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

Given Data

• Radius of mercury drop: \( r = 3.00 \,\text{mm} = 3.00 \times 10^{-3} \,\text{m} \)
• Surface tension of mercury: \( S = 4.65 \times 10^{-1} \,\text{N m}^{-1} \)
• Atmospheric pressure: \( P_{0} = 1.01 \times 10^{5} \,\text{Pa} \)

1. Excess pressure inside the drop

For a liquid drop (single interface), the excess pressure due to surface tension is \( \Delta P = \dfrac{2S}{r} \).

\( \Delta P = \dfrac{2 \times 4.65 \times 10^{-1}}{3.00 \times 10^{-3}} \) \( = \dfrac{9.30 \times 10^{-1}}{3.00 \times 10^{-3}} \) \( = \dfrac{0.93}{3.00 \times 10^{-3}} = 310 \,\text{Pa} \)

Excess pressure inside the mercury drop \( \Delta P = 310 \,\text{Pa} \).

2. Total pressure inside the drop

The pressure inside the drop is the sum of atmospheric pressure and excess pressure:

\( P_{\text{inside}} = P_{0} + \Delta P \) \( = 1.01 \times 10^{5} + 310 \,\text{Pa} \) \( = 1.0131 \times 10^{5} \,\text{Pa} \)

Given that the data are to three significant figures, this is written as

Pressure inside the drop \( P_{\text{inside}} \approx 1.01 \times 10^{5} \,\text{Pa} \); Excess pressure \( = 310 \,\text{Pa} \).

Physical note

• Surface tension acts to shrink the surface of the drop, creating extra pressure inside compared to the outside.
• Smaller drops (smaller radius) have larger excess pressure because \( \Delta P \propto 1/r \).