NEED HELP PLZ!!!

Question

Divers know that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa; at 20.4 m below the surface, the pressure is 301 kPa; and so forth. If the volume of a balloon is 2.2 L at STP and the temperature of the water remains the same, what is the volume 48.62 m below the water’s surface? Answer in units of L.

Answer 1

This question deals with the ideal gas law PV=nRT. There are multiple variations on this equation, including the one we will need to solve this problem. A good way of deciding which variation is needed is to simply look at the units being used in the question. In this case, we are dealing exclusively with pressure and volume, so we can use only those in our calculation. The starting equation we will use is:

P₁V₁ = P₂V₂

Pressure at STP is 101 kPa and volume is given at 2.2 L.

Pressure at a depth of 48.62 m is 101 kPa + [(48.62/10.2)*100 kPa] = 577.67 kPa

We are trying to find volume at this depth, so no value is added for this variable. Our new equation looks like this:

101 kPa * 2.2 L = 577.67 kPa * V₂

Simple algebra will rearrange this equation to give us our answer:

[(222.2 kPa * L)/577.67 kPa] = V₂

Cancel units and solve:

0.385 L = V₂