SATHEE: Properties of Matter 4 Question 13

Properties of Matter 4 Question 13

15. Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 {Nm}^{- 2}$ . The radii of bubbles $A$ and $B$ are $2 cm$ , respectively. Surface tension of the soap-water used to make bubbles is $0.04 {Nm}^{- 1}$ . 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]

(2009)

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Answer:

Correct Answer: 15. 6

Solution:

Although not given in the question, but we will have to assume that temperatures of $A$ and $B$ are same.

$\begin{aligned} \frac{n_{B}}{n_{A}} & = \frac{p_{B} V_{B} / R T}{p_{A} V_{A} / R T} = \frac{p_{B} V_{B}}{p_{A} V_{A}} \\ = \frac{p + 4 S / r_{A} \times 4 / 3 π {(r_{A})}^{3}}{(p + 4 S / r_{B}) \times 4 / 3 π {(r_{B})}^{3}} \end{aligned}$

$(S$ = surface tension $)$

Substituting the values, we get

$\frac{n_{B}}{n_{A}} = 6$

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Properties of Matter 4 Question 13

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