1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Diatomic Gas, Grand Canonical Ensemble

  1. Mar 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Part (a): Using grand canonical distribution, show ideal gas law ##P = nkT## holds, where ##n = \frac{\overline{N}}{V}##.

    Part (b): Find chemical potential of diatomic classical ideal gas in terms of ##P## and ##T##. The rotational levels are excited, but not the vibrational. Mass of each molecule is ##m##, separation is ##r##.

    2. Relevant equations



    3. The attempt at a solution

    Part(a)

    [tex]Z = \sum_{N=0}^{\infty} e^{\beta \mu N} Z_N = \sum_{N=0}^{\infty}\frac{1}{N!} \left( \frac{Ve^{\beta \mu}}{\lambda_{th}^3}\right)[/tex]

    Where ##Z_N = \frac{1}{N!} (\frac{V}{\lambda_{th}^3})^N## and ##\beta = \frac{1}{kT}##.

    [tex] Z = exp \left( \frac{Ve^{\beta \mu}}{\lambda_{th}^3}\right)[/tex]

    Starting with Gibb's Entropy, where probability ##P_i = \frac{e^{-\beta(E_i - \mu N_i)}}{Z}##

    [tex]S = -k \sum_i P_i ln(P_i) [/tex]
    [tex] S = -k \sum_i P_i \left[ -\beta (E_i - \mu N_i) - ln Z\right] [/tex]
    [tex] S = \frac{U}{T} - \frac{1}{T}\mu N + k ln Z[/tex]

    Grand Potential is defined as ##\Phi_G = -kT ln (Z)##.

    [tex]\Phi_G = U - TS - \mu N[/tex]
    [tex]d\Phi_G = -SdT - pdV - Nd\mu[/tex]

    To prove the ideal gas law, we must find a relation between P, V and T. We do this using grand potential.

    Starting:
    [tex]P = -\left( \frac{\partial \Phi_G}{\partial V} \right)_{T,\mu}[/tex]
    [tex]N = -\left( \frac{\partial \Phi_G}{\partial \mu} \right)_{V,T}[/tex]

    Then by differentiating, I showed that ##PV = NkT##.

    Part(b)

    Attempt #1: Using grand partition function

    We must first find the rotational partition function. Energies are given by ##E_a = \frac{\hbar ^2}{2I}j(j+1)##.

    [tex]Z_{rot} = \sum (2j+1) e^{-\beta(E_j -\mu N)} [/tex]

    Converting sum into integral:

    [tex] \int_0^{\infty} (2j+1) e^{-j(j+1)\frac{\theta_{rot}}{T}} dj \int_0^{\infty} e^{-\beta \mu N} dN[/tex]
    [tex] Z_{rot} = \frac{2I}{\hbar \mu} (kT)^2[/tex]

    Together:
    [tex]Z = Z_{rot}Z_{trans} = \frac{2I}{\hbar \mu} (\frac{1}{\beta})^2 exp \left( \frac{Ve^{\beta \mu}}{\lambda_{th}^3}\right) [/tex]

    [tex] ln Z = ln(\frac{2I}{\hbar \mu}) - 2ln (\frac{1}{\beta}) + Z_1 e^{\beta \mu}[/tex]


    Gibbs Potential ##\Phi_G = -kt ln (Z) = -\frac{1}{\beta} ln (Z)##

    [tex]\Phi_G = \frac{1}{\beta} ln \frac{2I}{\hbar \mu} - \frac{1}{\beta} ln(\frac{1}{\beta}) + \frac{1}{\beta} Z_1 e^{\beta \mu}[/tex]

    Using ##N = -(\frac{\partial \Phi_G}{\partial \mu})_{V,T}##:

    [tex]N = -\frac{1}{\beta \mu} + Z_1 e^{\beta \mu}[/tex]


    Attempt #2: Using single-particle partition function

    [tex]Z = Z_{rot}Z_{trans}[/tex]

    Translational one-particle Partition Function:
    [tex]Z_{trans} = \int e^{-\beta E} g_{(\vec{k})} d^3\vec{k} = \frac{V}{\lambda_{th}^3}[/tex]

    Rotational one-particle Partition Function:
    [tex]Z_{rot} = \sum (2j+1)e^{-j(j+1)\frac{\theta_{rot}}{T}}[/tex]

    where ##\theta_{rot} = \frac{\hbar^2}{2Ik}##

    Converting sum into integral:

    [tex]Z_{rot} = \frac{T}{\theta_{rot}}[/tex]

    Together, the total partition function is:

    [tex]Z = Z_{rot}Z_{trans} = \frac{V}{\lambda_{th}^3}\frac{T}{\theta_{rot}}[/tex]

    To find ##\mu = -\left(\frac{\partial F}{\partial N}\right)_{T,V}##, how do I find ##F## from partition function ##Z##?
     
    Last edited: Mar 22, 2014
  2. jcsd
  3. Mar 23, 2014 #2
  4. Mar 24, 2014 #3
    part (b) anyone?
     
  5. Mar 25, 2014 #4
    Part (b), Attempt #1: If I assume that the rotational partition function is simply ##Z_{rot} = \sum (2j+1)e^{-j(j+1)\frac{\theta_{rot}}{T}}##, the chemical expression comes out much nicer.
     
  6. Apr 2, 2014 #5
    bump part (b) anyone?
     
  7. Apr 6, 2014 #6
    bump part (b) anyone?
     
  8. Apr 8, 2014 #7
  9. Apr 12, 2014 #8
    bumpp part (b) anyone?
     
  10. Apr 14, 2014 #9
    bumppp part (b) anyone?
     
  11. Apr 18, 2014 #10
    bump part (b)??
     
  12. Apr 21, 2014 #11
    bumpp part (b)?
     
  13. Apr 24, 2014 #12
    bump part (b)?
     
  14. Apr 24, 2014 #13
    I think I have figured it out. If total energy is ##U_{tot} = U_{trans} + U_{rot}##, the chemical potential can be accounted for either in ##U_{trans}## or ##U_{rot}## but not both.

    Bearing this in mind, the partition function is ##Z = Z_{trans}Z_{rot}## where the ##e^{\beta \mu N}## only appears once in the ##Z_{trans}## and not in the ##Z_{rot}##. Hence the final equation should be less a ##-\frac{1}{\beta \mu}## term:

    [tex]N = Z_1 e^{\beta \mu}[/tex]

    So this gives chemical potential as:

    [tex]\mu = kT ln\left(\frac{N}{Z_1}\right)[/tex]

    Which is the same expression when derived using ##\mu = \left(\frac{\partial F}{\partial N}\right)_{T,V}##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Diatomic Gas, Grand Canonical Ensemble
Loading...