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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)`

Text Solution

Verified by Experts

The process of evaporation can be represented as
`H_(2)O(I) overset (" Vaporisation") to H_(2)O(g)`, ` Delta _(vap)H^(oplus) = 40.66 kJ mol^(-1)`
18 g 18 g
1 mol 1 mol
Assuming steam behaving as an ideal gas
`Delta _(vap)U ^(oplus) = Delta_(vap) H^(oplus) - pDelta V`
`= Delta _(vap) H^(oplus) - Delta n_(g) .RT`
Here `Delta n_(g) = 1-0 = 1 mol`
`Delta_(vap)U^(oplus) = 40.66 kJ mol^(-1) -(1 mol)(8.314 xx10^(3) kJ mol^(-1) K ^(-1))(373 K)`
` = 40.66 kJ mol^(-1) - 3.10`
`= 37.56 kJ mol^(-1)`
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