A swimmer coming out from a pool is covered with a film of water weighing about `80g`. How much heat must be supplied to evaporate this water ? If latent heat of evaporation for `H_(2)O` is `40.79 kJ mol^(-1)` at `100^(@)C`.

Options

A. `37.56 kJ mol^(-1)`
B. `43.76 kJ mol^(-1)`
C. `49.65 kJ mol^(-1)`
D. `32.65 kJ mol^(-1)`

Answer

A

Explanation

we have
`n_(H_(2)O) = (m_(H_(2) O))/("Molar mass"_(H_(2)O)) = (18 g)/(18 g mol^(-1)) = 1 mol`
We can represent the process of evaporation is
`underset(1 mol)(H_(2) O (1)) rarr underset(1 mol)(H_(2) O (g))`
According to Eq.
`Delta_(vap) H^(@) = Delta_(vap) U^(@) + (Delta n_(g)) RT`
or `Delta_(vap) U^(@) = Delta_(vap) H^(@) - (Delta n_(g)) RT`
`= (40.66 kJ mol^(-1)) - (1 mol) (8.314 k^(-1) mol^(-1)) (373 K) (10^(-3) kJ J^(-1))`
` = (40.66 kJ mol^(-1)) - (3.10 kJ mol^(-1))`
`= 37.56 kJ mol^(-1)`
Note that `Delta n_(g) = sum n_(P) (gas) - sum n_(R) (gas)`
`= (1 mol) - (0 mol) = 1 mol`
we have used the conversion factor to convert the value of `R` in kilojoules.