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 at100^(@)C. Delta_(vap) H^(@) for water at 373K = 40.66 kJ mol^(_1).

A swimmer coming out from a pool is covered a film of water weighing a

1.	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 at100^(@)C. Delta_(vap) H^(@) for water at 373K = 40.66 kJ mol^(_1).
Answer» Solution :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)`