Chemistry
posted by Angela .
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 3.2 L at STP and the temperature of the water remains the same, what is the
volume 40.32 m below the waterâ€™s surface? Answer in units of L

So the new pressure is 101 + (100*40.32/10.2) = ?. Then p1v1 = p2v2

Why do I need to add 101?

Read the problem.
"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;"
So the pressure 10.2m below is 100 + whatever it is at the surface (which is 101 kPa) so the total is 101 + 100 = 201. And at 20.4 it is 301 (that's 101 + 100+100 = 301). Right? The problem has given you the numbers to make a general formula and that's what I did. 
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