The vapour pressure of water is 0.04 atm at 27°C. The free energy change for the following process is
H₂O(g, 0.02 atm, 27°C) → H₂O(l, 0.02 atm, 27°C)

Question

The vapour pressure of water is 0.04 atm at 27°C. The free energy change for the following process is H₂O(g, 0.02 atm, 27°C) → H₂O(l, 0.02 atm, 27°C)

TestHub · Accepted Answer

Reversible path at 27°C (300 K): (i) compress the vapour isothermally from 0.02 atm to the equilibrium vapour pressure 0.04 atm: ΔG = RT ln(0.04/0.02) = R × 300 × ln 2; (ii) condense at 0.04 atm: ΔG = 0; (iii) changing the liquid's pressure back to 0.02 atm contributes negligibly. Total ΔG = R × 300 × ln 2 > 0, consistent with unsaturated vapour not condensing spontaneously. Matches (c).

Chemistry - Thermodynamics - II Question with Solution | TestHub

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