InterviewSolution
Saved Bookmarks
| 1. |
A swimmer coming out from a pool is covered with a film of water weighing about 18 g. how much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaperization at `100^(@)C`. `Delta_(vap)H^(Theta)` for water at 373 K = 40.66 kJ `mol^(-1)` |
|
Answer» The process of evaporation can be represented as `underset("1 mol")underset("18 g")(H_(2)O(I)) overset("Vaporisation") rarr underset("1 mol") underset("18 g")(H_(2)O(g)), Delta_(Vap)H^(Theta)=40.66 kJ" "mol^(-1)` Assumming steam behaving as an ideal gas `{:(Delta_(Vap)U^(Theta),=,Delta_(Vap)H^(Theta)-p Delta V),(,=,Delta_(Vap)H^(Theta)-Delta n_(g)RT),("Here" Delta n_(g),=,1-0 = "1 mol"),(Delta_(Vap)U^(Theta),=,"40.66 kJ mol"^(-1)-("1 mol")(8.314 xx 10^(-3)"kJ mol"^(-1))("373 K)),(,=,40.66 kJ mol^(-1)-3.10),(,=,37.56 kJ mol^(-1)):}` |
|