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Mechanical Properties of Fluids

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Question : 20 of 31
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What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50×10−2Nm−1?2.50 \times 10^{-2} \mathrm{N} \mathrm{m}^{-1}? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble?
(1 atmospheric pressure is 1.01×105Pa1.01 \times 10^{5} \mathrm{Pa}).
Solution:  
Here, surface tension of the soap solution, T=2.5×10−2Nm−1T = 2.5 \times 10^{-2} \mathrm{N} \mathrm{m}^{-1}
density of the soap solution, r=1.2×103kgm−3;r = 1.2 \times 10^{3} \mathrm{kg} \mathrm{m}^{-3};radius of the soap bubble,
R=5.0mm=5.0×10−3mR = 5.0 \mathrm{mm} = 5.0 \times 10^{-3} \mathrm{m}
Now, the excess pressure inside the soap bubble,
Pi−P0=4TRP_{i}-P_{0} = \frac{4 T}{R}
=4×2.5×10−25.0×10−3= \frac{4 \times 2.5 \times 10^{-2}}{5.0 \times 10^{-3}}
=20Pa= 20 \mathrm{Pa}
Also, the excess pressure inside the air bubble,
Pi−P0=2TRP_{i}-P_{0}= \frac{2 T}{R}
=2×2.5×10−25.0×10−3== \frac{2 \times 2.5 \times 10^{-2}}{5.0 \times 10^{-3}} =
10Pa10 \mathrm{Pa}
Now, the pressure outside the air bubble at a depth of 40cm i.e. 0.4m,40 \mathrm{cm} \text{ i.e. } 0.4 \mathrm{m},
P0=P_0 = atmospheric pressure + pressure due to 0.4 m column of soap solution
=1.01×105+0.4×1.2×103×9.8= 1.01 \times 10^{5}+ 0.4 \times 1.2 \times 10^{3} \times 9.8
=1.05704×105Pa= 1.05704 \times 10^{5} \mathrm{Pa}
Therefore, pressure inside the air bubble,
Pi=P0+2TRP_{i}=P_{0}+ \frac{2 T}{R}
=1.05704×105+10=1.05704 \times 10^{5}+10
=1.05714×105Pa=1.05714 \times 10^{5} \mathrm{Pa}.
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