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The Gibbs-Duhem equation - checking some work

  1. Apr 30, 2014 #1
    1. The problem statement, all variables and given/known data

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

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

    where [itex]\mu_a[/itex] and [itex]\mu_b[/itex] are chemical potentials and [itex]x=\frac{n_a}{n_b+n_a}[/itex]

    2. Relevant equations

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

    3. The attempt at a solution

    Looking at this I see that the sum of chemical potentials is equal to [itex]-SdT+VdP[/itex] (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:

    [itex]-SdT+VdP=n_a d\mu_a+n_b d\mu_b[/itex]

    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, [itex]x=\frac{n_a}{n_b+n_a}[/itex] so I would get

    [itex]-sdT+vdP=\frac{n_a}{n_b+n_a}d\mu_a + (1 - \frac{n_a}{n_b+n_a})d\mu_b[/itex]

    [itex]-sdT+vdP=xd\mu_a + (1 - x)d\mu_b[/itex]

    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

    [itex]-sT+vP=x \mu_a + (1 - x)\mu_b[/itex]

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

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

    do the same with μb w/r/t x:

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

    and since both equal zero they are equal to each other


    [itex]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[/itex]

    and moving everything around

    [itex]x \left(\frac{\partial \mu_a}{\partial x}\right)_{T,P} = (1-x) \left(\frac{\partial \mu_b}{\partial x}\right)_{T,P}[/itex]

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

    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:

    [itex]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[/itex] (this from the procedure above)

    and then I just straight up integrate both sides

    [itex]\frac{x^2}{2} \mu_a + (x-\frac{x^2}{2}) \mu_b= -ST + VP[/itex]

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

    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. jcsd
  3. Apr 30, 2014 #2
    It's much simpler than this. It looks like things got away from you. Starting with:

    [tex]-SdT+VdP=n_a d\mu_a+n_b d\mu_b[/tex]

    At constant temperature and pressure, this becomes:

    [tex]x_a d\mu_a+x_b d\mu_b=0[/tex]

    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
     
  4. Apr 30, 2014 #3
    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).
     
  5. Apr 30, 2014 #4
    The following looks mathematically incorrect to me:
     
  6. Apr 30, 2014 #5
    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?
     
  7. Apr 30, 2014 #6
    If dT and dP are zero, then: [itex]-sdT+vdP=0[/itex]

    Mathematically, the left side doesn't integrate to [itex]-sT+vP[/itex]

    Chet
     
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