1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Particle distribution in a box

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data
    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) ??

  2. jcsd
  3. May 13, 2009 #2


    User Avatar
    Homework Helper

    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)
  4. May 14, 2009 #3
    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!)
  5. May 14, 2009 #4


    User Avatar
    Homework Helper

    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)?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook