How Does Atmospheric Pressure Vary Between Micro and Macro Perspectives?

[Plep]

Hey all,

I have a question concerning atmospheric pressure. I understand that there are two ways to look at pressure, at micro scale (kinetic theory) and at macro scale (seeing pressure a a state variable).

Now i'd like to understand atmospheric pressure using both perspectives.

At micro scale i think it's pretty straightforward. Gravity causes an increase in air density closer to earth. More molecules means more collisions and thus a higher pressure.

At macro scale, i have been taught that atmospheric pressure at a point can be seen as the weight of the column of air over that point.

My problem arises when you slowly close a vessel and place it on the ground. Intuition tells me the pressure in the vessel will be the same as outside the vessel. Using the kinetic theory this can be explained since the density and temperature are the same inside and outside the vessel.
However, i am lost when ik try to use the macro scale perspective to explain the pressure in the vessel, since the weight of the air above is now carried by the walls of the vessel.

I've got the feeling I'm missing an important point here, can someone help me? Thanks!

 

[mgb_phys]

However, i am lost when ik try to use the macro scale perspective to explain the pressure in the vessel, since the weight of the air above is now carried by the walls of the vessel.

Not unless they are very strong it isn't.
It's like asking why the pressure indoors is the same as outside - it's because the forse excerted by the atmosphere on the walls/windows/roof is balanced by the same air pressure inside - there aren't many building that would survive you pumping all the air out of them.

 

[Lojzek]

You can determine a pressure in a material by:

A) determining the outside pressure
In this case the material will compress until its internal pressure will balance outside pressure.

B) determining the volume.

In this case you just force the material into a certain volume and it will respond with an internal pressure that depends on that volume.

The atmospheric pressure is determined by A), since each part of the gas is free to compress/expand as long as it can balance the weight of the air above.

A gas in an incompressible vessel is determined by B), so it is completely independent of
anything outside this vessel.

If you first prepare the gas by A), and then close a piece of that air in an incompressible vessel, then the pressure will of course stay the same, because it is already equal to internal pressure of this amount of gas at this temperature and volume.

In reality no vessel is completely incompressible, but this does not matter in this case, since
the pressures on both sides are equal and the vessel would not deform anyway.

 

[Robin07]

Not unless they are very strong it isn't.
It's like asking why the pressure indoors is the same as outside - it's because the forse excerted by the atmosphere on the walls/windows/roof is balanced by the same air pressure inside - there aren't many building that would survive you pumping all the air out of them.

Hi, so am I to understand that if you limit/reduce the opening to any building and have all ready housed a structure of extreme weight inside this building, the atmospheric pressure
would tend to drop?

I have often wondered of, the possible reason for constructing the kings' pyramid chamber c/w several tonne blocks above it as well as limiting the vents' sizes to the exterior. Perhaps this was the way they created a negative pressure inside this huge pressure pump, to suck water up from the Nile and irrigate their lands.

Interesting

 

[Danger]

Hi, so am I to understand that if you limit/reduce the opening to any building and have all ready housed a structure of extreme weight inside this building, the atmospheric pressure
would tend to drop?

I have often wondered of, the possible reason for constructing the kings' pyramid chamber c/w several tonne blocks above it as well as limiting the vents' sizes to the exterior. Perhaps this was the way they created a negative pressure inside this huge pressure pump, to suck water up from the Nile and irrigate their lands.

Interesting

I don't quite see why you would infer any of that. Something built under standard atmospheric pressure will remain at that same pressure unless air is either forcibly evacuated or pumped in. The Egyptians had no capability of that.

 

[Ranger Mike]

the reason the pyramid has small vents is to keep out grave robbers. lot of gold in them things!

 

[Crazy Tosser]

Pressure has to do with molecules' random motion and collisions, remember?
Now, air molecules inside the container when the container was sealed have pretty much the same average speed as the molecules outside.

