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Thermodyamics - Chemical Potential

  1. Nov 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Given [itex]dE=TdS-PdV+\mu dN[/itex], [itex]PV=NT[/itex] and [itex]Cv=\frac{3}{2}N[/itex]

    Find E

    2. Relevant equations
    [itex]dE=TdS-PdV+\mu dN[/itex], [itex]PV=NT[/itex] and [itex]Cv=\frac{3}{2}N[/itex]

    3. The attempt at a solution

    That part that is confusing me is the chemical potential, not sure what to do with it. Finding the energy for the ideal gas is pretty simple without it

    [itex]dE=TdS-PdV+\mu dN[/itex]
    [itex]dE=T\frac{\partial S}{\partial T}dT+(T\frac{\partial S}{\partial v}|_T-P)dV+\mu dN[/itex] expanding dS
    [itex]dE=C_v dT+(T\frac{\partial P}{\partial T}|_V-P)dV+\mu dN[/itex] by maxwell relation
    [itex]dE=C_v dT+(T\frac{N}{V}-P)dV+\mu dN[/itex] by taking derivative of ideal gas law
    [itex]dE=C_v dT+(P-P)dV+\mu dN[/itex]
    [itex]dE=\frac{3}{2}NdT+\mu dN[/itex]
    [itex]E=\int\frac{3}{2}NdT+\int\mu dN[/itex]

    Without the chemical potential its simply [itex]E=\frac{3}{2}NT[/itex] using [itex]E(t=0)=0[/itex] and pretty much no source I've looked at makes reference to the chemical potential when calculating the internal energy so I'm confused.

    Here are some comments by the person who set homework when someone asked him about N being constant
     
  2. jcsd
  3. Nov 7, 2014 #2
    What happened to the gas constant R in your equations?

    Chet
     
  4. Nov 8, 2014 #3
    It should be k for number of particles N, I guess the problem is in units where k=1
     
  5. Nov 8, 2014 #4

    BvU

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    Never heard of such a system of units. Help me there ... :)

    You make use of the ideal gas law. Is there something missing in the problem statement ? Like: "Find E for an ideal gas" instead of "find E" ?
     
  6. Nov 8, 2014 #5
    Please do BvU and me a favor and put the k back in. The equations, as they stand, hurt our eyes to look at them.

    If you change the number of molecules of the gas N at constant T and P, how does that affect the volume per molecule v, the internal energy per molecule e, and the entropy per molecule s?

    Chet
     
  7. Nov 8, 2014 #6
    http://en.wikipedia.org/wiki/Natural_units Pretty much all those unit conventions set k=1

    Well the equations of state are for an ideal gas so yes I am finding E for an ideal gas
     
  8. Nov 8, 2014 #7
    So from the ideal gas law with T and P constnat [itex]V \propto N[/itex] so Volume per molecule [itex]V/N \propto 1[/itex] so its a constant. I don't have a formula for the internal energy so i can't answer how changing N affects it, do I not need to know something about [itex]\mu[/itex]?
     
  9. Nov 8, 2014 #8
    If you have a container with N molecules of a gas at temperature T and pressure P (and internal energy E) and you have a second container with 2N molecules of exactly the same gas at the same temperature and pressure (i.e., twice the volume), what is the internal energy of the gas in the second container? How do the entropies of the gases in the two containers compare?

    Chet
     
  10. Nov 8, 2014 #9
    By extensivity of E it's 2E, similarly it's 2S
     
  11. Nov 8, 2014 #10
    Excellent. So we can write:

    V = N v(T,P)
    E = N e(T,P)
    S = N s(T,P)

    and thus,

    dV = N dv + v dN
    dE = N de + e dN
    dS = N ds + s dN

    Now, see what happens when you substitute these relations into your original equation for dE. (Make sure you collect terms that multiply dN, and also collect terms that multiply N).

    Chet
     
  12. Nov 8, 2014 #11
    I get [itex]N(de-Tds+Pdv)=(-e+Ts-Pv+\mu)dN[/itex]

    Still unsure what to do, there is something key I'm missing here...
     
  13. Nov 8, 2014 #12
    This equation contains a wealth of important information.

    First of all, since N and dN are arbitrary and independent of one another, the two terms in parenthesis in the equation must be equal to zero:
    [tex]\mu=e-Ts+Pv=g(T,P)=G/N[/tex]
    [tex]de=Tds-Pdv[/tex]

    The first relationship tells us that the chemical potential is equal to the Gibbs free energy per molecule g=G/N.

