Q. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 ${Nm}^{- 2}$. The radii of bubbles is $0.04{Nm}^{- 1}$. Find the ratio $\frac{n_{B}}{n_{A}}$, where $n_{A}$ and$n_{B}$ are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity]
A) 4
B) 6
C) 7
D) 8
Explanation
(b) Although not given in the question, but we will have to assume that temperatures of A and B are same.
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$\frac{n_{B}}{n_{A}} = \frac{p_{B}V_{B}/RT}{p_{A}V_{A}/RT} = \frac{p_{B}V_{B}}{p_{A}V_{A}}$
$= \frac{p + 4s/r_{A} \times 4/3\pi\left( r_{A} \right)^{3}}{p + 4s/r_{B} \times 4/3\pi\left( r_{B} \right)^{3}}$
$(s = surface\ tension)$
$Substituting\ the\ values,\ we\ get\ \frac{n_{B}}{n_{A}} = 6$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.