Question

A swimmer coming out from a pool is covered a film of water weighing about 18g. How much heat must be supplied to evaporate this water at `298K` ? Calculate the internal energy of vaporisation at`100^(@)C`. `Delta _(vap) H^(@)` for water at 373K `= 40.66 kJ mol^(_1)`.

Answer 1

The process of evaporation is `: 18 H_(2) O(l ) rarr 18 g H_(2)O(g) `
No. of moles of 18 g `H_(2) O = (18g )/(18 g mol^(-1))= 1 mol`
`Delta n_(g) = 1-0 =1 mol`
`:. Delta _(vap) U^(@) = V_(vap) H^(@) - Delta n_(g) RT = 40.66 k J mol^(-1) - (1 mol ) (8.314 xx 10^(-3) k J K^(-1) mol^(-1) ) ( 298 K) `
`= 40.66 k J mol^(-1) - 3.10 kJ mol^(-1) = 37. 56 kJ mol^(-1)`