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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 vaporization at $298 K . △_{v a p} H^{⊖}$ for water at $298 K = 44.01 k J m o l^{- 1}$

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Solution

Step 1: Moles of $H_{2} O (l)$

The process of evaporation represent as:

$H_{2} O (l) Vaporisation - ------ \to H_{2} O (g)$

We know, 𝑀𝑜𝑙𝑒 $= \frac{weight}{molar mass}$

Number of moles of $18 g$ 𝑜𝑓 $H_{2} O (l) = \frac{18 g}{18 g m o l^{- 1}} = 1 m o l$

Step 2: Heat supplied

Heat supplied to evaporate $18 g$ or $1 m o l$ of water at $298 K = n \times △_{v a p} H^{⊖}$

$= (1 m o l) \times (44.01 k J m o l^{- 1})$

$= 44.01 k J m o l^{- 1}$

Step 3: Internal energy

(Assuming steam behaving as an ideal gas).

$△_{v a p} H^{⊖} = △_{v a p} U^{⊖} + △ n_{g} R T$

Or $△_{v a p} U^{⊖} = △_{v a p} H^{⊖} - △ n_{g} R T$

Given $T = 298 K$

$△_{v a p} U^{⊖} = 44.01 (k J) - 1 (m o l) \times 8.314 \times 10^{- 3} (k J K^{- 1} m o l^{-}) \times 298 (K)$

$△_{v a p} U^{⊖} = 41.53 k J m o l^{- 1}$

Final answer:
Heat supplied, and internal energy are $44.01 k J$ and $41.53 k J / m o l$ respectively.

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