### Fugacity

We have already established that all of the property relations
that are used for the pure-species Gibbs free energy, $G$, also are
applicable for the partial molar Gibbs free energy. Specifically,
we are interested right now in the fact that:

$\left ( \frac{\partial \bar G_i}{\partial P} \right )_{T,
n_i} = \bar V_i$

If we have a closed, isothermal system then $T$ and $n_i$ are
*actually* constant (rather than being *held*
constant mathematically as in a partial differential) so that this
relation becomes:

$\frac{d \bar G_i}{d P} = \bar V_i$

This relation is useful because, in order to obtain a
*value* for $\bar G_i$ we need to calculate it relative to
some other value (i.e., a reference state).

If we consider the simplest case that we can think of, that is
an ideal gas, we can rearrange, substitute and integrate

$\int d \bar G_i = \int \bar V_i d P = \int \frac{RT}{P}
dP = RT \int \frac{dP}{P}$

$\bar G_i - \bar G_i^o = \mu_i - \mu_i^o = RT \ln \left
[\frac{P}{P^o} \right ]$

where the ${}^o$ values refer to whatever the reference state is
chosen to be.

There are two issues with this:

- there is not a simple choice of what the reference state should
be
- at low (zero) pressure the $\ln$ term goes to $-\infty$

To alleviate these problems, mixture equilibrium relations are
not built using the partial molar Gibbs free energy (or chemical
potential) but instead with a construct proposed by Lewis.

##### Definition:

The **fugacity** of species $i$ in a mixture,
$\hat f_i$, is defined in the following way:

$\bar G_i - \bar G_i^o = \mu_i - \mu_i^o = RT \ln \left
[\frac{\hat f_i}{\hat f_i^o} \right ]$

where the $\hat {}$ denotes the mixture value (as opposed to a
pure-species value which would not have the hat).

Note that the utility of this definition lies in the fact that
the reference state can be chosen somewhat arbitrarily and, in
practice, has very convenient "values" that differ by situation
(i.e., vapor, liquid, solid). Also, since $f$ is not
*actually* a pressure (even though it has units of
pressure), we no longer have the $\ln$ issues.

##### Note:

While we will not rigorously prove it, it is straight-forward to
show that the following equations are equivalent:

$\mu_L=\mu_v$

$\hat f_i^L=\hat f_i^v$

Henceforth, all phase equilibrium will be based on the fugacity
version of this relation.

##### Outcome:

Relate the fugacity and the chemical potential (or the partial
molar Gibbs free energy).