Posted by esmat on .
. Consider a large number, N(0)
, molecules contained in a volume V
(0)
. Assume that
there is no correlation between the locations of the molecules (ideal gas). Do not
use the partition function in this problem.
(a) Calculate the probability P (V; N) that an arbitrary region of volume V contains
exactly N molecules.
(b) Calculate the average value N¹
and the standard deviation of N.
(c) Show that if both V and V
(0) ¡ V are large, the function P (V; N) assumes a
Gaussian form for N close to N¹
.
(d) Show that if both V ¿ V
(0)
and N ¿ N(0)
, the function P (V; N) assumes a
Poisson form.

physics 
Count Iblis,
Probability for one molecule to be in that region is
q = V/V0
therefore:
P(V,N) =
N0!/(N! (N0N)!) q^N (1q)^(N  N0)
The average number N' is q N0 because the probability q for each molecule to be in the region is independent. You can also find this by direct computation:
N' = Sum from N = 0 to N0 of N P(V,N)
To perform the summation, consider the function
Q(V,N) =
N0!/(N! (N0N)!) q^N r^(N  N0)
where q and r are considered to be independent variables. Then we can compute the summation by differentiating w.r.t. q while keeping r constant and then we put r = 1q
So, we have:
N' = f(q,1q)
where
f(q,r) = Sum from N = 0 to N0 of
q d/dq Q(V,N) =
q d/dq Sum from N = 0 to N0 of Q(V,N) =
q d/dq (q+r)^N0 =
N0 q (q+r)^(N01)
We thus have:
N' = f(q,1q) = N0 q
The standard deviation can be evaluated in a similar way.
<N^2> = g(q,1q)
with
g(q,r) = q d/dq q d/dq (q+r)^N0 =
N0 q d/dq q (q+r)^(N01) =
N0 q (q+r)^(N01) +
N0 q^2 (N01)(q+r)^(N02)
So,
<N^2> = N0 q + N0(N01)q^2
<N^2>  <N>^2 =
N0 q + N0(N01)q^2  N0^2 q^2 =
N0 q  N0 q^2 =
N0 q (1q)
So, the standard deviation is:
sqrt[No q (1q)]
To derive the Gaussian and Poisson approximation in the appropriate limits, you need to expand
Log(P) using the Stirling approximation for log(N!) for large N. This only involves trivial manipulations.