 Solution

Use the pressure correction to a mineral reaction for the solids, with $P_0=1\mathrm{bar}$. The water contribution must instead use its chemical fugacity. For pure solid phases and water vapour treated as an ideal gas, $f_{{\mathrm{H}}_2\mathrm{O}}=P$, givingThe last term already accounts for the gas-pressure contribution; adding another gas-volume correction would double count it. The pressure ratio inside the logarithm is dimensionless.

With pressure expressed in bars, the solid correction is $-2.28(4500-1)=-10257.72\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. ThusUsing $4500$ rather than $4499$ for the solid pressure increment would silently shift the specified one-bar reference.