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