What is the free energy change for the process :
$\begin{aligned} H_{2} O (l, - 10^{\circ} C, 0.28 Pa, 1 mole) ⟶ H_{2} O \\ (s, - 10^{\circ} C, 0.26 Pa, 1 mole) \end{aligned}$
Given that vapour pressure of water and ice at -10°C is 0.28 and 0.26 Pa respectively.

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zero

$R \times 263 \times \ln \frac{14}{13}$

$R \times 263 \times \ln \frac{13}{14}$

$- R \times 10 \times \ln \frac{13}{14}$

answer is B.

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Detailed Solution

Gibbs energy,

$\begin{array}{l} d G = V d P - S d T \\ d G = V d P (T = Constant) \\ Δ G = n R T \ln (\frac{P_{2}}{P_{1}}) = 263 R \ln (\frac{0.26}{0.28}) \\ = 263 R \ln (\frac{13}{14}) \end{array}$

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