Question

What is the vapour pressure of benzene at ${25}^{\circ }C$ ? The normal boiling point of benzene is ${80}^{\circ }C$ and its molar heat of vaporization is $30.8kJ/mol$. [$e^{-1.94}=0.14$]

• A. $0.39atm$
• B. $0.14atm$
• C. $1.94atm$
• D. $0.45atm$

Answer 1

The correct option is B $0.14atm$

Given,
$\text{Temperature,}{T}_1={80}^{\circ }C=353K$
$\text{Temperature,}{T}_2={25}^{\circ }C=298K$
The normal boiling point of a liquid is the temperature at which its vapour pressure is equal to one atmosphere.
thus,
$\text{Vapour pressure at}353K,P_1=1atm$
$\text{Molar heat of vaporisation,}\Delta H=30.8kJ/mol$

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,
$ln\left(\frac{P_2}{1.00}\right)=\left(\frac{30800}{8.314}\right)(\frac{1}{353}-\frac{1}{298})$

$ln\left(\frac{P_2}{1.00}\right)=\left(3704.59\right)(-5.2\times {10}^{-4})ln{P}_2=-1.937P_2=0.144atm$
$\text{Vapour pressure at}{25}^{\circ }C,P_2=0.144atm$