Two spherical soap bubble coalesce. If `V` is the consequent change in volume of the contained air and `S` the change in total surface area, show that
`3PV+4ST=0`
where `T` is the surface tension of soap bubble and `P` is
Atmospheric pressure

Explanation

`PV=` constant after combining the two bubbles
`(P_(0)+(4T)/(R))V=` constant , `P_(0)V+(4T)/(R).(4)/(3)piR^(3)=C`
`P_(0)V+(4T)/(3)S=C` ..(1)
Before combining the two bubbles
`P_(0)(C_(1)+V_(2))+(4T)/(3)(S_(1)+S_(2))=C` ..(2)
According to Boyle's law `P_(1)V_(1)+P_(2)V_(2)=PV`
From equation (1) and (2)
`P_(0)(V_(1)+V_(2))+(4T)/(3)(S_(1)+S_(2))=P_(0)V+(4T)/(3)S`
`P_(0)DeltaV+(4T)/(3)DeltaS=0,3P_(0)(DeltaV)+4T(DeltaS)=0`