Question

A balloon’s internal pressure is 1.25 atm, and its volume is 2.50 L. If the balloon is taken to the bottom of the ocean (pressure = 95.0 atm), what would be its new volume? Assume that the temperature does not change. (Round to the nearest hundredth place) L

Answer 1

0.03 L

Explanation:

Since all other properties constant, we can use Boyle’s Law.

p₁V₁ = p₂V₂ Divide each side by p₂

V₂ = V₁ × p₁/p₂

Data:

p₁ = 1.25 atm; V₁ = 2.50 L

p₂ = 95.0 atm; V₂ = ?

Calculations:

V₂= 1.25 × (2.50/95.0)

V₂= 1.25 × 0.026 32

V₂ ≈ 0.03 L

Answer 2

Where temperature is constant P1V1 = P2V2 where P = pressure and V = volume.

so 1.25 * 2.5 = 95 * V2

V2 = (1.25 * 2.5) / 95

= 0.03 L to the nearest hundredth..