One mole of an ideal gas at 900 K , undergoes two reversible processes, I followed by II , as shown below. If the work done by the gas in the two processes are same, the value of ln V 3 V 2 is ( U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant) (Given: molar heat capacity at constant volume, C V , m of the gas is 5 2 R )

Step-by-step solution

Process -I: Adiabatic reversible process. (Since entropy is constant) W I = ΔU = 450 − 2250 R = - 1800   R Process II: Isothermal reversible process. (since internal energy is constant and entropy is increased) Work done: W II = − nRTln V f V i W II = − nRTln V 3 V 2 W II = − 900 5 Rln V 3 V 2 W II = 900 5 Rln V 3 V 2 U = 5 2 nRT 450   R = 5 2 nRT nRT = 900 5 R Given W I = W II 1800 R = 900 5 R   ln V 3 V 2 ln V 3 V 2 = 10