# The Binary System Review

### Azeotrope in Binary System Review

In this problem, we have a binary system wherewe've made a measurement of vaporliquid equilibrium at just one composition. So we have the liquidcomposition and the vapor composition, and the saturation pressures and the pressurewhere that vaporliquid equilibrium exists. And so the idea is, from that one measurement,can we predict whether this system has an azeotrope at this temperatureé And then, if so, what'sthe composition of the azeotropeé And we're going to assume, to simplify the calculations but the concept certainly holds in more general we use a simplified form of theMargules equation where there's only one parameter. And so we can calculate this one parameterfrom our vaporliquid equilibrium data. And

the idea, what we're looking at is, we have a system pressure versus mole fractions of component 1 and 2 saturation pressure here is 0.15and the saturation here is 0.35. And we know, when x1 is 0.2, that y1 is 0.5.So essentially, the question is, does this go through a maximum somewhere or noté Whatwe're going to do is first calculate the value of B. So we need the value of B, and we getthat from the vaporliquid equilibrium data. So this is the modified Raoult's law, assumingideal gas because it's low pressure, but clearly the liquid is not ideal. And so we can calculategamma 1 from the information we're given. And so if we look in a thermodynamics textbookfor this modified Raoult's law to save time

we're not going to derive this here thelog of gamma 1 is equal to this constant B times x2 squared. And so we can substitutevalues in here and determine the value of B. So we have the value of B it's dimensionless and in this equation, x1 was 0.2, then x2 of course is 0.8. So the question is, nowif we're going to use this model, is there anywhere in this region we have a maximumpressure azeotrope based on the data that we're looking at and that point, x1 equalsy1. So the question is, does that existé And so if we look at our modified Raoult's law,it means these terms would cancel. And we can write that for both component 1 and component2. Now we have the expression for gamma 1.

Gamma 1 is e times B x2 squared, times P1satequals pressure. And likewise for gamma 2 in terms of x1 squared, P2sat equals P. Sothat means these two terms are equal to each other. And we can substitute in now the numericalvalues. And now our only unknown is the mole fraction, of course we know x1's related tox2. So we can rearrange this and solve for x1. Take the log of both sides, get a littlecloser to our solution so this is log of 0.429. And again I'll substitute and calculatethe values. So there's multiple equations, now we can solve this. A nice way to do itis we can just factor this, then it's simplifies quite a bit because of course this is equalto 1. And so then x2 minus x1 is equal to

0.712. And then I can solve for x1. And sowe get an answer, a numerical value which says yes, we can have an azeotrope at x1 equals0.856 or perhaps, significant figures, 0.86. Now we can check this, we can then solve forthe pressure that's one thing we want to do. Let's do that first. We're going to solvefor the pressure by going back to our equation of x1 gamma 1 P1sat equals y1 times P. Andagain we're at the azeotrope so x1 is y1 and x2 is 0.144 times P1sat, that should giveus the pressure at the azeotrope. We can solve for that, 0.359 bar. This is one check becausethe pressure should be higher than either of the saturation pressures. In other words,if we look at our diagram, if we have an azeotrope,

this pressure should be higher than eitherof the saturation pressures. And indeed we calculate a value for pressure that's higher.We can check also, and I won't go through the calculations, but we can now go back andsubstitute in the azeotrope values to this equation to make sure we get the same pressure.We do that, we get the same pressure. So indeed we have an azeotrope and it corresponds tox1 of 0.86.