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Two identical soap bubbles each of radius r and of the same surface tension T combine to form a new soap bubble of radius R. The two bubbles contain air at the same temperature. If the atmospheric pressure is $p_{0}$ then find the surface tension T of the soap solution in terms of $p_{0}, r$ and R. Assume process is isothermal.

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$\frac{p_{o} (2 r^{3} - R^{3})}{4 (R^{2} + 2 r^{2})}$

$\frac{p_{o} (2 r^{3} - R^{3})}{2 (R^{2} + 2 r^{2})}$

$\frac{p_{o} (2 r^{3} + R^{3})}{4 (R^{2} - 2 r^{2})}$

$\frac{p_{o} (2 r^{3} - R^{3})}{4 (R^{2} - 2 r^{2})}$

answer is A.

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Detailed Solution

Total number of moles of air in the two soap bubbles = number of moles of air in the resulting bubble.

$\frac{2 pv}{RT} = \frac{p^{'} v^{'}}{RT}$

$2 pv = p^{'} v^{'}$

$2 (p_{o} + \frac{4 T}{r}) \frac{4}{3} π r^{3} = (p_{0} + \frac{4 T}{R}) \frac{4}{3} π R^{3}$

$2 (p_{o} + \frac{4 T}{r}) r^{3} = (p_{o} + \frac{4 T}{R}) R^{3}$

T = $\frac{p_{o} (R^{3} - 2 r^{3})}{8 r^{2} - 4 R^{2}} = \frac{p_{o} (R^{3} - 2 r^{3})}{4 (2 r^{2} - R^{2})}$

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