posted by esmat on .
. Consider a large number, N(0)
, molecules contained in a volume V
. 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
and N ¿ N(0)
, the function P (V; N) assumes a
Probability for one molecule to be in that region is
q = V/V0
N0!/(N! (N0-N)!) q^N (1-q)^(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
N0!/(N! (N0-N)!) 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 = 1-q
So, we have:
N' = f(q,1-q)
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)^(N0-1)
We thus have:
N' = f(q,1-q) = N0 q
The standard deviation can be evaluated in a similar way.
<N^2> = g(q,1-q)
g(q,r) = q d/dq q d/dq (q+r)^N0 =
N0 q d/dq q (q+r)^(N0-1) =
N0 q (q+r)^(N0-1) +
N0 q^2 (N0-1)(q+r)^(N0-2)
<N^2> = N0 q + N0(N0-1)q^2
<N^2> - <N>^2 =
N0 q + N0(N0-1)q^2 - N0^2 q^2 =
N0 q - N0 q^2 =
N0 q (1-q)
So, the standard deviation is:
sqrt[No q (1-q)]
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.