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

Options

A. `3P_(0)V + 4ST= 0`
B. `4P_(0)V + 3ST= 0`
C. `P_(0) V + 4TS= 0`
D. `4P_(0) V + ST= 0`

Answer

A

Explanation

Let a and b be the radii of two soap bubbles befores coalescing and c be the radius of bigger bubble.
`P_(a) = P_(0) + (4T)/(a), V_(a) = (4)/(3) pi a^(3)`
`P_(b) = P_(0) + (4T)/(b), V_(b) = (4)/(3) pi b^(3)`
`P_(c) = P_(0) + (4T)/(c ), V_(c ) = (4)/(3) pi c^(3)`
where `P_(0)` is the atmospheric pressure. Now, as mass is conserved,
`:. n_(a) + n_(b) = n_(c ) or (P_(a) V_(a))/(RT_(a)) + (P_(b) V_(b))/(RT_(b)) = (P_(c) V_(c))/(RT_(c))`
At constant temperature,
`P_(a) V_(a) + P_(b) V_(b) = P_(c) V_(c)`
`:. (P_(0) + (4T)/(a)) ((4)/(3) pi a^(3)) + (P_(0) + (4T)/(b)) ((4)/(3) pi b^(3)) = (P_(0) + (4T)/(c)) ((4)/(3) pi c^(3))`
or `4T (a^(2) + b^(2) -c^(2)) = P_(0) (c^(3) -a^(3) -b^(3))`
or `(4)/(3) T 4pi (a^(2) + b^(2) -c^(2)) = P_(0) (4)/(3) pi (c^(3) -a^(3) -b^(3))`
or `(4T)/(3) S = -P_(0)V or 3P_(0) V + 4ST = 0`