Question

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure `8N//m^2`.The radii of bubbles A and B are 2cm and 4cm, respectively. Surface tension of the soap-water used to make bubbles is `0.04N//m`. Find tha 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 1

Correct Answer - 6

Answer 2

For A, 

P_A- 8 = (4*0.04)/2*10²

P_A- 8 = 8

P_A = 16 N/m² 

P_AV_A = n_ART_A 

P_A * (4/3)π r₁² = n_ART_A ------(i)

For B,

P_B- 8 = (4*0.04)/4*10²

P_B- 8 = 4

P_B = 12 N/m² 

P_BV_B = n_BRT_B 

P_B* (4/3)π r₂² = n_BRT_B ------(ii) 

(ii)/(i)

T_A=T_B (given)

(P_B* (4/3)π r₂²)/(P_A * (4/3)π r₁²) = (n_BRT)/( n_ART)

n_B/n_A = [12*(4*10^-2)³]/[16*(2*10^-2)³] 

n_B/n_A = (3/4)*8

n_B/n_A = 6