A sample of an ideal gas at 1.00 atm and a volume of 1.95 L was placed in a weighted balloon and dropped into the ocean.

Question

(question continued) As the sample descended, the water pressure compressed the balloon and reduced its volume. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 80.0 atm, what was the volume of the sample? Assume that the temperature was held constant.

Answer 1

Use Boyle's Law to Solve.

P₁V₁ = P₂V₂

P₁ = 1.00 atm

P₂ = 80.0 atm

V₁ = 1.95 L

V₂ = ?

Rearrange to solve for V₂

V₂ = P₁V₁ / P₂ = (1.00 atm x 1.95 L) / 80.0 atm = 0.0244 L

Pressure and Volume are inversely proportional according to Boyle's Law. As the pressure is increased (1.00 atm → 80.0 atm), the volume decreases (1.95 L → 0.0244 L).