# 24.4: Ideal Solutions obey Raoult's Law

### Thermodynamic potentials of ideal solutions

As we have seen, an ideal solution obeys Raoult's law:

$P_j = x_jP^*_j$

When we substitute we find a logarithmic relationship between thermodynamic potential and mole fraction:

## Thermodynamics of ideal mixing

When we add $$n_1$$ moles of component $$1$$ and $$n_2$$ moles of component $$2$$ to form an ideal liquid solution this is generally a spontaneous process. Let us consider the Gibbs free energy change of that process:

And yes this is always negative, so mixing is spontaneous. However the expression looks suspiciously familiar. Apart from a factor -T is is equal to the entropy of mixing!

This leaves no room at all for an enthalpy effect:

Even though in the liquid there are strong interactions between neighboring particles there is no enthalpy change because it does not matter what this neighboring molecule is.

If we represent the average interaction energy between molecule i and j by Uij we are essentially assuming that

In practice this is seldom the case. It usually does matter and then the enthalpy term is not zero. As this affects the thermodynamics of the liquid solution it should als affect the vapor pressures that are in equilibrium with it.

### Molar volumes

From the change of $$G$$ in its natural variables we know that:

$\left (\dfrac{\partial G}{\partial P} \right)_T =V$

This means that if we take

$\left (\dfrac{\partial \Delta G}{\partial P} \right)_T =\Delta V_{mix}$

In the ideal case we get:

$\left (\dfrac{\partial \Delta G^{ideal}}{\partial P} \right)_T =\Delta V_{mix}^{ideal}$

$\left (\dfrac{\partial RT ( n_1 \ln x_1 + n_2\ln x_2)}{\partial P} \right)_T =\Delta V_{mix}^{ideal} =0$

In the ideal case volumes are simply additive and we need not worry about how the partial molar volumes change with composition.