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A sample of an ideal gas at 1.00 atm and a volume of 1.02 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 65.0 atm, what was the volume of the sample? assume that the temperature was held constant.

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0.0157 L. For this equation use the combined gas law, which states P1*V1/T1 = P2*V2/T2, but in this case we can remove the T, because T1 = T2, because the temperature is constant. Then we solve for V2 because it is the new volume. So P1*V1/P2 =V2 Plug in the variables, P1 = 100 atm, V1 = 1.02 L, P2 = 65.0 atm, and V2 = 0.0157 atm.

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