Engineering
Chemistry
Gibbs Free Energy Calculations and Third Law
Ideal Gas Equation
Gibbs Free Energy Entropy and Spontaneity
Question
The vapour pressure of liquid Hg at 433 K is 5 mm Hg. Calculate the free energy change accompanying the expansion of one mole of Hg vapour in equilibrium with liquid at 433 K to a pressure of 750 mm Hg at the same temperature assuming the vapour behaves like an ideal monoatomic gas in kJ mole–1. (Approximate integer and e5 = 150)
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Solution
Hg (ℓ) → Hg (g) at 433 K and 4.19 mm of Hg is reversible and ΔG = 0 but
ΔG for Hg (g) (4.19 mm of Hg) → Hg (g) (760 mm of Hg)