Two soap bubbles `A` and `B` are kept in a closed chamber where the air is maintained at pressure `8 N//m^(2)`. The radii of bubbles `A` and `B` are `2 cm` and `4 cm`, respectively. Surface tension of the soap. Water used to make bubbles is `0.04 N//m`. Find the ratio `n_(B)//n_(A)`, where `n_(A)` and `n_(B)` are the number of moles of air in bubbles `A` and `B` respectively. [Neglect the effect of gravity.]

Options

A. 2
B. 9
C. 8
D. 6

Answer

D

Explanation

Excess pressure inside the soap bubble `= (4S)/(r )`
So the pressure inside the soap bubble `= P _(atm) + (4S)/(r )`
From ideal gas equation `PV = nRT`
`(P _(A) V _(A))/( P _(B) V _(B))= (n _(A))/( n _(B)) implies ((8 + (4S)/( r _(A)))4/3 pi (r _(A)) ^(3))/((8 + (4S)/( r _(B)) ) 4/3 pi (r _(B)) ^(3)) = (n _(A))/( n _(B))`
Substituting `S = 0.04 N//m,r _(A) = 2cm, r _(B) =4cm.`
`(n _(A))/( n _(B)) = 1/6 therefore (n _(B))/( n _(A))=6.`