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 untill the radius of the bubble is halved. Calculate teh new pressure of air in the cylinder. Surface tension of soap solution is `0.08 Nm^(-1)`.

Explanation

Here, `r = 2.4 xx 10^(-4) m , P_(0)=10^(5)N//m^(2), P= ?, sigma = 0.08 Nm^(-1)`
Initial pressure inside the air bubble `P_(1)= P_(0)+(4 sigma)/(r)`
On compression radius of bubble becomes, `r//2`
Let the pressure in the cylinder becomes P.
`:.` Final pressure inside the air bubble = `P_(2) = P+(4 sigma)/(r//2) = P+(8 sigma)/(r)`
Initial vloume of bubble , `V_(1) = 4/3 pi r^(3)`, Final volume of bubble , `V_(2) = 4/3 pi (r//2)^(3) = 1/8 V_(1)`
As no change in temperature is involved in isothermal compression, therefore, applying Boyle's law,
`P_(2)V_(2)=P_(1)V_(1)`
`(P+(8 sigma)/(r)) 1/8 V_(1) = (P_(0)+(4 sigma)/(r))V_(1) or P+(8 sigma)/(r) = 8P_(0)+(32 sigma)/(r)`
`P=8P_(0)+(24 sigma)/(r) = 8xx10^(5) + (24 xx 0.08)/(2.4 xx 10^(-4)) = 8xx 10^(5)+0.08 xx 10^(5) = 8.08 xx 10^(5) Nm^(-2)`.