A soap bubble in vacuum has a radius of 3 , c | vedclass

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(C) When two soap bubbles coalesce in a vacuum under isothermal conditions,the total number of moles of air remains constant. Since the temperature is

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$5$

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(C) When two soap bubbles 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 given by $P_{ex} = \frac{4T}{r}$. Since the bubbles are in a vacuum,the internal pressure is $P = \frac{4T}{r}$.The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$.For the first bubble: $P_1 V_1 = \left(\frac{4T}{r_1}\right) \left(\frac{4}{3}\pi r_1^3\right) = \frac{16}{3}\pi T r_1^2$.For the second bubble: $P_2 V_2 = \left(\frac{4T}{r_2}\right) \left(\frac{4}{3}\pi r_2^3\right) = \frac{16}{3}\pi T r_2^2$.For the new bubble of radius $R$: $P V = \frac{16}{3}\pi T R^2$.Since the total amount of air is conserved: $P_1 V_1 + P_2 V_2 = PV$.$\frac{16}{3}\pi T r_1^2 + \frac{16}{3}\pi T r_2^2 = \frac{16}{3}\pi T R^2$.$r_1^2 + r_2^2 = R^2$.Given $r_1 = 3 \, cm$ and $r_2 = 4 \, cm$,we have $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, cm$.

$A$ soap bubble in vacuum has a radius of $3 \, cm$ and another soap bubble in vacuum has a radius of $4 \, cm$. If the two bubbles coalesce under isothermal condition,then the radius of the new bubble is ....... $cm$.

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