Homework Help: Chemistry

Posted by Jassia on Tuesday, January 29, 2013 at 10:42am.

Divers know that pressure exerted by the water increases about 100kPa with every 10.2 m of deptt. This means that at 10.2 m below the surface, the pressure is 201kPa; at 20.4 m, the pressure is 301kPa; and so forth. Given that the volume of a balloon is 3.5 L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water's surface. I know the answer 0.59 L, but I do not know how to work it.

Chemistry - DrBob222, Tuesday, January 29, 2013 at 2:33pm
100 kPa increase x 51m/10.2m = 500 kPa increase. Add that to STP pressure of 101.325 kPa + 500 kPa = 601.325 kPa.
Then P1V1 = P2V2
101.325*3.5L = 601.325*V2
Solve for V2.

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