The vapour pressures of water and ice at −5°C are 0.42 and 0.40 Pa, respectively. What is the free energy change for the process?
H₂O (l, −5°C, 0.42 Pa, 1 mole) → H₂O (s, −5°C, 0.40 Pa, 1 mole)

Question

The vapour pressures of water and ice at −5°C are 0.42 and 0.40 Pa, respectively. What is the free energy change for the process?

H₂O (l, −5°C, 0.42 Pa, 1 mole) → H₂O (s, −5°C, 0.40 Pa, 1 mole)

TestHub · Accepted Answer

Reversible path at −5°C (268 K): (i) l → g at 0.42 Pa, ΔG = 0; (ii) vapour expands 0.42 → 0.40 Pa, ΔG = RT ln(0.40/0.42) = R × 268 × ln(20/21); (iii) g → s at 0.40 Pa, ΔG = 0. Total ΔG = R × 268 × ln(20/21). Matches (c).

Chemistry - Thermodynamics - II Question with Solution | TestHub

Options:

Doubts & Discussion