The surface tension and vapour pressure of water at 20° C is 7.28× 10-2Nm-1 and 2.33× 103Pa respectively. The radius of the smallest spherical droplet which can form without evaporating is

Question

The surface tension and vapour pressure of water at 20° C is 7.28× 10-2Nm-1 and 2.33× 103Pa respectively. The radius of the smallest spherical droplet which can form without evaporating is (A)  6.25× 10-5m  (B)  6.25× 10-3m  (C)  7.45× 10+3m  (D)  8.9× 10-4m

Tardigrade Admin · Accepted Answer

Given T=7.28× 10-2N/m p=2.33× 103Pa The drop will evaporate if the water pressure greater than the vapour pressure p. i.e. p=(2T/R), where R = radius of drop ⇒ R=(2T/p)=(2× 7.28× 10-2/2.33× 103)=6.25× 10-5m

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

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

$6.25 \times 10^{- 3} m$

$7.45 \times 10^{+ 3} m$

$8.9 \times 10^{- 4} m$

Q. The surface tension and vapour pressure of water at 20∘C is 7.28×10−2Nm−1 and 2.33×103Pa respectively. The radius of the smallest spherical droplet which can form without evaporating is

6.25×10−5m

6.25×10−3m

7.45×10+3m

8.9×10−4m

Given T=7.28×10−2N/m p=2.33×103Pa The drop will evaporate if the water pressure greater than the vapour pressure p. i.e. p=R2T​, where R = radius of drop ⇒ R=p2T​=2.33×1032×7.28×10−2​=6.25×10−5m

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

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

$6.25 \times 10^{- 3} m$

$7.45 \times 10^{+ 3} m$

$8.9 \times 10^{- 4} m$