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.
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……………!
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.
Here, A1 =4 cm x 2.5 cm =10 cm2, A2 = 5.5 cm x 3.5 cm = 19.25 cm2,
T = 65 dyne/cm
∴W = T.dA
∴W = T.( A2 – A1)
∴W = 65 (19.25 – 10)
∴W = 65( 9.25)
∴W = 601.25 erg