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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2009 · Paper 2 · Physics · Thermodynamics

Two soap bubbles A and B are kept in a closed chamber where the air...

Two soap bubbles AA and BB are kept in a closed chamber where the air is maintained at pressure 8Nm−28 \mathrm{Nm}^{-2}. The radii of bubbles AA and BB are 2 cm2 \mathrm{~cm}, respectively. Surface tension of the soap-water used to make bubbles is 0.040.04 Nm−1\mathrm{Nm}^{-1}. Find the ratio nBnA\frac{n_B}{n_A}, where nAn_A and nBn_B are the number of moles of air in bubbles AA and BB, respectively. [Neglect the effect of gravity]

Correct answer

6

Step-by-step solution

Although not given in the question but we will have to assume that temperatures of A and Baresame.are same.nBnA=pBVB/RTpAVA/RT=pBVBpAVA=p+4s/rA×4/3π(rA)3(p+4s/rB)×4/3π(rB)3(s= surface tension) \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 \pi\left(r_A\right)^3}{\left(p+4 s / r_B\right) \times 4 / 3 \pi\left(r_B\right)^3} \\ &(s=\text{ surface tension) } \end{aligned}
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