Two soap bubbles of radii x and y coalesce to | vedclass

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(A) When two soap bubbles coalesce in a vacuum, the total number of moles of air remains constant. Assuming the process is isothermal, we use the idea

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$\sqrt{x^2+y^2}$

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(A) When two soap bubbles coalesce in a vacuum, the total number of moles of air remains constant. Assuming the process is isothermal, we use the ideal gas law $PV = nRT$. Since $n$ is constant and $T$ is constant, $PV$ must be constant.For a soap bubble, the excess pressure is given by $P_{ex} = \frac{4T}{r}$, where $T$ is the surface tension and $r$ is the radius. In a vacuum, the total pressure inside the bubble is $P = \frac{4T}{r}$.The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$.For the two initial bubbles, $P_1 V_1 = (\frac{4T}{x})(\frac{4}{3}\pi x^3) = \frac{16}{3}\pi T x^2$ and $P_2 V_2 = (\frac{4T}{y})(\frac{4}{3}\pi y^3) = \frac{16}{3}\pi T y^2$.For the final bubble, $P V = (\frac{4T}{z})(\frac{4}{3}\pi z^3) = \frac{16}{3}\pi T z^2$.Since the total amount of air is conserved, $P_1 V_1 + P_2 V_2 = PV$.$\frac{16}{3}\pi T x^2 + \frac{16}{3}\pi T y^2 = \frac{16}{3}\pi T z^2$.Dividing both sides by $\frac{16}{3}\pi T$, we get $x^2 + y^2 = z^2$, which implies $z = \sqrt{x^2 + y^2}$.

Two soap bubbles of radii $x$ and $y$ coalesce to form a single bubble of radius $z$. Then $z$ is equal to

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