apj
Sep2-08, 08:05 AM
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.
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.