Balloon inflated by external vacuum: why won't it burst?

In summary, the balloon in the bell jar is not inflated to its tensile strength limit, and the pressure difference between inside and outside the balloon is not large enough to cause the balloon to burst.
  • #1
Swamp Thing
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Why doesn't the balloon burst in this video? To me it seems that the differential (guage) pressure between inside and outside the balloon is what matters, and I would think it would be the same as a balloon inflated in the usual way.

https://youtube.com/shorts/3HZ0JkgtpDU?feature=share
 
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  • #2
Maybe the pressure difference is not large enough for the tensile strength of the balloon material to be exceeded?
 
  • #3
Isn't it the case that the inflated surface area of the balloon is a measure of the pressure differential? And the area seems to be such that the balloon would normally burst?
 
  • #4
Swamp Thing said:
Isn't it the case that the inflated surface area of the balloon is a measure of the pressure differential? And the area seems to be such that the balloon would normally burst?
There are competing effects at work. But no, the surface area is not a good gauge of the pressure differential. Paradoxically, increasing surface area is compatible with decreasing pressure differential.

If you've ever blown up a balloon by mouth, you may note that it is hard at first and easy after.

Rubber makes things more interesting. Rubber extends elastically only so far. Then it stops. A typical balloon is inflated approximately to this limit and has high tension as a result. One assumes that the balloon in the bell jar is not inflated that full. It has low tension as a result.
 
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  • #5
The curvature and tensile strength matter.
For a curved surface the Young-Laplace equation $$\Delta P=-\gamma (\frac 1 R_1 +\frac 1 R_2)$$ gives Pressure differential in terms the principal radii of curvature and the surface tension

Edit: I put in the minus sign
 
  • #6
I think this calls for a bit lower-level approach:
 
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  • #7
Bandersnatch said:
I think this calls for a bit lower-level approach:

So the vacuum setup is just a misleading distraction?
 
  • #8
I'd say so, yes. The problem is with the assumption that all balloons burst when pierced.
 
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  • #9
hutchphd said:
The curvature and tensile strength matter.
For a curved surface the Young-Laplace equation $$\Delta P=-\gamma (\frac 1 R_1 +\frac 1 R_2)$$ gives Pressure differential in terms the principal radii of curvature and the surface tension

Edit: I put in the minus sign
For a spherical balloon, the tensile stress in the rubber is $$\sigma=\frac{R}{2h}\Delta P$$where h is the present thickness of rubber. For incompressible rubber, h is related to the initial thickness and radius at low pressure by: $$h=\left(\frac{R_0}{R}\right)^2h_0$$
 
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  • #10
I think @Chestermiller's answer to the original question is the correct one. I have done this demonstration using what I called a "funny balloon" which the class quickly recognized as a condom. The darn thing filled the entire available volume without bursting until it was constrained from further expansion by the bell jar wall.
 
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  • #11
Swamp Thing said:
the differential (guage) pressure between inside and outside the balloon is what matters
It is and the maximum exterior 'suck' you can get is when ambient pressure is 0Bar. At that stage, the internal air will have expanded somewhat and what started as 1Bar will have reduced considerably so the pressure difference will be a lot less than -1Bar. Simple Boyle's Law would suggest that an increase in internal volume of 4 would produce a pressure difference of 0.25Bar - and so on. At some stage the envelope will be stressed but not enough to rupture it.
This is very different from the situation where you can use a pump to increase the internal pressure to however much the pump can provide (even your lungs) and that can easily (if you have the nerve) burst the balloon.
It's quite hard to measure the difference between a Good Vacuum and a Deep Vacuum.
kuruman said:
The darn thing filled the entire available volume without bursting until
The embarrassing question is why a condom should ever be required to stretch that far! Not by me, your honour.
 
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FAQ: Balloon inflated by external vacuum: why won't it burst?

How can a balloon be inflated by an external vacuum?

A balloon can be inflated by an external vacuum by placing the balloon inside a vacuum chamber and then reducing the air pressure inside the chamber. This causes the air inside the balloon to expand and inflate it.

Why won't the balloon burst when inflated by an external vacuum?

The balloon won't burst because the external vacuum is only reducing the air pressure inside the balloon, not increasing it. This means that the air molecules inside the balloon are still able to expand and fill the space, without causing the balloon to burst.

What is the difference between inflating a balloon with air and inflating it with an external vacuum?

When inflating a balloon with air, the pressure inside the balloon increases, causing it to expand and potentially burst if the pressure becomes too high. In contrast, when inflating a balloon with an external vacuum, the pressure inside the balloon decreases, allowing the air molecules to expand and inflate the balloon without causing it to burst.

Can any type of balloon be inflated by an external vacuum?

Yes, any type of balloon that is able to hold air can be inflated by an external vacuum. However, the material of the balloon may affect how much it can expand before bursting, so it is important to use a balloon made of a strong and flexible material.

Is inflating a balloon with an external vacuum safe?

Inflating a balloon with an external vacuum is generally considered safe, as long as the balloon is made of a strong and flexible material. However, it is important to follow proper safety precautions, such as using a vacuum chamber and not over-inflating the balloon, to prevent any potential hazards.

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