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Volume change with pressure

Saqib Husain, Syed
Saqib Husain, Syed asked
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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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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.
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.
How about a direct answer to my question? I am not looking for the theory or the working.
Mechanical Engineer
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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%.
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%
byundtMechanical Engineer
Most Valuable Expert 2013
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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%

Brad
thanks for clarifying