# Equation of state of a hard sphere gas

1. Sep 2, 2008

### apj

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 $$\sigma$$), the spatial distribution of the particles is no longer uncorrelated. Rougly speaking, the presence of $$n$$ particles in the system leaves only a volume $$(V-nv_0)$$ available for the (n+1)th particle; clearly, $$v_0$$ would be proportional to $$\sigma^3$$. Assuming that $$Nv_0 \ll V$$, determine the dependence of $$\Omega(N, V, E)$$ on V {For an ideal gas this would be $$\Omega \propto V^N$$} 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:
$$\Omega(N, E, V) \propto (V-Nv_0)^N$$
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 $$\sigma$$ in close contact. Together they occupy a space twice the volume of a sphere of diameter $$\sigma$$:

$$\frac{4}{3} \pi \left( \frac{\sigma}{2} \right)^3 = \frac{1}{3} \pi \sigma^3$$,
but they exclude a volume of a sphere of diameter $$2 \sigma$$:
$$\frac{4}{3} \pi \sigma^3$$
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. Nov 1, 2014

### CassiopeiaA

Hey can you please explain more elaborately how you got the factor of 4.