If we consider a balloon full of air submerged in water then we all know that it will rise rapidly. I am having trouble understanding this at the level of individual molecules of air and water. What is a molecular/microscopic explanation for this phenomenon?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
I will give it a try but remember I am new here.
Pressure in the end is brownian motion, therefore more pressure more motion, more total force. Therefore left and right forces are compensated, left and right recoils are compensated. What is not compensated is on the vertical axis therefore the water above the balloon will push it less downwards, than the water below the balloon. The total difference is gravity or better the weight of the water above the balloon minus the weight of the water below the balloon. Note that even if we consider an infinite quantity of water above and below, the total difference is substantially given by the size of the balloon, but in water. Therefore to recap the total difference is the weight in water of the same size of the baloon. Therefore there is less weight above and less molecules above that push down the baloon with their recoils.
Something similar is explained here http://en.wikipedia.org/wiki/Archimedes%27_principle
The subtraction of infinities may be an interesting curiosity for some readers.
-
$\begingroup$ Thank you. I am sure this is correct but I am trying to understand it at the level of the individual molecules. Imagine you were writing a computer simulation of the molecules of water and air. Why do the molecules below the balloon apply more force upwards than the molecules above the balloon apply downwards? $\endgroup$– SimdCommented Jun 8, 2014 at 9:44
-
$\begingroup$ @flyredeagle for brownian motion see Johannes's answer. $\endgroup$– user28737Commented Jun 8, 2014 at 14:16
-
$\begingroup$ @Waquar: "Pressure in the end is brownian motion": pressure is related to kinetic energy of molecules, more pressure means same average speed, but higher kinetic energy, i.e. higher variance of the speed. There are also corrections on top (e.g. the virial equation) which accounts substantially for compressibility and other second order effects (mentioned below). e.g. simulation models with hard spheres, sphere interactions etc. ( padding.awardspace.info/StatMech-fluids-Padding-part1.pdf ) $\endgroup$ Commented Jun 10, 2014 at 15:15
$\begingroup$
$\endgroup$
1
In simple terms, what is happening at a molecular level is that the water molecules rattling around and pushing against the bottom of the balloon do so slightly more frequently than those pushing against the top of the balloon. This is not because of a faster motion (if the water is at thermal equilibrium, anywhere in the water the molecules move at the same average thermal speed) but because at larger depths the water is slightly more compressed and therefore leaves less room per molecule. Conceptually you can think of the water molecules deeper down to rattle around at the same speed but in smaller 'cages'.
-
1$\begingroup$ It seems this argument relies on water being compressible? Would a liquid which was more/less comprehensible give a different effect? $\endgroup$– SimdCommented Jun 8, 2014 at 15:57