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 $100^{o}$ C.
$Δ_{v a p} H^{⊝}$ for water at $373$ K $= 40.66$ kJ $m o l^{- 1}$ .

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Solution

$H_{2} O 10 g (l) v a p o r i s a t i o n - ------ \to H_{2} O 10 g (g)$ $Δ_{v a p} H^{⊝} = 40.66 k J m o l^{- 1}$
$1 m o l$ $1 m o l$
As ideal gas, $Δ_{v a p} U^{⊝} = Δ_{v a p} H^{⊝} - p Δ V$
$= Δ_{v a p} H^{⊝} - Δ n_{g} R T$
Here $Δ n_{g} = 1 - 0 = 1 m o l$
$Δ_{v a p} U^{⊝} = 40.66 k J m o l^{1} - (1 m o l) (8.314 \times 10^{- 3} k J m o l^{- 1} k^{1}) \times (373 k)$
$= 40.66 k J m o l^{- 1} - 3.10$
$= 37.56 k J m o l^{- 1}$

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