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. `(P_(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)^(3)-r_(2)^(3)))/(4(r_(1)^(2)+r_(1)^(2)-r^(2)))`
D. `(P_(0)(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)^(2)-r_(2)^(2)-r^(2)))`

Answer

C

Explanation

the mass of the air is conserved.

`:. N_(1)+n_(2)=n` (as PV = nRT)
`:. (P_(1)V_(1))/(RT_(1))+(P_(2)V_(2))/(RT_(2)) =(PV)/(RT)`
As temperature is constant when two bubbles coalese, So, `T_(1)=T_(2)=T`
`:. P_(1)V_(1)+P_(2)V_(2)=PV`
or `(P_(0)+(4S)/(r_(1)))((4)/(3)pi r_(1)^(3))+(P_(0)+(4S)/(r_(2)))((4)/(3)pi r_(2)^(3)) = (P_(0)+(4S)/(r))((4)/(3)pi r^(3))`
On solving, we get
`S=(P_(0)(r^(3)-r_(1)^(3)-r_(2)^(3)))/(4(r_(1)^(2)+r_(2)^(2)-r^(2)))`