Question

Two soap bubbles $\text{A}$ and $\text{B}$ are kept in a closed chamber where air is maintained at pressure $8{\text{N/m}}^2$. The radii of bubbles $\text{A}$ and $\text{B}$ are $2\text{cm}$ and $4\text{cm}$ respectively. Surface tension of the soap - water solution used to make the bubbles is $0.04\text{N/m}$. The ratio $\frac{n_B}{n_A}$ is (where $n$ is the number of moles of air inside the bubble) [Neglect the effect of gravity and assume same temperature]

• A. $6:1$
• B. $5:1$
• C. $1:6$
• D. $1:5$

Answer 1

The correct option is A $6:1$


Given:
Pressure of air, $p=8{\text{N/m}}^2$
Radius of bubble $\text{A},$ $r_A=2\text{cm}=0.02\text{m}$
Radius of bubble $\text{B},$$r_B=4\text{cm}=0.04\text{m}$
Surface Tension of soap solution, $S=0.04\text{N/m}$

As we know,
Excess pressure inside soap bubble $=\frac{4S}{r}$
$\therefore$ Pressure inside the soap bubble $=p+\frac{4S}{r}$

Now, using the ideal gas equation,
$n=\frac{pV}{RT}$

So, $\frac{n_B}{n_A}=\frac{\frac{p_B{V}_B}{RT}}{\frac{p_A{V}_A}{RT}}=\frac{p_B{V}_B}{p_A{V}_A}$

$=\frac{(p+\frac{4S}{r_B})\times \frac{4}{3}\times \pi \times (r_B{)}^3}{(p+\frac{4S}{r_A})\times \frac{4}{3}\times \pi \times (r_A{)}^3}$

$=\frac{(8+\frac{4\times 0.04}{0.04})\times (0.04{)}^3}{(8+\frac{4\times 0.04}{0.02})\times (0.02{)}^3}=\frac{\frac{12}{15625}}{\frac{2}{15625}}$

$=\frac{6}{1}=6:1$

Hence, the correct option is $\left(a\right)$.