# If the pressure inside and outside a balloon balance, then why does air leave when it pops?

Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.

But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?

• The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon). – Hot Licks Jun 9 '19 at 2:59
• Surely you mean a rubber balloon, not a plastic one? – David Conrad Jun 9 '19 at 20:42
• @DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer. – shabby Jun 11 '19 at 6:07
• Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works? – Eric Lippert Jun 11 '19 at 20:37

For an inflated and tied balloon, the inner and outer pressures aren't equal. The inner pressure is higher by an amount $$2 \gamma |H|$$, where $$\gamma$$ is the inflated balloon's surface tension and $$H$$ is its mean curvature (which is $$-1/R$$ for a sphere). This is called the Young-Laplace equation.

After the balloon is untied and deflates, the pressures equalize and the surface tension becomes negligible.

• Not seen many balloons that are spherical... – user207455 Jun 8 '19 at 21:35
• @SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical. – tparker Jun 9 '19 at 13:56
• They are certainly more spherical than any cow I've ever seen! – Cort Ammon Jun 9 '19 at 15:16
• For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains. – dmckee --- ex-moderator kitten Jun 9 '19 at 17:48
• Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides. – Bilkokuya Jun 10 '19 at 10:05

But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?

You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.

Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.

Hope this helps.

• Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect. – Hagen von Eitzen Jun 9 '19 at 15:21
• @BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning. – mpasko256 Jun 10 '19 at 8:29
• As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/… – JMac Jun 10 '19 at 13:27
• @JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input. – Bob D Jun 10 '19 at 13:37
• @HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too. – Chronocidal Jun 11 '19 at 10:51

until its inner pressure equals to the atmospheric pressure

The inference that the balloon is not growing (or shrinking) because the pressure is the same is not correct.

The balloon is not growing because the effective force pushing the balloon out from inside is the same as the effective force pushing the balloon in from outside.

The force pushing outwards is indeed due to the pressure of the air inside the balloon.

But the force trying to collapse the balloon is the pressure of the air on the outside (atmospheric pressure) plus the elastics potential of the balloon trying to return to its original size and shape.

So, to counteract this additional force the pressure inside the balloon has to be higher than the air pressure outside the balloon.

Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.