# 24.3: Chemical Potential of Each Component Has the Same Value in Each Phase in Which the Component Appears

A lot of chemistry takes place in solution and therefore this topic is of prime interest for chemistry.

### Thermodynamic potentials of solutions

The Gibbs free energy of an ideal gas depends logarithmically on pressure

$G = G^o + RT \ln \dfrac{P}{P^o}$

Usually Po is taken to be 1 bar and is often dropped out of the formula. and we write.

$G = G^o + RT \ln P$

Notice however that although P and P have the same numerical value, the dimensions are different. P has a dimension as defined by what o means (usually bars), but P is dimensionless. (The value of Go still depends on what units we have chosen for P though.)

If we have a gas mixture we can hold the same logarithmic argument for each partial pressure as the gases do not notice each other. We do need to take into account the number of moles of each and work with (partial) molar values, i.e. the thermodynamic potential:

$μ_j = μ_j^o + RT \ln \dfrac{P_j}{P^o} \label{B}$

(Again Po is often dropped to confuse the enemy).

If we are dealing with an equilibrium over an ideal liquid solution the situation in the gas phase gives us a probe for the situation in the liquid. The equilibrium must hold for each of all components j (say two in binary mixture). That means that for each of them the thermodynamic potential in the liquid and in the gas must be equal:

$μ_j^{sln} = μ_j^{gas}$

for all $$j$$.

Consider what happens to a pure component, e.g. $$j=1$$ in equilibrium with its vapor. We can write:

$μ_1^{pure \,liq}= μ_1^{pure\, vapor}=μ_1^o + RT \ln \dfrac{P^*_1}{P^o}$

The asterisk in $$P^*_1$$ denotes the equilibrium vapor pressure of pure component 1 and we will use that to indicate the thermodynamic potential of pure compounds too:

$μ_1^{*liq}= μ_1^o + RT \ln \dfrac{P^*_1}{P^o} \label{A}$

Combining Equations $$\ref{A}$$ and $$\ref{B}$$ we find a relationship between the solution and the pure liquid :

$μ_j^{sln}=μ^*_j + RT \ln \dfrac{P_j}{P^*_j}$

Notice that the gas and its pressure is used to link the mixture and the pure compound.