Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 9 - Liquids, Solids, and Materials - Questions for Review and Thought - Topical Questions - Page 421b: 30


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, $\ln (\frac{2P_{1}}{P_{1}})=\frac{-\Delta_{vap}H}{8.314\,J\,mol^{-1}K^{-1}}[\frac{1}{353\,K}-\frac{1}{343\,K}]$ $\implies 0.693=-\Delta_{vap}H\times -9.934\times10^{-6}\,mol/J$ $\implies \Delta_{vap}H=\frac{0.693}{9.934\times10^{-6}\,mol/J}$ $=69800\,J/mol=69.8\,kJ/mol$
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