Ncert Exercises & Solutions

Q. 20 The surface tension and vapour pressure of water at $20 ° C$ $20 ° C$ are $7.28 \times 10^{- 2} N m^{- 1} a n d 2.33 \times 10^{3}$ $7.28 \times 10^{- 2} N m^{- 1} a n d 2.33 \times 10^{3}$ Pa respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20 ° C$ $20 ° C$ ?

Hint: The drop will evaporate if the water pressure is greater than the vapour pressure.

Step 1: Find the radius of the smallest droplet.

Given, the surface tension of water,

(S) = 7.28 \times 10^{- 2}

$(S) = 7.28 \times 10^{- 2}$ N/m

Vapour pressure, (P)=2.33 x

10^{3} P a

$10^{3} P a$

Let a water droplet of radius R can be formed without evaporating.

Vapour pressure = Excess pressure in drop

∴ ρ = \frac{2 S}{R}

$∴ ρ = \frac{2 S}{R}$

R = \frac{2 S}{ρ} = \frac{2 \times 7.28 \times 10^{- 2}}{2.33 \times 10^{3}}

$R = \frac{2 S}{ρ} = \frac{2 \times 7.28 \times 10^{- 2}}{2.33 \times 10^{3}}$

= 6.25 \times 10^{- 5} m

$= 6.25 \times 10^{- 5} m$

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