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I have a bit of misconception about weight which I want to clarify.

The air pressure is explained as the weight of the air column above our head acting per unit area. But since air is not continuous how can the weight of all the air molecules (above our head) be acting on our head? I mean we would only feel the weight of the molecules near the surface of our head (if not then why not?) but how do we feel the weight of molecules so far away?

I have edited this question and asked this follow up question (so that it remains specific).

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    $\begingroup$ You could equally ask why water pressure affects us because water isn't continuous. $\endgroup$ – Charlie Dec 8 '19 at 18:41
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    $\begingroup$ @Charlie I would love to ask that question but I think an answer for this question might answer that case! Would you mind answering why non-continuity isn't a problem? $\endgroup$ – user238497 Dec 8 '19 at 18:43
  • $\begingroup$ I've added an answer, I'll try and clear anything up if you have follow up question on it. $\endgroup$ – Charlie Dec 8 '19 at 19:34
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    $\begingroup$ @Christopher I have this view because I think that the molecules of air have large separation between them (an I thought this separation as vacuum) but as from your comment it seems that that isn't the case. Also it might be due to the fact that I haven't yet got any precise (or even rough) definition of continuous to serve the purpose of distinguishing the things. Would you be able to provide one? $\endgroup$ – user238497 Dec 10 '19 at 0:40
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    $\begingroup$ @TheLastAirbender The molecules of air does have large separation between them, and you could define the space between them as vacuum, if you so wish. What you are not considering is how humongously many there are of them. And how fast they on average move. It is a gigantic explosion of tiny collisions going on, every single microsecond - which we experience merely as... pressure. $\endgroup$ – Stian Yttervik Dec 10 '19 at 14:34
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Imagine that the air in the atmosphere was just somehow sitting there unpressurised. What would happen?

Well, Earth's gravity would be attracting all that air towards the centre. So the air would start to fall downwards.

The very bottom layer of air would be prevented from falling through the solid surface, as the air molecules rebound off the molecules of the surface. But the layer above that doesn't stop. So Earth's gravity forces the air in the lower part of the atmosphere to accumulate against the surface of the planet, becoming more and more dense.

As the air gets denser near the surface, it becomes more and more likely that air molecules collide. That's what air pressure is: the average force of all those air that would hit a surface you placed in the air. But the air pressure also acts on the air itself. So eventually the force of the air pressure at the bottom layer of air pushes up on the layer of air just above it enough to counteract the pull of Earth's gravity on that layer of air. And so you get another layer that is prevented from falling.

But the air above that is still being pulled down, and so more air is being squashed down into this second layer above the surface. This increases the force that the bottom layer needs to provide to the next layer; the air molecule collisions not only need to provide enough force to counteract the weight of the air immediately above it, but also to provide those molecules with enough momentum that when they in turn collide with the air in the third-bottom layer it can support the weight of that layer as well. So more air squeezes down to the surface until the pressure at the bottom layer is sufficient to support the weight of the 2 layers above that.

Obviously the atmosphere isn't actually split into discrete layers like this1, but hopefully it's a helpful way to think about it. You should be able to see how gravity squeezes the air down against the solid surface, until the pressure at the bottom is just enough to support the weight of all the air above it.

This is why air pressure drops off at higher altitude. As you go up, there is less air above squeezing down, so equilibrium with gravity is reached at a lower pressure.

So it's not literally that the air pressure you feel is the weight of the column of air above you. It's not that your head is somehow "holding up" a 100km column of air above it. But the air pressure of the air surrounding you must provide an equivalent force to the weight of all the air above it. If it did not then the weight of the air above would be partially unsupported, so gravity would squeeze it down further, increasing the pressure until it was equal to the weight of all of the air above.

This is also why the top of your head doesn't feel any difference in air pressure to the side of your body. Air pressure is the same in all directions, because the air molecules are really just zipping around in countless different directions, uncoordinated with each other. Those molecules colliding with things must supply enough average force in the upward direction to support the weight of the atmosphere, but when the pressure increases due to gravity it can't cause a coordinated force that is only upwards, so there is just as much force from air pressure on the side of your body as there is on your head.


1 And if you actually had the atmosphere of Earth spread out in a diffuse low pressure cloud and let it all fall under gravity the results would be much more exciting than I have described.

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  • $\begingroup$ Well, there are layers in the atmosphere, but they are pretty big, and not directly caused by the pressure gradient. And there is technically a minute, insignificant difference in pressure between your head and your feet. You don't feel it because it's just too small. And of course, wind and other air currents overwhelm it. $\endgroup$ – Spencer Dec 10 '19 at 21:30
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I think it is misleading to explain air pressure as the weight of the air column above our head acting per unit area.

Actually your head only feels those air molecules which are directly hitting the surface of your head. You don't feel any air molecules which are further away.

As explained in Wikipedia - Kinetic theory of gases - Pressure and kinetic energy:

... the pressure is equal to the force exerted by the atoms hitting and rebounding from a unit area of the gas container surface.

