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Is there a Conceptual problem here? (Pressure in the Kinetic theory of gases)

  1. Dec 13, 2011 #1
    I have reviewed recently the demonstration of the formula for the pressure of a gas by using the kinetic theory and it seems that there is a terrible flaw in it, which I will describe below. Can anyone comment on it, please?

    In all textbooks, the pressure of a gas is proven to be proportional to the mean square speed of its molecules. The demonstration starts with the observation that the pressure of a gas is due to the ellastic collision of molecules with the walls of the container. Therefore, it calculates the change in the momentum arising from the elastic collision of one molecule with the wall, which is
    Δp = 2mv (where v is the component of the molecule velocity perpendicular to that wall)
    Nothing wrong up to here.
    However, when calculating the force produced by this collision, instead of dividing this change in momentum to the time during which this collision takes place (Newton's second law), the authors divide the change in the momentum by the time taken by the particle to travel to the opposite wall and back 2L/v (where L is the length of the box perpendicular to the wall). They write
    F = change in momentum / time taken = 2mv / (2L/v)
    I contend that there is a serious problem here because the two time intervals obviously refer to different things.

    You can see this here http://en.wikipedia.org/wiki/Kinetic_theory but is in fact in every textbook.

    Don't you think that there is something wrong with this derivation?
    Waiting for your comments. Thanks
  2. jcsd
  3. Dec 13, 2011 #2


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    No, the derivation is fine.

    What you refer to is the force per collision, so to say. However, this is not the quantity of interest here. For calculating the pressure you want the mean (time-averaged) force of the particle on the walls. This is the force per collision divided by the waiting time between subsequent collisions and equivalent to the change in momentum per collision divided by the time the particle takes to reach the wall again.
  4. Dec 13, 2011 #3
    There still seems to be a problem... doesn't the equation treat all atoms/molecules simply by count and not distinguish them by their different masses, and thereby momentums?

    A box of H2 at the same pressure as a box of O2, both boxes with the same volume?
  5. Dec 13, 2011 #4


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    Now I am puzzled and do not know what you mean. How can a term poportional to 2mv and therefore directly proportional to mass not take mass into account?
  6. Dec 13, 2011 #5
    Once you get to the point change in momentum per collision = 2mv an alternative way to get an expression for the force on the wall is to say Force = change in momentum per second (ie force = rate of change of momentum)
    Therefore Force = 2mv x number of collisions with the wall per second.
    The time for the molecule to return to the wall is 2L/v seconds therefore the number of collisions per second by this molecule = 1/(2L/v) = v/2L

    Therefore force due to this molecule = 2mv x v/2L = mv^2/L

    Therefore the force due to N molecules = N/3 X mv^2/L = Nmv^2/3L

    Thereore the pressure = F/Area = F/(axb) (a, b are dimensions of end wall)
    P = Nmv^2/3(lab)

    hope this helps
  7. Dec 13, 2011 #6
    The derivation has stood the test of time.

    However, a more phenomenological way of looking at gas pressure is as the simple product of the mean number of molecular impacts per unit area per second and the mean impulse per impact; that is as the mean flux times the mean impulse. For non-standard notation, let's say, P=FI.

    The mean frequency of impact is the simple product of half the number density times the mean molecular speed along any single axis of movement. F=van/2.

    The mean impulse per impact is the simple product of the molecular mass and twice the mean impulse speed. I=m2vb

    The result is P = nmv2, the standard molecular definition of pressure.
  8. Dec 13, 2011 #7

