Question

The pressure inside a bubble of radius R and surface tension S is ______________.

Answer 1

Pressure and surface tension

1. Pressure- The force applied perpendicular to an object's surface per unit area across which that force is spread is known as pressure.
2. Surface Tension- Surface tension is the tendency of liquid surfaces to shrink to the smallest possible surface area.

Step 1: Given data

The radius of the bubble = $R$

Surface Tension = $S$

Step 2: Concept used

$W=S\times ∆A$

Where,

$W$ is work done.

$S$ is surface tension.

$∆A$ is the change in the area.

Step 3: Deriving the pressure

Pressure outside the bubble = $P_o$

Due to surface tension, the concave surface of a liquid will always have a higher pressure than the convex surface. Since the soap bubble's outer surface is convex and its interior surface is concave, the pressure on the interior surface is greater than that on the exterior surface.

Let the extra pressure inside the bubble be $P$

Total Pressure inside the bubble = $P_o+P$

Since the pressure inside is greater than the pressure inside so it will apply force toward the outside.

Let the change in radius is $dR$

Then work done can be taken as,

$W=S\times ∆A$

As there are two surfaces of the bubble,

$W=2\times S\times ∆A$

$Force\times InnerSurfacearea\times changeinradius=2\times S\times ∆A$

$P\times 4\pi R^2\times dR=2\times S[4\pi {(R+dR)}^2-4\pi R^2]$

$P·4\pi R^2·dR=2S[4\pi (R^2+2RdR)-4\pi R^2]$

as $d{R}^2$is very small so it will be neglected,

$P·4\pi R^{2}dR=2S·4\pi ·2RdR$

$$\begin{gathered} P=\frac{2S·4\pi ·2RdR}{4\pi R^{2}dR} \\ P=\frac{4S}{R} \end{gathered}$$

Therefore, the pressure inside a bubble of radius R and surface tension S is $\frac{4S}{R}$.