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(B) When two soap bubbles of radius $r$ coalesce in a vacuum to form a single bubble of radius $R$,the total number of moles of air inside the bubbles

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$2\sqrt{2} \ cm$

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(B) When two soap bubbles of radius $r$ coalesce in a vacuum to form a single bubble of radius $R$,the total number of moles of air inside the bubbles remains constant. Since the process is isothermal,the temperature $T$ is constant. The pressure inside a soap bubble of radius $r$ is $P = P_0 + \frac{4S}{r}$,where $P_0$ is the external pressure (which is $0$ in a vacuum) and $S$ is the surface tension. Thus,$P = \frac{4S}{r}$.Using the ideal gas law $PV = nRT$,for a soap bubble,$n = \frac{PV}{RT} = \frac{(4S/r) \cdot (4/3 \pi r^3)}{RT} = \frac{16 \pi S r^2}{3RT}$.Since the total number of moles is conserved,$n_{total} = n_1 + n_2$. Given $r_1 = r_2 = r = 2 \ cm$,we have $n_{total} = 2n = \frac{32 \pi S r^2}{3RT}$.For the new bubble of radius $R$,$n_{total} = \frac{16 \pi S R^2}{3RT}$.Equating the two,we get $R^2 = 2r^2$,which implies $R = r\sqrt{2}$.Substituting $r = 2 \ cm$,we get $R = 2\sqrt{2} \ cm$.

If two soap bubbles each of radius $2 \ cm$ combine in vacuum under isothermal conditions,then the radius of the new bubble formed is

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