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Mean Free Path of Nitrogen gas

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    At what pressure will the mean free path in room-temperature (20 degrees Celcius) nitrogen be 1.0 m? Answer in pascals.

    2. Relevant equations

    [tex]\lambda[/tex] = kBT/(4[tex]\sqrt{2}[/tex] [tex]\pi[/tex]pr2)

    where [tex]\lambda[/tex] = mean free path
    kB = Boltzmann Constant 1.38x10-23 J/K
    p = pressure
    r = diameter of gas molecule... in this case, since nitrogen is diatomic, my textbook tells me to use r=1.0x10-10m

    3. The attempt at a solution

    I have no idea what I'm doing wrong here... this question seems pretty straight forward. All I have to do is isolate for p, pressure, and substitute the numbers in but for some reason, the system is still telling me that my answer is wrong. After plugging everything in... I'm getting 2.28x10-11 Pascals. I MUST be wrong somewhere! I'm pretty sure my equation is correct since it is what my professor and textbook uses.

    EDIT: In the equation, square root 2 shouldn't be to the power of pi. I'm not sure why it looks like that..
  2. jcsd
  3. Feb 3, 2010 #2
    Nevermind. Figured it out.
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