Question
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.
This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.
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