Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8…
Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \mathrm{Nm}^{-2}$. The radii of bubbles $A$ and $B$ are $2 \mathrm{~cm}$ and $4 \mathrm{~cm}$, respectively. Surface tension of the soap-water used to make bubbles is $0.04$ $\mathrm{Nm}^{-1}$. 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]
Solution
Given:
Pressure of air, \(p=8 \mathrm{~N} / \mathrm{m}^2\)
Radius of bubble \(A, r_A=2 \mathrm{~cm}=0.02 \mathrm{~m}\)
Radius of bubble \(B, r_B=4 \mathrm{~cm}=0.04 \mathrm{~m}\)
Surface Tension of soap solution, \(\mathrm{S}=0.04 \mathrm{~N} / \mathrm{m}\)
As we know,
Excess pressure inside soap bubble \(=\frac{4 \mathrm{~S}}{r}\)
\(\therefore\) Pressure inside the soap bubble \(=p+\frac{4 S}{r}\)
Now, using the ideal gas equation,
\(n=\frac{p V}{R T}\)
\(\begin{aligned}
& \text {So, } \frac{n_B}{n_A}=\frac{\frac{p_B V_B}{R T}}{\frac{p_A V_A}{R T}}=\frac{p_B V_B}{p_A V_A} \\
& =\frac{\left(p+\frac{4 S}{r_B}\right) \times \frac{4}{3} \times \pi \times\left(r_B\right)^3}{\left(p+\frac{4 S}{r_A}\right) \times \frac{4}{3} \times \pi \times\left(r_A\right)^3} \\
& =\frac{\left(8+\frac{4 \times 0.04}{0.04}\right)^{\times(0.04)^3}}{\left(8+\frac{4 \times 0.04}{0.02}\right)^4 \times(0.02)^3}=\frac{\frac{12}{15625}}{\frac{2}{15625}} \\
& =\frac{6}{1}=6: 1
\end{aligned}\)
;