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How can you find this differential equation?

  1. Dec 10, 2014 #1
    This is actually a chemistry problem, but I feel it's more appropriate posted here, as I'm not having trouble with the chemistry but rather the mathematics. To avoid mixing subjects, I'll keep chemistry jargon out.

    We're given the formula for a component, and the problem request we give another component (which happens to be the derivative of the component given).

    Given formula: GT = C + RTln(xA)/xa
    where C, R, and T are constants for our purposes. (For clarity, A and B are different substances. Normal script T is temperature and subscript T is total, or the additive of components A and B for that variable)

    In the case of the problem, the component xA = nA/nT

    and μA = ∂GT/∂nA

    So, solving for μA, we get μA = RT(1-ln(xA / xA2nT)

    Now, onto my question. A certain equation says that

    nBB = -nAA

    And the problem requests I find a differential equation to solve for μB, but the required integration doesn't make sense to me. I mean, I assume I'm supposed to integrate both sides using the previously solved for μA. But I haven't taken a differential equations class, so I'm not sure how to produce a differential equation for this relation using the information given.

    Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Dec 11, 2014 #2

    Stephen Tashi

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    I don't see that. Is there some confusion in the subscripts between captial [itex] A [/itex] and lower case [itex] a [/itex] ?


    [itex] G_T = C + \frac{R\ T}{x_a} ln{(x_A)} [/itex]

    [itex] x_A = \frac{\eta_A}{\eta_T} [/itex]

    [itex] G_T = C + \frac{R\ T}{x_a} ln{( \frac{\eta_A}{\eta_T})} [/itex]

    [itex] \mu_A = \frac{\partial G_T}{\partial \eta_A} [/itex]

    [itex] = \frac{R\ T}{x_a} \frac{1}{ ( \frac{\eta_A}{\eta_T}) }\frac{1}{\eta_T} [/itex]

    [itex] = \frac{R\ T}{x_a}\frac{1}{\eta_A} [/itex]


    [itex] = \frac{R\ T}{x_a}\frac{1}{x_A \eta_T} [/itex]
     
  4. Dec 11, 2014 #3
    No, the lowercase and uppercase are the same. Merely a typo.

    Wouldn't there be a natural logarithm in the answer?

    [itex] \frac{d}{dn_{A}} \frac{RT ln(\frac{n_{A}}{n_{T}})}{\frac{n_{A}}{n_{T}}} [/itex]

    [itex] = RTn_{T} \frac{d}{dn_{A}}\frac{ln(\frac{n_{A}}{n_{T}})}{n_{A}}[/itex]

    [itex]= RTn_{T} \frac{d}{dn_{A}}[ln(\frac{n_{A}}{n_{T}})n^{-1}_{A} ][/itex]

    [itex]= RTn_{T} [\frac{n_{T}}{n_{A}}*n^{-1}_{A} + ln(\frac{n_{A}}{n_{T}})*n^{-2}_{A}][/itex]

    [itex]= RTn_{T} (\frac{n_{T}+ln(x_{A})}{n_{A}^{2}})[/itex]

    This is the same answer I have in the original post, with variable moved around.
     
  5. Dec 11, 2014 #4

    Stephen Tashi

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    Yes, if [itex] x_a = x_A [/itex]

    and I get:

    [itex]= R T n_{T} [ \frac{1}{n_A} * n^{-1}_A + ln( \frac{n_A}{n_T} )*(- n^{-2}_{A}) ][/itex]
     
  6. Dec 11, 2014 #5
    My question is about a differential equation, not about the derivative of the function. I just gave the information to aid in answering my question over the differential equation, since I don't know where to start.
     
  7. Dec 11, 2014 #6

    Stephen Tashi

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    It's actually very hard to say what the following symbolic manipulations do in a rigorous manner, but what your are expected to do is take an antiderivative of both sides.

    [itex] x dx = - y dy [/itex]

    [itex] \int x dx = - \int y dy [/itex]

    [itex] \frac{x^2}{2} + C_1 = -\frac{y^2}{2} + C_2 [/itex]

    [itex] x^2 + y^2 = 2(C_2 - C_1) = C [/itex] where [itex] C [/itex] is an arbitrary constant.
     
  8. Dec 11, 2014 #7
    The equation you are asking about is called the Gibbs Duhem equation. If you divide both sides of the equation by nT, you get:
    [tex]x_Bd\mu_B=-x_Ad\mu_A[/tex]
    For a binary system like the one you are looking at, xB=1-xA
    So, [tex]\frac{d\mu_B}{d\mu_A}=-\frac{x_A}{(1-x_A)}[/tex]
    Is that what you were looking for?

    Chet
     
    Last edited: Dec 11, 2014
  9. Dec 11, 2014 #8
    I'm not actually sure what I'm looking for. I asked my professor what form he wanted our answer in and he said he can't tell me, I suppose it gives the answer away.

    From my understanding, a differential equation is one where the left hand side is the derivative of a variable that is on the right side (i.e. y' = cy in the simplest form). But I haven't taken differential equations so my limited knowledge is only what was covered in calculus. In other words, I was trying to solve the differential you wrote above by somehow plugging in the previously solved for chemical potential of A. But if what your provided above is a differential equation in proper form, I suppose it will work. :P
     
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