Asample of an ideal gas at 1.00 atm and a volume of 1.91 l was placed in a weighted balloon and dropped into the ocean. as the sample descended, the water pressure compressed the balloon and reduced its volume. when the pressure had increased to 20.0 atm, what was the volume of the sample? assume that the temperature was held constant.

The answer to your question is:    V2 = 0.095 l          

Explanation:

Data

P1 = 1 atm

V1 = 1.91 l

P2 = 20 atm

V2 = ?

Formula

                       P1V1 = P2V2

                       V2 = (P1V1) /P2

Substitution

                       V2 = (1 x 1.91) / 20

                       V2 = 0.095 l                      

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