Repeat the previous problem for the opposite case where the liquid has a substantial negative mixing energy, so that its free energy curve dips |below the gas's free energy curve at a temperature higher than T_B. Construct the phase diagram and show that this system also has an azeotrope.

The liquid has a substantial negative mixing energy, so that its free energy curve dips |below the gas's free energy curve at a temperature higher than T_B.

Consider the following curve, which depicts the free energy of the gas and liquid at T_B. We can observe that the gas curve is more concave than the liquid curve, indicating that the two curves meet at two locations, indicating that the liquid and gas are stable in two different composition ranges. Because the liquid has a negative Gibbs energy, it dives below the gas curve.

Draw a tangent on a graph between x and T (the phase diagram) at the two intersection locations as indicated in the accompanying figure; this tangent intersects with the gas and liquid curves. Then draw perpendicular lines from the four intersection points on a graph between x and T (the phase diagram).

The Gibbs free energy is given by:

$G=U+PV-TS$

At constant volume and entropy, the change in Gibbs free energy is as follows:

$dG=dU+VdP-SdT$

By increasing the temperature, we get

$\frac{\partial G}{\partial T}=-S$

Because the entropy of the depends on the temperature and has a negative sign, the liquid curve moves downward as the temperature rises. Lowering the temperature causes the liquid curve to rise until it coincides with the gas curve at one point, forming an azeotrope combination.