[mps]

here's an interesting formula for gas diffusion from a container:

effusion rate (through pinhole) = (Area * dP(inside - outside)) / sqrt(2pi*MolWt * R * T)

I'm surprised effusion and temperature are inversely related-- as temperature increases, effusion rate slows.  And slower molecules (lower temp --> less KE) effuse faster. 

Can anyone explain?

Thanks.

[enahs]

While the rate of collision will depend on velocity, and increased velocity will result in more collision (higher temperature, more kinetic energy more velocity); the rate of effusion is directly proportional to the particle density,  Ñ.

The particle density of a container is given by the number of particles in that container divided by the volume of that container, n/V

PV=nRT

Solving for n/V:
n/V = P/RT

This might seem odd at first, but think about it. As you heat a gas up, add energy, the gas particles in the container move faster and are therefor more "spread out" on average, the more spread out, the lower the density. Pressure and density are not the same thing.


As effusion occurs, the pressure inside and outside changes, the number of molecules, the energy, the density, etc etc. They are all related, and it is a lot more complex then the explanation I gave with the Ideal Gas Equation.

The ideal gas equation is very poor to use in this case, and that is being generous. However, until you get into the advanced Kinetic Theory of Gas (you will not cover this in Gen Chem) it would be really hard to explain without writing a whole chapter of a book. That is, explain the concept of the particle density as a function of the temperature and pressure related to effusion, as it is a continually changing process.


I hope it helps; and grasping the concept of increasing the temperature (and thus pressure) decrease the particle density is the key here.

[Yggdrasil]

I don't think that this formula is correct.  The units do not match up.

However, when you think about the effect of temperature, note that pressure is also dependent on temperature.

[davidbbk]

Think of it this way...

When you put heat into a container with a gas the pressure increases. Pressure is the force of gas molecules colliging against the wall of a container.

If there is a pin hole leak in the case of effusion, the more kinetic energy the gas molecules possess, the faster they move, the more the pressure, the more the collisions of the gas molecules with the wall of the container.

Thus, the faster the molecules are moving the higher the Entropy of the system.

As temperature increases you increase the chances of an atom in gas phase to strike the hole of the container. Effusion thus occurs at a faster rate as temperature is increased. 

[Borek]

The particle density of a container is given by the number of particles in that container divided by the volume of that container, n/V

PV=nRT

Solving for n/V:
n/V = P/RT

First of all - define cotainer! Is it a baloon (V=nRT/P) or tank (V=const)?

[mps]

Thanks, all. 

Yes, Ygg-- you're right, the units don't add up.  There's no expression for time, which would be needed in a rate.  I found this formula in a reputable source, so I don't know what to make of it!

In any case, can someone give me the bottom line: is effusion rate directly or inversely proportional to temperature?  David's explanation makes intuitive sense:

"As temperature increases you increase the chances of an atom in gas phase to strike the hole of the container. Effusion thus occurs at a faster rate as temperature is increased."

Is this what we see in the lab?

Thanks again...

[enahs]

In any case, can someone give me the bottom line: is effusion rate directly or inversely proportional to temperature?  David's explanation makes intuitive sense:

Effusion rate is directly proportional to the particle density and directly proportional to the temperature (average kinetic energy).

However,
Particle density is inversely proportional to temperature.
The rate of the gas colliding with the appropriate area (the pin-hole) to escape is proportional to the temperature.


In effusion it is complicated, the gas is leaving the container, so the particle density, pressure, amount  of gas, and even temperature are all constantly changing. As a result, the type of equation you are using is only valid over a small change and in certain conditions.

[Borek]

In effusion it is complicated, the gas is leaving the container, so the particle density, pressure, amount  of gas, and even temperature are all constantly changing.

In general you are right. However, it is not that hard to set up experiment with tank loosing only very small part of the gas on the experiment timescale (mililiters out of liters), so that all parameters mentioned can be treated as constants. It all depends on what you are planning to research.

[Yggdrasil]

Alright, I looked up the formula for effusion in my p-chem notes.   For particles escaping from a rigid container into a vacuum and with the assumption that the size of the hole is small compared to the mean free path of the gas (i.e. no collisions while escaping), then

dN/dt = PA sqrt (ρ/2πm)

where P is pressure, A is the area of the hole, ρ is the number density (particles/volume), and m is the mass of the particle.

Note that P is dependent on temperature:

P = NkT/V = ρkT

So, dN/dt =  ρkT sqrt (ρ/2πm)

Since the container is rigid, number density does not change with temperature.  Therefore, effusion is directly proportional to temperature, and dN/dt α ρ^3/2.

[Edit: see Enahs post below]

[enahs]

I have some problems with what you say Yggdrasil, maybe I just do not see it, but they do not jive with what I have.

Let dN_c be the number of particles that hit the wall of the container. The collisiondNc rate, dN_c/dt is the number of collisions with the wall per unit time. This quantity will be be proportional to the area being struck, A. Also, the collision rate will depend on particle velocity, with increased velocity resulting in increased collision rate. The collision rate is also directly proportional to the particle density, Ñ.


dN_c/dt = Ñ*A*Integral( v_xf(v_x)dv_x)|0->infinity
This integral is the average particle velocity in the direction that will result in collision with the area of interest (taken in the positive x direction corresponding to limits from 0 to +infinity)


I will skip a few steps of the integral evaluation and substitutions, just because they are impossible to do justice in this form, but I can make them out and attach them if needed. Anyway, that integral evaluates to:

ÑA (kT/2 Pi m)^1/2

Using the maxwell speed distribution, to define the average speed, that becomes:
dN_c/dt = ÑA 1/4 v_ave

Defining the collisional flux, Z_c as the number of collisions per unit time per unit area, we have:

Z_c= (dN_cdt) /A = 1/4 Ñv_ave

To express collisional flux in terms of gas pressure:
Ñ = N/V = nN_a/v = P/kT

With this definition of Ñ, Z_c becomes

Z_c = P/(2Pi m kT)^1/2 = PN_A/(2 Pi MRT)^1/2
Where m is the mass of the particle in kg and M is the molar mass in kg mol^-1.


After evaluation Z_c for a given condition, you have to then multiply the collisional flux (Z_c) by the are of interest to get the collision rate.


I think maybe what you are calculating is the effusion time not rate, as asked by the original poster. The effusion time is inversely proportional to the effusion rate.

[Yggdrasil]

You're right.  I re-derived the formula and your answer is correct.  The professor I had in that p-chem class was notorious for writing the wrong formulas.