A sample of an ideal gas at 1.00 atm and a volume of 1.50 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 25.0 atm, what was the volume of the sample? Assume that the temperature was held constant.

Explanation:

Using Boyle's law

P₁ V₁ = P₂V₂ and temperature is constant

where P₁(pressure)  = 1.00atm, P₂ = 25 atm, V₁( volume) = 1.50L V₂ =

V₂ = ( P₁ V₁ ) / P₂ = ( 1 atm × 1.50L ) / 25 atm = 0.06 L

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explanation:

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