Under isothermal condition two soap bubbles of radii `r_(1)` and `r_(2)` coalesce to form a single bubble of radius r. The external pressure is `p_(0)`. Find the surface tension of the soap in terms of the given parameters.

Options

A. `(2p_(0)(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)^(2)+r_(2)^(2)-r^(2)))`
B. `(p_(0)(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)^(2)+r_(2)^(2)-r^(2)))`
C. `(p_(0)(r^(3)+r_(1)^(2)+r_(2)^(3)))/(4(r_(1)^(2)+r_(2)^(2)+r^(2)))`
D. None of these

Answer

B

Explanation

As mass of the air is conserved,

`therefore " " n_(1)+n_(2)=n " " ("as" pV=nRT)`
`therefore (p_(1)V_(1))/(RT_(1))+(p_(2)V_(2))/(RT_(2))=(pV)/(RT)`
As temperature is constant,
`T_(1)=T_(2)=T`
`therefore p_(1)V_(1)+p_(2)V_(2)=pV`
`therefore (p_(0)+(4S)/(r_(1)))((4)/(3)pir_(1)^(3))+(p_(0)+(4S)/(r_(2)))((4)/(3)pir_(2)^(3))=(p_(0)+(4S)/(r ))((4)/(3)pir^(3))`
Solving, this we get
`S=(p_(0)(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)r^(2)+r_(2)^(2)-r^(2)))`