Under isothermal conditions,two soap bubbles | vedclass

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Question

(C) For a soap bubble,the surface area is $A = 4\pi r^2$. Since a soap bubble has two surfaces (inner and outer),the total surface area is $8\pi r^2$.

Vedclass · Accepted Answer

$(r_1^2 + r_2^2)^{1/2}$

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(C) For a soap bubble,the surface area is $A = 4\pi r^2$. Since a soap bubble has two surfaces (inner and outer),the total surface area is $8\pi r^2$.Under isothermal conditions,the amount of air (number of moles) remains constant. Using the ideal gas law $PV = nRT$,where $P$ is the pressure inside the bubble,$V$ is the volume,and $T$ is constant.The pressure inside a soap bubble of radius $r$ is $P = P_0 + \frac{4S}{r}$,where $P_0$ is the atmospheric pressure and $S$ is the surface tension.The volume is $V = \frac{4}{3}\pi r^3$.Thus,$n = \frac{PV}{RT} = \frac{(P_0 + 4S/r)(4/3 \pi r^3)}{RT} = \frac{4\pi}{3RT} (P_0 r^3 + 4Sr^2)$.Since the total number of moles is conserved: $n_1 + n_2 = n_{final}$.Assuming $P_0$ is very large compared to the excess pressure,the volume remains constant: $\frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi R^3$,which gives $R = (r_1^3 + r_2^3)^{1/3}$.However,in standard physics problems of this type where surface area energy is considered or if the pressure is dominated by surface tension,the surface area is conserved: $8\pi r_1^2 + 8\pi r_2^2 = 8\pi R^2$.Therefore,$R^2 = r_1^2 + r_2^2$,which implies $R = (r_1^2 + r_2^2)^{1/2}$.

Under isothermal conditions,two soap bubbles of radii $r_1$ and $r_2$ coalesce to form a single soap bubble of radius $R$. The radius of the new bubble is

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