    The second equation is our basic first/second law equation, but without the chemical potential term present now. It tells us, from your previous analysis, that
    [tex]e = \frac{3}{2}kT[/tex]
    So, irrespective of whether the chemical potential term is present in your original equation, we conclude that, for an ideal gas, the internal energy is given by
    [tex]E = \frac{3}{2}NkT[/tex]

    Chet
     
  14. Nov 9, 2014 #13
    Sorry from my original analysis I end up with [itex]de=T\frac{\partial s}{\partial T}|_vdT[/itex]

    with s=S/N

    [itex]ds=\frac{\partial s}{\partial S}dS+\frac{\partial s}{\partial N}dN[/itex]

    [itex]\frac{\partial s}{\partial T}|_v=\frac{1}{N}\frac{\partial S}{\partial T}|_v-\frac{S}{N^2}\frac{\partial N}{\partial T}|_v[/itex]

    How do I show the term on the right is zero?
     
  15. Nov 9, 2014 #14
    Yes. This equation is correct. You also have that:
    [tex]\frac{\partial s}{\partial T}|_v=\frac{C_v}{T}[/tex]
    So,
    [itex]de=C_vdT[/itex]

    I wasn't able to make sense out of the rest of what you wrote.

    The key to this problem is recognizing that, for a single phase, single component, closed system, the chemical potential ##\mu## is not some arbotrary mysterious quantity, but, instead, is directly related to the system parameters E, S, T, P, V, and N by the equation:
    [tex]\mu=\frac{E+PV-TS}{N}[/tex]
     
  16. Nov 9, 2014 #15

    No I have [tex]\frac{\partial S}{\partial T}|_v=\frac{C_v}{T}[/tex] where S=sN
     
  17. Nov 9, 2014 #16
    It should be:
    [tex]\frac{\partial S}{\partial T}|_v=N\frac{C_v}{T}[/tex]
    where Cv is the heat capacity per molecule (or mole, depending on which units you are using).

    Chet
     
  18. Nov 9, 2014 #17
    I am really lost, I said at the top [itex]C_v=\frac{3}{2}N [/itex]

    If I have [itex]de=C_vdT[/itex] which I still don't understand how you got from [itex]\frac{\partial s}{\partial T}|_v[/itex]

    then [itex]e=C_vT[/itex]
    [itex]e=\frac{3}{2}NT[/itex]
    which gives [itex]E=\frac{3}{2}N^2T[/itex]
     
  19. Nov 9, 2014 #18
    The notation is driving both of us crazy. Yiiiiii.

    Let's follow the convention that lower case symbols are per molecule, and upper case symbols are for extensive total. Then:

    [tex]c_v=\frac{3}{2}k[/tex]
    [tex]C_v=\frac{3}{2}Nk[/tex]
    [tex]de=c_vdT=\frac{3}{2}kdT[/tex]
    [tex]dE=C_vdT=\frac{3}{2}NkdT[/tex]

    Chet
     
  20. Nov 9, 2014 #19
    Ok thats fine but what I have is [itex]de=T\frac{\partial s}{\partial T}|_vdT[/itex] and I was given [itex]T\frac{\partial S}{\partial T}|_v=C_v=\frac{3}{2}N[/itex]

    So I need to relate [itex]\frac{\partial s}{\partial T}|_v[/itex] and [itex]\frac{\partial S}{\partial T}|_v[/itex]
    which I tried to do by expanding ds(S,N) above and I didn't simply get [itex]\frac{\partial s}{\partial T}|_v=\frac{1}{N}\frac{\partial S}{\partial T}|_v[/itex] like you are saying, there is another term with [itex]\frac{\partial S}{\partial N}|_S\frac{\partial N}{\partial T}|_v[/itex] is this term simply zero?(This is in post 13)
     
    Last edited: Nov 9, 2014
  21. Nov 9, 2014 #20
    Hi Decerto,

    I need to apologize. I just haven't been able to figure out how to explain this in a way that works for you. Maybe someone with a background closer to yours could do better. Please don't take this personally and please don't give up. But I'm going to have to withdraw from contributing to this thread.

    Chet
     
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