Question

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.

Answer 1

1.01 × 10 5 Pa; 310 Pa
Radius of the mercury drop, r = 3.00 mm = 3 × 10 - 3 m
Surface tension of mercury, S = 4.65 × 10 - 1 N m - 1
Atmospheric pressure, P0 = 1.01 × 10 5 Pa

Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure

= $\frac{2S }{ r}$+ P0

$=\frac{ 2 × 4.65 × 10 ^{- 1}} {3 × 10 ^{- 3}} + 10.1 × 10^ 5$

= 1.0131 × 10 5

= 1.01 × 10 5 Pa

Excess pressure = $\frac{2S }{ r}$

$= \frac{2 × 4.65 × 10^{ - 1} }{ 3 × 10 ^{- 3} }$

= 310 Pa