Question

The molar enthalpy of vaporization of hexane $\left(C_6{H}_{14}\right)$ is $28.9kJ/mol$, and its normal boiling point is ${69}^{\circ }C$. What is the vapour pressure of hexane at ${25.00}^{\circ }C$ in $torr$? [$e^{-1.5}=0.222$]

• A. $169.5torr$
• B. $195.4torr$
• C. $134.6torr$
• D. $202.2torr$

Answer 1

The correct option is A $169.5torr$

Effect of temperature on the vapour​ pressure of a liquid is given by​ Clausius – Clapeyron equation,
$ln\left(\frac{P_2}{P_1}\right)=\frac{\Delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
The normal boiling point of a liquid is the temperature at which its vapour pressure is equal to one atmosphere.
thus,
$\text{Vapour pressure at}342K,P_1=760torr$
Given data:
$$\begin{array}{cc}{P}_1=xtorr& T_1={25.00}^{\circ }C=298K\\ P_2=760torr& T_2={69}^{\circ }C=342K\end{array}$$
$\text{Molar enthalpy of vaporisation,}\Delta H=28.9kJ/mol$

Substituting the above values, we get,
$ln\left(\frac{x}{760}\right)=-\left(\frac{28900}{8.314}\right)(\frac{1}{298}-\frac{1}{342})ln\left(\frac{x}{760}\right)=-3476.06\times 0.000432ln\left(\frac{x}{760}\right)\approx -1.500\frac{x}{760}=0.222x=169.5torr$
$\text{Vapour pressure at}{25}^{\circ }C,P_1=169.5torr$