One mole of an ideal gas undergoes two different cyclic processes I and II , as shown in the P - V diagrams below. In cycle I , processes a , b , c and d are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II , processes a ' , b ' , c ' and d ' are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is W I and that during cycle II is W I

Step-by-step solution

For first case: W I = W a + W b + W c + W d = 4 P 0 2 V 0 - V 0 + n R T   ln 4 V 0 2 V 0 + 2 P 0 V 0 - 4 V 0 + 0 = 4 P 0 V 0 + n R 8 P 0 V 0 n R ln 2 - 6 P 0 V 0 = 8 P 0 V 0   ln 2 - 2 P 0 V 0 For second case: W II = W ' a + W ' b + W ' c + W ' d = n R T   ln 2 V 0 V 0 + 0 + P 0 V 0 - 2 V 0 + 0 = n R 4 P 0 V 0 n R ln 2 - P 0 V 0 = 4 P 0 V 0   ln 2 - P 0 V 0 Therefore, required ratio is W I W II = 8 P 0 V 0 ln 2 - 2 P 0 V 0 4 P 0 V 0 ln 2 - P 0 V 0 = 2