Two identical spherical soap bubbles collapses. If `V` is consequent change in volume of the contained air, `S` is the chage in the total surface area and `T` is the surface tension of the soap solution. Then (if`p_(0)` is atmospheric pressure and assume temperature to remain same in all the bubbles).

Explanation

Let `r_(1)` and `r_(2)` be the radii of the two bubbles and r the radius of the single bubble. The temperature remains constant and there is no leakage of air. So the mass is constant. Hence, Boyle.s law may be applied.
`P_(1) V_(2) + P_(2) V_(2) = P_(0) V_(0)`
`P_(1)`, and `P_(2)` are the pressure inside the bubbles.
`P_(1) = P+( 4sigma)/( r_(1)) , P_(2) = P + ( 4 sigma)/( r_(2)) , P_(0) = P+ ( 4 sigma)/(r )`
`V_(1) = ( 4)/( 3) pi r_(1)^(3), V_(2) = ( 4)/( 3) pi r_(2) ^(3) , V_(0) = ( 4)/( 3) pi r^(3)`
Substituting in the above equation,
`( P + ( 4 sigma )/( r_(1)) ) ( 4)/(3) pi r_(1)^(3) + ( P + ( 4 sigma )/( r_(2))) (4)/(3) pi r_(2)^(3) = ( P + ( 4 sigma )/( r))( 4 )/( 3) pi r^(3)`
`P[ ( 4)/( 3) pi r_(1) ^(3)+( 4)/( 3) pi r_(2)^(3) - ( 4)/( 3) pi r^(3)] + ( 4 sigma )/( 3) [ 4 pi r_(1)^(2) + 4 pi r_(2)^(2) - 4 pi r^(2) ] =0`
` PV + ( 4)/( 3) sigma xx S = 0 ` or ` 3PV + 44 sigma S=0`
[ Note. Suppose S represents `8 pi r_(1)^(2) + 8 pi r_(2)^(2) - 8 pi r^(2)` then the result would have been `3PV + 2 sigma s = 0 `]