What is the value of the change in internal energy at 1 atm in the following process? H2O(l,323K)→H2O(g,423K) Given : Cv,m(H2O,1)=75.0JK−1;Cp,m(H2O,g)=33.314JK−1mol−1;ΔHvapat373K=40.7kJ/mol
A
42.910 kJ/mol
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B
43.086 kJ/mol
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C
42.600 kJ/mol
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D
49.600 kJ/mol
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Solution
The correct option is C42.600 kJ/mol We can break it down into three parts. (i) H2O(l,323K)→H2O(l,373K) (ii) H2O(l,373K)→H2O(g,373K) (iii) H2O(g,373K)→H2O(g,423K) now, in(i), q=nCv△T=1×75×50=3750J=3.75kJ w: negligible. ∴△E=3.75kJ in (ii), q=mL=△Hvap=40.7kJ w=−Pext△V=−105Pa(22.4×373273×10−3m3) =−3.06kJ △E=+37.64kJ
in (iii) q=nCp△T=1×33.314×50 =1.666kJ ω=−105Pa(22.4×423273−22.4×373273)×10−3 =−0.410kJ △E=+1.256kJ △Etotal=+42.646kJ