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

Let `a` and `b` be the radii of the two soap bubbles and `p_(a), p_(b) ` the pressure inside then before they coalesce. Let `c` be the radius and `p_(c)` the pressure inside the combined bubbles. Then
`P_(a)=P+(4T)/a, P_(b)=P+(4T)/b` and `P_(c)=P+_(4T)/c`
If `V_(a),V_(b)` and vector are the volumes of two soap bubbles and that of combined soap bubbles, respectively, then by Boyle's law under isothermal conditions.
`P_(a)V_(a)+P_(B)V_(b)=P_(c)V_(c)`
`P+(4T/a)4/3pia^(3)+(P_(4T)/b)4/3pib^(3)=(P+(4T)/c)4/3pic^(3)`
`P(4/3pia^(2)+44/3pib^(2)-4/3pic^(2))`
`+(4T)/3(4pia^(2)+4piab^(2)+4pic^(2))=0`
`(4/3pia^(2)+4/3pib^(3)-4/3pic^(3))=V`
`=` change in volume and `(4pia^(2)+4pib^(2)-4pic^(2))=S`
`=` Change in surface area
or `PV+(4T)/3S=0`
or `3PV+4ST=0`