How to Distribute Gas Atoms in Two Boxes?

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The discussion revolves around distributing N gas atoms between two boxes, A and B, with volumes V and 3V, respectively. It confirms that there are 2^N ways to divide the atoms, but emphasizes that the number of ways to choose n atoms for box A is given by the binomial coefficient N! / n!(N-n)!. The participants explore how to account for the volume difference when calculating microstates, noting that the number of spatial states in box A is proportional to its volume. They suggest dividing the volume into smaller units to determine the total microstates for both boxes. The conversation highlights the importance of understanding how to calculate microstates to derive the most likely distribution of atoms, which is identified as 25 in box A and 75 in box B when N equals 100.
villiami
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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
 
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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)?
 

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