What Happens When Air is Drained from a Hot Air Balloon with a Solid Shell?

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SUMMARY

This discussion centers on the physics of buoyancy and weight in different environments, specifically regarding hot air balloons and objects in vacuum chambers. It is established that a hot air balloon with a solid shell will rise as long as it is lighter than the air it displaces, regardless of the air inside it. Additionally, objects weighed in a vacuum will show different weights compared to when they are weighed in air due to the absence of buoyant forces acting on them. The conversation highlights the principles of buoyancy and Archimedes' Principle, emphasizing that buoyancy is dependent on the volume of air displaced.

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Zetan
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Hello to everyone.
My first question on this forum (it was quite tricky thinking of a heading and prefix for this thread!):
Imagine for a moment that there is a hot air balloon which has a strong shell forming the balloons shape rather than one made from material and it weighs the same as a normal hot air balloon, What would happen if air was gradually drained from the balloon, would it rise?
 
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It would rise as long as the balloon was lighter than the same volume of air, regardless of what's inside.
 
Thanks, that's cleared that one up.

Another question: If an object was placed on a set of scales in a vacuum chamber and the air was removed, does it weigh much less without the weight of any air pushing down on it?
 
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Zetan said:
Thanks, that's cleared that one up.

Another question: If an object was placed on a set of scales in a vacuum chamber and the air was removed, does it weigh much less without the weight of any air pushing down on it?
The air removed from the vacuum chamber I suppose you mean, and not from the object.

You might be able to answer that question yourself with a little help.

When dropped from a height in air, which will reach the ground first:
a blown up balloon, or not-blown up balloon. If you put them on a scale, which weighs more?

If you then put them in a vacuum chamber, and drop them,:
which one hits the ground first. On a scale in the vacuum chamber, which one weighs more?

( You can cheat by looking up something called bouyancy )
 
Zetan said:
Another question: If an object was placed on a set of scales in a vacuum chamber and the air was removed, does it weigh much less without the weight of any air pushing down on it?

I believe so.
 
Drakkith said:
I believe so.
Not so.
Do you weigh more on a scale when submerged in water than air.
Would you weigh more on a scale when submerged in air than a vacuum.
 
256bits said:
The air removed from the vacuum chamber I suppose you mean, and not from the object.

You might be able to answer that question yourself with a little help.

When dropped from a height in air, which will reach the ground first:
a blown up balloon, or not-blown up balloon. If you put them on a scale, which weighs more?

If you then put them in a vacuum chamber, and drop them,:
which one hits the ground first. On a scale in the vacuum chamber, which one weighs more?

( You can cheat by looking up something called bouyancy )

Yes, I mean the air being removed from the vacuum chamber and not the object.

Funnily enough I recently watched a feather and hammer dropped side by side in a vacuum chamber and they both fell at the same speed and hit the ground together. I haven't seen two objects weighed in a vacuum chamber yet, would the blown up balloon would weigh more because of the extra weight of the air inside it?
 
Zetan said:
Yes, I mean the air being removed from the vacuum chamber and not the object.

Funnily enough I recently watched a feather and hammer dropped side by side in a vacuum chamber and they both fell at the same speed and hit the ground together. I haven't seen two objects weighed in a vacuum chamber yet, would the blown up balloon would weigh more because of the extra weight of the air inside it?
That should tell you something.

In a vacuum, they both fall at the same rate.That's true.
And on the scale in the vacuum, the blown up balloon would weigh more due to the air inside.
So you are 2 for 2 - 100% so far.

( You didn't answer the first part when dropped in the air and weighed in the air, but since that's an everyday experience, so I'll give it to you. 4 for 4 )
( And no, we aren't going to worry about air friction. We will assume the friction is small and negligible )

So the question is :
Why does the blown up balloon weigh more than the non-blown up in a vacuum.
But weighs less than the non-blown up balloon , when weighed in air.

Have you looked up buoyancy yet?
 
