Question
Question: The excess pressure across a soap bubble of radius r is \({\text{p = }}\dfrac{{4\sigma }}{{\text{r}}...
The excess pressure across a soap bubble of radius r is p = r4σ
where σ is the surface tension of soap solution. What is the excess pressure across an air bubble of the same radius r formed inside a container of soap solution?
A) rσB) r2σC) r4σD) none of these
Solution
Hint : In this question, we can use the relation of pressures and surface tensions of two surfaces of a soap bubble. We can modify this relation as in the air, the soap bubble has two surfaces while in soap solution the air bubble has only one surface.
Complete step by step solution:
We know that the soap bubble in air has two surfaces. While the air bubble in soap solution has only one surface.
So the air bubble in the soap solution has half the excess pressure of the soap bubble in the air, that is to say
⇒P=21 × r4σ⇒P=r2σ∴ excess pressure across an air bubble isP=r2σ
Where P is the excess pressure across the air bubble and σ is the surface tension of the soap solution.
Hence option B is correct
Additional information: The surface tension of water provides the necessary wall tension for the formation of bubbles with water. The tendency to minimize wall stress causes the bubbles to assume spherical shapes.
The pressure difference between the inside and outside of a bubble depends on the surface tension and the radius of the bubble. The relationship can be obtained by visualizing the bubble as two hemispheres and observing that the internal pressure that tends to separate the hemispheres is counteracted by the surface tension acting around the circumference of the circle.
Note: Pascal's principle requires that the pressure be the same everywhere within the equilibrium balloon. But examination immediately reveals that there are large differences in wall tension in different parts of the globe. The variation is described by Laplace's Law.