1. Not finding help here? Sign up for a free 30min 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!

Stat Mech

  1. Apr 20, 2009 #1
    What is the probability that at one instant, in one mole of an ideal gas, all [itex]N=N_A[/itex] molecules will be ofund in a volume [itex]\frac{V}{2}[/itex]? Express your answer as a power of 10?

    usually this type of question is just a simple combinatorics question but i don't know how to set it up. i.e. how do i make it "something choose something"

    i thought there was some information missing to be honest but it's a past exam question so i'm assuming it's legit...
     
  2. jcsd
  3. Apr 20, 2009 #2

    dx

    User Avatar
    Homework Helper
    Gold Member

    Divide the volume in your mind into half. How many ways can you distribute N particles into these two halfs? How many of these ways have all particles in one half? The ratio of these two numbers is the required probability. If you have one mole of gas, then the number of molecules N is the Avogadro number.
     
  4. Apr 20, 2009 #3
    at a guess there's N! ways of arranging N distinguishable objects
    only 1 of these will have all in one half so the answer is just N! yeah?
     
  5. Apr 20, 2009 #4

    dx

    User Avatar
    Homework Helper
    Gold Member

    No it's not N! ways. N! is the number of ways of ordering N distinguishable objects in a line. We don't want to order the particles, but just to assign to each of them either "left side" or "right side". How many ways are there of doing that?
     
  6. Apr 20, 2009 #5
    [itex]\binom{N}{n_R} \binom{N-n_R}{n_L}[/itex] but the second coeffeicient will just be one will it not as [itex]n_L=N-n_R[/itex]
     
  7. Apr 21, 2009 #6
    is this right? i just found the same question again and still haven't really grasped what's going on.
     
  8. Apr 22, 2009 #7

    dx

    User Avatar
    Homework Helper
    Gold Member

    There are two ways to assign and L or an R to particle one, i.e. just L and R. Once you assign a side to particle one, there are two ways to assign a side to particle two, so for each of the two possible choices for particle one, there are two ways for particle two. This gives four ways for both: LL, LR, RL, RR. Can you see how to extend this reasoning to N particles?
     
  9. Apr 22, 2009 #8
    2N possible combinations

    i want 1 particular one (i.e. all on one side)

    therefore [itex]\binom{2N}{1}=2N[/itex]
     
  10. Apr 22, 2009 #9

    dx

    User Avatar
    Homework Helper
    Gold Member

    Check your result for three particles:

    LLR
    LLL
    LRR
    LRL
    RLR
    RLL
    RRR
    RRL

    There are 8 ways. But your formula 2N gives 2(3) = 6. Read post #7 again carefully and try to extend the logic to three particles, and you'll see the pattern.
     
  11. Apr 22, 2009 #10
    ok 2^N then but i still only want to choose one of those combinations (i.e. all on one side) so i have 2^N choose 1=2^N ways of doing this.
     
  12. Apr 22, 2009 #11

    dx

    User Avatar
    Homework Helper
    Gold Member

    There is only one way to assign "left" to all of them: (LLLL ... L). Be careful about what the "choose" operation means. N choose k is the number of ways of choosing a subset of k elements from N. Here, we don't want the number of ways to choose one element from 2^N, we just want the number of ways in which all of the N particles are on one side. They can all be either on the left side, or all on the right, so there are just two ways of doing it. The required probability is therefore 2/(2^N).
     
  13. Apr 23, 2009 #12
    ok. i kind of follow that argument. but still a little hazy about why we can't use the choose argument - clearly it gives a high probability which is unphysical but surely there are 2^Npossible configurations and we want to choose 1 of them?
     
  14. Apr 23, 2009 #13

    dx

    User Avatar
    Homework Helper
    Gold Member

    We want to count how many of them have all particles on one side. The technical meaning of "N choose k" is "the number of ways of selecting a subset of k elements from a set of N elements." Are we looking for the number of ways of choosing 1 element from a set of 2^N elements? No! We just want to count how many of them satisfy our condition "all on one side".
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stat Mech
  1. Stat Mech (Replies: 7)

  2. Stat Mech Question (Replies: 0)

Loading...