# FEC: Definition of Fugacity

### 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).