Rising Air Bubble in a River: Calculating Radius

AI Thread Summary
An air bubble with an initial radius of 2mm at the bottom of a 3.3m deep river experiences changes in pressure as it rises. The pressure at the bottom, P1, is calculated using the formula depth times density times gravitational acceleration, while the pressure at the surface, P2, equals atmospheric pressure. The volume of the bubble changes in proportion to the ratio of these pressures, P1/P2, demonstrating that pressure and volume are inversely related. This relationship is explained by the principle that, at constant temperature, an increase in pressure leads to a decrease in gas volume. The discussion clarifies that pressure is independent of surface area, as it is defined as force per unit area.
anand
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Homework Statement


An air bubble of radius 2mm if formed at the bottom of a 3.3m deep river.Calculate radius of bubble as it comes to the surface.
atmospheric pressure=10^5 pa and density of water=1000 kg/m^3
 
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Calculate the pressure at the bottom of the river(depth*density of liquid*graviational acceleration). Let it be P1 and the pressure above the surface of river (i.e equal to the atmosperic pressure) P2.
Then the volume enclosed by the buble change by the factor P1/P2 (pressure is inversely proportional to the volume).
Assuming the buble as a sphere you can easily work out the radius.
 
Why is the pressure inversely proportional to the volume and not the surface area?
 
anand said:
Why is the pressure inversely proportional to the volume and not the surface area?

At constant temperature, when pressure increases gas volume decreases(it is a general observation). Mathematically speaking, when pressure is doubled volume becomes halve of the original volume. So it is evident why volume and pressure are inversely related.

Pressure is independent of surface area. Because pressure is defined as force exerted perpendicularly on unit area(i.e P=F/A). So pressure must not be confused with force. :confused:
 
Thanks a lot!
 
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