 Solution

The fugacity coefficient gives $f_{{\mathrm{H}}_2\mathrm{O}}=\varphi P=0.82\times 4500=3690\mathrm{bar}$. Keep the solid pressure correction at the actual $4500\mathrm{bar}$; only the water activity uses the fugacity. ThusThe decrease from part (c) follows because $\varphi <1$ lowers the chemical potential of the water product. All logarithms in these thermodynamic formulae are natural logarithms.