Diatomic Gas, Grand Canonical Ensemble

  • #1
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Homework Statement



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##.

Homework Equations





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:

Answers and Replies

  • #3
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part (b) anyone?
 
  • #4
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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.
 
  • #13
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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}##
 

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