Two soap bubble of radii `r_(1)` and `r_(2)` combime to form a single bubble of radius r under isothermal conditions . If the external pressure is P, prove that surface tension of soap solution is given by `S=(P(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)^(2)+r_(2)^(2)-r^(2)))`.

Explanation

Excess of pressure in a soap bubble of radius `r_(1)` is
`P_(1)-P = (4S)/(r_1) or P_(1)=P+(4S)/(r_1)`
Similarly, pressure inside the bubble of radius r is
`P' =P+(4S)( r)`
Since the two bubbles combine under isothermal condition (i.e., temperature remains constant), so Boyle's law holds good during the combination of bubbles. so,
`P_(1)V_(1)+P_(2)V_(2)=P'V'`
or `(P+(4s)/(r_1)) 4/3 pi r_(1)^(3)+(P+(4S)/(r_2))4/3 pi r_(2)^(3)`
`=(P+(4S)/(r )) 4/3 pi r^(3)`
or `(P+(4s)/(r_1)) pi r_(1)^(3)+(P+(4S)/(r_2)) pi r_(2)^(3)`
`=(P+(4S)/(r )) pi r^(3)`
or ` 4S(r_(1)^(2)+r_(2)^(2)+r^(2))=P(r^(3)-r_(1)^(3)-r_(2)^(3))`
or `S=(P(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)^(2)+r_(2)^(2)-r^(2)))` Proved.