Two spherical soap bubbles of radii r1 and r2 coalesce in vacuum under isothermal conditions. What is the radius of the resulting single bubble?

r1*r2/(r1 + r2)
sqrt(r1*r2)
sqrt(r1² + r2²)
(r1 + r2)/2

Correct answer: sqrt(r1² + r2²)

Solution

Each soap bubble (two surfaces) has internal gas pressure equal to the excess pressure 4T/r (since outside is vacuum). Isothermal coalescence conserves the product of pressure and volume: P1V1 + P2V2 = PV. Substituting Pi = 4T/ri and Vi = (4/3)pi*ri³, the T/ri * ri³ = T*ri² terms give r1² + r2² = r², so r = sqrt(r1² + r2²).

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