Remember that the container walls are also composed of molecules that also move (Brownian motion). Now, if there are air molecules colliding with container molecules from both sides, the container's walls become a little thinner and denser, have a higher temperature, but generally maintain their shape. If you, however, have air on only on the outside of the container, air molecules will keep colliding with container's molecules, and keep pushing them inside. The increase of temperature of the walls would actually be less, but the container would have a really hard time maintaining it's shape - if it is not strong enough, it will break.

 

[Dale]

the weight of the air above is now carried by the walls of the vessel.

Is it? If you place a strain gauge on the walls of the vessel you can measure how much of a load the wall is carrying (many bathroom scales work on this principle).

So, how much strain do you think you would measure in the vessel wall in the situation you described? Does that help you answer your original question?

Under what conditions would you get a strain in the vessel wall and what would that imply about the pressure inside and out?

 

[Plep]

Wow, thanks for the replies, i pretty much thought this topic was dead.

So if i understand it correctly, the point that i was missing, was that no vessel is truly incompressible and because of that the vessel "transfers" the weight of the air above to the air inside the vessel, which causes a pressure inside the vessel.

Well then, as a thought experiment, what would the pressure be inside a truly incompressible vessel? The weight of the air above will now not be transferred to the air inside the vessel so by the same reasoning as above the pressure would be zero.

However, by the kinetic gas theory the molecules would still be moving since the temperature would be the same, which does give a pressure...

 

[Danger]

Well then, as a thought experiment, what would the pressure be inside a truly incompressible vessel? The weight of the air above will now not be transferred to the air inside the vessel so by the same reasoning as above the pressure would be zero.

The only way that you would have zero pressure inside would be if the vessel was constructed in a perfect vacuum. Otherwise, it would have atmospheric pressure inside. Likewise, if you constructed it on the bottom of the ocean, you would have several tonnes of pressure inside.

 

[Plep]

So how can you explain that pressure in terms of the weight of the air above it? Since now the (hypothetical) incompressible vessel is carrying all the weight.

Ever since i was taught the concept of pressure I've had trouble with it. On the one hand i imagine it as a force passed on through a fluid (like in Pascal's Principle), but on the other hand it is caused by molecules bouncing of the surface (kinetic theory). Do both explanations hold for every situation?

 

[mviswanathan]

Plep's question is very interesting.

Is it? If you place a strain gauge on the walls of the vessel you can measure how much of a load the wall is carrying ...

I don't think, this answers the question. The pressure developed inside due to the weight of air above the contaner will very much depend on the strength of the container walls. Even if there are not Really uncollapsible contaners, stronger the container more is the isolation between the outside pressure and that inside.

I am interested in the answer to Plep's question on the macro pressure based on weight of air column - how does it explain the pressure inside the closed vessel.

 

[mviswanathan]

In case of open atmosphere, the pressure is equal to weight of air column on unit area. This considers the weight air column upto the level above which there is vacuum.

In case of a closed container, there is no vacuum on top. Even on top of the container there is no gap (vacuum).

The weight per unit area is valid in case of say, mercury in a barometer where on top of the mercury column there is no more material. The pressure on top is zero due to presence of vacuum.

May be my explanation does not look very technical, but I am sure some one can explain better.

 

[Danger]

Forget about the container. What matters is under what conditions it is sealed in the first place.
Let's consider Mount Everest. That's a fairly incompressible structure. Now, drill a tunnel through its base. The air in that tunnel will be at atmospheric pressure. If you seal the tunnel, the pressure remains the same. A barometer inside will remain constant, even though one outside will vary with weather conditions.
You can also think of the Space Shuttle. The internal pressure doesn't change as it ascends.

 

[mviswanathan]

It is clear that in a closed space (container, space shuttle...) the pressure of the contained gas will remain constant, if not changed by heat etc. etc.

I was trying to analyse Plep's querry ...

However, i am lost when i try to use the macro scale perspective to explain the pressure in the vessel, since the weight of the air above is now carried by the walls of the vessel.

I've got the feeling I'm missing an important point here, can someone help me? Thanks!

