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Finding the state functions from the fundamental equation

  1. Mar 25, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I know that theoretically when one has the fundamental equation of a system, one can find the state equations and totally solve the system (if I understood well, I could make the analogy in classical mechanics of having the Lagrangian gives you the equations of motion). However I'm not sure how to do so. There are many problems asking you to do this and I'm stuck on this.
    From Callen's book (1st edition, page 34): 1)Find the three equations of state for a system with the fundamental equation [itex]u=\left ( \frac {\theta }{R} \right )s^2+ \left ( \frac {R \theta }{v_0 ^2} \right )v^2[/itex].
    2)Express [itex]\mu[/itex] as a function of T and P.

    2. Relevant equations
    [itex]du=Tds-Pdv[/itex] where [itex]u=U/N[/itex] (N is the number of moles), [itex]s=S/N[/itex] and [itex]v=V/N[/itex].


    3. The attempt at a solution
    I'm not exactly sure what are the 3 equations of state here. Apparently [itex]T(s,v)[/itex], [itex]P(s,v)[/itex] and [itex]\mu (s,v)[/itex] where mu is the chemical potential.
    Also I don't know what are theta and R. Constants? Maybe R is the gas constant?
    Can someone could get me started? I'm guessing the problem is very easy, a simple matter of differentiation/integration only and rearanging terms, etc.
    Thanks a lot for any help.
     
  2. jcsd
  3. Mar 25, 2012 #2
    In my opinion, the solution goes as follows:
    [itex]u=\frac{U}{N}[/itex]
    [itex]s=\frac{S}{N}[/itex]
    [itex]v=\frac{V}{N}[/itex]

    If we express the number of moles with the number of particles [itex]n[/itex]

    [itex]n=N \:N_{A}[/itex]


    where [itex]N_{A}[/itex] is the Avogadro's number, then you have

    [itex]U=\frac{N_{A}}{n}\:(\frac{\theta}{R}S^{2}+\frac{R \theta}{v_{0}^{2}}V^{2})[/itex]

    Then you can use
    [itex]dU=TdS-pdV+\mu dn[/itex]

    I think that you should treat [itex]R[/itex] and [itex]\theta[/itex] as constants.
    Best,
    francesco
     
    Last edited: Mar 25, 2012
  4. Mar 25, 2012 #3

    fluidistic

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    Thanks a lot. I reached [itex]T=\frac{2N_A \theta S}{nR}[/itex], [itex]P=-\frac{2R\theta V N_A}{nv_0^2}[/itex] and [itex]\mu =0[/itex].
    I doubt this is right.
     
  5. Mar 25, 2012 #4
    Why do you obtain [itex]\mu=0[/itex] ?
    In my opinion
    [itex]\mu=\frac{\partial U}{\partial n}=-\frac{N_A}{n}U(S,V,n)[/itex] ; then you substitute V(T,P,n) and S(T,P,n); maybe I don't see a macroscopic error in my computation.

    EDIT: sorry, I had to edit because i confused n with N
     
  6. Mar 25, 2012 #5

    fluidistic

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    Here is how I reached my result:
    Since [itex]U=\frac{N_{A}}{n}\:(\frac{\theta}{R}S^{2}+\frac{R \theta}{v_{0}^{2}}V^{2})[/itex], [itex]dU= \left ( \frac{N_A }{n} \right ) \left [ \left ( \frac{\theta }{R} \right ) 2SdS + \left ( \frac{R \theta }{v_0 ^2} \right ) 2VdV \right ][/itex]. But as you said, [itex]dU=TdS-pdV+\mu dn[/itex]. So I look for matching terms. Since in my expression I haven't any "dn", I conclude that [itex]\mu =0[/itex].
     
  7. Mar 25, 2012 #6
    In my opinion one should treat n as an independent variable, like S and V; the reason why it is not usually done is that n is considered constant; the chemical potential takes into account the fact that it can change.
    So
    [itex]dU=\frac{\partial U}{\partial S}dS+\frac{\partial U}{\partial V}dV+\frac{\partial U}{\partial n} dn[/itex] so that
    [itex]\frac{\partial U}{\partial n}=\mu[/itex]
     
  8. Mar 25, 2012 #7

    fluidistic

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    Thank you immensely. :smile:
    I get that [itex]\mu =-\frac{N_A}{n^2} \left [ \left ( \frac{\theta }{R} \right ) S^2 + \left ( \frac{R \theta }{v_0 ^2} \right ) V^2 \right ][/itex].
     
  9. Mar 25, 2012 #8
    You are welcome :)
     
  10. Mar 25, 2012 #9

    fluidistic

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    2)[itex]\mu (T,P)=-\frac{T^2}{4N_A \theta R^2}-\frac{P^2v_0^2}{4N_AR\theta }[/itex].
     
  11. Mar 25, 2012 #10

    fluidistic

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    Hmm I am confused now. My book says [itex]dU=TdS-pdV+\mu dN[/itex], not [itex]dU=TdS-pdV+\mu dn[/itex]. Where N is the number of moles and n is the number of particles.
     
  12. Mar 26, 2012 #11
    It is a matter of definitions, in my opinion; the two [itex]\mu[/itex]'s are related by multiplication of [itex]N_A[/itex]
     
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