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Consider the given cyclic process 1→2→3→4→5→6→7→8→9→10→11→12→1 if the net work done is KP0V0 find the value of K.

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answer is 5.

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Detailed Solution

Net work done W=Wcyclic clock+4Wcyclic anticlock⇒W=5P0−2P05V0−2V0+−4P0V0=5P0V0

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Consider the given cyclic process 1→2→3→4→5→6→7→8→9→10→11→12→1 if the net work done is KP0V0 find the value of K.