Question

The enthalpy of vaporization of a liquid is $30kJ{mol}^{-1}$ and entropy of vaporization is $75J{mol}^{-1}{K}^{-1}$. The boiling point of the liquid at $1$ atm is:

• A. $250K$
• B. $400K$
• C. $450K$
• D. $600K$

Answer 1

The correct option is B $400K$

$△H=$ enthalpy of vapourisation $=30$ kJ/mol $=30000$ J/mol

$△S=$ entropy of vaporisation $=75$ J/mol/K

At boiling point,

The reversible process liquid $\leftrightharpoons$ Vapour is in equilibrium at one atmospheric pressure.

$\therefore △G=0$

$△G=△H-T△S$

$\therefore T=\frac{△H}{△S}=\frac{30000Jmo{l}^{-1}}{75Jmo{l}^{-1}{K}^{-1}}=400K$