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Question

What is the value of the change in internal energy at 1 atm in the following process?
H2O(l,323 K)H2O(g,423 K)
Given : Cv,m(H2O,1)=75.0 JK1;Cp,m(H2O,g)=33.314JK1mol1;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 C 42.600 kJ/mol
We can break it down into three parts. 
(i) H2O(l,323 K)H2O(l,373 K)
(ii) H2O(l,373 K)H2O(g,373 K)
(iii) H2O(g,373 K)H2O(g,423 K)
now, 
in(i),
q=nCvT=1×75×50=3750 J=3.75 kJ
w: negligible.
E=3.75 kJ 
in (ii),
q=mL=Hvap=40.7 kJ
w=PextV=105 Pa(22.4×373273×103m3)
=3.06 kJ
E=+37.64 kJ

in (iii)
q=nCpT=1×33.314×50
=1.666 kJ
ω=105 Pa(22.4×42327322.4×373273)×103
=0.410 kJ
E=+1.256 kJ
Etotal=+42.646 kJ

Chemistry

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