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Thermodynamics, finding the fundamental equation of ideal gases

  1. Apr 6, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    An ideal monoatomic gas is characterized by the two equations [itex]PV=NRT[/itex] and [itex]U=\frac{3NRT}{2}[/itex] in which R is a constant.
    Find the fundamental equation corresponding to a monoatomic ideal gas.

    2. Relevant equations
    [itex]S=\left ( \frac{1}{T} \right ) U+\left ( \frac{P}{T} \right ) V- \left ( \frac{\mu }{T} \right ) N[/itex].


    3. The attempt at a solution
    I'm given 2 equations of state, one is missing in order to get the fundamental equation. In the entropy representation the variables of the fundamental equation are [itex]\frac{1}{T}[/itex], [itex]\frac{P}{T}[/itex] and [itex]\frac{\mu }{T}[/itex].
    The missing one is [itex]\frac{\mu }{T} (U,V,N)=\frac{\mu}{T} (u,v)[/itex] where u and v in lower script are the molar energy and volume respectively.
    So I have that [itex]\frac{1}{T}=\frac{3R}{2u}[/itex] and [itex]\frac{P}{T}=\frac{R}{v}[/itex].
    I use Gibbs-Duhem's relation [itex]d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )[/itex]. Replacing the variables in parenthesis and integrating both sides, I reach (like the book): [itex]\frac{\mu}{T}=-\frac{3}{2}R \ln \left ( \frac{u}{u_0} \right ) -R \ln \left ( \frac{v}{v_0} \right ) +\left ( \frac{\mu }{\mu _0} \right ) _0[/itex].
    When I replace this equation into the expression for S given above and using the fact that u=N/U, [itex]u_0=\frac{N}{U_0}[/itex], I reach that [itex]S =\frac{3U^2 R}{2N}+\frac{RV^2}{N}+ N \left [ \frac{3R}{2} \ln \left ( \frac{U_0}{U} \right ) +R\ln \left ( \frac{V_0}{V} \right ) - \left ( \frac{\mu }{T} \right ) _0 \right ][/itex]. Using some log property this simplifies to [itex]S=\frac{3U^2R}{2N} +\frac{RV^2}{N} +NR \ln \left [ \left ( \frac{U_0}{U} \right ) ^{3/2} \left ( \frac{V_0}{V} \right ) \right ] -N \left ( \frac{\mu }{\mu _0} \right ) _0[/itex].
    However this differs from the answer given in the book: [itex]S= \frac {5NR}{2}-N \left ( \frac {\mu }{T} \right ) _0 +NR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {N}{N_0} \right ) ^{-5/2} \right ][/itex].
    I do not see, for the life of me, what I've done wrong.
     
  2. jcsd
  3. Apr 6, 2012 #2

    I like Serena

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    Hi fluidistic.

    Let's start with your integration constant.
    It should be ##(\frac{\mu}{T})_0## instead of ##(\frac{\mu}{\mu_0})_0##.

    Then you write u=N/U, but that is not right.
    Since you defined u to be the molar volume, u=U/N.

    Perhaps you can redo it?
     
  4. Apr 7, 2012 #3

    fluidistic

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    Thanks a lot for helping me!
    You are absolutely right, I've redone everything with more care. I reach however [itex]S=\frac{5RN}{2}-N \left ( \frac{\mu}{T} \right ) _0+NR \ln \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \right ][/itex].
    Therefore it seems like I'm missing a term in the logarithm.
    It seems it comes from the integration of the Gibbs-Duhem relation: [itex]d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )[/itex]. It seems that when I integrate I should get an extra term like "[itex]-R \ln \left ( \frac{N}{N_0} \right )[/itex]. I don't really know how to get that.
     
  5. Apr 7, 2012 #4

    I like Serena

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    Maybe it's simpler than you think.
    What did you fill in for ##u_0##?
     
    Last edited: Apr 7, 2012
  6. Apr 7, 2012 #5

    fluidistic

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    Hmm I still don't see it. I took [itex]u_0=\frac{U_0}{N}[/itex].
     
  7. Apr 7, 2012 #6

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    Note that N is also a variable that changes.
    It has an initial value too, just like ##u_0## and ##U_0##.
     
  8. Apr 7, 2012 #7

    fluidistic

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    Yes I know this but when should I make use of this?
    Here are my steps:
    1)[itex]\frac{1}{T}=\frac{3NR}{2U}=\frac{3R}{2u}[/itex]
    2)[itex]\frac{P}{T}=\frac{NR}{V}=\frac{R}{v}[/itex]
    I use Gibbs-Duhem's relation [itex]d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )[/itex].
    3)[itex]ud \left ( \frac{1 }{T} \right ) =-\frac{3Rdu}{2u}[/itex]
    4)[itex]v d \left ( \frac{P }{T} \right ) =-R\frac{dv}{v}[/itex].
    Integrating the GD relations gives me [itex]\frac{\mu}{T}=-\frac{3}{2}R \ln \left ( \frac{u}{u_0} \right ) -R \ln \left ( \frac{v}{v_0} \right ) +\left ( \frac{\mu }{T} \right ) _0[/itex]. The book (Callen's page 52) reaches exactly the same result so far.
    He then says that inserting this into the Euler equation [itex]S=\left ( \frac{1}{T} \right ) U+\left ( \frac{P}{T} \right ) V- \left ( \frac{\mu }{T} \right ) N[/itex] one reaches [itex]S= \frac {5NR}{2}-N \left ( \frac {\mu }{T} \right ) _0 +NR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {N}{N_0} \right ) ^{-5/2} \right ][/itex].
    So my error is when I insert what I obtained into the Euler equation or it's in integrating GD relation (in which case the book would also be wrong) or it's in either of the 4 steps I denoted by 1), etc.?
    I really don't see it. :/
     
  9. Apr 8, 2012 #8

    I like Serena

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    I got to your book result substituting ##u_0=\frac{U_0}{N_0}## and ##v_0=\frac{V_0}{N_0}## in the integrated GD result.
     
  10. Apr 8, 2012 #9

    fluidistic

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    Ah!! That explains everything :biggrin: In my case those were variables, I didn't even notice. Now I reach the desired result... thank you very much for all.
     
  11. Apr 8, 2012 #10

    I like Serena

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    Just out of curiosity, were my previous hints too cryptic?
    It is not my intention to make a thread a cryptic puzzle...
    Perhaps you have an idea what kind of hint I might have given that would have helped you better?
     
  12. Apr 8, 2012 #11

    fluidistic

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    You mean post 6? No, not cryptic. In fact now I understand your "hint" that I should have used N_0 instead of N. I did not realize that with my definition of u_0, it wasn't even a constant.
    So in my case pointing out in post 6 that my definition of u_0 was wrong and that it's not even a constant would have made me understand where the problem lies. I did not realize it, because I already knew that N was a variable and I thought that was what you wanted me to know.
    But I'm maybe a "weird learner" :) Anyway, thank you very much for having went through all of this with me.
     
  13. Apr 8, 2012 #12

    I like Serena

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    Okay. Thanks.
     
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