The vapour pressures of water and ice at are 0.28 and 0.26 Pa, respectively. What is the free energy — Thermodynamics and Thermochemistry Chemistry Question
Question
The vapour pressures of water and ice at $-10^\circ\text{C}$ are 0.28 and 0.26 Pa, respectively. What is the free energy change for the process? $\text{H}_2\text{O (l, } -10^\circ\text{C, 0.28 Pa, 1 mole)} \to \text{H}_2\text{O (s, } -10^\circ\text{C, 0.26 Pa, 1 mole)}$
💡 Solution & Explanation
The initial and final states are respective equilibrium boundaries where $\Delta G_{\text{phase}} = 0$. The overall $\Delta G$ equates to the reversible isothermal expansion of the vapor from 0.28 Pa to 0.26 Pa. $\Delta G = nRT \ln(P_2/P_1) = 1 \times R \times 263 \times \ln(0.26/0.28) = R \times 263 \times \ln\left(\frac{13}{14}\right)$.