## Chemistry: The Molecular Science (5th Edition)

$39.4\,kJ/mol$
$T_{1}=(50.0+273)K= 323\,K$ $P_{1}= 233\, mmHg$ $T_{2}=(78.3+273)K=351.3\,K$ $P_{2}=1\,atm=760\, mmHg$ $\Delta_{vap}H$ is given by: $\ln (\frac{P_{2}}{P_{1}})=\frac{-\Delta_{vap}H}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$ That is, $\ln (\frac{760}{233})=\frac{-\Delta_{vap}H}{8.314\,J\,mol^{-1}K^{-1}}[\frac{1}{351.3\,K}-\frac{1}{323\,K}]$ $\implies 1.18228=-\Delta_{vap}H\times -2.9998\times10^{-5}\,mol/J$ $\implies \Delta_{vap}H=\frac{1.18228}{2.9998\times10^{-5}\,mol/J}$ $=39400\,J/mol=39.4\,kJ/mol$