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