# Question #73593

Feb 2, 2017

The Clausius-Clapeyron equation relates the vapor pressures of a liquid at two temperatures along with the enthalpy of vaporisation of liquid as follows.

$\textcolor{b l u e}{\ln \left({P}_{2} / {P}_{1}\right) = \frac{\Delta {H}_{\text{vap}}}{R} \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)}$

where

$\Delta {H}_{\text{vap"->"Enthalpy of vaporisation}}$

${P}_{1} \to \text{Vapor preessure at } {T}_{1} K$

${P}_{2} \to \text{Vapor preessure at } {T}_{2} K$

$R \to \text{Universal gas constant}$

In our problem for ether we have been given

${T}_{1} = \left(- 22.8 + 273\right) K = 250.2 K$

${T}_{2} = \left(25 + 273\right) K = 298 K$

${P}_{1} = 57.0 m m$

${P}_{2} = 534 m m$

$R = 8.314 J {K}^{-} 1 m o {l}^{-} 1$

Inserting these in the equation

$\textcolor{g r e e n}{\ln \left({P}_{2} / {P}_{1}\right) = \frac{\Delta {H}_{\text{vap}}}{R} \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)}$

$\implies \textcolor{b l u e}{\ln \left(\frac{534}{57}\right) = \frac{\Delta {H}_{\text{vap}}}{8.314} \left(\frac{1}{250.2} - \frac{1}{298}\right)}$

$\implies \Delta {H}_{\text{vap}} = \frac{298 \times 250.2 \times 8.314}{47.8} \ln \left(\frac{534}{57}\right) J m o {l}^{-} 1$

$\approx 29.0 k J$