Question

What is the free energy change $\Delta G$, when 1.0 mole of water at ${100}^{0}C$ and 1 atm pressure is converted into steam at ${100}^{0}C$ and 1 atm pressure?

• A. 540 cal
• B. -9800 cal
• C. 9800 cal
• D. 0 cal

Answer 1

The correct option is D 0 cal

Solution:- (D) $0cal$

As we know that,

$\Delta G=W=2.303nRT\log \left(\frac{P_f}{P_i}\right)$

$\Rightarrow \Delta G=2.303nRT\log \frac{P_f}{P_i}$

Given that

No. of mole of water $\left(n\right)=1$

$R=$ Gas constant $=2cal$

$T=100℃=(100+273)=373K$

$P_i=1atm$

$P_f=1atm$

$\therefore \Delta G=2.303\times 1\times 2\times 373\times \log \left(\frac{1}{1}\right)$

$\Rightarrow \Delta G=0[\log 1=0]$

Hence the value of free energy change is $0cal$.