∆Grxno=∆fG∘(vapour)−∆fG∘( liquid ) ∆Grxno=103−100.7=2.3kcal∕mol=2300cal∕mol ∆Grxn0=−RTlnKp(Kp=Pvap) 2300cal∕mol=−2cal∕mol∕K×500K×lnKp lnKp=−2.3log10Kp=−1 Kp= Antilog −1=0.1atm ∴ Vapour pressure of liquid ' S ' at 500K is approximately equal to 0.1atm.