Excess Pressure on Curved Surface of a Liquid and inside Liquid Drop and Shape of Liquid Surface Meniscus

Excess Pressure on Curved Surface of a Liquid

We know that the similar molecules of a liquid attract each other by a force of coheasion and force of surface tension, tangentially to the liquid surface at rest.

If the liquid meniscus is horizontal there is no net force normal to the surface on the molecules lying on the upper surface as shown in the figure (i)

If the meniscus of a liquid is convex, then there exists a net downward force due to surface tension normal to the meniscus in figure (ii)

Hence it can be concluded that excess pressure is less on a convex meniscus.

Excess Pressure inside a Liquid Drop

As we know that, the force due to surface tension act normal to the surface in inward direction on the molecules of a liquid drop. For an equilibrium of the drop, these must exist excess pressure insie it.

Let R be the radius of a drop and dR be the increase in its radius by small amount due to excess pressure.

$$\text {The work done by excess pressure} = \text {force} \times \text {displacement} $$

$$= P \times A \times dR$$

$$= P \times 4\pi R^2 \times dR \dots (i)$$

Increase in the area of the drop

$$= 4\pi (R +dR)^2 - 4\pi R^2 $$

$$= 4\pi (R^2 + 2RdR + dR^2 - R^2)$$

$$= 8\pi R dR \dots {2}$$ dR² can be neglected being small.

$$ \text {Increase in surface energy of the drop} = \text {Increase in area} \times \text {Surface tension}$$

$$= 8 \pi RdR \times T \dots (iii)$$

Equation (i) and (iii) represents same quantity

$$P \times 4\pi R^2 \times dR = 8\pi R dR \times T$$

$$P = \frac {2T}{R} \dots (iv)$$

$$ P_{\text {in}} - P_{\text{out}} = \frac {2T}{R}$$

Excess Pressure of Air Bubble

Since the air bubble has only free surface, so the excess pressure inside is given by

$$P = \frac {2T}{R} \dots (iv)$$

$$ P_{\text {in}} - P_{\text{out}} = \frac {2T}{R}$$

Excess Pressure Inside Liquid Bubble or Soap

Since a liquid bubble has two free surfaces so thee excess pressure is given by

$$P =2 \times \frac {2T}{R} =\frac {4T}{R} $$

$$ P_{\text {in}} - P_{\text{out}} = \frac {2T}{R}$$

Shape of Liquid Meniscus

The shape of a liquid meniscus depends on the result of cohesive and adhesive forces that exist on the molecules of the liquid which are in contact with the solid surface there may be three different cases of the resultant force which are described below

Let us consider a molecule A which is in contact with the solid surface and is acted upon by two forces i.e. theadhesive force F_a at right angle to the solid surface and cohesive force F_c which act along AQ at 45^o to the solid surface. The angle between Fa and Fc is always 135^o but their magnitude are different for different solid-liquidpairs.

Case I

If the resultant F_a and F_c lie along the solid surface then from the solid surface then from figure i

$$\sin 45^o = \frac {SQ}{AQ}$$

$$SQ = AQ \sin 45^o$$

$$F_a = F_c \frac {1}{\sqrt 2}$$

$$\sqrt 2FA = F_c$$

$$F_c = \sqrt 2 F_a$$

Case II

If the resultant of F_a and F_c lie outside the liquid and is represented by AS'

here SQ & S'Q

$$\frac {F_c}{\sqrt 2}< F_a$$

$$F_c <\sqrt 2 F_a \dots (ii)$$

and the meniscus will be concave

Case III

If the resultant of F_a and F_c lie within the liquid and is represented by AS'

from fig 3

$$SQ> S'Q$$

$$\frac {F_c}{\sqrt 2}> F_a$$

$$F_c >\sqrt 2 F_a \dots (iii)$$

The meniscus will be convex.