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? ,

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

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