Question

calculate the entropy change involved in the vaporization of water at 373 Kelvin to vapours at the same temperature latent heat of vaporization is equal to 2.2 75 kilojoule per gram

Answer 1

Molar enthalpy of vaporization

=2.275x18=40.95kj/mol

ΔvapS=ΔvapH/T

= 40.95x10^3/373.

= 109.78j/k/mol

Answer 2

Molar enthalpy=18× 2.275

= 40.626kj/mole

ΔS= ΔH/T

= 40.626× 10^-3

= 108.9J/k

This is correct...hope it help