Two spherical soap bubbles of radii r_1 and r | vedclass

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(D) The excess pressure inside a soap bubble of radius $r$ is given by $p = \frac{4T}{r}$,where $T$ is the surface tension.For the two initial bubbles

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$\sqrt{r_1^2+r_2^2}$

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(D) The excess pressure inside a soap bubble of radius $r$ is given by $p = \frac{4T}{r}$,where $T$ is the surface tension.For the two initial bubbles,the excess pressures are $p_1 = \frac{4T}{r_1}$ and $p_2 = \frac{4T}{r_2}$.The volumes of these bubbles are $V_1 = \frac{4}{3}\pi r_1^3$ and $V_2 = \frac{4}{3}\pi r_2^3$.Let the radius of the resulting bubble be $R$. The excess pressure inside it is $p = \frac{4T}{R}$ and its volume is $V = \frac{4}{3}\pi R^3$.Under isothermal conditions,the total number of moles of gas remains constant. Since $PV = nRT$,and $T$ is constant,the product $PV$ is conserved (assuming the external pressure is zero in vacuum).Thus,$PV = p_1V_1 + p_2V_2$.Substituting the values: $\frac{4T}{R} \left( \frac{4}{3}\pi R^3 \right) = \frac{4T}{r_1} \left( \frac{4}{3}\pi r_1^3 \right) + \frac{4T}{r_2} \left( \frac{4}{3}\pi r_2^3 \right)$.Simplifying this,we get $R^2 = r_1^2 + r_2^2$.Therefore,$R = \sqrt{r_1^2 + r_2^2}$.

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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