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Free fall in compressed air

  1. Sep 26, 2010 #1
    If air would be compressed to have equal pressure to that of water would objects which are buoyant in water be buoyant in such compressed air too or would they still fall down due to Earth gravity?

    Or asking differently, what would have to be the pressure of air (lower, equal or higher than that of water) to make, say wooden ball, stop falling in it and instead go up with the same speed as if it were in water?
    Last edited: Sep 26, 2010
  2. jcsd
  3. Sep 26, 2010 #2
    I don't think it's really a matter of pressure, but density. So for a wooden ball to float in air as it would in water, you would have to compress air (according to the ideal gas law) until it reaches the same density as water.

    [tex]p = \rho \cdot \frac{RT}{M}[/tex]

    p -- pressure (Pa)
    ρ -- density (kg/m^3) [= 1000]
    R -- specific gas constant (J/[kg·K]) [~ 286.9]
    M -- molar mass (kg/mol) [~ 28.97e-3]
    T -- temperature (K) [= 288.15]

    The answer would be approx. 2.854 GPa (more than 28,000 times the normal atmospheric pressure!).

    Someone correct me if I'm wrong.
    Last edited: Sep 26, 2010
  4. Sep 26, 2010 #3
    At the surface of any body of water, the pressure of the air is actually equal to the pressure of the water!

    What makes an object float in water has to do with the density. The high density of water causes a very large pressure gradient. It turns out that the pressure only a foot down is so much greater than the pressure at the surface that the resulting force (buoyancy) is enough to counteract the force of gravity.

    If you were to compress a gas to have the same density as water, then it would form the same pressure gradient, and any object which could float in water could float in the gas too. However, this would require huge amounts of pressure!

    Now, there's a stumbling block that would prevent you from getting air at exactly the same density as water. It would need to be in the liquid phase to be so dense, and that liquid would likely not be very compressible.

    EDIT: ninja'd!
  5. Sep 26, 2010 #4
    In relation I'd ask this...

    Aren't there submarines with the same internal pressure of air as the outside water pressure, so when deep under water a diver can go in/out of water through opening?

    Or is that solved in a different manner? So that pressure and density of internal air is lower than the pressure and density of external water, but if so, how is water prevented from filling up submarine with water?
  6. Sep 26, 2010 #5


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  7. Sep 26, 2010 #6
    And this...

    Imagine a tube in 'U' shape being closed on both ends. Left side of tube is filled with air, right side with water. To keep air and water separate exactly at the bottom-middle what would need to be the pressure / density of air?
  8. Sep 26, 2010 #7
    And how that works? Can you describe it shortly or point me to some website covering that? (I've googled it but no luck...)

    When a diver jumps in water, through that opening, is as much water as the diver's weight pushed inside the submarine and perhaps they just pump surplus of water out?
  9. Sep 26, 2010 #8
    Thought so, but increase in density can be easily done by increassing pressure.

    Interesting, especially since water is 'just' about 1000 times denser than the 'normal' air.
  10. Sep 26, 2010 #9
    Assuming a gravitational frame of reference, you would need to equate the pressure of the air and water column.

    [tex]p_{air} = p_{water}[/tex]

    [tex]\left( \frac{F}{A} \right)_{air} = \left( \frac{F}{A} \right)_{water}[/tex]

    [tex]a_{g} \left( \frac{m}{A} \right)_{air} = a_{g} \left( \frac{m}{A} \right)_{water}[/tex]

    [tex]a_{g} \left( \frac{\rho V}{A} \right)_{air} = a_{g} \left( \frac{\rho V}{A} \right)_{water}[/tex]

    [tex]a_{g} \left( \rho s \right)_{air} = a_{g} \left( \rho s \right)_{water}[/tex]

    [tex]a_{g} \left( \frac{pMs}{RT} \right)_{air} = a_{g} (\rho s)_{water}[/tex]

    [tex]p_{air} = \rho_{water} \cdot \frac{s_{water}}{s_{air}} \cdot \left( \frac{RT}{M} \right)_{air}[/tex]
    Last edited: Sep 26, 2010
  11. Sep 26, 2010 #10
    Thanks tuoni.

    At normal/usual conditions how many times would the pressure of that air have to be higher than normal atmospheric pressure?
  12. Sep 26, 2010 #11
    That depends on the amount of water you have. In this case, the U-shaped tube, the area is identical for both, so it only depends on the height of each column. If identical, it would be over 28,000 times the normal atmospheric pressure. If the water column was twice as high, it would be double that, over 56,000 times the normal atmospheric pressure.
  13. Sep 26, 2010 #12
    Is such pressure for air even possible?
  14. Sep 26, 2010 #13
    What I don't think tuoni is considering is that air will be far from an ideal gas at such high pressures. From my data the pressure at which air and water will have the same density is about 517 MPa or about 5100 atmospheres at 25C. Of course this would be a supercritical fluid, and I'm not sure how a mixture like air would actually behave under these conditions.
  15. Sep 26, 2010 #14
    Would a wooden object still free fall, due to Earth gravity, in such dense air (517 MPa)?

    Also, why such difference from other two posters who said that density of air, to equal that of water, would have to be 28000 atmospheres?
  16. Sep 26, 2010 #15


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    5100 and 28000 differ by less than one order of magnitude. Seeing as both are more than 4 orders of magnitude above ambient, do the details really matter?
  17. Sep 26, 2010 #16
    At that density things that would normally float in water would float on air - though as I said before it is supercritical and actually more dense than liquid air at most temperatures. The difference between my answer and the previous is that mine uses real gas information rather than ideal gas - as gases are compressed they become less ideal and the ideal gas law becomes less accurate.
  18. Sep 26, 2010 #17
    I see. Thanks to all for detailed explanations.
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