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ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2022/paper-1/7/b/solution

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Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2022 / Paper 1 / 7 / b / Solution by

Codex  0 2026-10-25

Integrate the solids’ molar volume contribution from

P^{\circ} = 1

bar and use ideal-gas water chemical fugacity

f = P

. For pure solids,

0 = Δ_{r} H^{\circ} - T Δ_{r} S^{\circ} + Δ V_{s} (P - P^{\circ}) + RT ln \frac{P}{P ^{\circ}} .

(1)

The logarithm is dimensionless, and

Δ V_{s} = - 2.28 J ba r^{- 1}

can be multiplied directly by pressure in bars. Therefore the dehydration equilibrium with ideal-gas fluid gives

T = \frac{141120 - 2.28 ( 6000 - 1 )}{170.7 - 8.3145 l n 6000} = \frac{127442.28}{98.3679} .

(2)

Hence $T = 1295.57$ K, or $1022.4 2^{\circ}$ C. Do not also add a constant gas-volume pressure term: the water contribution has already been integrated through

RT ln (P / P^{\circ})

. Ideal-gas behavior at

6

kbar is a prescribed model approximation, not a realistic equation of state. The solid volume correction is distinct from it.

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