# Particle distribution in a box

• villiami
For part 2, it says each atom is allocated a spatial state. In practice, these would be quantum states, but if you like you can imagine that the volume V is divided into little "miniboxes" of a constant size, say U (or whatever you want to call it), and that each atom goes into a minibox. The total number of miniboxes in box A is V/U - you can give this number a label, say C (again, call it whatever you want). Now, you have n atoms in box A which have to be distributed into the C miniboxes. Each way of distributing them is one microstate of box A. So

## Homework Statement

N atoms are in two boxes, box A: volume=V, and box B: volume=3V. (Assume a atoms make up a low density perfect gas).

1) How many ways can n be chosen, where n=(number of atoms in box A)?

2) Each atom if allocated a spatial state. Assuming that the number of available states within a volume V is proportional to that volume, write down an expression for the total number of microstates (Ω) present in the macrostate specified by n.

3) By optimising Ω determine the most likely number of gas atoms present in A and B. What are these values when N = 100?

4) Find the probability of the most likely distribution.

The attempt at a solution
1) 2^N right? (N binary choices)
2) ??
3) Clearly 25 in A, 75 in B. But need to answer 2) to derive it their way.
4) ??

Thanks,
Villiami

First of all, #1: yes there are 2^N ways of dividing the N atoms up between the two boxes. But how many of those ways put n of the atoms in box A? (Hint: it's not all of them)

diazona said:
First of all, #1: yes there are 2^N ways of dividing the N atoms up between the two boxes. But how many of those ways put n of the atoms in box A? (Hint: it's not all of them)

So #microstates(A) = N! / n!(N-n)! [Putting n in A]
but isn't num(B) = N! / (N-n)!n! as well? [putting N-n into B]

I still don't see how to weight them according to volumes..
(thanks for the help so far!)

villiami said:
So #microstates(A) = N! / n!(N-n)! [Putting n in A]
but isn't num(B) = N! / (N-n)!n! as well? [putting N-n into B]
Yep, that's exactly right... these numbers are not weighted by volume. Once you pick a particular set of particles to put in box A, all the rest have to go in box B - you don't have any choice left. So each way of putting n particles in box A corresponds to exactly one way of putting N-n particles in box B. Now do you see why you'd expect those two results to be the same?

For part 2, it says each atom is allocated a spatial state. In practice, these would be quantum states, but if you like you can imagine that the volume V is divided into little "miniboxes" of a constant size, say U (or whatever you want to call it), and that each atom goes into a minibox. The total number of miniboxes in box A is V/U - you can give this number a label, say C (again, call it whatever you want). Now, you have n atoms in box A which have to be distributed into the C miniboxes. Each way of distributing them is one microstate of box A. So how many microstates are there for box A alone? What about for box B alone? And once you've found those two numbers, what's the total number of microstates for the entire system (A+B)?