The radii of two soap bubbles are r_1 and r_2 | vedclass

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(C) When two soap bubbles of radii $r_1$ and $r_2$ coalesce in a vacuum under isothermal conditions,the total number of moles of air remains constant.

Vedclass · Accepted Answer

$\sqrt{r_{1}^{2}+r_{2}^{2}}$

Vedclass · Answer

(C) When two soap bubbles of radii $r_1$ and $r_2$ coalesce in a vacuum under isothermal conditions,the total number of moles of air remains constant. Since the temperature is constant,the product of pressure and volume $(PV)$ remains constant for the air inside the bubbles.For a soap bubble,the excess pressure is $P_{ex} = \frac{4T}{r}$,where $T$ is the surface tension. The absolute pressure inside is $P = P_{atm} + \frac{4T}{r}$. In a vacuum,$P_{atm} = 0$,so $P = \frac{4T}{r}$.The volume of a bubble is $V = \frac{4}{3}\pi r^3$.For the first bubble: $P_1 V_1 = (\frac{4T}{r_1})(\frac{4}{3}\pi r_1^3) = \frac{16}{3}\pi T r_1^2$.For the second bubble: $P_2 V_2 = (\frac{4T}{r_2})(\frac{4}{3}\pi r_2^3) = \frac{16}{3}\pi T r_2^2$.For the resultant bubble of radius $R$: $P_R V_R = (\frac{4T}{R})(\frac{4}{3}\pi R^3) = \frac{16}{3}\pi T R^2$.Since the total amount of air is conserved: $P_1 V_1 + P_2 V_2 = P_R V_R$.$\frac{16}{3}\pi T r_1^2 + \frac{16}{3}\pi T r_2^2 = \frac{16}{3}\pi T R^2$.Dividing by $\frac{16}{3}\pi T$,we get $R^2 = r_1^2 + r_2^2$,or $R = \sqrt{r_1^2 + r_2^2}$.

The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal conditions,they meet together in a vacuum. Then the radius of the resultant bubble is given by

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