= Solution
Integrate the solids’ <molar volume> contribution from $P^\circ=1$ bar and use ideal-gas water <chemical fugacity> $f=P$. For pure solids,
$$
0=\Delta_rH^\circ-T\Delta_rS^\circ+\Delta V_s(P-P^\circ)
+RT\ln\frac{P}{P^\circ}.
$$
The <logarithm> is dimensionless, and $\Delta V_s=-2.28\,\mathrm{J\,bar^{-1}}$ can be multiplied directly by <pressure> in bars. Therefore the <dehydration equilibrium with ideal-gas fluid> gives
$$
T=\frac{141120-2.28(6000-1)}{170.7-8.3145\ln6000}
=\frac{127442.28}{98.3679}.
$$
Hence \b[$T=1295.57$ K, or $1022.42^\circ$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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