    Thanks for your reply. But I am afraid that the problem still remains.
    Even if you calculate the average force, you will still have to divide the change in momentum in an elastic collision of a molecule (Δp = 2mv ) by the total time (Δt) taken until the next collision with the same wall takes place.
    This total time (Δt) is made up of: the time of the collision of the molecule with the initial wall (δt), the time taken by the molecule to travel to the opposite wall (L/v), the time of the collision with the opposite wall (δt), and the time taken by the molecule to arrive at the initial wall (L/v) ; only after the elapse of these intervals does the next collision begin. So the total time between collisions is not just 2L/v but Δt = δt + L/v + δt + L/v, that is Δt = 2(δt) + 2(L/v).
    Therefore the average force would be F = Δp / Δt = (2mv) / [2(δt) + 2L/v] and this becomes equal to (2mv) / (2L/v) only if you neglect δt and put it equal to zero.
    So now I came to think that probably the demonstration of the formula for pressure should include a reference to the fact that the collision time of the molecule with the wall is negligible. Still, I do not know if such an approximation is realistic.
  9. Dec 14, 2011 #8
    There are many assumptions in the kinetic theory.
    One assumption is that the time for a collision is zero.
    The fact that the theory is good for a wide range of pressures and temperatures means that the assumptions are good
  10. Mar 15, 2012 #9


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    The pressure on the wall is due to the force imparted by a population of particles. In steady-state with a uniform temperature and constant pressure, the particles (each of mass m) would collide with the wall at the same rate. Assuming also uniform density (no local concentration above the average/mean), some particles having recoiled from the wall will traverse back to the opposite wall and return in Δt=2L/v, during that period some N particles will have hit the wall at some rate. Summing up the total Δv from all the particles and divide by the period over which those particles are counted.

    So the force would be m Ʃi(Δvi)/Δt, i = 1, N, and Δv is the change in the velocity compoent normal to the surface being struck.
  11. Mar 15, 2012 #10
    Thanks, Astronuc.
    But having said that "some particles having recoiled from the wall will traverse back to the opposite wall and return in Δt=2L/v", you overlook the time spent by those particles in contact with the opposite wall (time of collision).
    It is probably insignificant, but it does no harm if aware of this approximation.
  12. Mar 15, 2012 #11

    Philip Wood

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    In my opinion the best kinetic theory derivation of the pressure formula is that used by Jeans in The Kinetic Theory of Gases. He doesn't consider the bouncing of molecules between opposite walls, but instead with the flux of molecules approaching a patch of wall at various angles. The method is simple and convincing.
  13. Mar 15, 2012 #12
    All explanations that match experimental observations are, by definition, ' good'
    The best is a subjective decision based, I would say, on experience and level of study.
    Kinetic theory implies dealing with a gas as a system of particles. At low temperatures this is not good enough and 'wave' properties need to be applied.
    This would not be called 'kinetic' theory
    Last edited: Mar 15, 2012
  14. Mar 16, 2012 #13

    Philip Wood

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    Agreed. That's why I said "in my opinion". My main criterion for a good derivation is how convincing I find it. Jeans's derivation doesn't rely on a special shape of box, nor on each molecule having an uninterrupted path across the box, nor on each collision with the wall being individually elastic.

    As you imply, the derivation based on molecules bouncing around in a cuboidal box delivers the right formula, and this isn't by accident; I just don't find it convincing. Put it this way: if I'd been clever enough to think of it myself, I would want someone to find a different way to do it, as a check.
  15. Mar 16, 2012 #14
    I think I go along with your thought Philip.
    When I taught kinetic theory the biggest problem I had was convincing the students that the derivation did not just apply to rectangular box.
    But there was no doubt that considering a rectangular box was the easiest way to get the idea across. The derivation considering components of velocity in x, y and z directions was usually too mathematically challenging for some students, especially those not studying maths at that level.
  16. Mar 16, 2012 #15

    Philip Wood

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    technician: Did you ever come across the derivation using a spherical container? It is a lot simpler than the one for the cuboidal container because you work with molecular speeds from the outset, rather than with velocity components, so you don't have to introduce
    [tex]\bar{c^{2}} = \bar{u^{2}} + \bar{v^{2}} + \bar{w^{2}}.[/tex]
    What is quite astonishing is the way the factor of 1/3 appears in this derivation, compared with the way it appears in the cuboidal box derivation!

    Using my how convincing? criterion, this derivation is at least as bad as the one for the cuboidal box, but I suppose if you take both together their joint impact is rather persuasive!
    Last edited: Mar 16, 2012
  17. Mar 16, 2012 #16
    I want to thank you all for your comments. They were very helpful and I think I have found the answer to my issue.
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