At 500 kbar and T K, the densities of graphite and diamond are 2.0 and 3.0 g/cm^3, respectively. The β Thermodynamics and Thermochemistry Chemistry Question
Question
At 500 kbar and T K, the densities of graphite and diamond are 2.0 and 3.0 g/cm^3, respectively. The value of (Ξ H - Ξ U) for the conversion of 1 mole of graphite into diamond at 500 kbar and T K is
Answer: B
π‘ Solution & Explanation
Ξ H - Ξ U = P * Ξ V = P * (V_diamond - V_graphite). Molar mass of carbon = 12 g/mol. V_graphite = 12 / 2 = 6 cm^3/mol. V_diamond = 12 / 3 = 4 cm^3/mol. Ξ V = -2 cm^3/mol = -2 * 10^-6 m^3/mol. Pressure = 500 kbar = 5 * 10^10 N/m^2. P * Ξ V = 5 * 10^10 * (-2 * 10^-6) = -10^5 Joules/mol = -100 kJ/mol.
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