# The Gibbs-Duhem equation - checking some work

1. Apr 30, 2014

### Emspak

1. The problem statement, all variables and given/known data

For a two-component system use the Gibbs-Duhem relation to show that

$$x \left(\frac{\partial \mu_a}{\partial x}\right)_{T,P} + (1-x)\left(\frac{\partial \mu_b}{\partial x}\right)_{T,P} = 0$$

where $\mu_a$ and $\mu_b$ are chemical potentials and $x=\frac{n_a}{n_b+n_a}$

2. Relevant equations

The Gibbs-Duhem equation is as follows: $-SdT+VdP- \sum_{i} n_i d\mu_i=0$

3. The attempt at a solution

Looking at this I see that the sum of chemical potentials is equal to $-SdT+VdP$ (otherwise it wouldn't be zero). I also note that the x is the molar fraction of the two substances, which is why the second term in the equation we are trying to show equals zero has a 1-x in it. (The two molar fractions have to add up to 1).

That would say to me that one way to approach this is:

$-SdT+VdP=n_a d\mu_a+n_b d\mu_b$

and since the capital S and V are the total S and V (not per mole) and na and nb are the moles of the two substances, if I want the per mole amounts I just need to convert those to molar fractions, $x=\frac{n_a}{n_b+n_a}$ so I would get

$-sdT+vdP=\frac{n_a}{n_b+n_a}d\mu_a + (1 - \frac{n_a}{n_b+n_a})d\mu_b$

$-sdT+vdP=xd\mu_a + (1 - x)d\mu_b$

Now, I can intuit that dT and dP ar both zero, because the original problem says T and P are kept constant. Integrating the above I should end up with

$-sT+vP=x \mu_a + (1 - x)\mu_b$

and taking a partial derivative of μa w/r/t x:

$x \left(\frac{\partial \mu_a}{\partial x}\right)_{T,P}+ \mu_a - \mu_b = 0$

do the same with μb w/r/t x:

$\mu_a + (1-x) \left(\frac{\partial \mu_b}{\partial x}\right)_{T,P}- \mu_b = 0$

and since both equal zero they are equal to each other

$x \left(\frac{\partial \mu_a}{\partial x}\right)_{T,P}+ \mu_a - \mu_b = \mu_a + (1-x) \left(\frac{\partial \mu_b}{\partial x}\right)_{T,P}- \mu_b$

and moving everything around

$x \left(\frac{\partial \mu_a}{\partial x}\right)_{T,P} = (1-x) \left(\frac{\partial \mu_b}{\partial x}\right)_{T,P}$

$x \left(\frac{\partial \mu_a}{\partial x}\right)_{T,P} - (1-x) \left(\frac{\partial \mu_b}{\partial x}\right)_{T,P} = 0$

But somehow I got a sign switch and I am not quite sure where. (I think I missed something dumb here). But more than that, the hint for the problem says to put μ in terms of P, T, and x and take the partial derivatives with respect to P etc. That seemed a much more complex way of going about it, and I also wondered if I was missing something about how they want you to do the problem. If I try that, what ends up happening is I try integrating to get μ and then re-deriving like this:

$x \left(\frac{\partial \mu_a}{\partial x}\right)_{T,P} + (1-x)\left(\frac{\partial \mu_b}{\partial x}\right)_{T,P}= -Sdt + VdP$ (this from the procedure above)

and then I just straight up integrate both sides

$\frac{x^2}{2} \mu_a + (x-\frac{x^2}{2}) \mu_b= -ST + VP$

and taking the partial derivatives $\frac{x^2}{2}\left(\frac{\partial \mu_a}{\partial P}\right)_{T,x} = V_a$, and $(x-\frac{x^2}{2})\left(\frac{\partial \mu_b}{\partial P}\right)_{T,x} = V_b$

But from there I feel like I am losing the plot, and I suspect I am approaching the math wrong, though I also suspect that if I try the second approach I have to get the partial derivatives of μ with respect to T as well, leaving P constant.

Anyhow, what I am interested in is a) is the first method I tried valid and b) how I messed up the second. (I will make a bet that I can't treat partial derivatives the way I did in the integration I tried).

Any assistance is welcome, and thanks.

2. Apr 30, 2014

### Staff: Mentor

It's much simpler than this. It looks like things got away from you. Starting with:

$$-SdT+VdP=n_a d\mu_a+n_b d\mu_b$$

At constant temperature and pressure, this becomes:

$$x_a d\mu_a+x_b d\mu_b=0$$

Let x = xa and (1-x) = xb. Now since there are only two components, at constant temperature and pressure, μaa(x) and μbb(x). So dμa=(dμa/dx)dx and dμb=(dμb/dx)dx

Chet

3. Apr 30, 2014

### Emspak

Yours is a lot simpler, but I was just suspecting that the text was looking for something else, is all. But hey, simple is good. But I am curious if the first method I used was OK. (I know it got away from me in the second part, which is why I left off that method -- i am not sure why the text proposed doing it that way).

4. Apr 30, 2014

### Staff: Mentor

The following looks mathematically incorrect to me:

5. Apr 30, 2014

### Emspak

Good to know, but is there something specific that is wrong? That is, I know partial derivatives were on one side and total derivatives on the other, so that can't just be integrated as I did it? Is that the case?

6. Apr 30, 2014

### Staff: Mentor

If dT and dP are zero, then: $-sdT+vdP=0$

Mathematically, the left side doesn't integrate to $-sT+vP$

Chet