In the following figures,in which case is hea | vedclass

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(C) For a cyclic process,the first law of thermodynamics is $\Delta U = Q - W = 0$,which implies $Q = W$. Heat absorbed $(Q > 0)$ corresponds to a pos

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(C) For a cyclic process,the first law of thermodynamics is $\Delta U = Q - W = 0$,which implies $Q = W$. Heat absorbed $(Q > 0)$ corresponds to a positive net work done $(W > 0)$ by the gas. In a $P-V$ diagram,work done is positive for a clockwise cycle and negative for an anticlockwise cycle. Figure $A$ is a $V-P$ diagram (note the axes). $A$ clockwise cycle in a $V-P$ diagram corresponds to a net work done $W Figure $B$ is a $P-V$ diagram with an anticlockwise cycle,so $W Figure $C$ is a $P-V$ diagram with a clockwise cycle,so $W > 0$. Figure $D$ is a $V-P$ diagram with a clockwise cycle,so $W Therefore,only in Figure $C$ is the net work done positive,meaning heat is absorbed by the gas.

In the following figures,in which case is heat absorbed by the gas?

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