A jar filled partially with water has an object floating in it. The jar is open. Now, if the jar is closed and a vacuum pump is used to pump the air out, what will happen to the floating object? will it rise up more or sink or ...? and why?

Buoyancy depends only on density.  If the object is more dense than water, it will sink to the bottom; if it is less dense, it will float to the top; if it has the same density as water, it will float within the water, moving randomly around due to Brownian motion.

Sucking the air out of the jar will create a vacuum that will eventually be replaced with water vapor as equilibrium is established. Sucking the air out of the jar does not change the density of either the water or the object, so the object will stay floating as before.  However, applying the vacuum will actually cause the water to "boil" as the water vapor fills the vacuum, and this boiling will certainly jostle the object around.

If you want to get hypertechnical, the energy needed to vaporize the water (that fills the vacuum left behind after the air is sucked out) will come from the liquid water, and therefore the liquid water will cool somewhat.  In general, the density of water increases as it cools (hence thermoclines in lakes).  Since the density of the object would not change, the object would then be less dense than the water and would therefore float to the top.  In a standard-sized mason jar, however, I don't think you'd notice this effect since the water temperature would equilibrate with the air temperature on the outside of the jar, which remains constant.  Maybe you'd notice if you had very little water in the jar.  Might be worth doing an experiment...

Posted by Snydley on 2005-10-18 21:22:35