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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

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The process of evaporation is
`18_((g))H_(2)O_((l))rarr18_((g))H_(2)O_((g))`
No. of moles in `18_((g))H_(2)O=(18_((g)))/(18_((g))mol^(-1))=1mol`
`Deltan_((g))=1-0=1mol`
`DeltaW_(vap)^(0)=DeltaH_(vap)^(0)-Deltan_((g))RT`
`=40.66KJmol^(-1)-(1)(8.314xx10^(-3)KJK^(-1)mol^(-1))(298K)`
`=40.66KJmol^(-1)-3.10KJmol^(-1)=37.56KJmol^(-1)`
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