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For a mineral reaction producing one mole of water, let $H^{\circ },S^{\circ }$ refer to the $P^{\circ }$ standard state and let the solids’ volume change $\Delta V_s$ be constant. If $A_s$ is the product/reactant solid-activity ratio, the Gibbs energy of reaction isFor temperature-independent specified activities and fugacity, equilibrium givesThe gas pressure work is already represented by the fugacity term; adding a second constant water-volume term double counts it. For an ideal gas, $f=P$; a given fugacity coefficient gives $f={\varphi }_{f}P$. If these quantities depend on temperature, the equilibrium equation must instead be solved implicitly. Pressure units in the solid volume correction must match those used for $\Delta V_s$.

With water fugacity coefficient ${\varphi }_f=0.8$, use $f={\varphi }_{f}P=4800$ bar. Retain the same ideal mineral-site activities, givingHenceThe lower fluid chemical fugacity stabilizes the water-bearing products of the forward dehydration relative to the ideal-gas calculation and lowers the required temperature by about $13.19$ K. Here the stated fugacity coefficient is treated as a given constant for evaluating the equilibrium.

The fugacity coefficient $\varphi =f/P=0.7$ gives $f=3500\mathrm{bar}$. Keep the same solid thermodynamic activities and pressure correction to a mineral reaction as in part (c), changing only the water chemical fugacity:The lower water chemical fugacity makes the dehydrated assemblage more favorable, so the equilibrium temperature decreases by about $29.0\mathrm{K}$. This direction is a useful independent check on the logarithm's sign.

For fugacity coefficient $\varphi =0.75$, use $f=\varphi P=3375\mathrm{bar}$ for the water activity. The solid pressure correction still uses the actual pressure $4500\mathrm{bar}$. ThusLowering the water chemical potential favours the product side, which lowers the equilibrium temperature relative to part (c) under the stated constant-property model.

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.