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

The process of evaporation can be represented as,
$H_{2} O (l) V a p o r i s a t i o n - ------- \to H_{2} O (g); Δ H_{v a p} = 40.66 K J m o l^{- 1}$
$18 g$ $18 g$
$1 m o l$ $1 m o l$
Assuming steam behaving as ideal gas,
${Δ U_{v a p}}^{-} = {Δ H_{v a p}}^{-} - p Δ V$
$= {Δ H_{v a p}}^{-} - Δ n g R T$ Here, $Δ n g = 1 - 0 = 1 m o l$
$∴ {Δ U_{v a p}}^{-} = 40.66 - 1 (8.314 \times 313)$
$= 40.66 - 3.10$
$= 37.56 K J m o l^{- 1}$ .

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