Question

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at the temperature (20° C) is 4.6 × 10⁻¹ N m⁻¹. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.

Answer 1

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, $P_e=\frac{2S}{r}$
Where S is the surface tension of the mercury.
r is the radius of the mercury drop.

Thus, P = $P_e+P_0$
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, $P_0$ = 1.01 × 10^5 Pa

so,  excess pressure , $P_e=\frac{2S}{r}$

= 2 × 4.65 × 10^-1/3 × 10^-3

= 2 × 1.55 × 10² Pa = 0.0031 × 10^5  Pa

Therefore,
P = 0.0031 × 10^5 Pa + 1.01 × 10^5 Pa
⇒ P = 1.0131 × 10^5Pa

The pressure inside the drop of mercury is 1.0131 × 10^5 Pa and the excess pressure inside the drop is 310 Pa.