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Concept of Force

Under isother...

Question

Under isothermal condition two soap bubbles of radii $r_{1} and r_{2}$ coalesce to form a single bubble of radius $r$ . The external pressure is $P_{0}$ . Find the surface tension of the soap in terms of the given parameters.

A

\frac{P_{0} (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}

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B

\frac{P_{0} (r^{3} - r^{3} - r_{2}^{3})}{2 (r_{1}^{2} + r_{2}^{2})}

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C

\frac{P_{0} (r^{3} - r_{1}^{3} - r_{2}^{3})}{(r_{1}^{2} + r_{2}^{2})}

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D

\frac{P_{0} (r^{3} - r_{1}^{3} - r_{2}^{3})}{(r_{1}^{2} + r_{2}^{2} - r^{2})}

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Solution

The correct option is A $\frac{P_{0} (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}$
As mass of the air inside the soap bubble is conserved.
$∴ n_{1} = n_{2} = n (as PV = nRT)$
Under isothermal condition, temperature is constant.

$\Rightarrow T_{1} = T_{2} = T$

$∴ P_{1} V_{1} + P_{2} V_{2} = P V$

$∴ (P_{0} + \frac{4 S}{r_{1}}) (\frac{4}{3} π r_{1}^{3}) + (P_{0} + \frac{4 S}{r_{2}}) (\frac{4}{3} π r_{2}^{3}) = (P_{0} + \frac{4 S}{r}) (\frac{4}{3} π r^{3})$

$∴ (P_{0} + \frac{4 S}{r_{1}}) (r_{1}^{3}) + (P_{0} + \frac{4 S}{r_{2}}) (r_{2}^{3}) = (P_{0} + \frac{4 S}{r}) (r^{3})$

After simplifying, we get,

$\Rightarrow S = \frac{P_{0} (r^{3} - r_{1}^{3} - r_{2}^{3})}{4 (r_{1}^{2} + r_{2}^{2} - r^{2})}$

Hence, $(A)$ is the correct answer.

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