $dQ = dW$.The work done $dW$ is

Vedclass · Accepted Answer

$4.17$

Vedclass · Answer

(A) For a cyclic process,the change in internal energy is zero,so the net heat supplied is equal to the net work done: $dQ = dW$.The work done $dW$ is equal to the area enclosed by the cycle $ABCA$ in the $V-P$ diagram.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Base $= (200 \, kPa - 100 \, kPa) = 100 	imes 10^3 \, Pa$Height $= (700 \, cc - 500 \, cc) = 200 \, cm^3 = 200 	imes 10^{-6} \, m^3$$dW = \frac{1}{2} 	imes (100 	imes 10^3 \, Pa) 	imes (200 	imes 10^{-6} \, m^3) = 10 \, J$.Given heat $dQ = 2.4 \, cal$.Since $dW = J 	imes dQ$,we have $J = \frac{dW}{dQ} = \frac{10 \, J}{2.4 \, cal} = 4.166... \approx 4.17 \, J/cal$.

$A$ gas is taken through a cyclic process $ABCA$ as shown in the figure. If $2.4 \, cal$ of heat is given in the process,what is the value of $J$ in $J/cal$ (mechanical equivalent of heat)?

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