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Effusion differential equation from Newtonian mechanics

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data

    If you poke a hole in a container full of gas: the gas will start leaking out. In this problem, you will make a rough estimate of the rate at which gas escapes through a hole: effusion. (This assumes the hole is sufficiently small).

    Consider such a hole of area "A". The molecules that would have collided with it will instead escape through the hole.

    Assume that nothing enters through the hole. Then: show that the number of molecules “N”, is governed by:

    [tex]\frac{{dN}}{{dt}} = - \frac{A}{{2V}}\sqrt {\frac{{kT}}{m}} N[/tex]

    2. Relevant equations

    [tex]2L = \Delta t \cdot \overline {{v_x}} [/tex] (round trip time for collision, but the factor of 2 coming from considering the walls of collision in just one dimension)
    [tex]{V = L \cdot A}[/tex] (volume = length times area)
    [tex]{PV = N{k_B}T}[/tex] (ideal gas law)
    [tex]\overline {{v_x}} \approx \sqrt {\overline {{v_x}^2} } [/tex] (root-mean-square/mean-velocity approximation)
    [tex]\sqrt {\frac{{kT}}{m}} = \sqrt {\overline {{v_x}^2} } [/tex] (I derived this result and know it to be true: it's from assuming (1/2)*mv^2 = (1/2)*kT: that is, thermal/kinetic energy equality in one dimension)

    3. The attempt at a solution

    work backwards: start reading from bottom up...sorry:

    \frac{{dN}}{{dt}} = - \frac{{{V^2}}}{{N \cdot \Delta t}} \\
    = {\left. { - \frac{V}{{\Delta t \cdot \overline {{v_x}} }}\overline {{v_x}} \frac{{{k_B}T}}{P}} \right|_{2L = \Delta t \cdot \overline {{v_x}} }} \\
    = {\left. { - \frac{{AL}}{{2L}}\overline {{v_x}} \frac{{{k_B}T}}{P}} \right|_{V = L \cdot A}} \\
    = {\left. { - \frac{A}{2}\overline {{v_x}} \frac{{{k_B}T}}{P}} \right|_{PV = N{k_B}T}} \\
    = {\left. { - \frac{A}{{2V}}\overline {{v_x}} N} \right|_{\overline {{v_x}} \approx \sqrt {\overline {{v_x}^2} } }} \\
    = {\left. { - \frac{A}{{2V}}\sqrt {\overline {{v_x}^2} } N} \right|_{\sqrt {\frac{{kT}}{m}} = \sqrt {\overline {{v_x}^2} } }} \\
    \frac{{dN}}{{dt}} = - \frac{A}{{2V}}\sqrt {\frac{{kT}}{m}} N \\

    Well…hard to say what I wanted d/dt to look like, so it's no wonder this just looks like algebraic junk. Well … I know A/V has units of inverse-length, and should be propotional to the volume of escaping air…

    Ansatz: the numer of particles striking the area, “A”, is a fraction of the total area, which is V/L, where “L” is some length of “gas” perpendicular to the area “A” related to the velocity.

    Is that a good ansatz? I'm looking for a "given" to start the derivation of this differential equation with. I'm sure I could make the quantities I wanted appear with the (2) Relevant Equations...
  2. jcsd
  3. Sep 17, 2010 #2
    Can I bump this?

    I've got a similar problem and I'm trying to derive it using my notes which derives the gas law using the pressure on an entire face of the box. It works out since multiplying the length L of the box against the area of the face gives the volume V of the box.

    But in this case it's a small area A and not an entire face, so I can't think of a way to manipulate L into V.
  4. Sep 17, 2010 #3
    ah, what does "bump" mean? :-|
  5. Sep 17, 2010 #4
  6. Sep 17, 2010 #5
    lol...okay. Anyway, I didn't get the problem solved. We didn't end up doing the problem for class anyway...but I may be more interested in the solution when test-time draws near.... El Professor-o doesn't have a test scheduled yet, though. Of course: you're quite free to think about it...
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