On average, a column of air one square centimeter in cross-section, measured from sea level to the top of the atmosphere, has a mass of about 1.03 kg and weight of about 10.1 N (2.28 lbf) (A column one square inch in cross-section would have a weight of about 14.7 lbs, or about 65.4 N).
The standard atmosphere (symbol: atm) is a unit of pressure equal to 101325 Pa or 1013.25 hectopascals or millibars. Equivalent to 760 mmHg (torr), 29.92 inHg, 14.696 psi. (The pascal is a newton per square meter or in terms of SI base units, kilogram per meter per second-squared.)
Therefore I conclude that Atmospheric pressure ≈ 1 bar.
The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by 14.5 psi (1 bar). In the deepest ocean, the pressure is equivalent to the weight of an elephant balanced on a postage stamp, or the equivalent of one person trying to support 50 jumbo jets!
Therefore I conclude that at 10 meters of depth in seawater, the pressure on an object is 2 bar.
Rigid, air-tight boxes
Imagine with have a rigid, air-tight box containing normal air at normal atmospheric pressure. With the assistance of weights, we lower the box to a steady 33 feet of depth. This is Box A.
Now imagine we take an identical box, but construct it in a vacuum chamber. As a result it contains no air. This box is then exposed to normal atmospheric pressure. This is Box B.
As Box A has 1 bar internal pressure (outwards) and 2 bar external pressure (inwards), does this result in a 1 bar pressure differential similar to Box B, but without the need of expensive vacuum chambers? Would this be a viable way to test a design for structural weak-points?