ThermodynamicsHard

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).

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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