Introduction:

  • Between any two molecules, there exists a force of attraction. This force is called the intermolecular force.
  • The attractive force between the two molecules of the same substance is called as a cohesive force and the attraction itself is called cohesion. e.g. attraction between water and water molecules.
  • The attractive force between the two molecules of the different substance is called an adhesive force and the attraction itself is called adhesion. e.g. attraction between water and glass molecules.

Range of molecular attraction (r) :

  • The maximum distance between two molecules up to which the intermolecular forces are effective is called the range of molecular attraction. It is of order 10-9 m.

Sphere of Molecular Influence :

  • A sphere drawn by taking the radius equal to the range of molecular attraction and centre as the centre of the molecule is called sphere of molecular influence.


Concept of Surface Tension:

  • Surface tension or force of surface tension is the force per unit length of an imaginary line drawn in any direction on the free surface of a liquid, the line of action of the force being on the surface and at right angles to the length of the imaginary line.
  • The dimensions of surface tension are [L0M1T-2].
  • S.I. unit of surface tension is N/m and c.g.s. unit is dyne/cm.

Explanation of Surface Tension:

 

Surface Tension Molecular Theory

  • The sphere of influence of the molecule A is well inside the liquid, and hence it will be acted upon by equal forces in all directions and these forces will balance one another and the net force acting on it is zero.
  • The part of the sphere of influence of molecule B is in the air, which contains air molecules.  The cohesive forces due to molecules in the liquid remain unbalanced and thus a net force in downward direction acts on the molecule.
  • Half of sphere of influence lies n the air. This molecule experiences the maximum possible unbalanced force in the downward direction.
  • Thus the molecules on the surface and in a surface experience net downward force.  The magnitude of force depends upon the distance of the molecule from the free surface.  The behaviour of this film is different from that of the rest of the liquid.  It is called the surface film.  This film behaves like a film which is under tension.  This phenomenon is known as surface tension.


Concept  of Surface Energy:

  • A work will have to be done to lift molecule from bulk of the liquid to the surface against the unbalanced downward forces of molecular attraction. This work will be stored in the molecule as potential energy.  This energy is called as surface energy.
  • Its S.I. unit is J and its dimensions are [L2M1T-2].

Relation Between Surface Tension and Surface Energy:

Surface Tension Surface Energy

  • Consider a rectangular frame ABCD in which side CD (of length l) is made of loose write and other three sides are fixed. The frame is immersed in a soap solution and taken out and held horizontally.  A force acting on the wire due to surface tension is directed towards AB, By definition of surface tension, we have

T = F / 2l

∴    F  =  T . 2 l

  • Imagine an external force is applied on CD which is equal and opposite to force F Let the wire at CD moves to C’D’ through small distance dx without acceleration.  Then the work done against the force of surface tension is given by

dW  =  F.dx  = T.2l. dx

 

But,  2l . dx = dA = increase in Area of both the surface of the film.

∴  dW   =  T.dA

∴   dU  =  T.dA

∴  T  =  dU / dA

  • This expression indicates that surface tension is equal to surface energy per unit area of the surface film.


Examples of Surface Tension (Due to the formation of the surface film):

  • When a glass rod is dipped in water and taken out it is found that the drop of water is sticking to the end of the rod.
  • When water falls slowly in drops from a tap, each drop gradually forms at the tip of the nozzle.  As the drop of water grows in size, it looks like a hanging elastic bag whose surface area is expanding as it fills up.
  • Water spiders are able to walk on the surface o the water because their feet produce dimples on the surface film without rupturing the film.
  • A safety razor blade, when placed gently with its flat surface on water floats on water.
  • A liquid tries to acquire minimum surface area on account of its surface tension. For a given volume sphere has the smallest surface area. Hence liquid drop falling down acquires spherical shape.

Factors Affecting Surface Tension:

Presence of  Impurity:
  • If the impurity added in a liquid is highly soluble, then the surface tension of liquid increases. and if the impurity added in a liquid is partly soluble, then the surface tension of liquid decreases.
Temperature:
  • Generally, the surface tension of liquid decreases with increase in temperature. it is due to increase in kinetic energy of liquid molecules. Exceptions:  cadmium and copper.
Contamination:
  • The presence of dust particles in liquid decreases its surface tension.


