Under isothermal conditions,two soap bubbles | vedclass

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(C) Under isothermal conditions,the temperature and surface tension $T$ remain constant. When two soap bubbles coalesce to form a larger bubble,the to

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$(r_{1}^{2}+r_{2}^{2})^{1/2}$

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(C) Under isothermal conditions,the temperature and surface tension $T$ remain constant. When two soap bubbles coalesce to form a larger bubble,the total surface area changes,but the total number of moles of air inside remains constant. Since the process is isothermal,the pressure $P$ inside a soap bubble of radius $r$ is given by $P = P_{0} + 4T/r$. The total number of moles $n$ is given by $PV = nRT$. For a soap bubble,the excess pressure is $4T/r$,so the total pressure is $P_{atm} + 4T/r$. The volume is $V = (4/3)\pi r^{3}$.Assuming the external pressure $P_{atm}$ is negligible or the process is governed by the conservation of surface energy in the context of bubble coalescence,the surface energy is $U = 2 	imes (4\pi r^{2}T) = 8\pi r^{2}T$ (since a soap bubble has two surfaces).Equating the initial and final surface energies: $8\pi r^{2}T = 8\pi r_{1}^{2}T + 8\pi r_{2}^{2}T$.Dividing by $8\pi T$,we get $r^{2} = r_{1}^{2} + r_{2}^{2}$.Therefore,the radius of the larger bubble is $r = (r_{1}^{2} + r_{2}^{2})^{1/2}$.

Under isothermal conditions,two soap bubbles of radii $r_{1}$ and $r_{2}$ coalesce to form a single larger bubble. The radius of the larger bubble is

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