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Deriving mean free path of ideal gas

  1. Jun 20, 2010 #1
    1. The problem statement, all variables and given/known data

    The mean free path for an ideal collisional gas can be calculated as shown on thishttp://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/menfre.html" [Broken]. I understand the derivation, except for one thing.

    2. Relevant equations

    3. The attempt at a solution

    The molecule is at one end of the cylinder at the beginning, and any of the particles could have collided with the molecule at that instant. The same consideration has to go when the molecule reaahes the other end of the cylinder. Does that not mean that we should attach hemispherical volumes to the ends of the cylinder, and take into account any collisions at the end.

    Please help!!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 20, 2010 #2
    I think your problem comes from misunderstanding the diagram. Unless i'm reading it wrong, that's not a cylinder, it's simply the volume which the molecules effective collision area sweeps out in a time t (moving at speed v).

    Of course it could collide with something at the start, it might not collide with anything for 10 minutes ! (Although that chance is about as much as England winning the world cup right now). But you're talking about the mean free path, which is the average distance it travels before colliding once.

    If you have any more questions feel free to ask.
  4. Jun 20, 2010 #3
    So, you're saying it's not a cylinder, then! Is it a hollow cylinder in that case?

    Secondly, in finding the number of particles that collide with the molecule, we have to use the

    1. particle number density

    2. the volume which is the sum of the volume traversed by the molecule and the envelope that surrounds it.

    In that case, we do need to include the hemispherical volms at th end, right??
  5. Jun 20, 2010 #4
    I think you misunderstand me

    That IS the "cylinder" in the picture, it's not a real cylinder, but it's just a picture of the volume which the particles effective collision area moves through in a time t.

    I'm not sure what you mean about the hemisphere's on the end. We're assuming a 2D collision area here, not a volume of collision or something like that.
  6. Jun 20, 2010 #5
    Well, I understand that it's not a real cylinder. If it were, the molecule could not have moved at all. What I mean is an imaginary cylinder that encloses the region over the collisions take place.

    If we were really assuming a 2d collision area, then the formula for the number of particles in that region would have a different form. But the formula is the volume of cylinder*number density, right??
  7. Jun 20, 2010 #6
    Why would it have a different form? We're assuming a 2D collision area which sweeps out a volume equal to the area* the velocity every second. Therefore we need to consider the number of particles in the given volume.

    Again, i'm not sure what you mean by this at all, but i repeat that the "cylinder" merely represents the volume which the molecules collision area moves through in a time t (going at a speed v).

    I'm also still not sure what you mean by hemispheres.
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