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)`.

Answer

8

Explanation

Here, `P=10^(5) N//m^(2), R =2.4xx10^(-4)m`,
`S=0.08 N//m`
Initial pressure inside the soap bubble in air cylinder, `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 after compression in the cylinder, then pressure inside the bubble is
`P_(2)=P' +(4S)/((R//2)) =(P'+(8S)/(R))`
Using Boyle's law, we have, `P_(1)V_(1) =P_(2)V_(2)`
or `(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) =8xx10^(5) +(24xx0.08)/(2.4xx10^(-4)) = 8.08xx10^(5)=nxx10^(5)`
`:. n=8.08 ~~8`