We know that To have minimum potential energy, molecules at liquid surface tend to reduces area to minimum value. Thus, liquid surface behave as a stretched membrane. This tendency of liquid surface is known as surface tension.

Surface Tension

*It is defined as the force per unit length.*

The force acting per unit length of imaginary line drawn on free liquid surface at right angles on either sides of the line.

If L is the length of line & F is the force acting on it, the surface tension (T) is given by,

∴ T = F/L

Unit of surface tension in SI system is newton per meter (N/m)

In this article we are going to discuss the relation between surface tension and surface energy.

**Let’s discuss the concept in details……………!**

**Surface tension**

The potential energy per unit area of the liquid surface is called the surface energy.

**Relation between surface energy & surface tension**:

Consider, ABCD as open rectangular frame of wire on which a wire PQ can slide without friction. Dip the frame into soap solution, take out and hold horizontally so that the soap film AB’C’D if formed.

Due to the surface tension, a force of F acts on the wire LM of length ‘L’ tending to pull it towards AB given as,

∴ F = 2TL

[2-indicates that the soap solution has 2 surface, both of which are in contact with the wire.]

Imagine that the wire is pulled by infinitesimal small distance ‘dx’, then the work done during against the force of surface tension is given as,

∴ W = Force x displacement

∴ W = F dx

∴ W = 2TL dx

Now, the change in the surface area of the film is.

∴ A = 2Ldx

Hence, the mechanical work done per unit surface area is.

∴W/A = 2TLdx/2Ldx = T

∴W = T.dA

This work is stored in the form of potential energy by the surface. This poetical energy of thee surface per unit area is surface energy.

∴U = T.dA

∴ Surface energy per unit area = surface tension

**Let’s solve some numerical for more illustration…..!**

Ex:1) Oil film forms on rectangular frame of wire area 4 cm x 2.5 cm increases to area 5.5 cm x 3.5 cm. Find the work done in increasing area of film if surface tension of oil is 65 dyne/cm.

Solution:

Here, A_{1} =4 cm x 2.5 cm =10 cm^{2}, A_{2} = 5.5 cm x 3.5 cm = 19.25 cm^{2},

T = 65 dyne/cm

We have,

∴W = T.dA

∴W = T.( A_{2} – A_{1})

∴W = 65 (19.25 – 10)

∴W = 65( 9.25)

∴W = 601.25 erg