8.3 Surface energy $(E)$

Physics > Fluid Mechanics > 8.0 Intermolecular forces

  Fluid Mechanics
    1.0 Introduction
    2.0 Fluid pressure
    3.0 Pascal's law
    4.0 Buoyant force
    5.0 Flow of fluids
    6.0 Viscosity
    7.0 Stoke's law
    8.0 Intermolecular forces
    9.0 Angle of contact
8.3 Surface energy $(E)$
It is defined as the amount of work done against the force of surface tension in increasing the liquid surface of a given area at a constant temperature. $$E = W = TdA$$ where,
$E$: Surface energy
$W$: Work done
$T$: Surface tension
$dA$: Increase in surface area

Derivation of surface energy $(E)$

Consider a wire frame equipped with sliding wire $AB$. It is dipped in soapy water. A film of liquid is formed on it.

A force $F$ is applied on the wire which moves it through a small distance $dx$ as shown in the figure.

Work done by the force is, $$\begin{equation} \begin{aligned} dW = Fdx \\ dW = (2Tl)dA\quad \left( {As,\;T = \frac{F}{{2l}}} \right) \\\end{aligned} \end{equation} $$ Also, $$2ldx = dA\quad \left( {{\text{increase in surface area of the film}}} \right)$$ So, $$dW = TdA$$
Surface tension can also be expressed as, $$T = \frac{{dW}}{{dA}}$$
So, surface tension is also defined as the work done in increasing the surface area by unit.

As there is no change in kinetic energy. So, the work done by the external force is stored as the potential energy of the new surface. $$T = \frac{{dU}}{{dA}}$$
The surface tension of a liquid is equal to the surface energy per unit area.

Note:
  • SI unit of surface energy is Joule $(erg)$ and CGS unit is $erg$
  • Work done in forming a liquid of drop of radius $R$ and surface tension $T$ is given by, $$W = 4\pi {R^2}T$$ Liquid drop has only one surface
  • Work done in forming a soap bubble of radius $R$ and surface tension $T$ is given by, $$\begin{equation} \begin{aligned} W = 2 \times 4\pi {R^2}T \\ W = 8\pi {R^2}T \\\end{aligned} \end{equation} $$ Soap bubble has two surfaces.
  • Work done in increasing the radius of liquid drop from radius $R_1$ to $R_2$ is given by, $$\begin{equation} \begin{aligned} {W_1} = 4\pi R_1^2T \\ {W_2} = 4\pi R_2^2T \\\end{aligned} \end{equation} $$ So, the work done is given by, $$\begin{equation} \begin{aligned} W = {W_2} - {W_1} \\ W = 4\pi \left( {R_2^2 - R_1^1} \right) \\\end{aligned} \end{equation} $$
  • Work done in breaking a liquid drop into $n$ equal small drops
Consider a liquid drop of radius $R$ which is breaked into $n$ smaller liquid drops each of radius $r$ as shown in the figure,

Volume of the bigger liquid drop, ${V_1} = \frac{4}{3}\pi {R^3}$

Volume of the $n$ smaller liquid drop, ${V_2} = n \times \frac{4}{3}\pi {r^3}$

According to the law of conservation of mass, $${V_1} = {V_2}$$ So, $$\begin{equation} \begin{aligned} \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {r^3}n \\ r = \frac{R}{{{n^{\frac{1}{3}}}}} \\\end{aligned} \end{equation} $$


Bigger bubbleSmaller bubble

Volume$${V_1} = \frac{4}{3}\pi {R^3}$$$${V_2} = n \times \frac{4}{3}\pi {r^3}$$
Surface energy$${E_1} = 4\pi {R^2}T$$$${E_2} = 4\pi {R^2}T$$
Conservation of mass

$$\begin{equation} \begin{aligned} {m_1} = {m_2} \\ \rho \left( {\frac{4}{3}\pi {R^3}} \right) = \rho \left( {\frac{4}{3}\pi {r^3}n} \right) \\ r = \frac{R}{{{n^{\frac{1}{3}}}}} \\\end{aligned} \end{equation} $$

Work done in breaking the soap bubble

$$\begin{equation} \begin{aligned} W = {E_1} - {E_2} \\ W = 4\pi {R^2}Tn - 4\pi {R^2}T \\ W = 4\pi {R^2}T\left( {{n^{\frac{1}{3}}} - 1} \right) \\\end{aligned} \end{equation} $$



Improve your JEE MAINS score
10 Mock Test
Increase JEE score
by 20 marks
Detailed Explanation results in better understanding
Exclusively for
JEE MAINS and ADVANCED
9 out of 10 got
selected in JEE MAINS
Lets start preparing
Read Complete Chapter
DIFFICULTY IN UNDERSTANDING CONCEPTS?
TAKE HELP FROM THINKMERIT DETAILED EXPLANATION..!!!
9 OUT OF 10 STUDENTS UNDERSTOOD
START LEARNING