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.
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.
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For smallest stable droplet, use \(r = 2 \gamma / P_{\text{vapour}}\). Ensure units of surface tension and pressure match.
Updated On: Jul 18, 2026
- 64 \(\mu \text{m}\)
- 30 \(\mu \text{m}\)
- 60 \(\mu \text{m}\)
- 32 \(\mu \text{m}\)
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The Correct Option is A
Solution and Explanation
Step 1: Recall the formula for equilibrium droplet radius.
The radius \(r\) is given by the balance between surface tension and vapour pressure:
\[ r = \frac{2 \gamma}{\Delta P} \]
where \(\gamma\) is surface tension and \(\Delta P\) is pressure difference.
Step 2: Identify pressure difference.
\(\Delta P = P_{\text{vapour}} = 2.5 \times 10^3 \, \text{Pa}\), \(\gamma = 8 \times 10^{-2} \, \text{N/m}\).
Step 3: Substitute values.
\[ r = \frac{2 \cdot 8 \times 10^{-2}}{2.5 \times 10^3} = \frac{0.16}{2500} \]
Step 4: Calculate radius.
\[ r = 6.4 \times 10^{-5} \, \text{m} = 64 \, \mu \text{m} \]
Step 5: Verify units.
\([r] = \text{N/m} / \text{Pa} = \text{m}\), units consistent.
Step 6: Final conclusion.
Hence, the radius of the smallest droplet is:
\[ \boxed{64 \, \mu \text{m}} \]
The radius \(r\) is given by the balance between surface tension and vapour pressure:
\[ r = \frac{2 \gamma}{\Delta P} \]
where \(\gamma\) is surface tension and \(\Delta P\) is pressure difference.
Step 2: Identify pressure difference.
\(\Delta P = P_{\text{vapour}} = 2.5 \times 10^3 \, \text{Pa}\), \(\gamma = 8 \times 10^{-2} \, \text{N/m}\).
Step 3: Substitute values.
\[ r = \frac{2 \cdot 8 \times 10^{-2}}{2.5 \times 10^3} = \frac{0.16}{2500} \]
Step 4: Calculate radius.
\[ r = 6.4 \times 10^{-5} \, \text{m} = 64 \, \mu \text{m} \]
Step 5: Verify units.
\([r] = \text{N/m} / \text{Pa} = \text{m}\), units consistent.
Step 6: Final conclusion.
Hence, the radius of the smallest droplet is:
\[ \boxed{64 \, \mu \text{m}} \]
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