ThermodynamicsHard
Question
The vapour pressures of water and ice at −10°C are 0.28 and 0.26 Pa, respectively. What is the molar free energy change for the process? H2O (l, −10°C, 0.28 Pa) → H2O (s, −10°C, 0.26 Pa)
Options
A.$R \times 263 \times \ln\frac{14}{13}$
B.$R \times 263 \times \ln\frac{13}{14}$
C.$R \times 10 \times \ln\frac{13}{14}$
D.$R \times 10 \times \ln\frac{14}{13}$
Solution
$H_{2}O\left( l, - 10^{o}C,0.28\text{ Pa} \right) \rightarrow H_{2}O\left( s, - 10^{o}C,0.26\text{ Pa} \right) $$${\downarrow \Delta G_{1} = 0 \uparrow \Delta G_{3} = 0 }{H_{2}O\left( g, - 10^{o}C,0.28\text{ Pa} \right)\overset{\quad\Delta G_{2}\quad}{\rightarrow}H_{2}O\left( g, - 10^{o}C,0.26\text{ Pa} \right)}$$
$\Delta G_{2} = nRT\ln\frac{P_{2}}{P_{1}} = 1 \times R \times 263 \times \ln\frac{0.26}{0.28} $$$\therefore\Delta G = \Delta G_{1} + \Delta G_{2} + \Delta G_{3} = 263R\ln\frac{13}{14}$$
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