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Homework Help: Mean collision time in a gas

  1. Dec 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Derive an expression for the mean collision time in a gas where the collision cross-section is σ and the number density is n.

    3. The attempt at a solution
    I'm just using my book to do this, and I can't get past the first bit...

    It says
    Consider a particular molecule moving at speed v with all other molecules in the gas stationary. In a time dt, the molecule sweeps out a volume σvdt, and if another molecule lies inside this volume, there will be a collision. With n molecules per unit volume, the probability of a collision in time dt is therefore nσvdt.

    I'm sure the quantity nσvdt is just the number of molecules in the volume it sweeps out in time dt, so how does this give a probability? And it can't be normalized can it?

    The remainder of the derivation is
    define P(t) as the probability of a molecule not colliding up to time t.
    Then P(t+dt)=P(t)+(dP/dt)dt
    However P(t+dt) is the probability of a molecule not colliding up to time t multiplied by not colliding in time dt, i.e
    P(t+dt)=P(t)(1-nσvdt)
    Then
    (1/P)dP/dt=-nσv
    P(t)=exp(-nσvt)
    The probability of not colliding up to time t, then colliding in the next dt is
    P(t)nσvdt=exp(-nσvt)nσvdt
    which is a normalized probability distribution with mean time 1/nσv which is the required result.

    This all makes sense assuming the first bit, but I just can't see why it is right, can anyone help please?
     
    Last edited: Dec 27, 2014
  2. jcsd
  3. Dec 27, 2014 #2

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    You see everything but the following, so far, so read it again, and think about what it says. And think about the definition of "number density."
     
  4. Dec 27, 2014 #3
    The number density is the number per unit volume, so we could write it as
    n=N/V where N is the total number and V is the volume, so the probability (according to my book) that a molecule has a collision in time dt is given by
    Nσvdt/V.

    Unfortunately that doesn't enlighten me in any way :(
     
  5. Dec 27, 2014 #4

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  6. Dec 27, 2014 #5
    Yup, but I'm multiplying it by a volume, namely σvdt, so it just gives a number, not a probability, right?
     
  7. Dec 27, 2014 #6

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  8. Dec 27, 2014 #7
    Hmm, the number that could collide in an infinitesimal amount of time? I don't think I can see what you're getting at, sorry...
     
  9. Dec 27, 2014 #8

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    I couldn't either. I didn't realize textbooks had become this bad. Sucked me under right along with you.
    Turn this expression into an equation; I'll do it since the book had me confused enough to try to make you follow its "mountain from a molehill" approach;
    dn = nσvdt .
     
  10. Dec 27, 2014 #9
    So nσvdt represents a differential number of molecule that our molecule moving with speed v could collide with in time dt, right? How does this translate into the probability of a molecule colliding in time dt?
     
  11. Dec 27, 2014 #10

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    Integrate this expression and see if it makes sense as far as probabilities go.
     
  12. Dec 27, 2014 #11
    Wait, is your n in dn the same as the n on the right hand side, because n is the number density, whilst dn is a differential number that will colide with our chosen molecule at speed v...

    Should we use dN=nσvdt?
     
  13. Dec 27, 2014 #12

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    Works for me. You do see where this is going?
     
  14. Dec 27, 2014 #13
    Nope, not sure where we're going.
     
  15. Dec 27, 2014 #14

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    See if I can get my brains unscrambled enough to make this work. Start over from the point where the book lost both of us.
    In time dt, the molecule will collide, or not collide with another molecule; the sum of probabilities for both outcomes is unity, 1, one. The number density n is the total number of molecules randomly distributed in the total volume. In some fraction of the total volume, say σvdt, the probability of finding a gas molecule is proportional to n times that volume fraction.
    Does this turn into a probability for you now?
     
  16. Dec 27, 2014 #15
    Well I suppose saying that
    nσvdt
    is proportional to the probability of a collision makes sense, however, in the step

    'however P(t+dt) is the probability of a molecule not colliding up to time t multiplied by not colliding in time dt, i.e
    P(t+dt)=P(t)(1-nσvdt)'

    we seem to require that it actually equals (i.e isn't just proportional to) this in order to say 1-nσvdt is the probability of not colliding in time dt.

    Page 9 here http://www.damtp.cam.ac.uk/user/tong/kintheory/kt.pdf derives the same thing but just without finding the form for the probability of collision in time dt, which might be useful in terms of understanding what's going on (although I haven't got anywhere with it).
     
    Last edited: Dec 27, 2014
  17. Dec 27, 2014 #16

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    I thought I had deleted "proportional" before I posted that last reply. Does it make sense to you without using the word "proportional?"
     
  18. Dec 27, 2014 #17
    No, sorry.

    So we're saying the probability of a chosen molecule colliding with other molecules in time dt = volume fraction in time dt * total number of molecules in volume?

    I can't even see why this can't exceed 1 for a start... The fraction of the volume is cleary between 0 and 1, but N is going to be huge because its the number of molecules...
     
  19. Dec 27, 2014 #18

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    Volume swept by one molecule (σvdt) over total volume, that fraction times total number of molecules in the total volume.
     
  20. Dec 27, 2014 #19
    Yep, bit ambiguous sorry, but I did mean total volume, so

    probability of a chosen molecule colliding with other molecules in time dt = volume fraction in time dt * total number of molecules in total volume

    So why must this be between zero and one?
     
  21. Dec 27, 2014 #20

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    You are absolutely correct --- it could hit every other molecule in the box Avogadro's number of times squared.
    You, take a break. I'm going to take a break. My preference is to make your book work for you rather than to subvert the course synopsis. Just need a couple hours to twist my mind around their way of thinking.
     
  22. Dec 27, 2014 #21
    Ha, thanks for persisting.
     
    Last edited: Dec 27, 2014
  23. Dec 27, 2014 #22
    So if dt and so vdt is small enough that our chosen molecule could only possibly collide with a maximum of one other molecule in this time, nσvdt would give us the probability of collision in time dt then. However, we can't just assume this about dt can we.
     
  24. Dec 27, 2014 #23

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    To continue, there's no obvious statement that the "scale" of "number density" is changing back and forth as the discussion proceeds, and this may be causing the difficulty. When number density is given as "n" for the problem, it's referring to a large scale, number of molecules in a syringe barrel, room, breath of air, party balloon, bubble rising in an aquarium, and is not measurably different for random samples of similar sizes. When number density is used to count collisions of one molecule moving through a volume of space σvdt, the variations in number density from one volume to another of equal size can be enormous; one volume could be completely unoccupied (nlocal,1 = 0), and another double or triple the large scale or macroscopic average (nlocal,2 = 3n). All that can be said is that the sum of the local number densities times their populations divided by the total number of local sample volumes will average to be "n" the large scale density. Then when looking at the number of collisions of a single molecule in the volume σvdt, all that is known is the overall probability of the actual number density the molecule experiences in that volume, or n.
     
  25. Dec 27, 2014 #24
    Yup, I see this. So I guess we just assume our macroscopic number density is valid on microscopic scales here.

    However, we're still not really any closer to understanding why nσvdt is the probability of a collision in time dt are we?
     
  26. Dec 28, 2014 #25

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    Toss a coin a million times, and it should come up half and half heads and tails. Probability of either for a single toss is half and half. You're not going to call it more than half the time.
    Sweep a volume of nσvdt a million times and you'll get one million times nσvdt collisions; sweep it once, and maybe you get nσvdt collisions, maybe you get more, maybe you get fewer.
     
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