Question

The surface tension and vapour pressure of a liquid at 25 °C are \(8 \times 10^{-2} \, \text{N/m}\) and \(2.5 \times 10^3 \, \text{Pa}\) respectively. Find the radius of the smallest spherical water droplet which can form without evaporating at 25 °C.

• A. 64 \(\mu \text{m}\)
• B. 30 \(\mu \text{m}\)
• C. 60 \(\mu \text{m}\)
• D. 32 \(\mu \text{m}\)

Hint:

For smallest stable droplet, use \(r = 2 \gamma / P_{\text{vapour}}\). Ensure units of surface tension and pressure match.

Answer 1

The Correct Option is A

Step 1: Derive the balance condition instead of just quoting the formula.
A curved liquid surface creates an excess pressure inside a droplet compared to outside, given by $\Delta P = \dfrac{2\gamma}{r}$ for a sphere, with the factor of 2 coming from the two equal principal curvatures of a sphere. For the droplet to survive without evaporating, this excess pressure must at least match the vapour pressure that is trying to push molecules off the surface.

Step 2: Set the excess pressure equal to the vapour pressure for the borderline (smallest) droplet.
\[ \frac{2\gamma}{r} = P_v \implies r = \frac{2\gamma}{P_v} \]

Step 3: Substitute $\gamma = 8\times10^{-2}$ N/m and $P_v = 2.5\times10^3$ Pa.
\[ r = \frac{2\times8\times10^{-2}}{2.5\times10^3} = \frac{0.16}{2500} \]

Step 4: Compute.
\[ r = 6.4\times10^{-5}\ \text{m} = 64\ \mu\text{m} \]

Step 5: Conclusion.
\[ \boxed{64\ \mu\text{m}} \]