. 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.
Probability for one molecule to be in that region is
q = V/V0
therefore:
P(V,N) =
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
Q(V,N) =
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)
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)^(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)
with
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)
So,
<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.
To answer these questions, we need to use concepts from statistical mechanics and probability theory. Let's go through each part step by step:
(a) To calculate the probability P(V; N) that an arbitrary region of volume V contains exactly N molecules, we can consider the probability of finding a single molecule in that region and then apply it to N molecules. Since there is no correlation between the locations of the molecules, each molecule has an equal probability of being in any part of the volume V(0).
Let's denote the total volume as V(0) and the volume of the arbitrary region as V. So, the probability of finding a single molecule in V is given by P = V/V(0).
Now, for N molecules, each molecule is independent, so the probability that exactly N molecules are found in V is calculated by multiplying the probabilities of finding a molecule in V for each molecule. Since each molecule has the same probability, we get P(V; N) = (V/V(0))^N.
(b) To calculate the average value N¹ and the standard deviation of N, we need to use the probability distribution function P(V; N) that we derived in part (a). The average value N¹ is the weighted sum of all possible values of N multiplied by their probabilities.
N¹ = Σ(N * P(V; N))
We need to sum over all possible values of N. The standard deviation σ of N is given by:
σ^2 = Σ((N - N¹)^2 * P(V; N))
Again, we sum over all possible values of N.
(c) To 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, we can use the concept of the large number limit and the Central Limit Theorem.
In the large number limit, when both V and V(0) - V are large, the probability distribution P(V; N) becomes continuous and approaches a Gaussian distribution. The Gaussian distribution is characterized by its mean value N¹ and its standard deviation σ.
(d) To show that if both V ≥ V(0) and N ≥ N(0), the function P(V; N) assumes a Poisson form, we can utilize the Poisson distribution. The Poisson distribution describes the probability of a given number of events occurring in a fixed interval when the events occur with a known average rate.
As V ≥ V(0) and N ≥ N(0), the molecules in the volume V become sufficiently spread out, and their distribution approaches the characteristics of a Poisson process. In this case, the probability distribution P(V; N) becomes consistent with the Poisson distribution.