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.]

Explanation

Data : ` R_(A) = 0.02 n, R_(B) = 0.04 m , p_(0) = 8 "N/m"^(2) ` , T = 0.4 N/m
` p - p_(0) = (4T)/R = pV = nRT`
` P_(A) -p_(0) = (4T)/R_(A) = (4(0.04))/0.02 = 8 " N/m"^(2)`
and ` p_(B) = p_(0) = (4T)/(R_(A) = (4(0.04))/(0.04) = 4 "N/m"^(2)`
` P_(A) = 8 + 8 = 16 "N/m"^(2)`
` and p_(B) = 8 + 4 = 12 "N/m"^(2)`
`n_(B)/n_(A) = (p_(B)V_(B))/(p_(A)V_(A)) = P_(B)/P_(A) ((R_(B))/R_(A))^(3)`
` 12/16 (0.04/0.02)^(3) = 3/4 xx 8 =6`
This is the required ratio .