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Under isothermal conditions two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If external pressure is P₀, the surface tension of soap solution is equal to:

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$\frac{P_{0} (a^{3} + b^{3} - c^{3})}{4 (a^{2} + b^{2} + c^{2})}$

$\frac{P_{0} (a^{3} + b^{3} - c^{3})}{(a^{2} + b^{2} + c^{2})}$

$\frac{P_{0} (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

$\frac{P_{0} (a^{3} + b^{3} - c^{3})}{4 (a^{2} - b^{2} - c^{2})}$

answer is C.

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Detailed Solution

$P_{a} = P + \frac{4 T}{a}; P_{b} = P + \frac{4 T}{b}; P_{c} = P + \frac{4 T}{c} P_{c} V_{c} = P_{a} V_{a} + P_{b} V_{b}$

$(P_{0} + \frac{4 T}{c}) \frac{4}{3} {πc}^{3} = (P_{0} + \frac{4 T}{a}) \frac{4}{3} {πa}^{3} + (P_{0} + \frac{4 T}{b}) \frac{4}{3} {πb}^{3}$

$P_{0} (c^{3^{}} - a^{3} - b^{3}) + 4 T (c^{2} - a^{2} - b^{2}) = 0$

$T = \frac{P_{0} (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

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