# Surface Tension

Surface Tension

Any molecule such as B, in the following figure, in the body of the liquid experiences equal attraction from all sides.

surface tension

Such that the resultant force on it is zero. But molecule such as A, on the surface layer experiences an unbalanced force, because it is attracted by molecules in the bulk of the solution. Due to this downward pull the molecule at the surface tends to go down if the surface is a stretched membrane. This force on the surface of the liquid is known as Surface Tension and may be defined as:

“The force per unit length of a line drawn on the surface and acting at right angle to the line, tending to pull the surface apart along the line”.

The dimension of surface tension is $MT^{-2}$ . It is expressed in dynes per centimeter. Generally stalagmometer is used for the determination of surface tension of the liquid.

Stalagmometer consists of a dropping tube with a capillary, the end of which is flattened. This flattened end helps to give a large dropping surface. This surface is already ground flat and polished. The other end of the capillary is sealed on a tube of wider bore containing a bulb. There are two marks etched on this stem, one above and other below the bulb.

Pyknometer is an apparatus for measuring the density or specific gravity of a liquid. A We know that a drop just held at a surface or just dropped from the capillary balances two forces:

(i) The gravity force exerted on the drop given by V.d.g where V is the volume of the drop, d its density and g the gravity, and

(ii)  The force tending to uphold the drop, given by 21cry, where 21:r is the circumference of the circular surface of radius r and 7 is surface tension.

When these two forces are balanced, then

$2 \pi r \gamma= V. d. g.$

But, if there are n drops in a volume, V of the liquid of density d, then the weight of each drop is Vd /n. Then the first equation may be arranged as

$2 \pi r \gamma = \dfrac{Vd}{n}.g$

If there are two liquids of densities $d_1$ and $d_2$ having the volume V and with surface tension $\gamma_1$and $\gamma_2$and if the number of drops counted is $n_1$and $n_2$ respectively in the same volume, then

$2 \pi r \gamma = \dfrac{Vd_1}{n_2}.g$ (for one liquid)

And $2 \pi r \gamma_2 = \dfrac{Vd_2}{n_2}.g$ (for another liquid)

Comparing these two, $\dfrac{\gamma_1}{\gamma_2} = \dfrac{n_2}{n_1}.\dfrac{d_1}{d_2}$

Thus, if densities of both the liquids and the number of drops in axe of each liquid are known and the surface Baku of any one liquid is also known then surface tension of its other may be calculated.

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