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Q. Entropy change involved in conversion of one mole of liquid water at $373 \,K$ to vapour at the same temperature (latent heat of vaporization of water $=2.257 \,kJ\, g ^{-1}$ ).

Thermodynamics

Solution:

$\Delta H_{\text {vap }}=2.257\, kJ / g =(2.257 \times 18) \,kJ \,mol ^{-1}=40.626 \,kJ \,mol ^{-1}$

$\Delta S_{\text {vap }}=\frac{40.626}{373}=0.1089\, kJ \,K ^{-1}\, mol ^{-1}$