Question

The latent heat of vapourization of water at ${100}^{o}C$ is $540cal.g^{-1}$. Calculate the entropy increase when one mole of water at ${100}^{o}C$ is evaporated.

• A. $26cal.K^{-1}mo{l}^{-1}$
• B. $1.45cal.K^{-1}mo{l}^{-1}$
• C. $367cal.K^{-1}mo{l}^{-1}$
• D. $1.82cal.K^{-1}mo{l}^{-1}$

Answer 1

The correct option is A $26cal.K^{-1}mo{l}^{-1}$

Solution:- (A) $26cal/mol-K$

As we know that,

$\Delta S=\frac{\Delta H}{T}$

$\Delta H=540cal/g$

Mol. wt. of water $=18g$

${\Delta H}_{\left(\text{per mole}\right)}={\Delta H}_{\left(\text{per gm}\right)}\times \text{Mol. wt.}$

${\Delta H}_{\left(\text{per mole}\right)}=540\times 18=9720cal/mol$

Now, as we know that,

$\Delta S=\frac{\Delta H}{T}$

Given $T=100℃=(273+100)=373K$

$\therefore \Delta S=\frac{9720}{373}=26cal/(mol.K)$