Question

What is excess pressure inside a bubble of soap solution of radius 5 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.5 × $10^{-2}\ Nm^{-1}$? If an air bubble of the same dimension were formed at a depth of 40 cm inside a container containing the soap solution (of relative density 1.2), what would be the pressure inside the bubble? (1 atm = 1.01 × $10^5$ Pa)

Answer 1

Answer:

1.05714 × 10^5 Pa

Explanation:

The excess pressure inside the soap bubble is given as - ∆P = 4S/R

where, S is the surface tension of the soap bubble . and R is the radius of soap bubble .

Surface tension of the soap bubble (S) = 2.5 × 10^-2 N/m (Given)

Radius of soap bubble (R) = 5mm = 5× 10^-3 m (Given)

Excess pressure inside the soap bubble is = 4S/R

= 4 × 2.5 × 10^-2 /5 × 10^-3

= 20 Pa

Excess pressure inside the air bubble is = 2S/R

= 2 × 2.5 × 10^-2 /5 × 10^-3

= 10 Pa

Therefore, pressure inside the bubble = atmospheric pressure + pressure due to soap solution + excess pressure in the bubble

= 1.01 × 10^5 Pa + dgh + 10 Pa

= 1.013 × 10^5 + (1.2 × 10³×9.8 × 0.4) + 10

= 1.05714 × 10^5 Pa

Thus, the pressure inside the bubble is 1.05714 × 10^5 Pa