Using this approach the quoted article derives this formula for the gas pressure $P$: $$ P = \frac{Nm\overline{v^2}}{3V}, $$ where $m$ is the mass of a single gas molecule, $\overline{v^2}$ is the average square velocity of the molecules, and $N$ is the number of molecules per volume $V$.

From this formula you can see quantitatively, that gas pressure is caused by the speed of the hitting molecules, and it increases with the density of the gas ($Nm/V$).

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  • $\begingroup$ I have edited my question, you can have a look. $\endgroup$ – user238497 Dec 8 '19 at 18:46
  • $\begingroup$ @ThomasFritsch said “I think it is misleading to explain air pressure as the weight of the air column”. I agree, and it seems that this misleading statement causes a lot of confusions. The weight of the air column gives the change in pressure, not the pressure. And even then only when it is in hydrostatic equilibrium. $\endgroup$ – Dale Dec 9 '19 at 0:40
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    $\begingroup$ "Actually your head only feels those air molecules which are directly hitting the surface of your head. You don't feel any air molecules which are further away." It might be unpedagogical to explain air pressure as the weight of the column, but I don't think that's a good explanation. If you pile a bunch of weight plates on top of your head, you'll only feel the molecules that are directly hitting the surface of your head, not any of them further away. So is it wrong to say that you feel the weight of the entire pile? $\endgroup$ – JiK Dec 9 '19 at 19:02
  • $\begingroup$ @JiK At the end the pressure force exerted by the bottom-most plate to your head is proportional to the weight of the whole pile of plates. So as a short-cut you are right to say, you feel the weight of the whole pile on your head. But this is due to a subtle interaction between gravity and the plates, and not due to some spooky action-at-a-distance of the higher plates to your head. $\endgroup$ – Thomas Fritsch Dec 9 '19 at 20:36
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Consider blocks of equal mass, stacked on each other.

The block at the top ‘feel’ the force of gravity and normal force from the block below. The block just below feels force of gravity (it’s weight) plus the force acting from the box above by Newton’s $3^\mathrm{rd}$ Law (if it ‘provides’ normal force on box A, box A acts on it with equal magnitude, but in opposite direction force). The result is the greater normal force.

If you continue with this stack and place yourself or any other ‘object’ below this stack, you will conclude that each block ‘contributes’ to the total force that is acting on you. Air molecules, although much smaller, must obey Newton’s Laws, too.

Your last assumption would be mostly correct if gravity field near air particles would be negligible.

Please note that atmospheric pressure varies due to various other factors, but this simple model should explain the principle.

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    $\begingroup$ Of course, this assumes that there is lots of interaction between the boxes. Is that a good assumption for air? $\endgroup$ – Luaan Dec 9 '19 at 8:30
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    $\begingroup$ This is a good assumption. If the air was not interacting, it would be pulled down by gravity until it was. $\endgroup$ – Oscar Smith Dec 9 '19 at 19:15
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You don't feel the air pressure of the air around you (and hence don't feel the "weight" of air above you) because the pressure inside your body is pushing out with the same force (assuming you're breathing normally).

This is different from, say, what you would experience if you held your breath and dove 20 feet down into the ocean -- then you'd be able to sense the water pressing on you from all directions. (But note that if you don a SCUBA mask and dive the pressure in your body will again be "equalized" with the pressure outside, and you will not sense the pressure (except perhaps in your ears).)

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  • $\begingroup$ Your ears are also connected to your breathing. Unless you have trouble with your eustachian tube, the pressure inside the ear should equalise with the outside pressure quickly enough (just like when you're driving down a hill, or dropping with an airplane). Even without breathing, there is some equalisation - our lungs change shape, though they cannot collapse completely. $\endgroup$ – Luaan Dec 9 '19 at 8:36
  • $\begingroup$ @Luaan - But, as you say, the eustacian tubes may be clogged, plus the cochlea may contain bubbles that compress. $\endgroup$ – Hot Licks Dec 9 '19 at 13:07
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Pressure is the consequence of gas particle colliding with the walls of the container, or in this case with you. Gas particles are encouraged to collide with you by other particles that do not have direct contact with you.

Particle hits you and bounces back, then the particle hits other particle, bouncess off it and it may hit you again. The higher pressure the more often you are hit by a has particle. The more gas is above you, the more likely it is that a gas particle will return to hit you again.

You are correct that you are not directly hit by particles by 10 km away, but those particles for other particles to hit you more often, hence higher pressure. Of course the gravity plays a role here.

