Two soap bubbles of radii r1 and r2 (r1 < r2) coalesce isothermally in vacuum to form a single soap bubble of radius r. What is the correct relation…

When two soap bubbles coalesce isothermally in vacuum, the total mass/amount of air remains constant, which implies the total volume of air remains constant: $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 \implies r^3 = r_1^3 + r_2^3$. Wait, if they are soap bubbles, the surface area is conserved because the total surface energy is conserved in an isothermal process in vacuum: $S \cdot 2(4\pi r^2) = S \cdot 2(4\pi r_1^2) + S \cdot 2(4\pi r_2^2)$. Thus $4\pi r^2 = 4\pi r_1^2 + 4\pi r_2^2 \implies r^2 = r_1^2 + r_2^2$. Note that soap bubbles have two free surfaces.