Two soap bubbles coalesce to form a single bu | vedclass

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Question

Let $P_i$ and $R_i$ be the inside pressure and radius of the ith soap bubble respectively.

$	herefore {P_1} = P + \frac{{4T}}{{{R_1}}}.\,\,\,{P_2}

Vedclass · Accepted Answer

$3PV+4ST = 0$

Vedclass · Answer

Let $P_i$ and $R_i$ be the inside pressure and radius of the ith soap bubble respectively.

$	herefore {P_1} = P + \frac{{4T}}{{{R_1}}}.\,\,\,{P_2} = P + \frac{{4T}}{{{R_2}}}\,\,and\,\,{P_3} = P + \frac{{4T}}{{{R_3}}}$

$Also\,{P_1}{V_1} + {P_2}{V_2} = {P_3}{V_3}$

$	herefore \left( {P + \frac{{4T}}{{{R_1}}}} ight)\frac{{4\pi }}{3}R_1^3 + \left( {P + \frac{{4T}}{{{R_2}}}} ight)\frac{{4\pi }}{3}R_2^3$

$ = \left( {p + \frac{{4T}}{{{R_3}}}} ight)\frac{{4\pi }}{3}R_3^3$

$P\left( {\frac{{4\pi }}{3}R_1^3 + \frac{{4\pi }}{3}R_2^3 - \frac{{4\pi }}{3}R_3^2} ight)$

$ + \frac{{4T}}{3}\left( {4\pi R_1^2 + 4\pi R_2^2 - 4\pi R_3^2} ight) = 0$

$P\left( {{V_1} + {V_2} - {V_3}} ight) + \frac{{4T}}{3}\left( {{S_1} - {S_2} - {S_3}} ight) = 0$

$PV + \frac{{4T}}{3}S = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,3PV + 4ST = 0$

Two soap bubbles coalesce to form a single bubble. If $V$ is the subsequent change in volume of contained air and $S$ change in total surface area, $T$ is the surface tension and $P$ atmospheric pressure, then which of the following relation is correct?

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