# Moving air up a column without lifting

Say we have n sealed tubes segments $$h$$ in length, capable of withstanding at least 1 atm of pressure (this means that at a pressure differential of 1 atm, the segment will not implode or explode) connected together with valves, also rated to hold back at least 1 atm of pressure, that can be open and closed via a computer between them and have this structure stand vertically. The ends also have valves to the outside.

Initially, all of the valves are open so the pressure in each segment would be the pressure that would be at a given height in a non-closed system ($$P = \rho h g$$ where $$\rho$$ is the density of the gas, $$h$$ is the height of the gas above it and $$g$$ is the gravitational force, note that this may only work for small $$h$$ where $$g$$ has a negligible difference between the top and bottom).

Now, if we close all of the valves, the gas in each section doesn't have the weight of the gas above it pushing down (or even the pressure of the atmosphere pushing in). This would mean that the pressure at a lower segment would be higher than the one above it for no reason other than that was it's initial state.

So if we then open a valve between two sections, since the entire column of gas above is no longer pushing down on the two sections, the sections should just equalize in pressure, averaging between the two. For small $$h$$, this would be $$PV = nRT$$. Once the pressure has equalized, the valve would be closed. This would result in an increase of the pressure in the section above and decrease the section below relative to the state before opening and closing the valve. This would effectively be moving the gas up the column.

This would indicate that we can move gas up the column without actually lifting it, which doesn't sound right. What am I missing?

• Closing the valves doesn't change the dependence of pressure on height, nor does it cause the pressure in a given chamber to equalize. – probably_someone Oct 27 '18 at 17:43
• @probably_someone, I thought that the pressure is based on the weight of the fluid above it. Hence $P = \rho h g$. – Adrian Oct 27 '18 at 17:45
• The air in each cell pushes on whatever is above and below. When the valves are open, that is air. When closed, it is the valve. – mmesser314 Oct 27 '18 at 18:10
• @mmesser314, if I have two tanks of gas where one has a lower pressure than the other, and they are connected by a valve. If I open that valve, the tanks will equalize to the same pressure, moving gas from the higher pressure tank to the lower pressure tank. How is this different? – Adrian Oct 27 '18 at 18:36
• You didn't start with an open cell at room pressure. Gas in a tank is compressed. Also, if one tank is above the other, they will be almost the same pressure. Pressure differences because of a foot or two of altitude is often ignored because it is so much small than the pressure difference inside and outside of a tank. – mmesser314 Oct 27 '18 at 18:39

Now, if we close all of the valves, the gas in each section doesn't have the weight of the gas above it pushing down (or even the pressure of the atmosphere pushing in). This would mean that the pressure at a lower segment would be higher than the one above it for no reason other than that was it's initial state.

Not true. The air has mass and therefore weight. So small differences in height produce small differences in pressure. We can see this with the formula $$P = \rho h g$$. For the atmosphere near sea level, $$\rho$$ is about $$1.225 \ \text{kg m}^{-3}$$. This means the pressure difference between the top and bottom of a 1m tall chamber is $$1.225 \text{kg m}^{-3} \cdot 1\text{m} \cdot 9.8\text{m s}^{-2} = 12\text{Pa}.$$

So even with the valves open, allowing the pressure to "equalize", you'll still expect a 12 pascal pressure drop for every meter of height (at least at first). Since this is the same drop that you have initially, there's no movement of air as you open and close the valves.

However, if you block of a section from the air above and below, you've created a closed system which is influenced by gravity, but not the weight of the air above it. So at most, the bottom of the section will be influenced by the height of the section, which is much less than the height of the air above it in the open system.

Depends on what you mean by "influenced". The pressure inside the vessel is the same as it was when the system was closed (assuming the vessel is perfectly rigid and we're not allowing heat transfer).

So you could certainly move the closed system to another altitude and it would have a pressure that differs from the atmosphere at that point. But that didn't seem to be your question.

You were asking if opening the valves where they started would cause air to move. It won't move because even though the section above might not be open to the atmosphere, the pressure inside is the same as the atmosphere at that altitude. Instead of the weight of the atmosphere, the rigidity of the chamber provides the pressure. The pressure at the bottom of the upper vessel is the same as the pressure at the top of the lower one. Therefore there is no pressure difference and the gas will not flow from one to the other.

I think one of the the problems I was having is that at the top of the closed off section, P=ρhg, but since h would be 0, that would imply that P=0 and was messing me up.

In this case $$P$$ is the increase in pressure due to the mass of the fluid above. At $$h=0$$, the mass of the fluid doesn't add anything to the pressure. But it doesn't mean the pressure is zero.

Ok, so then what is the formula for pressure of a closed vessel at different heights in a vessel. There needs to be some consistency or else you can rationalize anything.

First, lets imagine pressurizing a vessel in freefall. No weight to worry about. Unless it's very cold or very high pressure, we can get pretty close to the pressure by using the ideal gas law $$P = \frac{nRT}{V}$$. This pressure is created just by the motion of the molecules inside the vessel.

Now, we can place this vessel inside a gravitational field. If we make the assumption that it's not so big that the density at the top changes dramatically, then the pressure at the top will be just about the same.

But at the bottom, we have the same pressure as the top, plus the added weight of the gas within the vessel. That weight is supported by the bottom of the vessel, so the pressure is increased there.

Now consider a container of air on the ground a meter tall. We can consider the pressure in the vessel in a couple of equivalent ways:

• The pressure at the bottom of the vessel is due to the weight of 1m of air in the container, plus the overall pressurization of the vessel to 1atm.

