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InterviewSolution
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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)` |
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Answer» 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` |
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