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Is gas pressure caused by the weight of the gas above?

  1. Dec 19, 2013 #1
    I have two closely-related questions, which are probably just one question:

    1. On the National Weather Service website (http://www.srh.noaa.gov/jetstream/atmos/ll_airweight.htm),it [Broken] says

    "When we measure air pressure, we are measuring the weight of a column of air 15 miles (24 km) high directly over us."

    So what are we measuring when we measure the air pressure in the International Space Station?

    2. If I took an open screw-top container down a hundred meters below water and put a pressure-gauge inside it, the gauge would, presumably, measure the weight of the liquid above it.

    Would the reading then change if I screwed a strong top tightly shut onto the container?

    Presumably the liquid inside the container is no longer supporting the weight of the water above it (the top is doing that); and the liquid inside the container has not been compressed very much, so the pressure it is exerting on the gauge it is surrounding does not come from being compressed by the (previous) weight of the liquid above it (unlike, I assume, the identical case for a pressure gauge inside a sealed container of air, on the surface).
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  2. jcsd
  3. Dec 19, 2013 #2


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    The pressure of the atmosphere indicates the amount of gas located over a unit area. The pressure of a gas enclosed within a container, like a car tire or the ISS, is proportional to the amount of gas within the enclosure. According to the gas laws, pressure, volume, and absolute temperature are all related.

    Your screw top container example is more complicated. The pressure of any gas within the container will not change unless the hydrostatic pressure acting on the container causes it to compress and change the volume inside the container. See the explanation above about how gas pressure is influenced by amount of gas,volume of gas,and temperature of the gas.

    When we inflate a car tire, the change in pressure is caused by adding an additional amount of air to what is already inside the tire. If we add too much, then we lower the pressure by venting a small amount until we reach the correct pressure.
  4. Dec 19, 2013 #3


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    if you consider any volume of a fluidl (gas or liquid), the pressure in balanced by the forces acting on the surface that encloses the volume.

    If you imagine a small cube of air at ground level, one of the forces is the weight of the 15 miles of air above it. That is the basic reason for the air pressure at ground level, and the fact that the air presssure decreases at higher altitudes.

    But if you take a volume of air enclosed in a solid container, the forces on the air can be balanced by the stresses in the solid. If a car tire is pumped up to 30 psi above atmospheric pressure, the outside surface of the tire has a stress normal to the surface of 14.7 psi (balancing the atmospheric pressure) and the inside has a stress of 44.7 psi (balancing the internal pressure of 30 psi above atmospheric.).

    There are other stresses in the material of the tire as well to balance the change in stress through the thickness, but that's getting away from the question.

    So, the answer to question 2 gets complicated, because it depends how well the container can resist the stress gradients through it. If the container is completely full of liquid, any change of pressure will compress the liquid slightly, and also compress the container slightly. If would be fairly easy to work out the internal pressure for a simple case (e.g. a spherical container) but the answer would depend on the material properties of the container and the fluid, and the thickness of the container. For a more complicated shape like a screw top bottle, you would have to make a computer model to get the answer.

    But if there is some gas inside the container, the gas will probably be much more "flexible" than the container. If the container can withstand the pressure without collapsing in on itself (e.g. the hull of a submarine), the gas pressure inside won't change much. If the container does collapse, the internal and external pressures will probably be close to each other.

    The ISS is effectively a balloon, made out of metal, with a pressure difference between the air inside, and (almost) zero outside. I don't know about the interior of the ISS, but space suits are often filled with gas at lower than atmospheric pressure, so they can be made lighter and more flexiible than if they had to withstand the full 14.7 psi. The oxygen content of the gas inside is then increased proportionately, so the occupant's lungs are getting the same "partial pressure" of oxygen as they would be on earth, but there is less inert nitrogen.
  5. Dec 19, 2013 #4


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    It is not real clear what you are saying here.

    I work with oceanographic research instrumentation designed for deep water (2000-6000m) . If we have an instrument leak at depth, upon returning to the surface the cans, which are quit strong when the pressure is applied to the outside, can be severally damaged by internal pressure. Opening such a can is hazardous due to the stored pressure inside. Note that is is for a compromised can. If a container is opened at depth then closed, it will have that pressure inside upon returning to the surface. The only forces containing it are the tensile strength of the container.
  6. Dec 19, 2013 #5
    SteamKing: Thank you for the quick answer!

    I'm not sure it directly addresses my questions, especially the second one. In particular, by 'located over a unit area', do you mean 'directly impinging on it', i.e. the layer of molecules which at any given time are slamming into it? If this is the case, how is this the same as 'weighing the column of air over it', except perhaps in an indirect sense (because the 'density' of air molecules is a function of their location in the 'column of air'.

    What about the closed container of liquid? I didn't think that liquids got noticeably denser with depth, and thus the pressure in a liquid is a direct result of the weight of the liquid above it, not a function of some depth-related liquid density, as opposed to a gas -- so what happens when we remove this weight by putting a tight lid on the jar?

    AlephZero: What I am trying to understand is this: is there any difference in the causes of pressure exerted by gases, and pressure exerted by liquids? It seems to me that there is -- that's why it's the "Gas Law" and not the "Fluid Law".

    Gases exert pressure by slamming into a surface. The more of them there are per unit area, and the faster they're going, the greater the pressure. How many molecules are above them is only indirectly relevant: an individual molecule hitting a surface is not 'transmitting' the weight of molecules over it, the way a liquid molecule at depth is.

    And would I be right in saying that gravity has little or nothing to do with why the air gets less dense as you go up, except for its role in attracting air molecules towards a common center? If we envision starting with a large (but finite) volume of molecules, moving randomly, which are then attracted to a common point, isn't it just a matter of geometry that -- if they continue to move randomly -- they will be more closely packed around the center point than out at the edges?

