Question

A compound having molar heat of vaporization $8.314kJ/mol$ has a normal boiling point of ${127}^{\circ }C$
If the vapour pressure of same compound at ${27}^{\circ }C$ is denoted as $P_2$. Find the value of ${\log }_e\left(P_2\right)$.

• A. $\frac{-5}{6}$
• B. $\frac{+5}{6}$
• C. $\frac{-2}{3}$
• D. $\frac{+2}{3}$

Answer 1

The correct option is A $\frac{-5}{6}$

Given,
$\text{Temperature,}{T}_1={127}^{\circ }C=400K$
$\text{Temperature,}{T}_2={27}^{\circ }C=300K$
The normal boiling point of a liquid is the temperature at which its vapour pressure is equal to one atmosphere.
thus,
$\text{Vapour pressure at}400K,P_1=1atm$
$\text{Molar heat of vaporisation,}\Delta H=8.314kJ/mol$

Effect of temperature on the vapour​ pressure of a liquid is given by​ Clausius – Clapeyron equation,
$ln\left(\frac{P_2}{P_1}\right)=\frac{\Delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
we get,
$ln\left(\frac{P_2}{1}\right)=\left(\frac{8314}{8.314}\right)(\frac{1}{400}-\frac{1}{300})$

$ln{P}_2-ln1=\frac{-1000}{1200}ln{P}_2=\frac{-5}{6}$