A cylinder with a movable piston contains air under a pressure `p_(1)` and a soap bubble of radius 'r' . The pressure `p_(2)` to which the air should be compressed by slowly pushing the poston into the cylinder for the soap bubble to reduce its size by hald will be (The surface tension is `sigma` and the temperature T is maintained constant).

Answer

`[p_(2)=8p_(1)+(24sigma)/(r)]`

Explanation

by conservation of role :
`P_(A) xx v = P'_(A) = (P_(A)) xx (BV)` (`:.` doubling the radius, volume gots `8` times)
`rArr P'_(A) = (P_(A))/(8)`
Now for radius `R`:
`P_(A) - P_(0) = (4T)/(R)`
& `(P_(A))/(8) - P_(0) = (4T)/(2R) - (sigma^(2))/(2epsilon_(0))` {`:.` pressure by charge density `sigma` is given by `(sigma^(2))/(2epsilon_(0))`}
Now `Eq^(n)[(1) xx (1)/(8)] - Eq^(n)[2]` we get `((7P_(0))/(8) = - (3T)/(2R)) + (sigma^(2))/(2epsilon_(0))`
`rArr sigma^(2) = ((7P_(0))/(8) + (3T)/(2R)2epsilon_(0) rArr sigma = sqrt((7P_(0))/(8) + (3T)/(2R)2epsilon_(0))`
`Q = 4 pi (2R)^(2) sqrt((7P_(0))/(8) + (3T)/(2R)2epsilon_(0)) = 8piR sqrt(4R^(2)((7P_(0))/(8) + (3T)/(2R))2epsilon_(0))`
`= 8piR sqrt(epsilon_(0).R(7P_(0)R+12T))`.
Hence proved.