--%>

Liquid surfaces

The surface between a liquid and a vapour distinguishes these fluids.


The surface tension of liquids can be looked upon as that the property which draws a liquid together and forms a liquid vapour interface, therefore, distinguishing liquids from gases.

The molecular basis for this property is suggested, where the unbalanced attractions experienced by the surface molecules are shown to lead to the amount of free liquid will pull it together to form a less spherical drop. The surface layer can be expected to have properties that differ from those of the bulk of the liquid.

The surface tension of the liquid can be defined with reference to where it is most easily pictured is a wire frame, arranged as a piston, used to expand a soap film. The definition also applies to the mechanically more difficult systems where the film is replaced by a layer of liquid of appreciable thickness. The force required stretching the film or liquid vapour is proportional to the length l of the piston. Since there are two surfaces of the film, the total length of the film is 2l, and the proportionality equation:

ƒ = γ(2l) can be written.

The proportionality constant γ is known as the surface tension, and according to the above equation it can be looked upon as the force by a surface of unit length.
Of more general use is the relation between surface tension and surface energy. The mechanical energy required to expand the surfaces by moving the piston a distance dx is f dx, or 2l dx. Since the area of new surface is 2l dx, the result:

Mechanical energy/change of surface area = 2lγ dx/ 2l dx = γ, can be obtained. This expression shows that the surface tension can be interpreted as the energy per unit surface area and that it is a mechanical rather than thermal energy. In these terms, the tendency of a surface to reduce its area is just another example of a system tending toward an arrangement of low free energy.

Surface tension of some liquids, N m-1:

Liquid 20°C 60°C 100°C Liquid t, °C Surface tension
H2O 0.07275 0.06618 0.05885 Hg 0 0.480
C2H5OH 0.0223 0.0223 0.0190 Ag 970 0.800
C6H6 0.0289 0.0237   NaCl 1080 0.094
(C2H5)2O 0.0170   0.0080 AgCl 452 0.125


Example: compare the heights to which water and carbon tetrachloride will rise as a result of capillary action in a tube with an internal diameter of 0.1 mm. at 20°C the surface tensions of water and carbon tetrachloride, respectively, are 0.0727 and 0.0268 N m-1, and their densities are 0.998 and 1.595 g mL-1.

Solution: 
we use to obtain:

L = 2 γ/rpg


The radius of the cube is 0.5 mm = 0.5 × 10-4 m, and the densities are 9.98 × 103 and 1.598 × 103 kg m-3.

For water: l = 2 (0.0727 N m-1)/(0.5 × 10-4 m) (9.98 × 103 kg m-3) (9.81 m s-1)

= 0.0297 m = 29.7 mm

For CCl4: l = 2 (0.0268 N m-1)/(0.5 × 10-4 m) (1.595 × 103 kg m-3) (9.81 m s-2)

= 0.00685 m = 6.85 mm.

   Related Questions in Chemistry

  • Q : Relative lowering of the vapour pressure

    Choose the right answer from following.The relative lowering of the vapour pressure is equal to the ratio between the number of: (a) Solute moleules and solvent molecules (b) Solute molecules and the total molecules in the solution (c) Solvent molecules and the tota

  • Q : Problem on vapour pressure Choose the

    Choose the right answer from following. If P and P are the vapour pressure of a solvent and its solution respectively N1 and N2 and are the mole fractions of the solvent and solute respectively, then correct relation is: (a) P= PoN1 (b) P= Po N2 (c)P0= N2 (d)

  • Q : Question relatede to calculate molarity

    Select the right answer of the question. What is molarity of a solution of HCl that contains 49% by weight of solute and whose specific gravity is 1.41 : (a) 15.25 (b) 16.75 (c) 18.92 (d) 20.08

  • Q : What are condensation polymers? Give

    These types of polymers are formed as a result of condensation reaction between monomer units. Some common examples are being discussed here: 1. Polyesters 2047_condensat</span></p>
                                        </div>
                                        <!-- /comment-box -->
                                    </li>
   
   </td>
	</tr><tr>
		<td>
       
      <li>
                                        <div class=

    Q : Molar mass of solute The boiling point

    The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is: (a) 5.8g mol-1  (b)

  • Q : Problem based on molality of glucose

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Problem on vapor-liquid equilibrium Two

    Two tanks which contain water are connected to each other through a valve. The initial conditions are as shown (at equilibrium): 683_tank question.jpg

  • Q : Explain Ionic Bond with examples. The

    The bonding in ionic molecules can be described with a coulombic attractive term.For some diatomic molecules we take quite a different approach from that used in preceding sections to describe the bonding. Ionic bonds are interpreted in terms of the coulom

  • Q : Analytical chemistry 37% weight of HCl

    37% weight of HCl and density is 1.1g/ml. find molarity of HCl

  • Q : Group Cations Explain how dissolving

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately