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Question: The surface tension and vapour pressure of water at\(20 ^ { \circ } \mathrm { C }\) is \(7.20 \times...

The surface tension and vapour pressure of water at20∘C20 ^ { \circ } \mathrm { C } is 7.20×10−2 N m−17.20 \times 10 ^ { - 2 } \mathrm {~N} \mathrm {~m} ^ { - 1 } and 2.33×1032.33 \times 10 ^ { 3 } Pa respectively. The radius of the smallest spherical water droplet which can form without evaporating at 25°C is

A

(a) 1.25×10−5 m1.25 \times 10 ^ { - 5 } \mathrm {~m}

A

(b) 6.25×10−5 m6.25 \times 10 ^ { - 5 } \mathrm {~m}

A

(c) 4.3×108 m4.3 \times 10 ^ { 8 } \mathrm {~m}

A

(d) 3.4×103 m3.4 \times 10 ^ { 3 } \mathrm {~m}

Explanation

Solution

(b) Here, S=7.28×10−2Nm−1,P=2.33×103 Pa\mathrm { S } = 7.28 \times 10 ^ { - 2 } \mathrm { Nm } ^ { - 1 } , \mathrm { P } = 2.33 \times 10 ^ { 3 } \mathrm {~Pa}

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

As ,Where R is the radius of drop

∴\therefore R=2 SP=2×7.28×10−22.33×103\mathrm { R } = \frac { 2 \mathrm {~S} } { \mathrm { P } } = \frac { 2 \times 7.28 \times 10 ^ { - 2 } } { 2.33 \times 10 ^ { 3 } }

=6.25×10−5 m= 6.25 \times 10 ^ { - 5 } \mathrm {~m}