Problem Statement
The surface tension of water in contact with air is 0.0725 N/m. The pressure outside the droplet of water with a diameter of 0.02 mm is atmospheric (1.032 × 105 N/m2). Calculate the pressure within the droplet of water.
Solution
Given:
- Diameter of droplet = 0.02 × 10-3 m
- Radius of droplet (r) = 0.01 × 10-3 m
- Surface tension (σ) = 0.0725 N/m
- Pressure outside droplet (Po) = 1.032 × 105 N/m2
- Pressure within droplet (Pd) = ?
Formula:
Excess pressure inside the droplet:
P = 2σ / r
Calculations:
Substitute the given values:
P = 2 × 0.0725 / (0.01 × 10-3)
P = 14500 N/m2
Now calculate the total pressure inside the droplet:
Pd = P + Po
Substitute the values:
Pd = 14500 + 1.032 × 105
Pd = 117700 N/m2
Result:
The pressure within the droplet is 117700 N/m2.
Explanation
This problem illustrates the calculation of the pressure within a droplet of water using the principles of surface tension:
- Excess pressure: The pressure inside the droplet is greater than the atmospheric pressure outside, due to surface tension.
- Formula: The excess pressure is given by the formula P = 2σ / r, which relates the surface tension, radius, and pressure difference.
- Total pressure: Adding the atmospheric pressure to the excess pressure gives the total pressure inside the droplet.
This is a practical application of fluid mechanics to calculate the effects of surface tension in small-scale systems.
