Two soap bubbles of the same radius $r$ coalesce isothermally to form a bigger bubble of radius $R$ . If the atmospheric pressure is $P_{o}$ , find the surface tension of the soap solution.

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Solution

The mass of air inside the bubble must be conserved
$n_{1} + n_{2} = n_{0}$
$P V = n R T$ _____(1)
$\frac{P_{1} V_{1}}{R T_{1}} + \frac{P_{2} V_{2}}{R T_{2}} = \frac{P V}{R T}$
$T_{1} = T_{2} = T$ ______(ii)
$V_{1} = V_{2} = \frac{4}{3} π r^{2}$
$P_{1} V_{1} + P_{2} V_{2} = P V$
Let $S$ Surface tension then
$(P_{0} + \frac{4 s}{r}) \frac{4}{3} π r^{3} + (P_{0} + \frac{4 s}{r}) \frac{4}{3} π r^{3} = (P_{0} + \frac{4 s}{r}) \frac{4}{3} π R^{3}$
$2 (P_{0} + \frac{4 s}{r}) \frac{4}{3} π r^{3} = (P_{0} + \frac{4 s}{R}) \frac{4}{3} π R^{3}$
$2 (P_{0} + \frac{4 s}{r}) r^{3} = (P_{0} + \frac{4 s}{R}) R^{3}$
$P_{0} r^{3} + \frac{8 r^{3} s}{r} = P_{0} R^{3} + \frac{4 S R^{3}}{R}$
$2 P_{0} r^{3} + 8 r^{2} s = P_{0} R^{3} + 4 S R^{2}$
$2 P_{0} r^{3} - P_{0} R^{3} = 4 S R^{2} - 8 S r^{2}$
$P_{0} (2 r^{3} - R^{3}) = 4 s (R^{2} - 2 r^{2})$
$S = \frac{P_{0} (2 r^{3} - R^{3})}{4 (R^{2} - 2 r^{2})}$
$= \frac{- P_{0} (R^{3} - 2 r^{3})}{- 4 (2 r^{2} - R^{2})}$
$= \frac{P_{0} (R^{3} - 2 r^{3})}{4 (2 r^{2} - R^{2})}$

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