Question

derive an expression for laplaces law of spehiracal membrane ​

Answer 1

Answer:

The excess of pressure is Pi– Po. Laplace's Law of a Spherical Membrane for a Liquid Drop: The excess of pressure is Pi– Po. Let the radius of the drop increases from r to r + Δr, where Δr is very very small, hence the inside pressure is assumed to be constant.

Answer 2

According to this law the difference between the pressure of inner and outer surfaces of a spherical membrane is expressed as

P in - P out = 2T/r

under the equilibrium conditions

Force due to excess pressure = force due to surface tension

P in - P out (pie r )² = T (2pie/r)

(P in - P out) r = 2T

P in - P out= 2T/r

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