# Effusion differential equation from Newtonian mechanics

• bjnartowt
In summary, this conversation involves a problem about estimating the rate of gas escaping through a hole in a container. The conversation discusses the number of molecules escaping, the ideal gas law, and the root-mean-square/mean-velocity approximation. The conversation ends with a question about manipulating the length of the box into the volume.
bjnartowt

## 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:

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

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...

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...

## 1. What is an effusion differential equation?

An effusion differential equation is a mathematical equation used in Newtonian mechanics to describe the flow of a fluid through a small opening. It relates the rate of effusion (flow) to the pressure difference across the opening and other physical properties of the fluid.

## 2. How is the effusion differential equation derived?

The effusion differential equation is derived from the laws of motion and conservation of mass in Newtonian mechanics. It involves using the Navier-Stokes equations, which describe the motion of a fluid, and applying them to a small opening or orifice.

## 3. What is the significance of the effusion differential equation?

The effusion differential equation is significant because it allows us to understand and predict the flow of fluids through small openings, which is important in many practical applications such as gas and liquid pipelines, chemical reactions, and ventilation systems.

## 4. What are some examples of effusion in everyday life?

Effusion can be observed in many everyday scenarios, such as the release of gas from a pressurized container, the flow of water through a faucet, and the movement of air through a small crack in a window. It is also relevant in industrial processes, such as in the production of gases, liquids, and powders.

## 5. How does the effusion differential equation relate to other equations in physics?

The effusion differential equation is closely related to other equations in physics, including the ideal gas law, Bernoulli's equation, and the continuity equation. It is also related to the concept of viscosity, which is a measure of a fluid's resistance to flow.

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