  • #10
Buckleymanor said:
Not so.
Do you weigh more on a scale when submerged in water than air.
Would you weigh more on a scale when submerged in air than a vacuum.

Does that mean every object has a certain amount of buoyancy whilst it is submerged in air, depending on the amount of air it displaces, and in a vacuum it would weigh more without any buoyancy?
 
  • #11
Zetan said:
Does that mean every object has a certain amount of buoyancy whilst it is submerged in air
If there is a pressure gradient in the air, for example due to gravity.
 
  • #12
256bits said:
That should tell you something.

In a vacuum, they both fall at the same rate.That's true.
And on the scale in the vacuum, the blown up balloon would weigh more due to the air inside.
So you are 2 for 2 - 100% so far.

( You didn't answer the first part when dropped in the air and weighed in the air, but since that's an everyday experience, so I'll give it to you. 4 for 4 )
( And no, we aren't going to worry about air friction. We will assume the friction is small and negligible )

So the question is :
Why does the blown up balloon weigh more than the non-blown up in a vacuum.
But weighs less than the non-blown up balloon , when weighed in air.

Have you looked up buoyancy yet?

Thanks, yes I've looked up buoyancy and displacement, it's forming a picture in my head gradually!

Hmm, in a vacuum, there isn't any buoyancy at all, there is nothing to "squeeze" the balloon upward through layers of less dense atmosphere. There is more density to the balloon with the air inside it, so it would weigh more on the scales in the vacuum?

Would the blown up balloon weigh less when weighed in air because it is displacing more air? I'm not sure if can see that yet because I think the inflated balloon would contain more air pressure than the outside atmosphere, which means it wouldn't be displacing enough air to make it buoyant?
 
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  • #13
Zetan said:
because I think the inflated balloon would contain more air pressure than the outside atmosphere, which means it wouldn't be displacing enough air to make it buoyant?
Somewhat more air pressure, but most party balloons are quite buoyant you will have to admit.
 
  • #14
Zetan said:
I think the inflated balloon would contain more air pressure than the outside atmosphere,
The content is irrelevant for the force of buoyancy.
 
  • #15
256bits said:
Somewhat more air pressure, but most party balloons are quite buoyant you will have to admit.

Yes. So the material of the balloon when inflated\spread out, is displacing more air than the non inflated one, making it more buoyant?

Am I right in thinking that as an object is stretched out, the closer the physical matter of the object becomes to the density of air, until it eventually matches the weight of air and achieves neutral buoyancy. Then if it is stretched out further it becomes lighter than surrounding air and is squeezed upward until it reaches air density that matches it?
 
  • #16
Zetan said:
Yes. So the material of the balloon when inflated\spread out, is displacing more air than the non inflated one, making it more buoyant?
The volume of the object as a whole matters for the force of buoyancy.
 
  • #17
Zetan said:
Does that mean every object has a certain amount of buoyancy whilst it is submerged in air, depending on the amount of air it displaces, and in a vacuum it would weigh more without any buoyancy?
Yes.
 
  • #18
Right, think I've got it now! As a balloon is inflated and the skin stretches out, the actual skin itself displaces more and more of the surrounding air until eventually the air it has displaced weighs more than the balloons skin.

If the balloon skin were stretched out far enough and it displaced enough air, it would begin to be squeezed upwards towards thinner layers of air by the weight of the air around it, until it matched the weight of the air it was displacing and achieved neutral buoyancy?
 
  • #19
Zetan said:
Right, think I've got it now! As a balloon is inflated and the skin stretches out, the actual skin itself displaces more and more of the surrounding air until eventually the air it has displaced weighs more than the balloons skin.
No. The balloon as a whole displaces more and more air.

Zetan said:
If the balloon skin were stretched out far enough and it displaced enough air, it would begin to be squeezed upwards towards thinner layers of air by the weight of the air around it, until it matched the weight of the air it was displacing and achieved neutral buoyancy?
No. Balloons float when filled with gases of lower density than the surrounding air.
 
