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Homework Help: [Doubt] Chemical Potential of an Ideal Gas

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Basically, find the chemical potential of an ideal gas knowing its heat capacities.

    2. Relevant equations
    [itex]P V = n R T \ \ \ \ (1)[/itex]
    [itex]U = n c_V T + U_0 \ \ \ \ (2)[/itex]
    [itex]S = S_0 + n c_V ln (T) + nR ln (V) = S_0 + n c_V ln (T) + nR ln \left ( \frac{nRT}{P} \right ) \ \ \ \ (3)[/itex]
    [itex] \mu = \left ( \frac {\partial G}{\partial n} \right )|_{T,P} \ \ \ \ (4)[/itex]
    [itex] G = U - TS + PV \ \ \ \ (5)[/itex]

    3. The attempt at a solution

    Mixing (1), (2) and (3) into (5) I get

    [itex] G = n c_V T + U_0 - T \left (S_0 + n c_V ln (T) + nR ln \left ( \frac{nRT}{P} \right ) \right ) + nRT [/itex]

    Then differentiating with n, while treating P and T as constants

    [itex] \mu (P, T, n) = c_V T - T \left (c_V ln (T) + R ln \left ( \frac{nRT}{P} \right ) + R \right ) + RT [/itex]

    Which has no constants, but I suppouse that the chemical potential, as every good classical potential, must be defined beggining at some constant [itex]\mu_0[/itex].

    What I am doing wrong?

    Thank you for your time.
    Last edited: Nov 27, 2012
  2. jcsd
  3. Nov 27, 2012 #2
    In the third term for S, the V should just be RT/P. Then, re-express μ as

    μ = μ0(T) + RT ln P

    μ is an intensive property and cannot depend on the number of moles.

    Determine what μ0 is as a function of T.
  4. Nov 27, 2012 #3
    Good catch on the molar dependance, hadn't looked at that. The excercise I was looking at says [itex] \mu (P,T,n) [/itex]. I have some doubts on that being on purpose, but anyway...
    Then I have

    [itex] G = n c_v T + U_0 - T \left (S_0 + n c_v ln (T) + nR ln \left ( \frac{RT}{P} \right ) \right ) + nRT [/itex]

    Then the differentiation is

    [itex] \mu (P, T,) = c_v T - T \left (c_v ln (T) + R ln \left ( \frac{RT}{P} \right ) \right ) + RT [/itex]

    Which is the same as

    [itex] \mu (P,T) = c_v T - T c_v ln(T) + TR ln (RT) - TR ln (P) + RT [/itex]

    On a side note, I understand your approach, but I'm concerned about what's wrong with my derivation rather than getting the actual answer. I don't mean to be rude and I greatly appreciate your help on that other way to make the derivation, but I'm afraid I might have some wrong concepts and that's why what I've done is wrong.

    Thanks :)
  5. Nov 27, 2012 #4
    There are some sign errors in your derivation. Watch the algebra. Also, you can use Cp = Cv + R

    to make the final result more concise.

    I might also mention that the chemical potential is usually regarded as the Gibbs free energy per mole. You also need to understand that U0 and S0 are extensive properties, and thus are proportional to the number of moles. Thus, they can be expressed as U0 = n u0 and S0 = n s0, where u0 is the internal energy per mole in some reference state. With these substitutions, you can divide G by n and get μ.
    Last edited: Nov 27, 2012
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