Gas Law and Hollow Spheres
Date: Fall 2011
If we have two hollow spheres of the same weight, made out of a non-expandable material, yet of different volumes: the second being 1/2 the volume of the first (i.e. one vessel holds 1 cup of atmospheric air, and the other holds 1/2 cup atmospheric air), what happens when we pressurize the 1/2 cup vessel to 2x atmospheric pressure, (for a total of 1 cup of natural atmospheric air). Which one will be more buoyant, or will they be equal because the density/mass will be the same? Or will the smaller one sink because of the difference of surface pressure?
If the two spheres have the same mass but have different volumes, how
can you say that they have the same density? There seems to be a major flaw in your argument.
Remember that buoyancy is a function of mass and displaced volume, since the two objects will have the same mass (the vessels have the same mass and the air inside have equal mass [pressurizing to 2x the pressure means having 2x the moles and 2x the mass]), then only the displaced volume is different between the two cases. So this is like saying the difference between a solid steel bar of the same weight as a ship made of the same steel, the ship floats but the solid bar sinks. The bigger sphere has a lower density and is going to be more buoyant.
Greg (Roberto Gregorius)
The missing information is the mass and density of the spheres. If the mass of the smaller sphere is half that of the first, then there should be no difference in the buoyancy. Density is the key to this answer. Doubling the air pressure with no change in temperature would also also double the density.
Ray Tedder, NBCT
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Update: June 2012