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

The excess of pressure above atmospheric pressure, due to surface tension in a bubble ` = ( 4 T ) /( r ) `
The surrounding pressure ` P_ 0 = 8 N//m ^ 2 . " " therefore P_A ` for 1st bubble ` = P_0 + ( 4 T ) /(r _A ) = 8 + (4 xx 0.04 ) /( 0.02 ) `
` P_A = 16 N//m ^ 2 , P_B = P_0 + (4T ) /( r _B ) = 8 + (4 xx 0.04 ) /(0.04 ) = 12N//m ^ 2 `
` PV = nR T `
For ` A : 16 (4)/(3) pi ( 0.02) ^3 = n _ A RT, ` for ` B : (12)((4)/(3) pi (0.04) ^3) = n_B RT rArr (n_B)/(n_A) = 6 `