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Partition Function Help Please

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data

    An ideal monatomic gas at the temperature T is confined in a spherical container of radius R. There are N molecules of mass m in the gas. Molecules move in a spherically symmetric potential V(r) where r is the distance from the center of the container. The potential V(r) is given by a piecewise function:
    V(r) = -Uo*r/R for r<Ro
    = 0 for Ro<r<R
    = ∞ for r>R

    U0 is a positive constant (i.e., “attractive well”). Thus, there is a potential well (a trap) of radius R0 inside the container. Notice also that the interaction between the molecule and the cavity wall is hard-core like.

    a) Calculate the partition function Z.

    b) What is the probability distribution function (probability per unit volume), P(r), that a particle is located at distance r from the center of spherical container?

    c) What is the probability, Ptrap(r), that a particle is located inside the potential well (i.e. at a distance r smaller than R0 from the center of the spherical container).

    d) What is the average number of molecules inside the trap, Ntrap?

    e) Calculate the internal energy E=<H(r,p)> and the heat capacity Cv of this gas.

    f) Calculate the pressure p exerted by the molecules on the wall of the spherical container.

    2. Relevant equations

    H = p^2/(2*m) + V(r)

    3. The attempt at a solution

    ℤ =1/(N!*h^(3N)) ∫ exp(-β*p^2/(2*m)∫exp(-β*V(r))

    I get stuck with the integration of the potential term. I don't know if you would need to integrate each part of the piecewise separately?


    If someone could help me out with this, I would be extremely grateful!
     
    Last edited: Dec 10, 2014
  2. jcsd
  3. Dec 11, 2014 #2

    stevendaryl

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    Let's simplify to the case [itex]N=1[/itex]. Then [itex]Z = \frac{1}{h^3} \int d^3 p \int d^3 r e^{-\beta (\frac{p^2}{2m} + V(r))}[/itex]. The integral can be factored as:

    [itex]Z = \frac{1}{h^3} (\int d^3 p e^{-\beta \frac{p^2}{2m}}) (\int d^3 r e^{-\beta V(r)})[/itex].
     
  4. Dec 11, 2014 #3
    I don't understand the ∞ in the piecewise. I feel like that would just cause the whole thing to go to ∞.

    I don't know how to generate equations in this forum, so I will just try to explain my thought process.

    The container is spherical, thus need to use spherical coordinates for V(r).

    The first integral involving the momentum is well known, so that doesn't need to be "solved" by me. The second integral (the piecewise) needs to be separated into the three pieces of the function. The first going 0<r<Ro, which is the region closest to the center of the sphere (the trap), the second pieces has Ro<r<R, which goes from the potential trap to the edge of the sphere, and then R<r<∞. But, putting an exp(∞) into the integral causes it to go to infinity, and adding infinity to anything is causing the whole thing to go to ∞... What am I missing here?
     
  5. Dec 11, 2014 #4

    stevendaryl

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    Well, if [itex]V(r) \rightarrow \infty[/itex], then [itex]e^{-\beta V(r)} \rightarrow 0[/itex]. So that means that the integral over [itex]d^3 r[/itex] only needs to cover the region where [itex]V(r) < \infty[/itex]
     
  6. Dec 11, 2014 #5
    Ahh, I was forgetting the (-) in front of the ∞. That would explain a lot! This is still a huge ℤ, so I thought that I was doing it wrong but it sounds like I was on the right track. Thanks!
     
  7. Dec 11, 2014 #6
    One more quick question: Would there be a V^N term in front of everything? I know that in the classical monatomic gas solution, there is. And , considering that the pressure is a partial wrt V, I assume it needs to be there again?
     
  8. Dec 11, 2014 #7

    stevendaryl

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    No. The [itex]V^N[/itex] term is from integrating [itex]d^3 r[/itex] (there is one such integral for each particle). If you integrate [itex]d^3 r[/itex] over a region, you get the volume of that region. But in this case, you're not integrating [itex]d^3 r[/itex], you're integrating [itex]e^{-\beta V(r)} d^3 r[/itex]. That's not going to give you the volume. (We have two different [itex]V[/itex]s here: the volume and the potential).
     
  9. Dec 11, 2014 #8
    Hmm. Then I don't get how to determine the pressure without having a volume term?
     
  10. Dec 11, 2014 #9

    stevendaryl

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    The integral [itex]\int e^{-\beta V(r)} d^3 r[/itex] will involve [itex]R[/itex]. [itex]R[/itex] is related to the volume [itex]V[/itex] through [itex]V = \frac{4}{3} \pi R^3[/itex], or [itex]R = (\frac{3}{4\pi} V)^{\frac{1}{3}}[/itex]
     
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