The last message was the explanation based on what I understand.

 

[Lojzek]

I think the original poster is trying to overgeneralize a formula that connects air pressure with the weight of the air column.

Formula p=rho*g*h
or more accurately dp=rho*g*dh
can only tell us the difference in pressure in a gas or fluid. It is of no use when one puts
a substance into a small volume at a known temperature. In this case pressure is determined by the equation of state for that substance (for example ideal gas equation).

Imagine a gas in a closed box in space: there is no gravity and no weight, but the gas will
still exert pressure on the box walls.

 

[Dale]

Well then, as a thought experiment, what would the pressure be inside a truly incompressible vessel? The weight of the air above will now not be transferred to the air inside the vessel so by the same reasoning as above the pressure would be zero.

I don't think, this answers the question. The pressure developed inside due to the weight of air above the contaner will very much depend on the strength of the container walls. Even if there are not Really uncollapsible contaners, stronger the container more is the isolation between the outside pressure and that inside.

You are both being distracted by unimportant details. The compressibility of the container walls is irrelevant. The same concepts apply if your container is made of adamantium or plastic.

There are two theories that you need to understand and use together here: Newton's laws, and the kinetic theory of gasses. When you take the kinetic theory of gasses and apply Newton's 2nd and 3rd laws then you can derive the ideal gas law, this is the "microscopic" view you were talking about.

Now, consider a volume of gas. If there is no bulk movement then, by Newton's first law, you know that all of the forces on that volume are balanced. By symmetry, the pressure forces on the sides balance, but what about the pressure forces on the top and the bottom? In order for the net force to be 0, the pressure force on the bottom must be greater than the pressure force on the top by an amount exactly equal to the weight of the gas. This is the "macroscopic" view you were talking about.

Note that the macroscopic view is not a general way of determining pressure, it is simply a result of applying Newton's first law to a volume of gas. Because the air in the atmosphere is often accelerating it does not even apply in most realistic cases. However, assuming that you have a column of air that is perfectly still from the ground up to space, since the pressure force on the top is 0 the pressure force on the bottom is equal to the weight of the column of air.

Now, let's consider a sealed vessel. The pressure inside the vessel is completely determined by the ideal gas law: the volume of the container, the temperature of the gas, and the amount of the gas. It does not matter if the vessel is at the bottom of the ocean, in the air, or in space, the pressure inside the vessel is not affected by any of that. The only relevance of the strength of the vessel walls is how the volume of the vessel changes, but the ideal gas law always holds.

Now, if the vessel is not moving we can again apply Newton's first law. Let's say that we sealed the vessel as you described above and then sunk it down to the bottom of the ocean, and for the analysis let's draw our free-body diagram by chopping the vessel in half and applying Newton's first law to the top half. Now, we see that the pressure force of the water pushing down is much greater than the pressure force of the air pushing up, but since we know that the vessel is not moving we know that there is another force, the stress in the vessel wall, which is pushing up to balance the otherwise unbalanced pressure forces. We can carry this same analysis out in other situations, in space the walls are in tension in order to balance out the pressure forces, and if the pressure forces are equal then the walls are unstressed, all by Newton's first law, which is the macroscopic view.

You really need both Newton's laws and the kinetic theory, neither will do alone. But you need to properly understand the way that Newton's first law is used here.

 

[rcgldr]

The Earth and gravity act as a container, with gravity preventing the outside layer of air from escaping and the surface of the Earth opposing the weight of the air with an equal and opposite force.

The weight of the air decreases with altitude, both inside and outside the container. The pressure within a sealed container decreases with altitude, more pressure at the bottom, less at the top, and the net downforce due to this pressure is exactly equal to the weight of the air inside the container.

For example, imagine a 1937.5 foot tall column of air with 1 in^2 cross sectional area with it's base at sea level. The weight of the air in this column is 1 lb. The pressure at the bottom of the column is 14.7 psi and the pressure at the top of this column is 13.7 psi. It doesn't matter if this column of air is contained or not.