# Marble like atoms in a jar

1. Jan 19, 2012

### jfy4

1. The problem statement, all variables and given/known data
In a classical gas of hard spheres (of diameter D), the spatial distribution of the particles is no
longer uncorrelated. Roughly 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 $D^3$. Assuming that $Nv_0 \ll V$, determine the dependence of $\Omega(N,V,E)$ on V (compare to equation (1.4.1)) and show that, as a result of this, V in the ideal-gas law (1.4.3) gets replaced by (V − b), where b is four times the actual volume occupied by the particles.

2. Relevant equations
(1.4.1) $\Omega (N,V,E) \propto V^N$
(1.4.3) $PV=NkT=nRT$

3. The attempt at a solution
Well I took
$$(V-Nv_0)^N \approx V^N - N^2 v_0 V^{N-1}+....$$
Then I tried to do something similar as to the construction of the ideal gas law by trying
$$\frac{P}{T}=k\frac{\partial \ln (\Omega)}{\partial V}\frac{\partial \Omega}{\partial V}$$
assuming $\Omega \propto V^N - N^2 v_0 V^{N-1}$ similar to the original derivation.

Thanks,

2. Jan 22, 2012

### gomboc

You got to $\Omega \propto (V-Nv_0)^N$, but when you think about it, this doesn't properly represent the situation - this actually represents putting a single particle into the remaining volume of the jar N times.

What you have to do is put one particle in, and then progressively add particles, one by one, with each successive particle having less and less space to occupy. Thus, the expression you need to begin with is $$\Omega \propto \prod_{n=0}^{N-1} (V-nv_0)$$

You can then use a logarithm to turn the product into a sum, make any needed simplifying assumptions, do all necessary manipulations to express the sum algebraically, which then lets you get a simple expression for $\Omega$.