Question

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 $0.04 \,N/m$. The ratio of $n_{B}/n_{A}$ is (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. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (C)

The excess of pressure above atmospheric pressure, due to surface tension in a bubble $=\frac{4T}{r}$ .
The surrounding pressure $P_{0}=8 N /m^{2}$.
$\therefore P_{A}$ for $1^{st}$ bubble $= P_{0}+\frac{4T}{r_{A}}=8+\frac{4\times0.04}{0.02}$
$P_{A}=16\, N /m^{2}$
$P_{B}=P_{0}+\frac{4T}{r_{B}}
=8+\frac{4\times0.04}{0.04}
=12\, N/ m^{2}$
$PV = nRT$
$\left(16\right)\frac{4}{3}\pi\left(0.02\right)^{3}=n_{A}RT$
$\left(12\right) \left(\frac{4}{3}\pi \left(0.04\right)^{3}\right)$
$=n_{B}RT$
$\frac{n_{B}}{n_{A}}=6$