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Question: A liquid drop of diameter \(4mm\) breaks into 1000 droplets of equal size. Calculate the resultant c...

A liquid drop of diameter 4mm4mm breaks into 1000 droplets of equal size. Calculate the resultant change in Surface energy. The surface tension is 0.07 Nm−10.07\,N{m^{ - 1}} .

Explanation

Solution

The disruption of intermolecular bonds that occurs when a surface is formed is measured by surface energy. Surface free energy or interfacial free energy are other names for it. In layman's terms, surface energy is the work performed per unit area by the force that produces the new surface.

Complete answer:
Surface tension is the phenomenon that occurs when the surface of a liquid comes into contact with another phase, according to the description (it can be a liquid as well). Liquids prefer to have as little surface area as possible. The liquid's surface behaves like an elastic sheet.
Let us come to the problem:
Since the diameter of drop=4mm = 4mmRadius of drop=2mm=2×10−3 = 2mm = 2 \times {10^{ - 3}}
S=Surface tension=0.07 Nm−10.07\,N{m^{ - 1}}
a large drop in volume=1000×= 1000 \times volume of each droplet
Let R be the major drop's radius and r be the small droplet's radius.
43πR3=1000×43πr3\dfrac{4}{3}\pi {R^3} = 1000 \times \dfrac{4}{3}\pi {r^3}
R3=(10r)3 R=10r  {R^3} = {(10r)^3} \\\ R = 10r \\\
liquid drop's initial energy=E1=T.A = {E_1} = T.A
E1=T×4πR2{E_1} = T \times 4\pi {R^2} T= surface tension, A=area
Final energy of 1000 droplets =E2=1000T.a = {E_2} = 1000T.a
E2=1000T×4πr2{E_2} = 1000T \times 4\pi {r^2} T= surface tension, a=area
Now, change in energy=E2−E1 = {E_2} - {E_1}
=1000T×4πr2= 1000T \times 4\pi {r^2}-T×4πR2T \times 4\pi {R^2}

=T×4π[1000r2−R2] =T×4π[1000(R10)2−R2]∵R=10r =T×4π[10R2−R2] =T×4π×9R2 =36πTR2  = T \times 4\pi [1000{r^2} - {R^2}] \\\ = T \times 4\pi [1000{\left( {\dfrac{R}{{10}}} \right)^2} - {R^2}]\\{ \because R = 10r\\} \\\ = T \times 4\pi [10{R^2} - {R^2}] \\\ = T \times 4\pi \times 9{R^2} \\\ = 36\pi T{R^2} \\\

Now, let us put the value of T and R
=36×3.14×0.07  Jm−2×(2×10−3)= 36 \times 3.14 \times 0.07\,\,J{m^{ - 2}} \times (2 \times {10^{ - 3}})
=3.165×10−5  Jm−2= 3.165 \times {10^{ - 5}}\,\,J{m^{ - 2}}

Note:
The surface tension of a liquid refers to the attractive force that the molecules present at its surface exert on each other. The equivalent attractive force present between the molecules at the surface of a solid material is known as surface energy.