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)
Best Answer
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: