Two spherical soap bubbles of radii r_{1} and | vedclass

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(B) When two soap bubbles combine in a vacuum under isothermal conditions,the total number of moles of gas remains conserved.Using the ideal gas law $

Vedclass · Accepted Answer

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

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(B) When two soap bubbles combine in a vacuum under isothermal conditions,the total number of moles of gas remains conserved.Using the ideal gas law $PV = nRT$,and since the temperature $T$ is constant,we have $n = \frac{PV}{RT}$.Conservation of moles: $n_1 + n_2 = n_3 \Rightarrow P_1 V_1 + P_2 V_2 = P_3 V_3$.The excess pressure inside a soap bubble is given by $P = \frac{4S}{r}$,where $S$ is the surface tension.The volume of a spherical bubble is $V = \frac{4}{3} \pi r^3$.Substituting these into the conservation equation:$\left(\frac{4S}{r_1}\right) \left(\frac{4}{3} \pi r_1^3\right) + \left(\frac{4S}{r_2}\right) \left(\frac{4}{3} \pi r_2^3\right) = \left(\frac{4S}{r_3}\right) \left(\frac{4}{3} \pi r_3^3\right)$.Simplifying the equation:$16 \pi S \left(\frac{r_1^2}{3} + \frac{r_2^2}{3}\right) = 16 \pi S \left(\frac{r_3^2}{3}\right)$.Therefore,$r_1^2 + r_2^2 = r_3^2$,which gives $r_3 = \sqrt{r_1^2 + r_2^2}$.

Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to -

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