Basic Maths and Units and DimensionsHard
Question
Standard enthalpy of vaporisation ᐃvapHo for water at 100oC is 40.66 kJ mol-1. The internal energy of vaporisation of water at 100oC (in kJ mol-1) is
(Assume water vapour to behave like an ideal gas).
(Assume water vapour to behave like an ideal gas).
Options
A.+ 37.56
B.- 43.76
C.+ 43.76
D.- 40.66
Solution
For the above reqaction,
ᐃ ng = np - nr = 1 - 0 = 1
∴ 40.66 kJ mol-1 =
+ 1 × 8.314 × 10-3 × 373 = 40.66 kJ mol-1 - 3.1 kJ mol-1= + 37.56 kJ mol-1
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