Two soap bubbles and are kept in a closed chamber where the air is maintained at pressure . The radii of bubbles and are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is . Find the ratio where and are the numb

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 $0.04 N/m$ . 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.] [2009]

Smart Achievers · Accepted Answer

(6)

$Pressure,$

$P_{A} = P_{air} + \frac{4 T}{R_{A}}$

$(∵ Excess pressure due to surface tension (T) in soap bubble = \frac{4 T}{r})$

$B$

$Similarly, P_{B} = 8 + \frac{4 T}{R_{B}} = 8 + \frac{4 \times 0.04}{0.04} = 12 {N/m}^{2}$

$According to ideal gas equation, P_{V} = n R T$

$P_{A} V_{A} = n_{A} R T_{B} \Rightarrow 16 \times \frac{4}{3} π {(0.02)}^{3} = n_{A} R T_{A} ...(i)$

$P_{B} V_{B} = n_{B} R T_{B} \Rightarrow 12 \times \frac{4}{3} π {(0.04)}^{3} = n_{B} R T_{B} ...(ii)$

$Dividing eq. (ii) by (i)$

$\frac{12 \times \frac{4}{3} π {(0.04)}^{3}}{16 \times \frac{4}{3} π {(0.02)}^{3}} = \frac{n_{B}}{n_{A}} [∵ T_{A} = T_{B}]$

$∴$ $\frac{n_{B}}{n_{A}} = 6$

Solution

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