There is a soap bubble of radius `2.4xx10^(-4)m` in air cylinder which is originally at a pressure of `10^(5)(N)/(m^(2))`. The air in the cylinder is now compressed isothermally untill the radius of the bubble is halved. (the surface tension of the soap film is `0.08Nm^(-1))`. The pressure of air in the cylinder is found to be `8.08xx10^(n)(N)/(m^(2))`. What is the value of n?

Answer

5

Explanation

`r_(i)=r=2.4xx10^(-4)m,P_(i)=10^(5)Nm^(-2)`
`r_(f)=(r)/(2),P_(i)=?,T=0.08Nm^(-1)`
Initial pressure inside the bubble, `P_(1)=P_(1)+(4T)/(r)`
When the air in the cylinder is compressed isothermally, final pressure inside the bubble
`P_(1)=P_(2)+(4T)/((r)/(2))`
Initial volume `V_(1)=(4)/(3)pir^(3),V_(2)=(V_(1))/(8)`
Since number of moles in bubble remains same,
`P_(1)V_(1)=P_(2)V_(2)`
`(P_(1)+(4T)/(r))r^(3)=(P_(2)+(8T)/(r))(r^(3))/(8)implies8(P_(1)+(4T)/(r))=P_(2)+(8T)/(r)`
Final pressure outside the bubble.
`P_(2)=8P_(1)+(32T)/(r)-(8T)/(r)=8P_(1)+(24T)/(r)`
`=8xx10^(5)+(24xx0.08)/(2.4xx10^(-4))=8.08xx10^(5)(N)/(m^(2))` (given), hence `n=5`.