Problems Based on Surface Tension

  1. A needle 5 cm long can just rest on the surface of water without wetting. What is its weight? Surface tension of water = 0.07 N/m. (Ans:  0.007 N)
  2. A thin and light ring of a material of radius 3 cm is rested flat on the liquid surface. When slowly raised, it is found that the pull required is 0.03 N more before the film breaks than after. Find the surface tension of the liquid.(Ans:  0.08 N/m)
  3. A light square wire frame each side of which is 10 cm long hangs vertically in the water with one side just touching the water surface. Find the additional force necessary to pull the frame clear of water (T=0.074 N/m)(Ans:  0.0148 N)
  4. A thin wire is bent in the form of a rectangle of length 4 cm and breadth 3 cm. What force due to the surface tension does the side experience when a soap film is formed in the frame? S.T. of soap solution = 0.030 N/m. (Ans: 8.4 × 10-3 N)
  5. By dipping a U-shaped wire in a soap solution, a film is formed between it and a light sliding wire resting on it. The sliding wire supports a weight of 0.01 N when its length is 20 cm. Find the surface tension of the liquid. (Ans: 0.025 N/m).
  6. A glass tube of internal diameter 3.5 cm and thickness 0.5 cm is held vertically with its lower end immersed in water. Find the downward pull on the tube due to surface tension. The surface tension of water is 0.074 N/m. (Ans: 0.0186 N)
  7. A ring of glass is cut from a tube 7.4 internal and 7.8 cm external diameter. This ring with its lower side horizontal is suspended from one arm of a balance so that the lower edge is just immersed in a vessel of water. It is found that an additional weight of 3.62 g must be placed in the other scale pan to compensate for the pull of surface tension on the ring. Calculate the S.T. of water. g = 9.8 m/s2. (Ans: 0.074 N/m.)
  8. A rectangular glass plane (height 10 cm, breadth 8 cm and thickness = 0.2 cm) rests with the largest face flat on the surface of the water. Calculate the additional force required to pull the plane clear of the water. What is the force due to surface tension acting on the plane it is held vertically with the edge of the longest side just touching the water surface? Surface Tension of water = 0.070 N/m. (Ans: 0.0252 N, 0.0143 N)
  9. A square glass plate 10 cm long and 1 mm in thickness is suspended vertically with the lower edge horizontal from one arm of a balance and counterpoised so that the beam is horizontal. When the glass plate is allowed to touch the surface of a soap solution, an additional mass of 0.72g has to be added to the other pan to keep the beam horizontal. Find the S.T. of the liquid. g = 9.8 m/s2  (Ans: 0.036 N/m.)
  10. Calculate the force required to take away a flat circular plate of radius 0.01 m from the surface of the water. The surface tension of water is 0.075 N/m. (Ans:  4.713  × 10-3 N) 
  11. A rectangular film of liquid is formed in a frame of wire and a movable rod of length 4 cm. What force must be applied to the rod to keep it in equilibrium if the surface tension of the liquid is  40  × 10-3 N/m? (Ans:  3.2  × 10-3 N)
  12. A horizontal circular loop of wire of radius 0.02 m is lowered into crude oil form film. The force due to the surface tension of the liquid is 0.0113 N. Calculate the surface tension of crude oil. (Ans:  0.04496 N/m)
  13. The surface tension of water at 0 °C is 70 dynes/cm. Find the surface tension of water at 25 °C. Given α for water  = 0.0027/°C. (Ans:  65.28 dyne/cm)

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Problems Based on Surface Energy

  1. A water film is formed between two straight parallel wires of 10 cm each with a separation of 0.1 cm. If the distance between the wires is increased by 0.1 cm, how much work is done? T = 0.072 N/m (Ans:  1.44 × 10-5  J)
  2. A liquid film is formed in a rectangular frame one side of which is a light and thin movable wire of length 4 cm. The frame is held in a vertical plane with the movable wire in the downward position. Find what weight required to hold the movable wire in equilibrium. How much work is done if the film is stretched by pulling the movable wire downwards by 2 cm? T = 0.025 N/m. (Ans:  4 × 10-5  J)
  3. There is a soap film on a rectangular wire frame of area 4cm × 4 cm. If the area of the frame is increased to 4cm x 5 cm, find work done in the process. The surface tension of soap film is 3 × 10-2 N/m. (Ans: 2.4 × 10-5  J)
  4. Find the work done in blowing a soap bubble of radius 5 cm. S.T. of soap solution = 0.035 N/m.  (Ans: 2.2 × 10-3  J)
  5. Find the work done in blowing a soap bubble of radius 10 cm. S.T. of soap solution = 30 dyne/cm. (Ans: 7.54 × 10-3  J)
  6. Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of a soap solution is 30 dyne/cm. (Ans: 2.262 × 10-4  J)
  7. Two mercury drops one of radius 2 mm and the other of radius 1 mm coalesce to form a single drop. Find the change in the free surface energy. S.T. of mercury = 0.544 N/m. (Ans:  4.61 × 10-6  J)
  8. How much energy would be liberated if 103 water droplets each 10-8 m in diameter coalesce to produce a single drop? S.T. of water = 0.072 N/m.Assume drops are spherical. (Ans: 2.036  ×  10-14 J)|
  9. A drop of mercury of radius 0.1 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 540 dyne/cm.(Ans: 6.786 × 10-6 J)
  10. A mercury drop of radius 10-3 m breaks up into 125 small droplets. Calculate the change in energy assuming that the drops are spherical and S.T. of the mercury is 0.55 N/m. (Ans: 2.76 × 10-5 J)
  11. A mercury drop of radius 0.1 cm breaks up into 27 small droplets. Calculate the work done assuming that the drops are spherical and S.T. of the mercury is 540 dyne/cm. (Ans: 1.357 × 10-5 J)
  12. Eight droplets of water each of radius 0.2 mm coalesce together to form a single drop, Find the change in total surface energy, Given: surface tension of water = 0.072 N/m. (Ans:  1.448 × 10-7 J)
  13. A mercury drop of radius 0.5 cm falls from a height on a glass plate and breaks up into million droplets, all of the same sizes, Find the height from which the drop must have fallen. The density of mercury 13600 kg/m3. Surface tension of mercury = 0.465 N/m (Ans:  0.2072 m.)
  14. The total energy of the free surface of a liquid drop is 2π times the surface tension of the liquid. What is the diameter of the drop? Assume all terms in S.I. unit.(Ans:  1.414 m)
  15. n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4π R²T (n1/3   – 1)
  16. n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4πr² T. ( n   –  n2/3)

For Solutions to Problems Click Here