General Chemistry

Question

What amount of energy is required to change a spherical drop of water with a diameter of  to four smaller spherical drops of equal size? The surface tension,  , of water at room temperature is  .

Answer 1

This problem seems more appropriate for physics than chemistry, but I'll give it a try.....

Surface area of sphere = 4π r²

Volume of sphere = (4/3)π r³

radius = 1/2 diameter = 1/2 2.20 mm = 1.10 mm

Surface area = 4 x 3.14 x (1.10)² = 15.20 mm² = 0.00001520 m²

Volume = (4/3) x 3.14 x (1.10)³ = 5.57 mm³

Surface tension of original droplet = γ = 72.0 mJ/m² x 0.00001520 m² = 0.00114 mJ

Volume of the 4 new droplets must have the same total volume as the original droplet, thus...

volume of 1 of the new droplets = 5.57 mm³ / 4 = 1.39 mm³

From this we can find the radius of the new droplet: (4/3)π r³ = 1.39 mm² and r = 0.692 mm

Surface area of each new droplet = 4π r2 = 4 x 3.14 x (0.692 mm)² = 6.01 mm² = 0.00000601 m²

Surface tension of each new droplet = γ = 72.0 mJ/m² x 0.00000601 m² = 0.000433 mJ

Total surface tension of all new droplets = 4 x 0.000433 mJ = 0.00173 mJ

Difference in surface tension of the 4 new droplets from the original droplet is

0.00173 mJ - 0.00114 mJ = 0.00059 mJ = 5.9x10^-4 mJ

and this would be the energy required to perform the desired task.

(NOTE: I just noticed the "comment" by Brian L, and this answer is indeed very close to his answer of

6.4x10^-4 mJ).