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Molecule collisions in an Ideal Gas

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Molecules in an ideal gas collide with each other at random times. The probability distribution governing the time between collisions is P(t) = Ae^(-bt).
    (a) Find the value of A so that P(t) is correctly normalized.
    (b) Find the average time between collisions, t. This time is traditionally called tau. Now re-write P(t) in terms of tau, without the original parameters A and b.
    (c) Find the standard deviation of the collision times, σ_t.


    2. Relevant equations
    ∫(x^n)e^(-x/a)dx = n!a^(n+1) from 0 to ∞


    3. The attempt at a solution
    I believe I have the right answer to (a), I normalized it to obtain P(t) = -be^(-bt), but I can't figure out what to do for part (b) on.
    My attempt was to take the average time t to equal (ƩP(t))/n where n is the number of collisions and the sum goes from 0 to ∞, but that thought process got me no where. Any help would be appreciated.
     
  2. jcsd
  3. Feb 5, 2013 #2

    Dick

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    I don't think the first part is right. The integral of P(t) from t=0 to t=infinity should be +1.
     
  4. Feb 5, 2013 #3

    haruspex

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    If you have the probability density function p(x) for some continuous variable X, what expression gives the average value of X?
     
  5. Feb 5, 2013 #4
    For the first response, I rechecked my work on part (a) and realized my signs were off, using A = b the ∫Ae^(-bt) from 0 to ∞ gives me a value of +1.

    For the second response, the only solution I can think of would be the Expected value E(x) which would give E(t) = ƩtP(t) = Ʃt(Ae^(-bt)) from t = 0 to ∞ which gives 0 + be^(-b) + 2be^(-2b) +....+ 0 which I don't know how to generalize or even how to apply to the next part. Quite frankly, I am lost on this problem after part (a).
     
  6. Feb 5, 2013 #5

    Dick

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    The mean time is an expectation value. It's going to be the integral of t*P(t) from 0 to infinity.
     
  7. Feb 5, 2013 #6

    haruspex

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    That's for a discrete distribution. As Dick says, the equivalent for a continuous distribution is integration.
     
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