# Two-phase systems

In which we...

- Figure out precisely how a vdW gas separates into separate phases,
- Show how coexisting phases are a key engineering solution to make machines with a high efficiency--that is, more nearly reversible.
- Calculate the slope of the coexistence curve for two phases.

### Van der Waals phase transition re-visited

The van der Waals equation of state resulted in isotherms with a kink below a critical temperature $T_c$ and some phase separation.

The Helmholz energy, $F$, can be used to precisely locate these phases on a $P-V$ diagram:

If **two** phases are **in equilibrium** with each other, (in equilibrium both thermally and mechanically) then they ought to have the same pressure and temperature. But a line at constant pressure intersects **three** points on any isotherm below $T_c$.

For a system at constant temperature and volume, the condition for equilibrium is that the Helmholz energy for the system, $F$, should be a minimum.

Since $[dF]_T = - W_r$, we can calculate the free energy **of a single phase** by integrating the reversible work, $PdV$, along an isotherm:
$$\Delta F = \int (-P\,dV).$$

We can sketch the integration visually as shown at the right (solid line) for a homogeneous phase. $f$ is a continuously decreasing function of $v$, since both $P$ and $v$ are always positive.

The orange dashed line segment on the Helmholz energy part of the graph is the free energy of a linear combination of the two phases at $V_1$ and $V_2$.

Since the Helmholz energy of any linear combination of 1 and 2 is always lower than the single-phase energy between 1 and 2, the system will have, for example, a lower free energy by breaking up into a combination of two phases, 1 and 2, (b) rather than staying in the single, homogeneous phase (a).

The pressure is also constant for all the linear combinations between 1 and 2, because the single phases 1 and 2 are at the same pressure.

#### Equal area construction

But what pressure are the two phases at? We will see that pressure line should be drawn such that the areas $A$ and $B$ shown are equal, by using the "Maxwell construction":

The conditions that determine 1 and 2 are:

- In order for the pressures to be the same, the slopes $\partial F/\partial V$ must be the same, because: $$\frac{\partial F}{\partial V_1} = -P_1 = -P_2=\frac{\partial F}{\partial V_2}.$$
- But points 1' and 2' also have the same pressure: Why not them?
The blue line segment shows the free energy of a linear combination of 1' and 2' which is always higher than the orange line.

$\Rightarrow$ The points for the two phases must lie on the *same* tangent line in order to get the lowest possible Helmholz energy for linear combinations.

Using the condition that the points lie on the *same* tangent line to $F$: $$\frac{F_2-F_1}{V_2-V_1}= \left(\frac{\partial F}{\partial V}\right)_{\text{at }V_1\text{ or }V_2}.$$

Rearranging a bit... $$-(F_2-F_1) = -\frac{\partial F}{\partial V}(V_2-V_1).$$

This is: $$\int_{V_1}^{V_2}P\,dV = P_1(V_2-V_1).$$

The terms on either side of this equation correspond to these areas on the graph:

$$P_1(V_2-V_1)$$

$$\int_{V_1}^{V_2}P\,dV$$

#### Image credits

Matthieu Marechal, Spakovszky, figure 8.6, Andrew Ainsworth, Mahan, Elementary Chemical Thermodynamics, 1963, cited in Spakovszky