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# Volume change with pressure

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If a certain volume of air at atmospheric pressure is trapped under water and the head of water is 34 m above the air pocket by what factor would the volume of air be reduced?

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Commented:
Let your original volume of air be 1 cubic meter (1×1×1) at 14 psi of the metric equivalent.

Now put a tower of water (1×1×34 meters) on top of it.

Calculate the weight of the water for the added pressure.
Retired

Commented:
Hydrostatic Equilibrium and the Homogeneous Atmosphere

Hydrostatic equilibrium

Basic ideas
The principle of hydrostatic equilibrium is that the pressure at any point in a fluid at rest (whence, “hydrostatic”) is just due to the weight of the overlying fluid.

As pressure is just force per unit area, the pressure at the bottom of a fluid is just the weight of a column of the fluid, one unit of area in cross-section.

This principle is simple to apply to incompressible fluids, such as most liquids (e.g., water). [Note that water and other common liquids are not strictly incompressible; but very high pressures are required to change their densities appreciably.] If the fluid is incompressible, so that the density is independent of the pressure, the weight of a column of liquid is just proportional to the height of the liquid above the level where the pressure is measured. In fact, the mass of a unit-area column of height h and density ρ is just ρh; and the weight of the column is its mass times the acceleration of gravity, g. But the weight of the unit-area column is the force it exerts per unit area at its base — i.e., the pressure. So

P  =  g ρ h .

Examples
For example, the density of water is 1000 kilograms per cubic meter (in SI units), so the weight of a cubic meter of water is 1000 kg times g, the acceleration of gravity (9.8 m/sec2), or 9800 newtons. This force is exerted over 1 m2, so the pressure produced by a 1-meter depth of water is 9800 pascals (the Pa is the SI unit of pressure, equal to 1 newton per square meter).

The unit of pressure used in atmospheric work on Earth is the hectopascal; 1 hPa = 100 Pa. So the pressure 1 m below the surface of water (ignoring the pressure exerted by the atmosphere on top of it) is 98 hPa. Standard atmospheric pressure is 1013.25 hPa, so it takes 1013.25/98 = 10.33 meters of water to produce a pressure of 1 atmosphere. (That's about 34 feet, for those who like obsolete units.)

The pressure in the ocean increases by about 1 atmosphere for every 10 meters of depth. The average depth of the ocean is about 4 km, so the pressure on the sea floor is about 400 atmospheres.

Commented:
How about a direct answer to my question? I am not looking for the theory or the working.
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Commented:
Atmospheric pressure is 1.033 kg/cm2

The weight of a 34 m column of water 1 cm2 in cross-section is 3.4 kg.

So the pressure difference between the trapped air and atmosphere pressure is 3.4 kg/cm2, and the absolute pressure of the trapped air is 1.033 + 3.4 = 4.433 kg/cm2.

Ideal gas law gives the relationship P1*V1 = P2*V2

So V2/V1 = P1/P2 = 1.033/4.433 = 23.3%. The reduction in volume is 76.7%.
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Commented:
i don't get the same answer as @byundt

p1 = 1 atm at (sea level)

p2 = 3.29 atm at 34 meters

p1/p2= 1/3.9 = ‭25.64%
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Commented:
Paul,
Your p2 is a gauge pressure, not absolute.

The calculation needs to be V2/V1 = p1/p2 = 1/(1+3.29) = 23.3%