= Solution
Apply the <pressure correction to a mineral reaction> to the solids and use the <chemical fugacity> of the water product. For pure solids and <water vapour> treated as an <ideal gas>,
$$
0=\Delta_rH^\circ-T\Delta_rS^\circ+\Delta V_{\rm solids}(P-P_0)+RT\ln(P/P_0),
\qquad P_0=1\,\mathrm{bar}.
$$
The gas contribution is already included by the logarithm and must not be counted again as an independent volume correction. With pressures in bars,
$$
\boxed{T=\frac{141120-2.28(4500-1)}{170.7-8.3145\ln4500}
=1298.755\,\mathrm K\simeq1025.605\,{}^\circ\mathrm C.}
$$
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