NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
Diamonds are formed from graphite under high pressure in coal mines. Calculate the equilibrium pressure (in atm) at which graphite is converted to diamonds at 25 o C (assumed constant) given densities of ρ graphite = 2 g / cc & ρ diamond = 3 g / cc Δ G f o for diamonds is 3 kJ m - 1 from graphite
Correct answer
15001
Step-by-step solution
Differential equation of free energy dG = VdP - SdT at constant temp. with change in allotropic modification there is a change in molar volume ∴ d Δ G = Δ V dP . Δ V = M Diamond ρ Diamond - M graphite ρ graphite = 1 2 3 - 1 2 2 Δ V = - 2 cm 3 / mole = - 2 × 1 0 - 6 m 3 mole -1 ∫ d Δ G Δ G f o at 2 9 8 K, 1 atm Δ G f o at 2 9 8 K, p atm = 0 = - 2 × 1 0 - 6 m 3 mole - 1 ∫ dP P = 1 atm P = p atm Because at equilibrium when diamonds are formed