The surface tension of water in contact with air is given as 0.0725 N/m. The pressure outside a droplet of water of diameter 0.02 mm is atmospheric (10.32 N/cm²). Calculate the pressure within the droplet of water.

Pressure Inside a Water Droplet

Problem Statement

The surface tension of water in contact with air is given as 0.0725 N/m. The pressure outside a droplet of water of diameter 0.02 mm is atmospheric (10.32 N/cm²). Calculate the pressure within the droplet of water.

Given Data

  • Surface Tension, \(\sigma = 0.0725 \, \text{N/m}\)
  • Droplet Diameter, \(d = 0.02 \, \text{mm}\)
  • Outside Pressure, \(P_{out} = 10.32 \, \text{N/cm}^2\)

Solution

1. Convert All Units to SI

For consistency, we convert the diameter and outside pressure to their standard SI units (meters and N/m²).

$$ d = 0.02 \, \text{mm} = 0.02 \times 10^{-3} \, \text{m} $$
$$ P_{out} = 10.32 \, \frac{\text{N}}{\text{cm}^2} \times \frac{10000 \, \text{cm}^2}{1 \, \text{m}^2} $$ $$ P_{out} = 103200 \, \text{N/m}^2 $$

2. Calculate Excess Pressure (p)

The excess pressure inside a spherical droplet due to surface tension is given by the Young-Laplace equation.

$$ p = \frac{4\sigma}{d} $$ $$ p = \frac{4 \times 0.0725 \, \text{N/m}}{0.02 \times 10^{-3} \, \text{m}} $$ $$ p = \frac{0.29}{0.00002} \, \text{N/m}^2 $$ $$ p = 14500 \, \text{N/m}^2 $$

3. Calculate Total Pressure Inside the Droplet ($P_{in}$)

The total pressure inside is the sum of the outside pressure and the excess pressure.

$$ P_{in} = P_{out} + p $$ $$ P_{in} = 103200 \, \text{N/m}^2 + 14500 \, \text{N/m}^2 $$ $$ P_{in} = 117700 \, \text{N/m}^2 $$

4. Convert Final Pressure to N/cm²

For easier comparison, we convert the final pressure back to N/cm².

$$ P_{in} = 117700 \, \frac{\text{N}}{\text{m}^2} \times \frac{1 \, \text{m}^2}{10000 \, \text{cm}^2} $$ $$ P_{in} = 11.77 \, \text{N/cm}^2 $$
Final Result:

The pressure inside the droplet of water is \( P_{in} = 11.77 \, \text{N/cm}^2 \).

Explanation of Excess Pressure

Surface tension is a property of a liquid’s surface that causes it to behave like a stretched elastic membrane. For a droplet to maintain its curved, spherical shape, the pressure inside must be higher than the pressure outside. This pressure difference, known as excess pressure or Laplace pressure, is created by the cohesive forces of the liquid molecules pulling inward.

Physical Meaning

This problem highlights the inverse relationship between droplet size and internal pressure. Because the droplet is extremely small (0.02 mm diameter), the effect of surface tension is very strong, creating a significant excess pressure of 14500 N/m².

This excess pressure is about 14% of the surrounding atmospheric pressure, a substantial increase. This demonstrates that the smaller a droplet is, the higher the internal pressure must be to counteract the powerful surface tension forces trying to collapse it. This principle is fundamental in understanding phenomena like cloud formation, atomization in fuel injectors, and the stability of emulsions.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top