What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Total pressure inside a mercury drop is given as ‘P’,


P = Excess pressure inside the mercury drop + Atmospheric pressure


Excess pressure inside the mercury drop, P1 =


Where,


‘S’ is the surface tension of the mercury


‘r’ is the radius of the mercury drop.


Thus,


P = P1 + P0


Given,


Surface tension of the mercury, S = 4.65 × 10-1 N


Radius of the mercury drop, r = 3.0 mm = 3.0 × 10-3 m


Atmospheric pressure, P0 = 1.01 × 105 Pa


⇒ P1 =


⇒ P1 = 3.1 × 102 Pa = 0.0031 × 105 Pa


Therefore,


P = 0.0031 × 105 Pa + 1.01 × 105 Pa


⇒ P = 1.0131 × 105 Pa


The pressure inside the drop of mercury is 1.0131 × 105 Pa and the excess pressure inside the drop is 310 Pa.


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