Question

A swimmer coming out from a pool is covered with a film of water weighing about $18g.$ How much heat must be supplied to evaporate this water at $298K?$ Calculate the internal energy of vaporization at $298K.{△}_{vap}{H}^{\ominus }$ for water at $298K=44.01kJmo{l}^{-1}$

Answer 1

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

The process of evaporation represent as:

$H_{2}O\left(l\right)\stackrel{\text{Vaporisation}}{\to }{H}_{2}O\left(g\right)$

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

Number of moles of $18g$ 𝑜𝑓 $H_{2}O\left(l\right)=\frac{18g}{18gmo{l}^{-1}}=1mol$

Step 2: Heat supplied

Heat supplied to evaporate $18g$ or $1mol$ of water at $298K=n\times {△}_{vap}{H}^{\ominus }$

$=\left(1mol\right)\times \left(44.01kJmo{l}^{-1}\right)$

$=44.01kJmo{l}^{-1}$

Step 3: Internal energy

(Assuming steam behaving as an ideal gas).

${△}_{vap}{H}^{\ominus }={△}_{vap}{U}^{\ominus }+△n_{g}RT$

Or ${△}_{vap}{U}^{\ominus }={△}_{vap}{H}^{\ominus }–△n_{g}RT$

Given $T=298K$

${△}_{vap}{U}^{\ominus }=44.01\left(kJ\right)–1\left(mol\right)\times 8.314\times {10}^{-3}\left(kJ{K}^{-1}mo{l}^{-}\right)\times 298\left(K\right)$

${△}_{vap}{U}^{\ominus }=41.53kJmo{l}^{-1}$

Final answer:
Heat supplied, and internal energy are $44.01kJ$ and $41.53kJ/mol$ respectively.