I am not sure I understood your question correctly, but as I understand it, the answer is no.
I assume you are talking about the effect on air spaces.
33 feet of sea water is the same pressure as air pressure at sea level. The change in pressure is actually determined by a simple equation: V(1) * P(1) = V(2) * (P2)
In simple terms, if you are at 33 feet, you are at 2 atmospheres of pressure--one for the air and one for the water. If you have 1 liter of air in a flexible container at the surface and take it to 33 feet, the formula tells you what will happen:
1 * 1 = V(2) * 2, so 0.5 = V(2). You container will be half its original size.
If you want to figure out the effect at intermediate depths, you have to figure out the atmospheres at each depth. At 10 feet, it will be 10/33 +1 (for the air pressure), or 1.3 atmospheres. At 20 feet it will be 20/33 +1, or 1.6 atmospheres.
At 10 feet the formula will be 1 * 1 = V(2) * 1.3, so V(2) = 0.77. The container will be more than 3/4 its original size.
At 20 feet the formula will be 1 * 1 = V(2) * 1.6, so V(2) = 0.63. The container will be about 5/8 its original size.
Generally, the closer you are to the surface, the greater the effect changes in depth have on volume.