• The pressure at the bottom of the vessel is due to the weight of 1m of air in the container, plus the weight of the air above, extending to the edge of the atmosphere.

When the valves are open, the added pressure in the container is directly caused by the column of atmosphere above. When the valves are closed, this connection is broken and the two values can diverge.

If, by closing the valves, you don't have the weight of the gas above the vessel pressing down, then the pressure at the bottom of the vessel should now be less.

Imagine you have a stack of 50 heavy springs. The springs at the bottom are quite compressed due to the weight of the springs above. Now you take a rigid box and slide it so that it exactly holds the bottom spring. If you now take the other springs off the box, the bottom spring doesn't expand, it stays just exactly as compressed. It's being held by the box now, not the other springs.

This is the same situation as your scenario. If the valves are open, the box has no lid. The air pressure is created by the weight of the air above. If the valves are closed, the box has a lid. The air pressure or spring pressure is created by the rigid container. If you move either to a new location, the contents remain at the same internal pressure. If you open the container at a new location, the pressure will equalize to whatever is outside.

Simply closing the valves, or sliding the box on top of the spring doesn't change the pressure. You'd have to compress or expand the container to do that.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Nov 1 '18 at 16:07
• What happened to the messages? The chat is also gone. :( – Adrian Nov 12 '18 at 19:54

Air acts something like a lot of little boxes made out of springs. Each box presses outward against whatever is next to it. Often this is nearby boxes of air. Sometimes it is the walls of a container. If nothing is moving, something is pressing back on each little box with the same force as the box presses outward.

If you cram more boxes into a container, the springs are compressed. Each box shrinks and presses harder. The volume of each cube goes down. The pressure goes up.

Each box has a small mass. At atmospheric pressure, it is just about $$1000$$ times less dense than water. A liter of water is $$1kg$$. $$1 m^3$$ is $$1000$$ kg, or a metric ton. For air, the mass of a liter is about $$1 g$$ and $$1 m^3$$ is about $$1 kg$$.

If you tried to hold up a trash bag full of water, you would have a hard time. But it is effortless underwater. Water is slightly different than air (Imagine very stiff springs that barely compress or stretch at all.) But the trash bag is held up by forces from the water around it. If you were a fish, you wouldn't really be aware that water has mass. Likewise, a trash bag full of air is held up by the air around it and we are not very aware of the mass of air.

Air pressure at sea level is from the weight of all the little boxes above. The mass of all the boxes above a $$1 cm^2$$ patch of ground is about $$1 kg$$. The mass above $$1 m^2$$ is about $$10^4 kg$$.

This weight presses on the outside of an oil drum. It would be easily enough to crush the oil drum, except that air inside presses equally on the inside of the drum. It doesn't matter if you close the cap of the drum or not. The air inside pushes outward with the same force either way.

The volume of a box with a $$1 cm^2$$ base and a height of $$1m$$ is 1 liter. The mass of all the air above the base is about $$1 kg$$. The mass above the top is about $$1 g$$ less. This difference is small enough that it is easy to ignore. But it is real.

It is noticeable if you go up a mountain. Almost half the air is below the top of Mt. Whitney. Because of the decreased weight of air above, the springs in each box of air are less compressed. A box has about twice the volume as it would at sea level. Conversely, a 1 liter box has about $$1/2 g$$ of air at the top of Mt. Whitney.

Suppose you had an open tube of air. You want to increase the pressure at the bottom, so you push more air in the bottom. The new air pushes upward on the air above it, and all the air in the tube moves up. Some air flows out the top. As it does, each little box of air has less air above it. It expands a little and pushes outward a little less. It becomes like the air that was at that height.

Suppose you have a rigid closed tank of air at atmospheric pressure. The walls of the tank press inward on the air in the tank just as hard as surrounding air would have if the tank wasn't there. The force on the bottom is the force on the top + the weight of the air in between.

Suppose you push some more air in the tank. The volume of the tank doesn't change. So each box of air is compressed and presses outward harder. Pressure goes up. The force on the bottom is still the force on the top + the weight of the air in between. But now the force on the top is larger, and so is the weight of air between the top and bottom.

For a normal size tank of air, the weight of the air in the tank is generally much smaller than the weight of the air above it. The differences in forces and pressures in a tank of air are usually ignored. A tank of water is different. The weight of water in a big tank can exceed the weight of the air above it.

I have glossed over some properties of air. For example, if you heat air, the springs press harder. If air expands, it cools.

• Suppose you had an open tube of air... Some air flows out the top. unless the tube goes higher than the atmosphere, then it should just continue to go higher. Suppose you have a rigid closed tank of air at atmospheric pressure. If the tank can separate 1 atm from a vacuum without exploding, then by definition the force couldn't be transmitted through the wall/valve. – Adrian Oct 28 '18 at 20:04
• True, I wasn't thinking of a tube that tall. And a rigid tank is strong enough to withstand a pressure difference without deforming. The forces come from the strength of the wall. An oil drum is rigid when the pressure difference between inside and outside is small. Not so much when it is 1 atm. – mmesser314 Oct 28 '18 at 21:51
• So if each container and valve is capable of withstanding at least 1 atm, then the pressure should only be determined by $PV=nRT$, regardless of $r$ assuming that $h$ is small. So does that change your answer? – Adrian Oct 28 '18 at 22:56