$A$ 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 vaporisation at $298 \ K$. Given: $\Delta_{vap} H^{\ominus}$ for water at $298 \ K = 44.01 \ kJ \ mol^{-1}$.

(N/A) The process of evaporation is represented as: $H_2O_{(l)} \rightarrow H_2O_{(g)}$.
Number of moles in $18 \ g$ of $H_2O$ is $n = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
Heat supplied to evaporate $18 \ g$ of water at $298 \ K$ is $q = n \times \Delta_{vap} H^{\ominus} = 1 \ mol \times 44.01 \ kJ \ mol^{-1} = 44.01 \ kJ$.
To calculate the internal energy of vaporisation $(\Delta_{vap} U)$,we use the relation: $\Delta_{vap} U = \Delta_{vap} H^{\ominus} - \Delta n_g RT$.
Here,$\Delta n_g = 1$ (since $1 \ mol$ of gas is produced from $1 \ mol$ of liquid).
$\Delta_{vap} U = 44.01 \ kJ - (1 \ mol) \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times (298 \ K)$.
$\Delta_{vap} U = 44.01 \ kJ - 2.48 \ kJ = 41.53 \ kJ$.