Hot Air Balloon Rises due to Kinetic Energy of Molecules

[Ntstanch]

When you heat the gases inside a hot air balloon it makes more sense (for myself at least) to say that it rises due to the velocity and kinetic energy of the molecules in their relation to the gases outside.

Outside the balloon the molecules are obviously cooler, more dense, and moving slower (less kinetic energy) than inside. The way I imagine it is to think of how a bullet is less affected by gravity until it loses it's energy over time. The more energy in the bullets motion, the less gravity can pull it down in relation to someone throwing the bullet... or maybe just a weaker caliber :tongue2:... So with all these molecules being heated and moving so quickly inside the balloon, outside of it there is far less motion and energy, but the gravity remains a constant, leaving the balloon to essentially be pushed upwards by the 'pile up' of slower molecules forming a fluid like cushion below.

Once higher into the atmosphere it levels out due to there being fewer molecules outside the hot air balloon. Making the heated molecules motion inside of the balloon less significant concerning the constant pull of gravity... given that there are, at greater heights, fewer molecules outside the balloon (to my knowledge) to present that 'pushing upwards by settling below' effect on the balloon at those higher altitudes. So, regardless of the outside molecules being significantly cooler, the balloon will eventually reach the point where the gases inside it can't be heated enough to override that difference.

It also seems like this logic makes sense with gases (or liquids) in their natural state being trapped in basic rubber balloons. They just don't require the manually applied energy. Lastly, I apologize if this is sort of confusing or unclear. I've never actually taken a physics class. High School or College. I've taken some college chemistry at a tech school, however nothing ever goes into great detail.

 

[elegysix]

convection currents and density are the main factors.

with gravity it does not matter how fast you go, though its effects may not be as obvious. It always applies a force of GMm/r^2 downwards, no matter what it is, how high it is, or how fast its going.

 

[Ntstanch]

convection currents and density are the main factors.

with gravity it does not matter how fast you go, though its effects may not be as obvious. It always applies a force of GMm/r^2 downwards, no matter what it is, how high it is, or how fast its going.

Though what I am thinking is that, when the density changes once you heat the gas contained inside the balloon to a less dense and therefore more buoyant state, causing it to rise like a piece of wood or a bubble in water, you are creating that buoyant factor through the quicker motion of the particles being less effected by gravity.

Like how hot air always rises over cold, but without it being contained (inside the balloon) the thermodynamic equilibrium balances it out as quickly as possible. With the piece of wood in water it appears that the wood is forced up by the settled and more dense water underneath it. So does the same apply for the hot air balloon to some degree?

Maybe a better question is to ask why the introduction of heat into the balloon makes the space inside the balloon lighter? I mean, if you somehow increased my heat or overall energy (without setting me on fire or electrocuting me) I'm not so sure I'd become much lighter... Then again there is that video of that very large electromagnet levitating a frog. :tongue2:

 

[Janus]

Maybe a better question is to ask why the introduction of heat into the balloon makes the space inside the balloon lighter? I mean, if you somehow increased my heat or overall energy (without setting me on fire or electrocuting me) I'm not so sure I'd become much lighter... Then again there is that video of that very large electromagnet levitating a frog. :tongue2:

The kinetic energy of the gas molecules are a factor. Faster moving particles exert more pressure per particle than slower moving ones. A balloon holds its shape due to the pressure exerted by the gas it contains. Thus a hotter gas can keep the balloon inflated with fewer particles in the same volume, and the balloon will weight less per cubic cm of volume.

If you heat the air in a balloon, one of three things will happen: If the balloon is sealed and made of an elastic material, the balloon will expand, increasing its volume while maintaining the same weight. If the balloon is not elastic but not sealed, (typical for hot air balloons) some of the gas will be forced out of the opening, leaving less gas inside the same volume. In both these cases the balloon will decrease in density.

The third possibility is that the balloon is sealed and not elastic, in which case the pressure inside the balloon will increase without an increase in volume. The density will remain the same. Such a balloon will not be any more buoyant after heating than before. This is something like when you heat a solid or liquid. They generally don't change volume very much when heated.

Buoyancy itself is caused by the fact that pressure increases with depth. Thus the air pressure at the top of a balloon is less than that at the bottom, and the air pushes up on bottom of the balloon harder than it pushes down on the top, resulting in a net upward force. In this way, everything is a bit buoyant. However, the weight of most objects exerts a downward force that exceeds this buoyant force.

To get a balloon to rise, the downward force due to its weight must be less than the upward buoyant force. This happens when the density of the balloon (mass/volume) is less than the density of the same volume of air were it is located.

Thus getting a balloon to rise means filling it with something less dense than the surrounding air, either a gas like hydrogen, which is less dense even at the same temp, or hot air (which is less dense for the reasons given above.)

 

[Ntstanch]

If you heat the air in a balloon, one of three things will happen: If the balloon is sealed and made of an elastic material, the balloon will expand, increasing its volume while maintaining the same weight. If the balloon is not elastic but not sealed, (typical for hot air balloons) some of the gas will be forced out of the opening, leaving less gas inside the same volume. In both these cases the balloon will decrease in density.

The third possibility is that the balloon is sealed and not elastic, in which case the pressure inside the balloon will increase without an increase in volume. The density will remain the same. Such a balloon will not be any more buoyant after heating than before. This is something like when you heat a solid or liquid. They generally don't change volume very much when heated.

