Question

Give the expression for the excess pressure in a soap bubble.

Answer 1

Hint: Here we will use the concept of the surface tension. It is the property of the surface of the liquid which allows resisting an external force due to the cohesive nature of the water molecules and the molecules on the surface of the water experiences an inward pull. Surface Tension is denoted by “T” and is measured in Newton per metre.

Complete step by step answer:

Let us consider a soap bubble with radius of curvature R and Surface Tension T. A soap bubble has two free surfaces. As a result, due to surface tension the molecule on the surface experiences the net force in the inward direction normal to the surface.
Let the pressure inside the liquid drop be $ = {p_i}$
And the pressure outside the liquid drop be $ = {p_o}$
There net pressure is $p = {p_i} - {p_o}$
As a result, the drop will expand due to excess pressure inside the liquid drop and the free surface of the drop will experience the net force in the outward direction.
Under an isothermal conditions, let the free surfaces be displaced be $ = dR$
Hence, the excess pressure of the soap bubble does the work in the form of displacement of the surface and that the work will be stored in the form of the potential energy.
Now, work done by the excess pressure is
$dW = F.d$
Where F is the excess pressure into the surface area of the spherical bubble
D is the displacement of the surface
$ \Rightarrow dW = p \times 4\pi {R^2} \times dR\;{\text{ }}.......{\text{ (a)}}$
Increase in the potential energy is –
$dU = T \times \Delta A$
$\Delta A = $ Increase in area of the free surface
$dU = T[2\{ 4\pi {(R + dR)^2} - 4\pi {R^2}\} ]$
Simplify the above equation –
$dU = T[2\{ 4\pi (2RdR)\}] \\$
$\implies dU = T(8\pi RdR){\text{ }}.......{\text{ (b)}} \\ $
Work done is equal to the increase in the potential energy.
By using equations (a) and (b)
$p \times 4\pi {R^2} \times dR = T(8\pi RdR) $
Take common factors from both the sides of the equation and remove them.
$\Rightarrow p \times R = 4T \\$
$\Rightarrow p = \dfrac{{4T}}{R} \\$

Hence, the required answer is - the excess pressure in a soap bubble is $\dfrac{{4T}}{R}$

Note:
The above problem can be solved by using other method as below-
As the excessive pressure inside the layer is \[P = \dfrac{{2T}}{R}\] where R is the radius of curvature and T is the surface tension of the liquid.
Hence, a soap bubble having two spherical surfaces,
The excess pressure inside the bubble is $P = 2 \times \dfrac{{2T}}{R} = \dfrac{{4T}}{R}$