    Integral: So you are saying that a container which has had liquid under great pressure admitted to it, as in the two examples you gave, will -- if sealed again -- have that same (high) pressure when it is brought up to the surface? I am assuming that dissolved gases play no significant role here, and I am assuming that water is not significantly compressible, so ... where is this extra force coming from?

    Everyone: thank you for taking the time to try to enlighten a non-scientist.
    Last edited: Dec 19, 2013
  7. Dec 19, 2013 #6
    If the container were perfectly rigid, the reading on the gauge would not change. In fact, it would not even change if you brought the container back up to the surface. The water within the container would remain slightly compressed (enough for the pressure to be higher). Once the lid is opened at the surface, however, the liquid will expand slightly, and the pressure will return to the atmospheric value.
  8. Dec 19, 2013 #7


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    Where the hazards arise if the container has trapped air. This, due to the rigid container, will be at surface pressure. Slowly leaking high pressure water in will compress that air. Now when returned to the surface that trapped air will be at high pressure. This can be very dangerous. If the container is opened and completely filled with water at depth, then the only pressure will be due to the differential expansion of the water and container.

    When I conduct a pressure test to 6000m I pump about 1l of water into our already full pressure bom. (Left off the last "b" to avoid NSA snoops) Our pressure test vessel has a 40l capacity.
  9. Dec 19, 2013 #8
    Here’s an interesting thought problem on this subject:

    Postulate a tower whose top story contains a structure whose horizontal cross-section is in the shape of an airfoil. This structure can be rotated so that the leading edge of the airfoil faces directly into the wind. The structure contains four rooms that are airtight except for a single opening to the outside. Each room contains a calibrated barometer. The room openings are at the leading edge of the airfoil (the room containing barometer A), the trailing edge of the airfoil (barometer B), the point of maximum curvature on the curved length of the airfoil (barometer C) and the midpoint of the straight side of the airfoil (barometer D).

    There is a 25 meter-per-second wind blowing, and the structure has been rotated to face the leading edge of the structure into the wind. At the same time, a balloon is drifting past the tower bearing a basket whose elevation is the same as the tower rooms. This basket also contains a barometer (barometer E). All of the barometers may be read remotely and at the same time by an observer.

    I maintain that each barometer will simultaneously measure a different atmospheric pressure reading.

    Question: Which of these five barometers (if any) measures “the weight of the overlying air”?
  10. Dec 19, 2013 #9
    So, the increased pressure in a rigid container which was completely filled with water (no air to compress), then sealed up, and then brought back to the surface, would be entirely due to the compression of the water?

    I found this description of bringing up an early (unoccupied) bathysphere very interesting, although it seems that the enormous pressure may have been due at least in part to compressed air: http://www.pmel.noaa.gov/eoi/nemo1998/education/pressure.html

    However, what puzzles me is that we are normally taught that water -- like all liquids -- is incompressible:

    Here is what another government site says:


    "Water is essentially incompressible, especially under normal conditions. ... But, squeeze hard enough and water will compress—shrink in size and become more dense ... but not by very much. Envision the water a mile deep in the ocean. At that depth, the weight of the water above, pushing downwards, is about 150 times normal atmospheric pressure .... Even with this much pressure, water only compresses less than one percent."

    Yet it seems counter-intuitive that you could get such huge effects, opening the sealed container, from the water inside expanding by one percent.
  11. Dec 19, 2013 #10


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    What is counter-intuitive about this? This is exactly what "practically incompressible" means: huge pressure for little volume reduction.
  12. Dec 19, 2013 #11


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    Did you actually read any of the article? The Bathysphere was indeed filled with water and highly compressed air.
  13. Dec 19, 2013 #12
    Yes. The Bathysphere was indeed filled with water *and highly compressed air*.

    Recall that I'm trying to establish if there is any difference between the cause(s) of pressure in liquids, and the cause(s) of pressure in gases.

    So my question is: had the Bathysphere been filled with water alone, what would have happened?

    The consensus here seems to be that the 1% or so compression of water would have been sufficient to produce the dramatic effects described, without any air being involved. For example, 1600 litres of depth-compressed water [ my back-of-the-envelope calculation of the internal volume of the original Bathysphere, described in the link above] expanding to 1616 litres would blow the hatch bolt across the deck and produce 'a solid cylinder of water' shooting through the hole in the door.

  14. Dec 19, 2013 #13


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    Do not forget that the bathysphere itself will contract with depth then expand on decreasing depth. I will bet that the internal volume will increase MORE then the expansion of the water, meaning that if the container if filled at depth, there will be a DECREASE in internal pressure with decrease in depth.
  15. Dec 19, 2013 #14
    Ok. I think Chestermiller (above) has answered one of my questions: the weight of water (or, presumably, any liquid) above a certain depth compresses, even if just a little, the liquid below it. If this liquid is inside a sufficiently-strong sealed container, it will retain that increased pressure. This is not in contradiction to the statement that the pressure increase as one descends below the surface of a liquid is due to the weight of the liquid above.

    Now: what about the pressure in a gas? Is it correct to say that the pressure increase in a gas as one descends towards the point towards which it is being pulled (by gravity, say) is due to the weight of the gas above it, as opposed to the increasing density of the gas?

    The reason I ask is that I find it hard to visualize weight being transmitted 'downwards' by particles which are only in intermittent contact as they bounce off each other (unlike the case for a liquid or a solid, where I visualize the particles as in contact with each other, although with a liquid it is a sliding sort of contact).

    Or is it the case that the pull of gravity accelerates particles in a downwards direction, and thus increases the force they exert on the ones they hit below them, and decreases it on the ones they hit above them, and that this is how the weight of the ones above is transmitted?

    Again, I thank everyone who has taken the time to respond.
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