# 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 [itex]V_{out}=3 m^3[/itex].

The moment the box touches the sea surface the air inside it gets trapped and has a volume at [itex]V_0=2.5 m^3[/itex] and a pressure at [itex]p_0=101 kPa[/itex]. As the box is lowered the temperature is constant and the mass of the box and air is [itex]m=4000 kg[/itex].

Calculate the volume of the air in the box as it is lowered to [itex]h=19 m[/itex] below the sea surface.

**2. Relevant equations**

[itex]p=p_0+ρgh[/itex]

[itex]pV=nRT[/itex]

**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:

[itex]p=101000 Pa+9.8\frac{m}{s^2}\cdot 19 m=287000 Pa[/itex]

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.

http://ift.tt/OCzIT3

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