  • #20
A.T. said:
No. The balloon as a whole displaces more and more air.No. Balloons float when filled with gases of lower density than the surrounding air.

Thank you.

Supposing a sheet of rubber were stretched out until it is volume displaced more air than the rubber itself weighed, would it float?
 
  • #21
Zetan said:
Thank you.

Supposing a sheet of rubber were stretched out until it is volume displaced more air than the rubber itself weighed, would it float?

Being "lighter than air" and buoyancy are not the same thing, even though related.

To float, the total weight of the balloon and the gas enclosed within the balloon together have to weigh as much as the air it displaces.
If the total weight is heavier, the balloon will sink to the ground.
If the total weight is lighter, the balloon will rise.
 
  • #22
Zetan said:
Supposing a sheet of rubber were stretched out until it is volume displaced more air than the rubber itself weighed, would it float?
If you could actually increase the volume of the rubber by several orders of magnitude, so it's mass density becomes less than air's, then yes.

But this not what happens when you stretch rubber sheets. The stretched rubber's surface area increases, while the thickness decreases. So the rubber volume doesn't change much.
 
  • #23
Zetan said:
Right, think I've got it now! As a balloon is inflated and the skin stretches out, the actual skin itself displaces more and more of the surrounding air until eventually the air it has displaced weighs more than the balloons skin.

If the balloon skin were stretched out far enough and it displaced enough air, it would begin to be squeezed upwards towards thinner layers of air by the weight of the air around it, until it matched the weight of the air it was displacing and achieved neutral buoyancy?
Not
A.T. said:
If you could actually increase the volume of the rubber by several orders of magnitude, so it's mass density becomes less than air's, then yes.

But this not what happens when you stretch rubber sheets. The stretched rubber's surface area increases, while the thickness decreases. So the rubber volume doesn't change much.
Don't imagine that even if you increased the volume by several orders of magnitude it would.The density of air within a balloon is always more than the surrounding air when inflated because the skin stretches so it pushes the air it surrounds into a smaller volume.
 
  • #24
Buckleymanor said:
Don't imagine that even if you increased the volume by several orders of magnitude it would.The density of air within a balloon is always more than the surrounding air when inflated because the skin stretches so it pushes the air it surrounds into a smaller volume.
He was talking about stretching a rubber sheet there, but of course it's completely unfeasible anyway.
 
  • #26
Zetan said:
Many thanks for all of your replies, very helpful.

I came across this useful description whilst searching for information:
http://www.pbs.org/wgbh/nova/lasalle/buoybasics.html
Yes it's basically Archimedes Principle which states a floating body when partially submerged in water displaces it's own weight.
So if you weighed the volume of water displaced it would be equal to the piece of wood or whatever else was floating.
In air though most objects don't float and if they do they don't normally stay part submerged.
What does happen to an object in air is that is denser than the air around it is that whatever volume of air it displaces there is an up thrust which is equal to the weight of air the object occupies.
So if you were able to remove the air which the object had displaced (occupied) and weighed it, that weight would equal the amount of up thrust or buoyant force the object experienced.
 
  • #27
So, you have a folded up Mylar balloon. It weights X. You inflate it to 1m in diameter such that the internal pressure is still 1 atm. What does it weigh now?
 
  • #28
insightful said:
So, you have a folded up Mylar balloon. It weights X. You inflate it to 1m in diameter such that the internal pressure is still 1 atm. What does it weigh now?
You can't inflate the Mylar balloon to have an internal pressure of still 1 atmosphere at best it would be 1 atmosphere + the weight of the Mylar balloon which surrounds the air when inflated.
 
  • #29
What will you get when you add pressure to weight? :confused:
 
  • #30
(Whew, tough crowd.)

Okay, you hang a deflated Mylar balloon from a scale that reads 100g. You blow it up with ambient air to 1.001 atm to produce a roughly 1m diameter sphere. What does the scale read?
 

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