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Pressure from Grand Partition

  1. Apr 11, 2014 #1
    Can someone take a look at picture and show me how to derive the pressure from the grand partition function ?
     

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  3. Apr 12, 2014 #2
    Grand potential ##\Phi_G = -PV##
    Grand potential ##Z = e^{-\beta \Phi_G}##

    Thus, ##PV = kT ln Z##
     
  4. Apr 12, 2014 #3
    Unscietific,

    I cant use the fact that the grand potential equals -PV because my goal is to prove that the grand potential in terms of the partition function is equivalent to (-PV).

    I know that those sums on the left side must equal (PV/KT) but I don't know the details of how to show it.
     
  5. Apr 13, 2014 #4
    The grand partition function is sum of all states ##Z_G = \sum_i e^{\beta(\mu N_i - E_i)}## and Probability is i-th state over all possible states: ##P_i = \frac{e^{\beta(\mu N_i - E_i)}}{Z_G}##.
    [tex]S = -k\sum_i P_i ln P_i = \frac{U - \mu N + kT ln (Z_G)}{T}[/tex]
    Rearranging,
    [tex]-kT ln (Z_G) = \Phi_G = U - TS - \mu N = F - \mu N[/tex]

    Now we must find ##F## and ##\mu##.

    Starting with partition function of an ideal gas: ##Z_N = \frac{1}{N!}(\frac{V}{\lambda_{th}^3})^N##, what is ##F##?

    Using the below relation, how do you find ##\mu##?

    [tex]dF = -pdV - SdT + \mu dN[/tex]



    Putting these together, can you find ##\Phi_G##?
     
    Last edited: Apr 13, 2014
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