Question

What is the free energy changes $\Delta G$, when $1.0$ mole of water at ${100}^o$C and $1$ atm pressure is converted into steam at ${100}^o$C and $1$ atm pressure?

• A. $540$ Cal
• B. $-9800$ Cal
• C. $9800$ Cal
• D. $0$ Cal

Answer 1

Answer: (D)

$0$ Cal

Solution:- (D) $0cal$

As we know that,

$\Delta G=nRT\ln \frac{P_2}{P_1}$

$\because P_1=P_2=1atm$

$\therefore \Delta G=nRT\ln \frac{1}{1}=0(\because \ln 1=0)$

Hence the free energy change will be $0$.