There is a soap bubble of radius `2.4 xx 10^(-4)m` in air cylinder at a pressure of `10^(5) N//m^(2)`. The air in the cylinder is compressed isothermal until the radius of the bubble is halved. Calculate the new pressure of air in the cylinder. Surface tension of soap solution is `0.08 Nm^(-1)`.

Explanation

Here,` P = 10^(5) Nm^(-2), R = 2.4 xx 10^(4)m`,
`S = 0.08 Nm^(-1)`
Initial pressure inside the soap bubble in air cylinder is,
`P_(1)=P+(4S)/(R )`
Let `V_(1)` be the initial volume of the soap bubble
`V_(1) = 4/3 pi R^(3)`
After compression, volume of the soap bubble
`V_(2) = 4/3 pi (R/2)^(3) = (V_1)/(8)`
If P' is the air pressure in teh cylinder after compression, then pressure, indside the bubble is
`P_(2) = P' + (4S)/((R//2)) = P' + (8S)/(R)`
Using Boyal's law, we have
`(P+(4S)/(R ))V_(1) = (P' + (8S)/(R))(V_1)/(8)`
or `P +(4s)/(R ) = 1/8 (P' + (8S)/(R))`
or `P' = 8P +(24S)/(R )= 8 xx 10^(5) + (24 xx 0.08)/(2.4 xx 10^(-4))`
`=8.08 xx 10^(5)Nm^(-2)`.