4
$\begingroup$

I see numerous websites talking about how a small balloon is at higher pressure than a big balloon. This is a fun counter-intuitive factoid... unless it isn't a fact at all.

Young-Laplace gives

$$T = \Delta p \frac R2 $$

Which means that for a given tension, increasing the radius does decrease the pressure. But why would the tension be constant? A latex balloon is like a 2D spring, right? So the tension in a inflated spherical balloon should be proportional to the surface area, correct?

$$T \propto R^2 $$

But this gives

$$\Delta p \propto R$$

Which means the bigger the balloon, the higher the internal pressure. Which, again, would be the normal intuition - but it would burst the bubble of some simplified explanations. (har har)

Improve this question
$\endgroup$
4
  • $\begingroup$ By definition, a factoid isn't a fact. The "-oid" ending means "like something, but not something". humanoid = like a human, but not a human. factoid = like a fact, but not a fact. $\endgroup$
    – 410 gone
    Commented Oct 1, 2012 at 15:35
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/10372/2451 $\endgroup$
    – Qmechanic ♦
    Commented Oct 1, 2012 at 15:42
  • $\begingroup$ @Qmechanic, I saw that question. Unfortunately, it has no answer and the link it references is dead. $\endgroup$
    – weymouth
    Commented Oct 1, 2012 at 23:00
  • $\begingroup$ FYI the statement is correct. Not sure about the explanation, though. $\endgroup$
    – Vorac
    Commented Oct 2, 2012 at 9:41

2 Answers 2

Reset to default
4
$\begingroup$

You're quite correct that in a rubber balloon the tension in the rubber increases as it stretches. The pressure is only lower at larger radii when the tension is constant. Soap bubbles are an excellent example of this because the surface tension depends only on surfactant concentration and not on the bubble size.

The pressure in a rubber balloon increases gradually as it is inflated then usually rises more rapidly just before the burst as you reach the elastic limit of the rubber. See http://www.youtube.com/watch?v=fwh-i0WB_bQ for experimental measurements of the pressure, or I'm sure some Googling would find many more examples.

In case the link dies in the future, here's the final graph from the video I mentioned above:

Balloon

Improve this answer
$\endgroup$
3
  • $\begingroup$ That is a nice link. To summarize (in case of a future dead link) it does not show a decrease in pressure at larger radii. It shows an increase in pressure with increasing time and size until the pop. $\endgroup$
    – weymouth
    Commented Oct 1, 2012 at 23:05
  • $\begingroup$ Good point about the link. I'll grab a copy of the pressure:time graph. $\endgroup$
    – John Rennie
    Commented Oct 2, 2012 at 5:44
  • $\begingroup$ The phenomenon is the spike and decline on the left. This has come up again. I would like to have an intuitive understanding rather than just copy the derivation from Wikipedia. $\endgroup$
    – Keith McClary
    Commented Nov 23, 2016 at 22:55
0
$\begingroup$

Take it from someone who has blown up a lot of party balloons, it’s hard to get started and gets much easier until it pops. So the surface tension may increase in the latex but not enough to offset the young Laplace equation. That radius is the largest factor.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.