# Chemistry question

## Homework Statement

A diving bell has an air space of 3.0 m2 when on the deck of a boat. What is the volume of the air space when the bell has been lowered to a depth of 50 m? Take the mean density of sea water to be 1.025 g cm-3 and assume that the temperature is the same as on the surface.

PV = nRT

## The Attempt at a Solution

I'm thinking I should use the ideal gas law to solve this problem.

V1P1 = nRT

V2P2 = nRT

V1P1 = V2P2

P1 = 1 atm (at surface of water)

P2 = ? (would I use the density of sea water and surface area of diving bell somehow?)

Thanks!

$$P_{2}=P_{1}+\rho g \Delta z$$
$$\Delta z$$= difference in depth