Question
The sufrace tension and vapour pressure of water at 20°C is $7.28 \times 10^{-2}N$ $m^{-1}$ and $2.33 \times 10^3$ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?
✓
Answer
The drop will evaporate if the water pressure on liquid, is greater than vapour pressure above the surface of liquid. Let a water droplet of radius R be formed without evaporation then. Vapour pressure = Excess pressure in a drop$\rho=\frac{2\sigma}{\text{R}}$ (onlt one surface in drop)
$\text{R}=\frac{2\times7.28\times10^{-2}}{\text{Vapour preessure}}$
$=\frac{2\times7.28\times10^{-2}}{2.33\times10^{3}}=\frac{1456\times10}{233\times10^5}$
$\text{R}=6.25\times10^{-5}\text{m}$
$\text{R}=\frac{2\times7.28\times10^{-2}}{\text{Vapour preessure}}$
$=\frac{2\times7.28\times10^{-2}}{2.33\times10^{3}}=\frac{1456\times10}{233\times10^5}$
$\text{R}=6.25\times10^{-5}\text{m}$
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