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For the transformation, $H_{2} O (l, 1 atm)$ to $H_{2} O (g, 0.1 atm); Δ H_{vap} = 40.668 kJ {mol}^{- 1}$ The change in entropy $({JK}^{- 1} {mol}^{- 1})$ is :

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19.14

89.89

109.03

128.17

answer is D.

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Detailed Solution

$\begin{array}{l} H_{2} O (L, 1 atm) ⟶ H_{2} O (g, 1 atm) \\ \begin{aligned} Δ S & = \frac{40.688 \times 10^{3}}{373} + 1 \times 8.314 \ln (\frac{1}{0.1}) \\ = 128.2 {JK}^{- 1} {mol}^{- 1} O (g, 0.1 atm) \end{aligned} \end{array}$

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