Link between atm. pressure by air weight and brownian motion

[wj02]

Dear list,

Imagine a table of upper surface area S sitting in an open field, with nothing on it. We know that S is subjected to a downward atmospheric pressure P due to a cylindrical column of air of volume V extending vertically from S to the end of the terrestrial atmosphere. Assume this column contains n air molecules each of mass m. We know from the kinetic theory of gases that these zillions of molecules are moving around in a Brownian motion at an average speed v. To simplify further, assume that earth’s gravity g is independent of altitude. Then:
1-From a hydrostatic point of view, we know that P = nmg/S; here pressure is due to the weight nmg of the air column above the table.
2-From a kinetic point of view (after many simplifying assumptions), P = nmv^2/3V; here pressure is due to collisions at a speed v between air molecules and S.
Since both point of views are correct, they should both yield the same pressure.

Now assume we cool down the air column to very low temperatures such that the molecules become nearly still (v approaches 0), then theory 1 would still predict the same pressure P= nmg/S. Theory 2 however predicts a pressure P approaching 0!

For the life of me, I cannot answer the following questions:
-How can both theories be correct, yet totally disagree?
-How can one phenomenon of nature (pressure) be the result of two totally unrelated natural phenomena (static molecular weight and dynamic molecular collisions)?
-Why can’t we derive one formula from the other (even when we make some assumptions)?

I am puzzled. Can someone explain?
You can answer if you want offlist to wasjal@gmail.com

thx
wassim

 

[Bill_K]

Consider the cubic foot of air at the bottom of that column. It has all the air above it pressing down on it, right? So it has to push back with an equal and opposite force. Newton' Law of Action and Reaction. How does it push back? With the pressure you calculated in (2) due to the thermal motion of its molecules.

What happens when you cool the gas? As you lower T, the weight pressing down remains the same. So in your expression P = nmv^2/3V, as v gets smaller, so does V. The cooler air occupies less and less volume. But P remains the same.

 

[wj02]

Thank you dear Bill.
You are right, v and V both get smaller and the pressure remains constant. Very good point indeed.

But by what concrete physical mechanism does my original air column exert force on the table surface? at zero temperature (for the whole column), all molecules are sitting at the column's bottom exerting pressure P on the table by virtue of their pure static weight only. Now raise the temperature sowly, some moving molecules are no longer in contact with the table surface and cannot exert static weight on it. This is presumably compensated by the growing collisional energy of the moving molecules colliding with the table's surface. So as the direct static weight effect on the table is reduced, the growing collisional forces compensate. And so P should be a mathematical expression with 2 terms in it: one term accounting for the statically induced weight pressure, the other for the dynamically induced collisional pressure. One term goes down, tho other goes up and visa versa. P remains constant.
Where is this equation supposed to link both effects?


Regards

 

[wj02]

Dear forum,

How can we state that the table surface is subjected to the total weight of all air molecules above it (P=nmg/S) when these molecules are known to be bouncing in all directions, and thus at any specific instant a good fraction of these molecules is probably traveling upwards or sideways thus exerting no force nor impulse on the surface nor in its direction (downward)?
P=nmg/S makes perfect sense if molecules are at rest on top of the table as their weight is exerted on the table by direct contact. How can the equation still hold water when no one knows the average number of air molecules contributing to downward pressure at any instant?

thanks

 

[russ_watters]

I think you're just seeing the cause-effect relationship backwards. Atmospheric pressure is due to the weight of the column of air, period. Brownian motion determines the density of this column of air based on what is required for the bouncing-around to exactly cancel the pressure. If you heat the atmosphere, the bouncing around increases and the atmosphere expands until it goes back to the equilibrium.

How can we state that the table surface is subjected to the total weight of all air molecules above it (P=nmg/S) when these molecules are known to be bouncing in all directions, and thus at any specific instant a good fraction of these molecules is probably traveling upwards or sideways thus exerting no force nor impulse on the surface nor in its direction (downward)?

By adding them all together.

How can the equation still hold water when no one knows the average number of air molecules contributing to downward pressure at any instant?

But we do know. The surface area of the Earth is constant and the number of molecules in the atmosphere is (essentially) constant, so the average number of molecules over any given piece must also be constant.

 

[wj02]

Dear All:
Thank you for your input.
Kindly consider (macroscopically) a closed vertical column standing 10m high with vacuum inside it with 1m x1m square cross section.
Suppose it has inside it 100 rubber balls (engaging in elastic ever lasting collisions) with initial individual speed v bouncing around inside the column in all directions.
Can you say that the pressure on the column's base is the weight of the 100 balls divided by 1m^2? Probably not because not all 100 balls are continuously contacting the column's base (they are not at rest sitting down), so at any instant the only force on the base is the fraction of balls colliding with the base, not all 100 balls.
Even if we assumed (statistical averaging effect), for example, that 30 of them are colliding with the base any instant, we still cannot say that the pressure is the weight of 30 balls divided by the base area because these 30 balls are not resting on the base, they are colliding with it, thus it is their momentum that is now relevant not just their mass (weight).
Now suppose instead of the balls, you had air molecules, which is the real world case. Why in this case should you be able to say that the pressure is so simply equal to the air molecules weight divided by the base?
This is the question I am trying to illustrate? I know that weight of air divided by area is the pressure, but I am questioning how it is that this formula works so well when the molecules are not at rest.

Your comments?

thx

 

[russ_watters]

The pressure inside a closed vessel is not related to the weight of the air inside, but the atmosphere is not a closed vessel. In your closed vessel, what keeps the molecules from flying out the top of the vessel is the top of the vessel. In the atmosphere, what keeps them from flying out the top is their weight.

If you poured a bag of sand into your vessel, you'd certainly agree that the pressure is equal to the weight divided by the area, right? The bouncing-around of your rubber balls or the atmosphere acts to counter that weight: they are in equilibrium with each other.

 

[wj02]

thanks dear Russ.

You say "the pressure inside a closed vessel is not related to the weight of the air inside ..."

So if my vessel is 80km high (say about the same hight as the atmospheric layer) and open at the top, naturally filled with air, and I hovered over that top and ever so gently put a sealing lid on it so that the vessel became closed, can I say that the pressure onto the bottom area is no longer related to the air weight?

you say
"The bouncing-around of your rubber balls or the atmosphere acts to counter that weight"

do you mean that the bouncing of the molecules acts to counter or decrease their weight effect? if yes, then how can we still equate their full weight to the pressure?

Thank you for your advise, I am still trying to understand this.

 

[russ_watters]

thanks dear Russ.

You say "the pressure inside a closed vessel is not related to the weight of the air inside ..."

So if my vessel is 80km high...

Let me be more precise: the pressure inside a small closed vessel is not related to the weight of the air inside. Strictly speaking, the weight is always a factor that causes the pressure at the bottom to be greater than the pressure at the top, but for a small vessel that difference is a tiny fraction of the total pressure and can safely be ignored.

you say
"The bouncing-around of your rubber balls or the atmosphere acts to counter that weight"

do you mean that the bouncing of the molecules acts to counter or decrease their weight effect?

No, I already answered that in the beginning of post #5: it counters the weight in that it pushes particles up, which doesn't decrease the pressure due to the weight, but rather only decreases the density of the gas.