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Surface Energy

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_{1}^{3} - r_{2}^{3})}{2 (r_{1}^{2} + r_{2}^{2})}

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C

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

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D

\frac{2 P_{0} r_{1} r_{2}}{(r - r_{1} - 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 the mass (number of moles) of the air is conserved,
$n_{1} + n_{2} = n$

Using ideal gas equation $P V = n R T$ ,

$\frac{P_{1} V_{1}}{R T_{1}}$ + $\frac{P_{2} V_{2}}{R T_{2}}$ = $\frac{P V}{R T}$

Under isothermal condition,

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

$\Rightarrow P_{1} V_{1} + P_{2} V_{2} = P V$

Pressure inside the soap bubble is the sum of external pressure and excess pressure inside the soap bubble,

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

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

Hence, option $(A)$ is correct.

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