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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)`

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`Delta_(vap)H^(@) = 40.66 kJ mol^(-1)`
`H_(2)O(l) overset "(Vaporise)" to H_(2)O(g)`
Molar mass of water = 18 g "`mol^(-1)`
Mass of water = 18g
Moles of water = `(18)/(18) = 1` mol
Heat needed for vaporisation = `n Delta_(vap)H^(@)`
` = 1 mol xx 40.66 kJ mol^(-1)`
= 40.66 kJ
Now `Delta_(vap)U = Delta_(vap) H^(@) - p DeltaV`
= `Delta_(vap)H^(@) - Deltan_(g)RT`
`Deltan_(g) = 1 -0 = 1 mol`
Assuming steam behaving as an ideal gas
`Delta_(vap)U = Delta_(vap)H^(@) - Deltan_(g)RT`
` = 40.66 - (1) 8.314 xx 10^(-3) xx 373`
= ` 40.66 - 3.10`
` = 37.56 kJ mol^(-1)`
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