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    $\begingroup$ But this argument would also work if you have a horizontally long container, but we know that in such a case the pressure is the same $\endgroup$ – Wolphram jonny Dec 8 '19 at 15:35
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    $\begingroup$ Yes, but there is a difference: if you add more particles horizontally they do not change anything, as there is no difference whether say 10 m away there is a wall or there is more gas. If you add layers vertically it matters, because of the gravity: gravity acts on all particles. So if gravity forces a "high altitude" particle down, it bumps into "lower altitude particle" and kicks it even more down. $\endgroup$ – Lacek Dec 8 '19 at 16:57
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    $\begingroup$ In the vertical arrangement the effects of gravity add up, in the horizontal they do not. $\endgroup$ – Lacek Dec 8 '19 at 16:58
  • $\begingroup$ so there is a preferred motion down,? this must also increase the particle speed of those going down, So more temp and pressure because it exchanges potential energy as it moves down? $\endgroup$ – Wolphram jonny Dec 8 '19 at 17:16
  • $\begingroup$ I have edited my question, you can have a look. $\endgroup$ – user238497 Dec 8 '19 at 18:47
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Imagine

  • A very tall square bottomed container of 1 unit "floor" area.

  • A weighing device in the bottom of the container that weighs objects placed into the container.

  • The availability of a number of 1 unit a side cubes each of 1 unit mass.

Add one cube to the container - the weighing device will show one unit weight.
Add a total of 10 cubes -> 10 units weight
Add N cubes - 100 or 1000 or .... - and weight will increase proportionally.

Now provide an array of such containers in a 100 x 100 array, or 1000 x 1000 or .... . Placed adjacent to each other, if each container has say 50 cubes in it you'll get 50 units of weight on each weighing device - = 50 units per unit area.
Overall if there are say 1000 containers there will be 50,000 units of weight spread over 1000 weighing devices - so you'll still get 50 units of weight per unit floor area in this example.

No remove the container sides - as long as you have perfectly balanced the cubes (good luck with that :-) ) the weights shown will not change.

Now consider replacing the cubes with cubes of air of finite weight (about 1.2 kg per cubic metre at sea level). Or use 1 cm^3 containers. Or individual molecules.

The containers and cubes analogy is a reasonably good fit to the air pressure situation. It's not perfect - "All models are wrong. Some models are useful" - but good enough to steer you on the path to a complete answer

*-George Box.

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The very top air molecule of our atmosphere is attracted by gravity. So it has a weight.

  • The second-top-most molecule thus must carry the weight of the top-most molecule.

  • The third-top-most molecule must hold back against both the force on the second-top-most molecule as well as the weight of it - so it must carry two air molecule weights.

  • The forth-top-most molecule must carry the force on the third-top-most molecule plus the weight of it - so it carries three air molecule weights.

And so on. In this way, the weight that must be carried - the force that is applied - increases the closer you are to the ground.

Your head, shoulders and body carries/withstands the weight of the air molecules directly above as well as the force that is pushing down on them. So, the weights if all air molecules above you all the way up. Just like you hand carries both the tray and the cups that are weighing down on the tray.

Why we technically can think of it as the simplee vertical column of air above us and all the way up, can be a technical discussion. But the fact that we indeed are carrying more air molecules than only those that touch us, should now be clear.

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Answers in words have already been posted, but perhaps an answer with equations may also help, so I will add this.

Consider a thin slice of fluid, whether liquid or gas, in a uniform gravitational field. Say the bottom of the slice is at height $z$ and the top at height $z + w$ where the width $w$ will be small (you can call it $\delta z$ if you like). The upwards force on the bottom of this slice is $p(z) A$ where $p(z)$ is the pressure at height $z$ and $A$ is the cross-sectional area of the slice. The weight of the slice is $m g = \rho A w g$ where $\rho$ is its density. The downwards force on the top of the slice from the material above it is $p(z+w) A$. In equilibrium the forces are balanced so $$ p(z) A = p(z+w) A + \rho A w g . $$ This equation is saying that the pressure at $z$ is just enough to balance the pressure at the next region up, plus the contribution of the weight of the slice. The equation applies to all the slices in the air or other fluid; each one thus supports the material above it. In a gas this pressure is transmitted from one layer to another through collisions of the molecules. It is not necessary for the molecules at the very top to hit the ones at the very bottom; it is enough for each layer to hit the ones next to it. In the case of a gas the result is that the density increases as you go to lower heights, and this is why you feel a substantial air pressure at sea level.

If you are happy to do a little differential calculus then you can write $$ p(z + w) \simeq p(z) + w \frac{dp}{dz} $$ where the approximation becomes exact in the limit where $w$ tends to zero. Using this in the above equation we have $$ \frac{dp}{dz}= - \rho g $$ Thus the pressure gradient goes as the density times gravity. In an incompressible fluid such as water, this is easy to solve because the density is the same everywhere. Then you get $$p(z) = p(0) - \rho g z. $$ Since $z$ here is the height, this is saying the pressure increases with depth in proportion to the depth (and density and gravity).

In a gas the density is itself a function of height so the result is more complicated. One easy case is when the temperature is the same at all heights (not true for Earth's atmosphere by the way). In that case $\rho$ is proportion to $p$ and one finds the solution is an exponential function.

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The reason air pressure acts on us as humans really has nothing to do with mathematics, its science. Our bodies are more water than mass, so the pressure exerted on us by air pressure acts on our blood and other body fluids, making increases and decreases in our own pressures. Its really only common sense - this is why air pressure changes cause headaches! I should know!

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