Effusion differential equation from Newtonian mechanics

AI Thread Summary
The discussion revolves around deriving the effusion differential equation for gas escaping through a small hole in a container. The key equation presented is dN/dt = - (A/2V)√(kT/m) N, which describes the rate of gas escape based on the area of the hole and the properties of the gas. Participants express difficulty in manipulating the volume and area relationships to derive this equation, particularly when considering a small hole versus an entire face of a container. There is also a mention of a lack of resolution on the problem, with one participant expressing interest in revisiting it closer to exam time. The conversation highlights the challenges in applying theoretical concepts to specific scenarios in gas dynamics.
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Homework Statement



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:

\frac{{dN}}{{dt}} = - \frac{A}{{2V}}\sqrt {\frac{{kT}}{m}} N

Homework Equations



2L = \Delta t \cdot \overline {{v_x}} (round trip time for collision, but the factor of 2 coming from considering the walls of collision in just one dimension)
{V = L \cdot A} (volume = length times area)
{PV = N{k_B}T} (ideal gas law)
\overline {{v_x}} \approx \sqrt {\overline {{v_x}^2} } (root-mean-square/mean-velocity approximation)
\sqrt {\frac{{kT}}{m}} = \sqrt {\overline {{v_x}^2} } (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)


The Attempt at a Solution



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

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

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...
 
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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.
 
ehhh said:
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.

ah, what does "bump" mean? :-|
 
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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