Question

What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10¯² N m¯¹ ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 105 Pa).

Answer 1

Given, the surface tension of the soap solution at temperature at 20° C is 2.5× 10 −2   Nm -1 , the radius of the soap bubble is 5.00 mm=5× 10 −3  m , an air bubble is formed inside the container at depth of 40 cm=0.4 m and the relative density of the soap is 1.2 .

Let P be the excess pressure inside the soap bubble, then

P= P i − P 0 = 4T r

Here, p i is the pressure inside the soap bubble, P o is the pressure outside the soap bubble, T is the surface tension and r is the radius of the bubble.

Substitute the values in the above expression.

P= 4×2.5× 10 −2 5× 10 −3 =20 Pa

Hence, the excess pressure of the soap bubble is 20 Pa.

When an air bubble is formed inside the container at depth of 40 cm=0.4 m , which is filled with the soap solution of relative density 1.2 or density ρ=1.2× 10 3  kg  m -3 , then excess pressure of the air bubble is given as,

P air bubble = 2T r P ai − P ao = 2T r …… (1)

Here P ai is the pressure inside the bubble and P ao is pressure outside of the bubble.

The pressure outside the air bubble is,

P ao = P a +ρhg

Here, P a is the atmospheric pressure, h is the depth and g is the acceleration due to gravity.

Substitute the values in the equation (1).

P ai −( P a +ρhg )= 2T r P ai =( P a +ρhg )+ 2T r

Substitute the values,

P ai =( 1.01× 10 5 +0.4×1.2× 10 3 ×9.8 )+ 2×2.5× 10 −2 5× 10 −3 =1.01× 10 5 +4.704× 10 3 +10 ≈1.06× 10 5  Pa

Hence, the pressure inside the bubble is 1 .06×10 5  Pa .