# Finding the volume of air in a box when it’s lowered into water?

1. The problem statement, all variables and given/known data
A box that is open at the bottom is lowered into the sea (density like water). The outer volume of the box and the air inside it is $V_{out}=3 m^3$.
The moment the box touches the sea surface the air inside it gets trapped and has a volume at $V_0=2.5 m^3$ and a pressure at $p_0=101 kPa$. As the box is lowered the temperature is constant and the mass of the box and air is $m=4000 kg$.
Calculate the volume of the air in the box as it is lowered to $h=19 m$ below the sea surface.

2. Relevant equations
$p=p_0+ρgh$

$pV=nRT$

3. The attempt at a solution
I was thinking about using the first equation to calculate the pressure at 19 m below the sea surface since I know the pressure at the sea surface and I know the density of water:
$p=101000 Pa+9.8\frac{m}{s^2}\cdot 19 m=287000 Pa$

I’m supposed to use the ideal-gas equation to calcuate the volume of the air trapped in the box at 19 m below the sea surface… So I just calculated the pressure at this level but I’m completely lost on what else to do now.

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