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Question: A drop of mercury of radius 2 mm is split into 8 identical drops. Find the increase in surface energ...

A drop of mercury of radius 2 mm is split into 8 identical drops. Find the increase in surface energy. Surface tension of mercury =0.465 Jm−20.465{\text{ }}J{m^{ - 2}}

Explanation

Solution

Surface tension is the propensity of liquid surfaces to shrink to the smallest possible surface area while they are at rest. Surface tension is what permits things with a higher density than water, such as razor blades and insects, to float on a water surface without submerging. The stronger attraction of liquid molecules to each other than the molecules in the air causes surface tension at liquid–air contacts.

Complete answer:
There are two main processes at work here. One causes the liquid to compress due to an inward push on the surface molecules. A tangential force parallel to the liquid's surface is the second. Surface tension is the name given to this tangential force (per unit length). As a result, the liquid acts as if it were surrounded by a stretched elastic membrane. However, this parallel must be used with caution since the tension in an elastic membrane is determined by the degree of deformation, whereas surface tension is a characteristic of the liquid–air or liquid–vapour interface.
The breakdown of intermolecular interactions that happens when a surface is produced is measured by surface free energy, also known as interfacial free energy or surface energy. Surfaces must be fundamentally less energetically advantageous than the bulk of a material in solids physics (the molecules on the surface have more energy than the molecules in the bulk of the substance), otherwise there would be a driving force for surfaces to form, eliminating the bulk of the material (see sublimation). As a result, surface energy may be described as the extra energy at a material's surface relative to the bulk, or as the effort necessary to construct a specific surface area.
From the question
R=2  mm⇒2×103  m{\text{R}} = 2\;{\text{mm}} \Rightarrow 2 \times {10^3}\;{\text{m}}
The volume of drop here is given as:
43πR3=8×43×r3\dfrac{4}{3}\pi {{\text{R}}^3} = 8 \times \dfrac{4}{3} \times {{\text{r}}^3}
∴R3=8r3\therefore {{\text{R}}^3} = 8{{\text{r}}^3}
R38=r3⇒(R2)3=r3\dfrac{{{{\text{R}}^3}}}{8} = {{\text{r}}^3} \Rightarrow {\left( {\dfrac{{\text{R}}}{2}} \right)^3} = {{\text{r}}^3}
∴r=1  mm=10−3  m\therefore {\text{r}} = 1\;{\text{mm}} = {10^{ - 3}}\;{\text{m}}
We know that
A1=4πR2{{\text{A}}_1} = 4\pi {{\text{R}}^2}
Upon substituting we get
⇒4π(2×10−2)2=4π×4×10−6=16π×10−6\Rightarrow 4\pi {\left( {2 \times {{10}^{ - 2}}} \right)^2} = 4\pi \times 4 \times {10^{ - 6}} = 16\pi \times {10^{ - 6}}
also
A2=8×4πr2=32π(10−2)3=32π10−6{{\text{A}}_2} = 8 \times 4\pi {{\text{r}}^2} = 32\pi {\left( {{{10}^{ - 2}}} \right)^3} = 32\pi {10^{ - 6}}
Difference in area can be explained as
ΔA=A1−A2\Delta {\text{A}} = {{\text{A}}_1} - {{\text{A}}_2}
(32π−16π)10−6=16π×10−6(32\pi - 16\pi ){10^{ - 6}} = 16\pi \times {10^{ - 6}}
So, The increase in surface energy will be:
Es=TΔA=16π×10−6(0.465)=23.376{{\text{E}}_{\text{s}}} = {\text{T}}\Delta {\text{A}} = 16\pi \times {10^{ - 6}}(0.465) = 23.376
⇒Es=23.376μJ\Rightarrow {{\text{E}}_{\text{s}}} = 23.376\mu J

Note:
A molecule is pushed equally in every direction by neighbouring liquid molecules due to cohesive forces, resulting in a net force of zero. Because the molecules near the surface are not surrounded by the identical molecules on all sides, they are pushed inward. This produces internal pressure, causing liquid surfaces to shrink to the smallest possible area.