A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at `100^@C`. `DeltaH_(vap)^@` for water at `373K = 40.66 kJ mol^(-1)`

Explanation

We can represent the process of evaporation as
`18gH_2O(l) overset("vaporisation")to 18gH_2O(g)`
No, of moles in `18gH_2O(g) `
`=(18 g)/("18 g mol"^(-1))="1 mol"`
`triangle_("vap") U= triangleU_("vap") -p triangleV=triangle_("vap") H^@=-trianglen_(g)RT` (assuming steam behaving as an ideal gas).
`triangle_("vap")H^@-trianglen_(g)RT="40.66kJ mol"^(-1)`
`-(1) (8.314 JK^(-1)" mol"^(-1)) (373K) (10^(-3)kJ J^(-1))`
`triangle_(vap) U^@= "40.66kJ mol"^(-1) -"3.10kJ mol"^(-1).`
= 37.56kJ mol`""^(-1)`