 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2024/paper-3/9/b/solution

0 = Δ_{r} H^{\circ} - T Δ_{r} S^{\circ} + Δ V_{solids} (P - P_{0}) + RT ln (P / P_{0}), P_{0} = 1 bar .

The gas contribution is already included by the logarithm and must not be counted again as an independent volume correction. With pressures in bars,

T = \frac{141120 - 2.28 ( 4500 - 1 )}{170.7 - 8.3145 ln 4500} = 1298.755 K ≃ 1025.605^{\circ} C .

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