Divers know that the pressure exerted buy the water increases about 100kpa 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 , the pressure is 301 Kpa; and so forth. Given that the volume of a balloon is 3.5L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water's surface?

asked by Natasha

11 years ago

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1 answer

So the new pressure is
101 + (100 x 51/10.2) = ?
Then P1V1 = P2V2

answered by DrBob222

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Question

Divers know that the pressure exerted buy the water increases about 100kpa 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 , the pressure is 301 Kpa; and so forth. Given that the volume of a balloon is 3.5L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water's surface?

1 answer

So the new pressure is
101 + (100 x 51/10.2) = ?
Then P1V1 = P2V2

Ask a New Question or answer this question.

DrBob222 · Accepted Answer

So the new pressure is 101 + (100 x 51/10.2) = ? Then P1V1 = P2V2