# How to calculate the pressure of a trapped, bubble of air 1m below water surface

Ahoy hive mind!

The rough scenario I'm looking for some help over is;

• Picture a tub of $$\mathrm{10 \ (l) × 5 \ (w) × 2 \ (h)}$$ floating in a body of water
• The full mass is $$50,000 \ \mathrm{kg}$$, so the displaced $$50\ \mathrm{m^3}$$ of water reaches half way up the tub sides
• Add a mirrored cavity, equivalent to the top, except with only $$\mathrm{1\ m}$$ vertical walls (structure included in the $$50,000\ \mathrm{kg}$$ of mass)
• This occurs in freshwater, and at seal level.

Now bubble air into the inverted cavity, filling the entire $$50\ \mathrm{m^3}$$ and raising the double-tub $$1\ \mathrm{m}$$ higher in the water, to the point where the top of the trapped air pocket is level with the water surface.

My question :

How would one go about calculating the pressure of the trapped air?

Any help appreciated! The idea is to hypothesise how vast a submerged tank would be needed to qualify as a realistic, low-pressure, energy stoage reservoir (any further thoughts would also be welcome!)

The pressure of the air at the bottom is the pressure on top plus the pressure due to the change in height: $$p=W/A+\rho_{air}gh$$. You can also calculate that pressure by considering the pressure of the water there: $$p=p_{atmosphere}+\rho_{water}gh$$.
From these two you can get both, $$p$$ and $$h$$