How to Distribute Gas Atoms in Two Boxes?

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Homework Help Overview

The problem involves distributing N gas atoms between two boxes with different volumes, specifically box A with volume V and box B with volume 3V. The discussion centers around calculating the number of ways to choose atoms for each box, determining the total number of microstates, and finding the most likely distribution of atoms based on these calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the number of ways to distribute N atoms between the two boxes, with some confirming the calculation of 2^N for binary choices. Questions arise regarding how to calculate the number of microstates and the implications of volume on these calculations.

Discussion Status

The discussion is ongoing, with participants providing insights into the calculations for microstates and questioning the assumptions about volume weighting. There is a collaborative effort to clarify the relationship between the number of atoms in each box and the corresponding microstates.

Contextual Notes

Participants are navigating the constraints of the problem, including the assumption of low density and perfect gas behavior, as well as the need for further clarification on how to approach the calculation of microstates based on the distribution of atoms.

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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