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.