Derive an expression for exess pressure inside a liquid drop
Asked by shobhit_dc | 18th Jan, 2012, 07:24: PM
Excess pressure in a liquid drop. Let us consider a drop of liquid of surface tension T and radius R. Let Pi and P0 be the values of pressure inside and outside the drop of the liquid . Then,
Excess pressure inside the liquid drop = pi po
Suppose that the radius of the drop is increased from R to R + R under the pressure difference (pi p0).
The outward force acting on the drop = excess of pressure × surface area = (pi p0) × 4 R2
The small amount of work done to increase its radius by (R),
W = (pi p0) × 4 R2 × (R) or W = 4 R2 (pi p0) (R)
This work (W) is done by the excess of pressure against the force of surface tension and is stored inside the liquid drop in the form of its potential energy (U).
Also, increase in the potential energy of the liquid drop,
U = surface tension × increase in surface area
=T × (4 (R + (R))2 4 R2) = T × (4 R2 + 8 R (R) + 4 (R)2 4 R2)
As R is small, the term containing R2 can be neglected.
Therefore, increase in the potential energy of drop, U = 8 T R (R)
From the equations (i) and (ii), we have
4 R2(pi p0) (R) = 8 T R (R) or pi po = 2T/R
Answered by | 19th Jan, 2012, 10:34: AM
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