Two spherical soap bubble coalese. If V be the change in volume of the contained air, A is the change in total surface area then show that `3PV + 4AT = 0` Where T is the surface tension and P is atmospheric pressure.

Explanation

Let `P_(1) - P = (4T)/(r_(1)) rArr P_(1) = P + (4T)/(r_(1))`
Similarly,
`P_(2) - P = (4T)/(r_(2)) rArr P_(2) = P + (4T)/(r_(2))`
When bubbles coalesce
`P_(1)V_(1) + P_(2)V_(2) = PV` …(1)
`:.` The pressure inside the new bubble,
Substituting for `P, P_(1)` and `P_(2)` in equation (1),
`P_(1)V_(1) + P_(2)V_(2) = PV`
`(P + (4T)/(r_(1)))V_(1) + (P + (4T)/(r_(2))) V_(2) = (P + (4T)/(r ))V`
`(P + (4T)/(r_(1)))(4)/(3)pi r_(1)^(3) + (P + (4T)/(r_(2)))(4)/(3)pi r_(2)^(3) = (P + (4T)/(r ))(4)/(3)pi r^(3)`
[`because` Vlume of the sphere `V = (4)/(3)pi r^(3)`]
or `(4)/(3) pi P(r_(1)^(3) + r_(2)^(3) - r^(3)) + (16pi T)/(3)((r_(1)^(3))/(r_(1)) + (r_(2)^(3))/(r_(2)) - (r^(3))/(r )) = 0`
`(4)/(3)pi P(r_(1)^(3) + r_(2)^(3) - r^(3)) + (16piT)/(3)(r_(1)^(2) + r_(2)^(2) - r^(2)) = 0` ...(2)
Given change in Volume,
`V = (4)/(3)pi r_(1)^(3) + (4)/(3)pi r_(2)^(3) - (4)/(3)pi r^(3)` ...(3)
Change in Area, `A = 4pir_(1)^(2) + 4pi r_(2)^(2) - 4pi r^(2)` ...(4)
Using equations (3) and (4) in equation (2) we get,
`P((4)/(3)pi r_(1)^(3) + (4)/(3) pi r_(2)^(3) - (4)/(3)pi r^(3)) + (4)/(3)(4pi r_(1)^(2) + 4pi r_(2)^(2) - 4pi r^(2)) = 0`
`P(V) + (4T)/(3)(A) = 0`
`PV + (4T)/(3)A = 0 rArr 3PV + 4TA = 0`
`:. 3PV + 4TA = 0`.