Question

The surface tension of water is

72.8 dynes {cm}^{- 1}

. Calculate the energy required to disperse one spherical drop of radius

3.0 mm

to

3.0 \times 10^{- 3} mm

.

Accepted Answer

According to the definition of surface tension,

$γ = \frac{F}{l} γ = \frac{72.8 dyne}{1 cm} = \frac{72.8 \times 10^{- 5} N}{10^{- 2} m} = 72.8 \times 10^{- 3} {Nm}^{- 1}$

Now, the surface area of spherical drop having radius $3.0 mm$ is,

$4 {πr}_{1}^{2} = 4 \times 3.14 \times {(\frac{3}{1000})}^{2} = 1.13 \times 10^{- 4} m^{2}$

and the surface area of spherical drop having radius $3.0 \times 10^{- 3} mm$ is,

$4 {πr}_{2}^{2} = 4 \times 3.14 \times {(\frac{3.0 \times 10^{- 3}}{10^{3}})}^{2} = 1.13 \times 10^{- 10} m^{2}$

Now, the number of small-sized drops $= \frac{Volume of bigger drops}{Volume of small ‐ sized drops}$

$= \frac{\frac{4}{3} π {(\frac{3}{1000})}^{3}}{\frac{4}{3} π {(\frac{3 \times 10^{- 3}}{1000})}^{3}} = 10^{9}$

Now, the total surface area of small-sized drop

$= 1.13 \times 10^{- 10} \times 10^{9} = 1.13 \times 10^{- 1} m^{2}$

Increase in surface area

$= (1.13 \times 10^{- 1} - 1.13 \times 10^{- 4}) m^{2} = (1.13 \times 10^{- 4} \times 999) m^{2}$

Hence, work done in dispersion= $surface tension \times increase in surface area$

$= 72.8 \times 10^{- 3} {Nm}^{- 1} \times (1.13 \times 10^{- 4} \times 999) m^{2} = 82181 \times 10^{- 7} Nm = 82181 \times 10^{- 7} J$

Seema Saini Solutions for Chapter: States of Matter, Exercise 7: DPP 4.7

Seema Saini Chemistry Solutions for Exercise - Seema Saini Solutions for Chapter: States of Matter, Exercise 7: DPP 4.7

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