Calculate the entropy change when $36 g$ of liquid water evaporate at $373 K$ (latent heat of vaporization of water is $0.63 k J m o l^{- 1}$ ).

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Solution

Number of moles of water $= \frac{36}{18} = 2$ $m o l e s$
$E n t r o p y = \frac{Δ H_{v a p o r i z a t i o n}}{T}$ (during phase change)
$⟹ E n t r o p y = \frac{0.63 \times 1000}{373}$
$⟹ E n t r o p y = 1.689$ $J m o l^{- 1} T^{- 1}$
For $2$ $m o l e s$ ,
$E n t r o p y = 2 \times 1.689 = 3.378$ $J T^{- 1}$

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