= Solution
The <fugacity coefficient> gives $f_{\mathrm{H_2O}}=\phi P=0.82\times4500=3690\,\mathrm{bar}$. Keep the solid pressure correction at the actual $4500\,\mathrm{bar}$; only the <water activity> uses the fugacity. Thus
$$
\boxed{T=\frac{130862.28}
{170.7-8.3145\ln\!\left[3690(0.23/0.85)^3\right]}
=969.243\,\mathrm K\simeq696.093\,{}^\circ\mathrm C.}
$$
The decrease from part (c) follows because $\phi<1$ lowers the chemical potential of the water product. All logarithms in these thermodynamic formulae are natural <logarithms>.
Back to article page
