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 vaporization at 100°C. Delta_("vap") H^( Theta ) for water at 373K = 40.66 "kJ mol"^(-1)

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

1.	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 vaporization at 100°C. Delta_("vap") H^( Theta ) for water at 373K = 40.66 "kJ mol"^(-1)
Answer» Solution :We can represent the process of EVAPORATION as `18g H_(2) O_((l)) overset("vaprisation")(to) 18g H_(2) O_((g))` No. of moles in `18 g, H_(2) O_((l))` is `= (18g)/( 18 "g mol"^(-1) )=1` mol `Delta_("vap") U= Delta_("vap") H^( Theta ) - PDELTAV = Delta_("vap") H^( Theta ) - Deltan_(g) RT` (assuming steam behaving as an ideal gas) `Delta_("vap") H^( Theta ) - Delta n_(g) RT = 40.66 "kJ mol"^(-1) -(1)` `(8.314 "JK"^(-1) "mol"^(-1) ) (373 K) (10^(-3) "kJ J"^(-1))` `Delta_("vap") U^( Theta ) = 40.66 "kJ mol"^(-1) -3.10 "kJ mol"^(-1)` `=37.56 "kJ mol"^(-1)`