Excess pressure inside a Liquid Drop
Consider a drop of liquid R as shown in the figure. The molecules lying on the surface of the liquid drop due to surface tension will experience a resultant force acting inwards perpendicular to the surface. As a result, the pressure inside the drop must be greater than the pressure outside it. The excess pressure inside the drop will provide a force acting outwards perpendicular to the surface, to balance the resultant force due to surface tension.
Let T be the surface tension and P be the excess pressure inside the drop. Suppose due to excess pressure, there be an increase in the radius of the drop by quantity dR. In such case we can write, work done by excess pressure,
W = Force * displacement = (Excess pressure * area) * displacement
or, W = P * 4πR2 * dR …….. (i)
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