question_answer

Two soap bubbles A and B are kept in a dased chamber where the air is maintained at pressure\[8\text{ }N/{{m}^{2}}\]. The radii of bubbles A and B are 2cm and 4cm, respectively. Surface tension of the soap-water used to make bubbles is\[0.04\text{ }N/m\]. Find the ratio \[{{\eta }_{B}}/{{\eta }_{A}},\] where \[{{\eta }_{A}}\] and \[{{\eta }_{B}}\] are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity]

A) 4         

B) 6       

C) 5                     

D) 9

Correct Answer: B

Solution :

For bubble A:

If \[{{P}_{A}}\] is the pressure inside the bubble then

\[{{P}_{A}}-8=\frac{4T}{{{R}_{A}}}=\frac{4\times 0.04}{0.02}=8\,\,\Rightarrow {{P}_{A}}=16N/{{m}^{2}}\]

According to ideal gas equation,

\[{{P}_{A}}{{V}_{A}}={{n}_{A}}R{{T}_{A}}\,\,\Rightarrow \,\,16\times \frac{4}{3}\pi {{(0.02)}^{3}}={{n}_{A}}R{{T}_{A}}\]   ...(i)

For bubble B:

If \[{{P}_{B}}\] is the pressure inside the bubble then

\[{{P}_{B}}-8=\frac{4T}{{{R}_{B}}}=\frac{4\times 0.04}{0.04}=4\]  \[\Rightarrow \,\,{{P}_{B}}=12N/{{m}^{2}}\]

According to ideal gas equation,

\[{{P}_{B}}{{V}_{B}}={{n}_{B}}R{{T}_{B}}\Rightarrow 12\times \frac{4}{3}\pi {{(0.04)}^{3}}={{n}_{B}}R{{T}_{B}}\]    ...(ii)

Dividing (ii) by (i) we get

\[\frac{12\times \frac{4}{3}\pi {{(0.04)}^{3}}}{16\times \frac{4}{3}\pi {{(0.02)}^{3}}}=\frac{{{n}_{B}}}{{{n}_{A}}}\]     \[\left[ \because \,\,{{T}_{A}}={{T}_{B}} \right]\]

\[\therefore \,\,\,\frac{{{n}_{B}}}{{{n}_{A}}}=6\]