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Chemical potential and fixed number of particles

  1. Aug 22, 2013 #1
    Can we talk about the chemical potential of a system with fixed number of particles? Is this physically meaningful? Why/why not?

    P.s: I know that chemical potential is the partial derivative of free energy with respect to number of particles. But in the formulation of grand canonical ensemble, we write N=Ʃf (for example f would be fermi-dirac distribution function) and can't we fix N in this formula, and solve chemical potential μ by changing energy ε?
     
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  3. Aug 22, 2013 #2

    DrClaude

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    Staff: Mentor

    Yes, for the same reason you can talk about the temperature of a system with a fixed entropy or the pressure of a system with a fixed volume.
     
  4. Aug 22, 2013 #3

    jfizzix

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    You can indeed solve for the chemical potential [itex]\mu(N,V,T)[/itex] in the way you suggest.

    Consider the ideal gas as an example. Here, the occupancy comes from the classical maxwell-Boltzmann statistics [itex]f_{MB}(\epsilon) = e^{-(\frac{\epsilon-\mu}{k_{B}T})}[/itex].

    The number of particles N can be expressed as

    [itex]N=\sum_{j}g_{j}f_{MB}(\epsilon_{j})[/itex]
    where [itex]g_{j}[/itex] is the number of states at energy level [itex]\epsilon_{j}[/itex]

    The partition function for a single particle [itex]Z_{1}[/itex] is

    [itex]Z_{1} = \sum_{j}g_{j} e^{-(\frac{\epsilon_{j}}{k_{B}T})}=\eta_{q}V : \eta_{q}=(\frac{m k_{B}T}{2 \pi \hbar^{2}})^{\frac{3}{2}}[/itex]

    Here [itex]\eta_{q}[/itex] is a characteristic quantum concentration (you can see it has dimensions of [itex]\frac{N}{V}[/itex]). It's a large concentration indicating when quantum effects (bose/Fermi statistics) must be taken into account. For ideal gases, the real concentration [itex]\eta=\frac{N}{V}[/itex] is much less than [itex]\eta_{q}[/itex].

    We can relate the total particle number [itex]N[/itex] to the partition function [itex]Z_{1}[/itex], giving us

    [itex]N=Z_{1}e^{-\frac{\mu}{k_{B}T}}[/itex]

    We can solve for [itex]\mu[/itex] to find

    [itex]\mu =k_{B}T \ln (\frac{N}{Z_{1}}) [/itex]

    Then, substituting out expression for [itex]Z_{1}[/itex], we arrive at the final result

    [itex]\mu =k_{B}T \ln (\frac{\eta}{\eta_{q}}) :\eta=\frac{N}{V}[/itex].

    It's a total non-sequitur, but it seemed worth explaining since I had already written lecture notes on the subject.
     
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