Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the balloon in an airplane, where the pressure is 0.87 atm. If the temperature is unchanged, what's the balloon's new radius?,

Answer: 17.8 cm

Explanation:

1) Since temperature is constant, you use Boyle's law:

PV = constant = > P₁V₁ = P₂V₂

=> V₁/V₂ = P₂/P₁

2) Since the ballon is spherical:

V = (4/3) π (r) ³

Therefore, V₁/V₂ = (r₁) ³ / (r₂) ³

3) Replacing in the equation V₁/V₂ = P₂/P₁:

(r₁) ³ / (r₂) ³ = P₂/P₁

And you can solve for r₂: (r₂) ³ = (P₁/P₂) x (r₁) ³

(r₂) ³ = (1.0 atm / 0.87 atm) x (17 cm) ³ = 5,647.13 cm³

r₂ = 17.8 cm