Answer
69.8 kJ/mol
Work Step by Step
$T_{1}=(70.0+273)K= 343\,K$
$P_{2}= 2P_{1}$
$T_{2}=(80.0+273)K=353\,K$
$\Delta_{vap}H$ is given by:
$\ln (\frac{P_{2}}{P_{1}})=\frac{-\Delta_{vap}H}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$
That is,
$\ln (\frac{2P_{1}}{P_{1}})=\frac{-\Delta_{vap}H}{8.314\,J\,mol^{-1}K^{-1}}[\frac{1}{353\,K}-\frac{1}{343\,K}]$
$\implies 0.693=-\Delta_{vap}H\times
-9.934\times10^{-6}\,mol/J$
$\implies \Delta_{vap}H=\frac{0.693}{9.934\times10^{-6}\,mol/J}$
$=69800\,J/mol=69.8\,kJ/mol$
