Question

What is the pressure within a 1 mm diameter spherical droplet of water relative to the atmospheric pressure outside? The surface tension of water is 0.07 N/m.

• A. 140 Pa
• B. 280 Pa
• C. 420 Pa
• D. 560 Pa

Hint:

For a soap bubble (which has two free surfaces), the excess pressure is given by \( \Delta P = \frac{4\sigma}{R} \). Be careful to use the correct formula based on whether it's a droplet (one surface) or a bubble (two surfaces). Always ensure consistent units.

Answer 1

The Correct Option is B

Step 1: Understand the concept of excess pressure inside a liquid droplet due to surface tension.
Due to surface tension, the pressure inside a curved liquid surface (like a droplet or bubble) is higher than the pressure outside. This excess pressure is what balances the forces due to surface tension. Step 2: Recall the formula for excess pressure inside a spherical droplet.
For a spherical droplet of radius \( R \) and surface tension \( \sigma \), the excess pressure \( \Delta P \) inside the droplet relative to the outside pressure is given by: $$\Delta P = P_{inside} - P_{outside} = \frac{2\sigma}{R}$$ where:
\( \Delta P \) is the excess pressure
\( \sigma \) is the surface tension of the liquid
\( R \) is the radius of the droplet
Step 3: Identify the given parameters and convert units if necessary.
We are given:
Diameter of the droplet \( d = 1 \) mm = \( 1 \times 10^{-3} \) m
Radius of the droplet \( R = \frac{d}{2} = \frac{1 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \) m
Surface tension of water \( \sigma = 0.07 \) N/m
Step 4: Substitute the values into the formula for excess pressure.
$$\Delta P = \frac{2 \times 0.07 \text{ N/m}}{0.5 \times 10^{-3} \text{ m}}$$$$\Delta P = \frac{0.14 \text{ N/m}}{0.5 \times 10^{-3} \text{ m}}$$$$\Delta P = 0.28 \times 10^{3} \text{ N/m}^2$$ $$\Delta P = 280 \text{ N/m}^2$$ Since 1 Pa (Pascal) = 1 N/m\(^2\), the excess pressure is 280 Pa. This is the pressure within the droplet relative to the atmospheric pressure outside.