Derive an expression for excess pressure inside a liquid drop and bubble.

Excess PressureDue to the property of the surface tension, a liquid surface always tends to have minimum surface area. The small drops and bubbles are found to possess spherical shapes. However, a big drop of liquid is not spherical in shape. The reason is that in case of a small drop of liquid, the effect of gravity is negligible. As the small drops and bubbles have spherical shapes, it implies that their spherical surfaces must be possessing minimum surface area. Further, as a drop or a bubble does not collapse under the effect of the force of surface tension (which tends to minimize its surface area), it indicates that the pressure inside the drop or the bubble must be greater than that outside it. a) Excess pressure in a liquid drop.Let us consider a drop of liquid of surface tension T and radius R. Let P_iand P₀be the values of pressure inside and outside the drop of the liquid (Fig.) Then,Excess pressureinside the liquid drop = p_i p_oSuppose that the radius of the drop is increased from R to R +R under the pressure difference (p_i p₀).The outward force acting on the drop = excess of pressure×surface area=(p_i p₀)×4R²The small amount of work done to increase its radius by (R),W = (p_i p₀)×4R²×(R) orW = 4R²(p_i p₀) (R) (i)This work (W) is done by the excess of pressure against the force of surface tension and is stored inside the liquid drop in the form of its potential energy (U).Also, increase in the potential energy of the liquid drop,U = surface tension×increase in surface area=T × (4(R + (R))² 4R²)= T×(4R²+ 8R (R) + 4(R)² 4R²)AsR is small, the term containingR²can be neglected.Therefore, increase in the potential energy of drop,U = 8T R (R) (ii)From the equations (i) and (ii), we have 4R²(p_i p₀) (R) = 8T R (R) or p_i p_o= 2T/R (iii)b) Excess pressure in a liquid bubble.A liquid bubble has air both inside and outside it and therefore it has two free surfaces. If a liquid bubble increases in size from radius R to R +dR, then area of its inner surface as well as that of outer surface will increase.Therefore, increase in potential energy of the bubble,U = 2 × T [4(R + (R))2 4R²] orU =16TR (R) .... (iv)As obtained above, the small work done in increasing the radius of the liquid bubble from R to R +R,W = 4R²(pi po)(R) ..... (v)From the equations (iv) and (v), we have 4R²(p_i p₀) (R) = 16T R (R) or p_i p_o= 4T/R .... (vi)

c) Excess pressure in an air bubble.

In a drop of liquid, air is outside the drop and the liquid is inside it. In an air bubble, liquid is outside the spherical core of air. This, as in the case of a drop, an air bubble also has one free surface. As before, it can be easily proved that excess of pressure inside an air bubble,p_i p_o= 2T/R

Solved example 1:

What is theexcess pressureinside a bubble of soap solution of radius 5.0 mm given that the surface tension of soap solution is 2.5×10²N m¹? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.2), what would be the pressure inside the bubble? Given that 1 atm = 1.01×10⁵Pa.

Solution:

Here, surface tension of the soap solution, T = 2.5×10²N m¹;

Density of the soap solution,= 1.2×10³kg m³; radius of the soap bubble, R = 5.0 mm = 5.0×10³m

Now, theexcess pressureinside the soap bubble,

p_i p_o= 4T/R = 4 × 2.5 × 10²/5.0 × 10³= 20 Pa

Also, theexcess pressureinside the air bubble,

Now, the pressure outside the air bubble at a depth of 40 cm i.e. 0.4 m,

p_i p_o= 2T/R = 2 × 2.5 × 10²/5.0 × 10³= 10 Pa

P₀= atmospherically pressure + pressure due to 0.4 m column of soap solution

= 1.01 ×10⁺⁵+ 0.4×1.2 × 10³×9.8 = 1.05704×10⁵Pa

Therefore, pressure inside the air bubble,

p_i p_o= 2T/R = 1.05704 × 10⁵+ 10. = 1.05714 × 10⁵Pa