4 SolutionAlong ac path Work done W 35 J Heat added Q 63

Solution

Along ac path :

Work done W = -35 J

Heat added Q = -63 J

Change in internal energy U = Q - W
                                        = -63 -(-35)

                                        = -28 J

Along abc path :

Work done W \' = -48 J

We know W \' = Pb(Vb-Va)

      Pb(Vb-Va) = -48 -------( 1)

Along cda path :

Work done W \" = Pc ( Vd-Vc)

                       = (Pb/2)[-(Vb-Va)]

                       = (1/2) (-W \')

                       = -24 J

(a). Work done for path cda is W \" = -24 J

Change in internal energy for cda U \" = -U

                                                      = 28 J

Q for path cda is Q \" = U \" + W \"

                             = 28 J -24 J

                            = 4 J