Going to look more into what you said, but for now I'm sort of immediately curious about the first paragraph I have in quotes, and why (in as much layman detail as you can, if you have the time) they become less dense compared to the example in the second paragraph.

Off the top of my head, it sounds like the expansion in volume decreases the density on account of your other point about pressure on the bottom being higher than on the top. So the expansion of the volume allows for that difference to create buoyancy. For the elastic but not sealed balloon the faster moving particles would escape out of the top and dissipate while the volume inside stays less so than outside... but there is no increase in volume so where I am most confused is why the second example allows for increased buoyancy, but the third does not... Does the decrease in density and increase in buoyancy have something to do with the escaping of gas in the non-elastic and non-sealed (hot air balloon) example?

 

[Antiphon]

Don't forget also that buoyancy is a matter of regions of differing density. If you had a rigid spherical balloon of a very light material holding a vacuum (no kinetic energy at all of gas molecules in a vacuum) it would still rise.

 

[Ntstanch]

Don't forget also that buoyancy is a matter of regions of differing density. If you had a rigid spherical balloon of a very light material holding a vacuum (no kinetic energy at all of gas molecules in a vacuum) it would still rise.

If you had the means to heat the gas molecules inside that vacuum would it rise any quicker? Density, as far as fluids go... in this case gases, seems to hold a similar relation to gravity as a rock shooting past the gravitational field of earth. If it was traveling very fast, the pull of gravity would have less effect on the current path of the rock than if it were traveling far slower. What I'm curious about is if this still applies to excited molecules trapped inside the hot air balloon, or inside your very light/spherical balloon example.

 

[Antiphon]

There are no gas molecules in the vacuum I posited so you can't heat them up.

If however I place a small amount of gas inside my rigid ballon, then the cold balloon will rise ever so slightly faster than the hot one. This is because of the extra mass equivalent (very tiny) of the heat in the hot balloon.

 

[Ntstanch]

There are no gas molecules in the vacuum I posited so you can't heat them up.

If however I place a small amount of gas inside my rigid ballon, then the cold balloon will rise ever so slightly faster than the hot one. This is because of the extra mass equivalent (very tiny) of the heat in the hot balloon.

So the cold vacuum will always be lighter than one containing any substance, even if the contents are heated to near plasma levels (in the hot balloon)? In other words, the kinetic energy and velocity of the contents inside the rigid balloon vacuum will have no impact on the mass as far as being heavier or lighter while the atmosphere and gravity are normal on the outside of the vacuums... if you placed the cold (empty) and hot (high energy) balloons next to one another?

 

[Janus]

Going to look more into what you said, but for now I'm sort of immediately curious about the first paragraph I have in quotes, and why (in as much layman detail as you can, if you have the time) they become less dense compared to the example in the second paragraph.

With an elastic balloon, it is the interior pressure exerted by the gas on the inside of the balloon. This outward acting force is countered by the both the air pressure outside the balloon and force of the balloon material itself as it tries to return to non-stretched state. The more you inflate the balloon and stretch its skin, the harder it squeezes back. It is the balance of these forces that determine the size of the balloon.

Air pressure is caused by the collision of the gas molecules against the skin of the balloon. The faster the molecules are moving the more force per collision. If you heat the gas in the balloon, you increase the outward pressure exerted by the gas. This causes the balloon to expand until it finds a new balance point were the increased force of the stretched skin cancels out the increase in internal pressure.
Density is just mass divided by volume. The mass of the gas has not changed (same number of molecules, each with the same mass as before), but since the volume it is enclosed by has increased due to the balloon expanding, the over all density of the gas decreases. (the average spacing between molecules has increased)

In the second case, we have an opening at the bottom of the balloon. We heat the air, increasing the pressure. The balloon does not expand but the force of the particles trying to get out of the balloon at the hole is greater than that of those trying to get in from outside. The result is that the heated air will escape out the hole. As it does, the interior pressure decreases. (fewer air molecules in the balloon to create the pressure. When the inside and outside air pressure equalizes you have less air molecules and less mass inside the balloon than when you started. less mass and same volume equals lower density.

[/quote]

Off the top of my head, it sounds like the expansion in volume decreases the density on account of your other point about pressure on the bottom being higher than on the top.
[/quote]These two effects are unrelated. The difference in outside air pressure from top to bottom is due to the air pressure is due to the weight of the air above a point. the higher you go, the less air above you, thus less weight and lower pressure. The decrease in density due to expansion is simply due to the same mass being enclosed in a larger volume.

So the expansion of the volume allows for that difference to create buoyancy. For the elastic but not sealed balloon the faster moving particles would escape out of the top and dissipate while the volume inside stays less so than outside... but there is no increase in volume so where I am most confused is why the second example allows for increased buoyancy, but the third does not... Does the decrease in density and increase in buoyancy have something to do with the escaping of gas in the non-elastic and non-sealed (hot air balloon) example?

The second example you end up with fewer particles inside the balloon exerting the same pressure due to their increased energy. Fewer particles and same volume means lower density.

In the third example, no molecules are allowed to escape nor is the balloon allowed to expand. Same mass and same volume means the same density and no increase in buoyancy.

Buoyancy it simple due to the difference in overall density of the balloon vs the outside air.
The heating of the air allows this by either increasing the volume of the balloon or letting some of the air out and maintaining the same volume. This is the only effect heating the gas has.

 

[Ntstanch]

Thanks for the information. Janus, you were very helpful. Thanks for the detailed explanations.