In this example, I'd like to demonstrate how we can get partial molar properties when we're
looking at a binary mixture from just a graphical plot of the molar property of the mixture
versus the mole fraction of one of the components. So I've plotted on the black line some partial
molar properties. Could be enthalpy, entropy, Gibbs free energy, volume. And then I've indicated
M1 which is the pure component. So this is the pure component value for component 1,
the enthalpy for example of component 1, and likewise over here for component 2, this is
the value for the solution. So at any mole fraction, we are given a value which is not
the same as the pure component values, is not the same as an average. And then these
are the partial molar quantities. So again, partial molar enthalpy of component 1 and
over here component 2. So what I'd like to demonstrate by calculations is that when I
draw a tangent line. If I pick this point, this molar property at this mole fraction,
then I can draw a tangent line to that point and where it intersects the axes gives me
the partial molar properties. And we're going to start off with basically how we define
these partial molar quantities and we're going to take advantage of some equations and some
manipulations to get to a relation that demonstrates that indeed these intersections of this tangent
are the partial molar properties. So I'm starting with the definition so the property is some
state function. It's a property of the mixture is equal to the mole fraction of component
1 times the partial molar property of component 1 plus the mole fraction of component 2 and
the partial molar properties. So we're saying we take partial molar properties and weight
them by the mole fraction to get the property of the mixture. So what we're going to do
next is take a differential of both sides of this equation. So here I've used the product
rule to get two terms. First I hold x1 constant and take the differential of the partial molar
property then I hold the partial molar property and take the differential of x1. So it's going
to turn out this is going to simplify because this term plus this term when added together
equal to zero. And this sit eh result of something called the Gibbs-Duhem equation. So the Gibbs-Duhem
equation is at constant temperature and pressure, x1 times the differential of the partial molar
quantity for component 1 plus x2 the differential of the partial molar quantity for component
2, that equals zero. So the two terms circled in red add to zero and as a result, the equation
reducers to just 2 terms. Now we're going to simplify this because we know x1+x2 adds
to 1. Binary systems, the two mole fractions add to 1. If we take the differentials of
both sides, right sides a constant. So this means dx1 is equal to -dx2. So we can simplify
this equation and replace dx2 by minus dx1. And so I make that assumption if I then divide
through by that differential. So my objective here is to get an equation to solve for the
partial molar quantity for M1. That partial molar quantity in terms of M. So I want to
replace this, and what I'm going to do is go back to our original starting equation
where M, the overall property for the mixture is x1 times partial molar quantity and x2
times the partial molar quantity. And I can use this to substitute for M2. So here I've
solved for partial molar quantity of component 2 in the equation on the right. I'm going
to substitute that back into here and I end up then with this equation. And so now I've
made the substitution and multiplied through by x2 and then i'm going to rearrange by bringing
the solution property to the other side. So I have the equation in terms of the solution
property and mole fractions on one side. The other side I have x1+x2 times partial molar
quantity for property 1 which of course since x1 plus x2 adds to 1 is equal to partial molar
quantity. So keep in mind this equation, we need a bar over this symbol here to represent
the partial molar quantity. And if I rearrange this equation to just solve for the derivative
which would be the same as the slope at that point. If I pick a point here. So the property
M and I'll just draw a line so that's M at this mole fraction. So this is a value for
x1 corresponding to M. So this would be x2 at that point. So I can see the difference.
M1, partial molar quantity, minus M that's right here, this difference, divided by x2,
that's the slope at this point, and therefore, that's equal to the derivative of M with respect
to x1 which means this intercept here must be the partial molar quantity because that's
what this equation indicates. And the same thing for partial molar quantity M2 at the
other end. So we can look at one of these graphical representations and see how the
partial molar quantity changes as we get closer to component 1, for example a point here the
partial molar quantity is going to get close to the molar quantity as we expect but the
partial molar quantity for a component that's very dilute can be quite different from the
molar quantity. Partial molar quantity if I take a point here and draw a straight line,
partial molar quantity is way down here for component 2. And so for binary solutions,
we can just look at the graph and very quickly see how the partial molar quantity changes
as we change the mole fraction.