Question

Derive an expression for excess of pressure inside a liquid drop.

Solution

Consider a liquid drop of radius R and surface tension T.
Due to surface tension the molecules on the surface film experience the net force in inward direction normal to the surface.

Therefore there is more pressure inside than outside.
Let p_i be the pressure inside the liquid drop and p_o be the pressure outside the drop.
Therefore excess of pressure inside the liquid drop is,

p = p₁– p₀

Due to excess of pressure inside the liquid drop the free surface of the drop will experience the net force in outward direction due to which the drop will expand.
Let the free surface displace by dR under isothermal conditions.
Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.
The work done by excess of pressure in displacing the surface is,
dW = Force x displacement
= (Excess of pressure x surface area) x displacement of surface

$equals straight p cross times 4 πR squared xdR space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis$
Increase in the potential energy is,
dU = surface tension x increase in area of the free surface
$equals straight T open square brackets 4 straight pi open parentheses straight R plus dR close parentheses squared minus 4 πR squared close square brackets equals straight T open square brackets 4 straight pi open curly brackets 2 RdR close curly brackets close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis From space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis straight p cross times 4 πR squared cross times dR equals straight T left square bracket 4 straight pi open curly brackets 2 RdR close curly brackets right square bracket rightwards double arrow space space space space space space space space space space space space space space space space space space space straight p equals fraction numerator 2 straight T over denominator straight R end fraction$
The above expression gives us the pressure inside a liquid drop.

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Mechanical Properties Of Fluids

Derive an expression for excess of pressure inside a liquid drop.

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