For a substance A, vapour pressure of liquid and solid state at temperature 400 K is 0.1 bar and 0.16 bar respectively. What will be Gibbs free energy change for the following process.
$A (s) [0.1 bar, 400 K, n = 1] ⟶ A (l) [0.16 bar, 400 K, n = 1]$

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400 R In $\frac{1}{1.6}$

400 R ln 1.6

400 R [0.16- 0.1]

answer is C.

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Detailed Solution

Gibbs energy,

$\begin{aligned} Δ G = - R T \ln (\frac{P_{2}}{P_{1}}) \\ = - R \times 400 \ln (\frac{0.16}{0.1}) \\ = 400 R \ln \frac{0.1}{0.16} \\ = 400 R \ln \frac{1}{1.6} \end{aligned}$

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