If S and V are the change in surface area and volume- when two soap bubbles coalesce under isothermal condition- then - is surface tension- P 0 is atmospheric pressureA- 3 P0 V-4 S -0B- 4 P0 V-3 - S-0C- P0 V-4 - S-0D- 4 P0 V- S-0

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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

If S and V ar...

Question

If S and V are the change in surface area and volume, when two soap bubbles coalesce under isothermal condition, then ( $σ$ is surface tension, $P_{o}$ is atmospheric pressure)

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Solution

The correct option is A

Let a and b be the radius of two soap bubbles forming a soap bubble of radius c. Then
$(P_{o} + \frac{4 σ}{a}) \frac{4}{3} π a^{3} + \frac{4}{3} π b^{3} (P_{o} + \frac{4 σ}{b}) = (P_{o} + \frac{4 σ}{c}) \frac{4}{3} π c^{3}$
i.e., $P_{o} [\frac{4}{3} π (a^{3} + b^{3} - c^{3})] = 4 σ [\frac{4}{3} π (c^{2} - a^{2} - b^{2})]$
$P_{o} V = - \frac{4 σ S}{3}$
$3 P_{o} V + 4 σ S = 0$
So choice (1) is correct and the rest are wrong

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Similar questions

Q. Two soaps bubbles (surface tension

T

) coalesce to form a big bubble under isothermal condition. If in this process the change in volume be

V

and the surface area be

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Q. Two spherical soap bubbles collapses. If

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Q. Two spherical bubbles coalesce isothermally.

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V

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If S and V are the change in surface area and volume, when two soap bubbles coalesce under isothermal condition, then (σ is surface tension, Po is atmospheric pressure)

The correct option is A Let a and b be the radius of two soap bubbles forming a soap bubble of radius c. Then (Po+4σa)43πa3+43πb3(Po+4σb)=(Po+4σc)43πc3 i.e., Po[43π(a3+b3−c3)]=4σ[43π(c2−a2−b2)] PoV=−4σS3 3PoV+4σS=0 So choice (1) is correct and the rest are wrong

If S and V are the change in surface area and volume, when two soap bubbles coalesce under isothermal condition, then ( $σ$ is surface tension, $P_{o}$ is atmospheric pressure)