The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm⁻³, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol⁻¹, the pressure at which graphite will be transformed into diamond at 298 K is:
11.08 × 10⁸ Pa
9.92 × 10⁷ Pa
9.92 × 10⁶ Pa
11.08 × 10⁵ Pa
At constant temperature, the change in Gibbs free energy with pressure is given by the relation: dG = VdP. For a macroscopic change, \Delta G_P - \Delta G_1 = \Delta V(P - 1). At equilibrium, the free energy of graphite and diamond must be equal, so \Delta G_P = 0. Given, \Delta Gº = 1895 J mol⁻¹. The molar volume of graphite, V_graphite = Mass / Density = 12 / 2.25 = 5.333 cm³ mol⁻¹. The molar volume of diamond, V_diamond = Mass / Density = 12 / 3.31 = 3.625 cm³ mol⁻¹. Change in molar volume, \Delta V = V_diamond - V_graphite = 3.625 - 5.333 = -1.708 cm³ mol⁻¹ = -1.708 × 10⁻⁶ m³ mol⁻¹. Now, substituting the values into the equation: 0 - 1895 = (-1.708 × 10⁻⁶) × (P - 10⁵) P - 10⁵ = -1895 / (-1.708 × 10⁻⁶) = 1109.48 × 10⁶ Pa = 11.09 × 10⁸ Pa. Since 10⁵ Pa is negligible compared to 11.09 × 10⁸ Pa, the required pressure P ≈ 11.08 × 10⁸ Pa.
Join thousands of students and practice with AI-generated mock tests.