Derive an expression for exess pressure inside a liquid drop
Asked by shobhit_dc | 18th Jan, 2012, 07:24: PM
Expert Answer:
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)
(i)
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)
(ii)
From the equations (i) and (ii), we have
4
R2(pi p0) (
R) = 8
T R (
R) or pi po = 2T/R
(iii)
Answered by | 19th Jan, 2012, 10:34: AM
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