# Question b39e8

Jul 20, 2017

By Clausius - Clapeyron equation weknow

color(red)(ln(P_1/P_2) = (DeltaH_"vap")/R xx (1/T_2 - 1/T_1)

Where

${P}_{1}$ is the vapor pressure of the liquid at ${T}_{1} K$

${P}_{2}$ is the vapor pressure of the liquid at ${T}_{2} K$

$R$ is the universal gas constant, equal to ${\text{8.314J mol"^(-1)"K}}^{- 1}$

$\Delta {H}_{\text{vap"= "enthalpy of vaporization of the liquid}}$

Given

${T}_{1} = 273 + 5 = 278 K$

${P}_{1} = 8.54 m m$ of Hg

${T}_{2} = 273 + 120 = 393 K$

$\Delta {H}_{\text{vap"=41.8kJmol^-1}} = 41.8 \times {10}^{3} J m o {l}^{-} 1$

P_2=?

Inserting these in above equation

color(blue)(ln(P_1/P_2) = (DeltaH_"vap")/R xx (1/T_2 - 1/T_1)

=>color(blue)(ln(8.54/P_2) = (41.8xx10^3)/8.314xx (1/393 - 1/278)#

$\implies \textcolor{b l u e}{\ln \left(8.54\right) - \ln {P}_{2} = \frac{41.8 \times {10}^{3}}{8.314} \times \left(\frac{1}{393} - \frac{1}{278}\right)}$

$\implies \textcolor{b l u e}{\ln {P}_{2} = \ln \left(8.54\right) - \frac{41.8 \times {10}^{3}}{8.314} \times \left(\frac{1}{393} - \frac{1}{278}\right)}$

$\implies \textcolor{b l u e}{\ln {P}_{2} = 7.437}$

$\implies \textcolor{b l u e}{{P}_{2} = {e}^{7.437} \approx 1698 m m}$