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.]

Answer

6

Explanation

`P_(A)=P+(4T)/(r_(A))" "implies" "P_(A)=16N//m^(2)`
`P_(B)=P+(4T)/(r_(B))" "implies" "P_(B)=12N//m^(2)`
Applying gas equation
`P_(A)B_(A)=n_(A)RT`
`P_(B)V_(B)=n_(B)RT`
`(P_(B))/(P_(A))xx((V_(B))/(V_(A)))=(n_(B))/(n_(A))=(12)/(16)xx((r_(B))/(r_(A)))^(3)=(n_(B))/(n_(A))implies(12)/(16)xx(2)^(3)=(n_(B))/(n_(A))implies(n_(B))/(n_(A))=6`