: A small air bubble of $0.1mm$ diameter is formed just below the surface of water. If surface tension of water is $0.072N{m^{ - 1}}$ , the pressure inside the air bubble in kilo pascal is (Atmospheric pressure $ = 1.01 \times {10^5}Pa$ )
1)28.9
2) 0.289
3) 0.0289
4) 103.88
Answer
265.2k+ views
Hint The excess pressure inside the bubble is Pressure inside the bubble would be the sum of pressure outside and the excess pressure inside the bubble. So by adding both we will get the desired answer.
Complete step by step solution
For the bubble to be stable and not to collapse, the pressure inside the bubble must be greater than the pressure on the outside. The force due to the pressure difference must balance the force from the surface tension. The outward force due to the pressure difference between the inside and the outside air is balanced by the surface tension. We have two surfaces, the inner and the outer surface of the bubble both having different pressure.
Given:
Radius of air bubble $R = \dfrac{{0.1}}{2}mm = 0.05 \times {10^{ - 3}}m$ (As $1mm = {10^{ - 3}}m$)
Surface tension of water $T = 7.2 \times {10^{ - 2}}N{m^{ - 1}}$
The excess pressure inside an air bubble would be :
${P_2} - {P_1} = \dfrac{{2T}}{R}$
Where
${P_1}$ is Atmospheric Pressure (${P_0}$)
And ${P_2}$ is pressure inside the air bubble (${P_i}$)
So
${P_i} = {P_0} + \dfrac{{2T}}{R}$
${P_i} = 1.01 \times {10^5} + \dfrac{{2 \times 0.072}}{{0.05 \times {{10}^{ - 3}}}}$
${P_i} = 1.01 \times {10^5} + \dfrac{{2 \times 72}}{5} \times {10^2}$
${P_i} = 1.01 \times {10^5} + 28.8 \times {10^2}$
${P_i} = 1.01 \times {10^5} + 0.288 \times {10^5}$
${P_i} = 1.0388 \times {10^5}Pa$
As we were asked to give answer in kilopascal So,
${P_i} = 103.88kPa$
Note The soap film has less surface tension as compared to the pure water, which would pull itself into the tiny droplets. Surface tension of the water provides the necessary surface tension for the formation of bubbles with water. And also the Soap bubbles and raindrops are spherical because the sphere is the shape with the smallest surface to volume ratio.
Complete step by step solution
For the bubble to be stable and not to collapse, the pressure inside the bubble must be greater than the pressure on the outside. The force due to the pressure difference must balance the force from the surface tension. The outward force due to the pressure difference between the inside and the outside air is balanced by the surface tension. We have two surfaces, the inner and the outer surface of the bubble both having different pressure.
Given:
Radius of air bubble $R = \dfrac{{0.1}}{2}mm = 0.05 \times {10^{ - 3}}m$ (As $1mm = {10^{ - 3}}m$)
Surface tension of water $T = 7.2 \times {10^{ - 2}}N{m^{ - 1}}$
The excess pressure inside an air bubble would be :
${P_2} - {P_1} = \dfrac{{2T}}{R}$
Where
${P_1}$ is Atmospheric Pressure (${P_0}$)
And ${P_2}$ is pressure inside the air bubble (${P_i}$)
So
${P_i} = {P_0} + \dfrac{{2T}}{R}$
${P_i} = 1.01 \times {10^5} + \dfrac{{2 \times 0.072}}{{0.05 \times {{10}^{ - 3}}}}$
${P_i} = 1.01 \times {10^5} + \dfrac{{2 \times 72}}{5} \times {10^2}$
${P_i} = 1.01 \times {10^5} + 28.8 \times {10^2}$
${P_i} = 1.01 \times {10^5} + 0.288 \times {10^5}$
${P_i} = 1.0388 \times {10^5}Pa$
As we were asked to give answer in kilopascal So,
${P_i} = 103.88kPa$
Note The soap film has less surface tension as compared to the pure water, which would pull itself into the tiny droplets. Surface tension of the water provides the necessary surface tension for the formation of bubbles with water. And also the Soap bubbles and raindrops are spherical because the sphere is the shape with the smallest surface to volume ratio.
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