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Q.

If water vapour is assumed to be a perfect gas, enthalpy change for vaporisation of 1 mole of water at 1 bar and 373 k is 41 kJ per mole. Change in internal energy for 1 mole of water to be vapourised at 1 bar and 100∘C is

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a

44.1 kJ per mole

b

42.3 kJ per mole

c

31.5 kJ per mole

d

37.9 kJ per mole

answer is D.

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Detailed Solution

H2O(l)→H2O(g)

Δng=1−0=1

ΔH=Δ u+ΔngRT

+41 kJ=Δu+(1)(8.314×10−3)(373)

Δu=+37.9kJ

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