Posted by **Maci** on Wednesday, September 28, 2011 at 2:08pm.

If the pressure on a balloon were quadrupled what would happen to the balloons volume?

- Chemistry -
**DrBob222**, Wednesday, September 28, 2011 at 2:36pm
P1V1 = P2V2

Since no initial P is given, just make up a value, then quadruple it for P2.

- Chemistry -
**Maci**, Wednesday, September 28, 2011 at 4:12pm
So just any random number I can quadruple it to?

- Chemistry -
**DrBob222**, Wednesday, September 28, 2011 at 10:28pm
right. Why not go the easy route and make P1 = 1 and P2 = 4.

Then you need to make V1 = 4 and solve for V2.

Another way to do it is use

P1V1 = P2V2 and set P2 = 4P1

So P1V1 = 4P1V2 and solve for

V1/V2 = 4P1/P1

So V1/V2 = 1P1/4P1, the P1 cancels to obtain

V1/V2 = 4

V1 = 4V2 or

V2 = (V1/4) = 1/4 V1

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