Two spherical soap bubble collapses. If `V` is the consequent change in volume of the contained air and `S` in the change in the total surface area and `T` is the surface tension of the soap solution, then it relation between `P_(0), V, S` and `T` are `lambdaP_(0)V + 4ST = 0`, then find `lambda` ? (if `P_(0)` is atmospheric pressure) : Assume temperature of the air remain same in all the bubbles

Answer

`3`

Explanation

Let `(a)` and `(b)` coalesce to form `(c)`.
By mole conservation :
`P_(a) . a^(3) + P_(b) . b^(3) = P_(c) . c^(3) …..(i)`
Also `P_(a) = P_(0) + (4gamma)/(a) …..(ii)`
`P_(b) = P_(0) + (4gamma)/(b) ….(iii)`
`P_(c) = P_(0) + (4gamma)/(c)...... (iv)`

Putting there values :
`(P_(0) + (4gamma)/(a))a^(3) + (P_(0) + (4gamma)/(b))b^(3) = (P_(0) + (4gamma)/(c))c^(3)`
`rArr P_(0)|a^(3) + b^(3) - c^(3)|+ 4gamma|a^(2) + b^(2) - c^(2)| = 0`
also `c^(3) - (b^(3) - a^(3)) = (3V)/(4pi)` and `c^(2) - (a^(2) + b^(2)) = (s)/(4 pi)`.
Putting there values
`P_(0)((-3v)/(4pi)) + 4T((-S)/(4pi)) = 0 rArr 3P_(0)V + 4ST = 0`