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 $0.04N/m$. 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 $P_0=8N/m^2$.
$\therefore P_A$ for $1^{st}$ bubble $=P_0+\frac{4T}{r_A}=8+\frac{4\times 0.04}{0.02}$
$P_A=16N/m^2$
$P_B=P_0+\frac{4T}{r_B}<br>=8+\frac{4\times 0.04}{0.04}<br>=12N/m^2$
$PV=nRT$
$\left(16\right)\frac{4}{3}\pi {\left(0.02\right)}^3=n_{A}RT$
$\left(12\right)\left(\frac{4}{3}\pi {\left(0.04\right)}^3\right)$
$=n_{B}RT$
$\frac{n_B}{n_A}=6$