The pressure outside a water droplet of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if the surface tension of water is 0.0725 N/m.

Pressure Inside a Water Droplet

Problem Statement

Given Data

Diameter of droplet, $d = 0.04 \, \text{mm}$
Pressure outside the droplet, $p_{out} = 10.32 \, \text{N/cm}^2$
Surface Tension, $\sigma = 0.0725 \, \text{N/m}$

Solution

1. Convert All Units to SI

To ensure consistency in calculations, we convert all given values to standard SI units (meters and N/m²).

$$ d = 0.04 \, \text{mm} = 0.04 \times 10^{-3} \, \text{m} $$ $$ p_{out} = 10.32 \, \frac{\text{N}}{\text{cm}^2} \times \left(\frac{100 \, \text{cm}}{1 \, \text{m}}\right)^2 = 10.32 \times 10^4 \, \text{N/m}^2 = 103200 \, \text{N/m}^2 $$

2. Calculate the Excess Pressure ($p_{excess}$)

The excess pressure $p_{excess}$ inside a droplet is caused by surface tension and is calculated using the formula:

$$ p_{excess} = \frac{4\sigma}{d} $$

Substituting the given values:

$$ p_{excess} = \frac{4 \times 0.0725 \, \text{N/m}}{0.04 \times 10^{-3} \, \text{m}} = \frac{0.29}{0.00004} \, \text{N/m}^2 $$ $$ p_{excess} = 7250 \, \text{N/m}^2 $$

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

The total pressure inside the droplet is the sum of the outside atmospheric pressure and the excess pressure from surface tension.

$$ p_{in} = p_{out} + p_{excess} $$ $$ p_{in} = 103200 \, \text{N/m}^2 + 7250 \, \text{N/m}^2 = 110450 \, \text{N/m}^2 $$

4. Convert Final Pressure to N/cm²

For easier comparison with the given outside pressure, we convert the result back to N/cm².

$$ p_{in} = 110450 \, \frac{\text{N}}{\text{m}^2} \times \left(\frac{1 \, \text{m}}{100 \, \text{cm}}\right)^2 = \frac{110450}{10000} \, \frac{\text{N}}{\text{cm}^2} $$ $$ p_{in} = 11.045 \, \text{N/cm}^2 $$

Final Result:

The pressure inside the droplet is $ p_{in} = 11.045 \, \text{N/cm}^2 $.

Explanation of Excess Pressure

Surface Tension ($\sigma$) creates a net inward force on the surface molecules of a liquid, causing the liquid to behave as if it's covered by a thin, stretched membrane. To maintain the equilibrium of a curved surface like a droplet, the pressure inside must be greater than the pressure outside. This difference is the excess pressure.

The total pressure inside the droplet is therefore the sum of the external pressure (e.g., atmospheric pressure) and this additional pressure generated by surface tension forces.

Physical Meaning

This calculation demonstrates that even for a tiny droplet, the internal pressure is noticeably higher than the surrounding atmospheric pressure. The excess pressure, $p_{excess}$, is inversely proportional to the droplet's diameter ($p_{excess} \propto 1/d$).

Smaller Droplets, Higher Pressure: This means that as a droplet gets smaller, the pressure inside it increases dramatically. This is a key principle in the physics of aerosols, emulsions, and cloud formation.
Magnitude: In this case, the excess pressure (7250 N/m²) is about 7% of the atmospheric pressure (103200 N/m²), a significant increase caused solely by surface tension acting on a very small surface.

The pressure outside a water droplet of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if the surface tension of water is 0.0725 N/m.

Problem Statement

Given Data

Solution

1. Convert All Units to SI

2. Calculate the Excess Pressure ($p_{excess}$)

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

4. Convert Final Pressure to N/cm²

Explanation of Excess Pressure

Physical Meaning

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