Friday

August 29, 2014

August 29, 2014

Posted by **Angela** on Sunday, March 2, 2014 at 6:10pm.

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 water’s surface? Answer in units of L

- Chemistry -
**DrBob222**, Sunday, March 2, 2014 at 6:19pmSo the new pressure is 101 + (100*40.32/10.2) = ?. Then p1v1 = p2v2

- Chemistry -
**Angela**, Sunday, March 2, 2014 at 6:27pmWhy do I need to add 101?

- Chemistry -
**DrBob222**, Sunday, March 2, 2014 at 6:57pmRead 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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