Question

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 298K.
`Delta_("vap")H^(Ө)` for water
at 298K =44.01 kJ `"mol"^(-1)`

Answer 1

We can represent the process of evaporation as
`underset(1"mol")(H_(2)O)(1)overset("vaporisation")rarrunderset(1"mol")(H_(2)O)(g)`
No. of moles in 18 g`H_(2)O(1)` is
`=(18g)/(18g"mol"^(-1))=1"mol"`
Heat supplied to evaporate18g water at 298 K
`" " nxxDelta_("vap")H^(Ө)`
`" " =(1"mol")xx(44.01 kJ "mol"^(-1))`
`" " =44.01 kJ`
(assuming steam behaving as an ideal gas).
`Delta_("vap")U=Delta_("vap")H^(Ө)-pDeltaV=Delta_("vap")H^(Ө)-Deltan_(g)RT`
`Delta_("vap")H^(v)-Deltan_(g)RT=44.01kJ`
`-(1)(8.314kJ^(-1)"mol"^(-1))(298K)(10^(-3)kJ J^(-1))`
`Delta_("vap")U^(v)=44.01 kJ-2.48kJ`
`" "41.53 kJ`