Effusion differential equation from Newtonian mechanics

Click For Summary
SUMMARY

The discussion focuses on deriving the effusion differential equation from Newtonian mechanics, specifically the rate at which gas escapes through a small hole in a container. The key equation derived is dN/dt = - (A/2V) * sqrt(kT/m) * N, where N is the number of molecules, A is the hole area, V is the volume, k is the Boltzmann constant, T is the temperature, and m is the molecular mass. The discussion also touches on the relationship between collision time, velocity, and the ideal gas law, emphasizing the importance of understanding these concepts for accurate derivation.

PREREQUISITES
  • Understanding of Newtonian mechanics
  • Familiarity with the ideal gas law (PV = NkBT)
  • Knowledge of root-mean-square velocity calculations
  • Basic differential equations
NEXT STEPS
  • Study the derivation of the ideal gas law in detail
  • Learn about the kinetic theory of gases and its implications for effusion
  • Explore the concept of mean free path in gas dynamics
  • Investigate applications of the effusion equation in real-world scenarios
USEFUL FOR

Students in physics, particularly those studying thermodynamics and gas dynamics, as well as educators looking for practical examples of effusion in gas behavior.

bjnartowt
Messages
265
Reaction score
3

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...
 
Physics news on Phys.org
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...
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
5K
Replies
19
Views
3K
Replies
4
Views
2K