Question

A soap bubble has two surfaces of equal surface area i.e. the outer and the inner but pressure inside is
different from the pressure outside. Obtain the expression for the difference in pressure inside a soap
bubble floating in air

Answer 1

Answer:

Consider a soap bubble of radius R surface tension T.Due to surface tension the molecules on the surface film experience the net force inward direction normal to the surface. Therefore there is more pressure inside than outside.

pi= pressure inside liquid drop

po=pressure outside drop

excess pressure inside liquid drop

p= pi- po

work done by excess pressure in displacing surface

dw=force ×displacement

= excess pressure × surface area× displacement of surface

= p×4

$p \times 4\pi \: r ^{2} \times dr$

increase in potential energy is

du= surface tension × increase in area of the free surface.

=

$t \: (2 {4\pi{r + dr} }^{2} - 4\pi \: r {}^{2} )$

=

$t \: (2 { 4\pi{2rdr})$

1, 2 we get

.

. . p= 4T/R

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