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Homework Help: Equation of state of a hard sphere gas

  1. Sep 2, 2008 #1


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    1. The problem statement, all variables and given/known data
    I am trying to solve problem 1.4 from Statistical Mechanics by R.K. Pathria 2nd edition. This is the problem:
    In a classical gas of hard spheres (of diameter [tex]\sigma[/tex]), the spatial distribution of the particles is no longer uncorrelated. Rougly speaking, the presence of [tex]n[/tex] particles in the system leaves only a volume [tex](V-nv_0)[/tex] available for the (n+1)th particle; clearly, [tex]v_0[/tex] would be proportional to [tex]\sigma^3[/tex]. Assuming that [tex]Nv_0 \ll V[/tex], determine the dependence of [tex]\Omega(N, V, E)[/tex] on V {For an ideal gas this would be [tex]\Omega \propto V^N[/tex]} and show that, as a result of this, V in the gas law (PV=nRT) gets replaced by (V-b), where b is four times the actual space occupied by the particles.

    2. The attempt at a solution
    I first tried:
    [tex]\Omega(N, E, V) \propto (V-Nv_0)^N[/tex]
    and then:
    \frac{P}{T} = \left( \frac{\partial S}{\partial V} \right)_{N,E} = \left( \frac{\partial}{\partial V} k_B \ln \Omega(N, E, V) \right)_{N,E} = k_B \frac{N}{(V-Nv_0)}
    rearranging yields
    P(V-Nv_0) = k_B N T
    which looks a lot like what I need to prove, however I did not prove the factor 4.

    A second guess was more like a hand waving argument. Suppose two hard spheres of diameter [tex]\sigma[/tex] in close contact. Together they occupy a space twice the volume of a sphere of diameter [tex]\sigma[/tex]:

    \frac{4}{3} \pi \left( \frac{\sigma}{2} \right)^3 = \frac{1}{3} \pi \sigma^3
    but they exclude a volume of a sphere of diameter [tex]2 \sigma[/tex]:
    \frac{4}{3} \pi \sigma^3
    From here we find the factor 4

    Third try:
    Probably the best thing to do is to assume the following:
    \Omega \propto \prod_{i=0}^{N-1} (V-iv_0)
    and then continue from thereon, but I don't know how to do this properly.

    Any help would be appreciated greatly.
  2. jcsd
  3. Nov 1, 2014 #2
    Hey can you please explain more elaborately how you got the factor of 4.
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