When I heat up a balloon, does the air inside the balloon increase in pressure as well as volume? I thought pressure and volume were inversely proportional? Or does pressure and volume increase as temperature increases?

  • $\begingroup$ What exactly do you mean by pressure? I think your notion of pressure is quite different. $\endgroup$ – Asker123 May 24 '15 at 19:46
  • $\begingroup$ The pressure of a gas is the force that the gas exerts on the walls of its container, or in this case the balloon. $\endgroup$ – John May 24 '15 at 19:49
  • $\begingroup$ Yes, the vapor pressure increases. $\endgroup$ – Asker123 May 24 '15 at 19:51
  • $\begingroup$ What is the difference between pressure and volume in this scenario? Is there one? Thats where I am getting so confused. $\endgroup$ – John May 24 '15 at 19:53
  • 1
    $\begingroup$ As you said, pressure and volume are inversely proportional but I believe that you are getting confused in a different matter which I will go over in the answer. $\endgroup$ – Asker123 May 24 '15 at 19:55

If the balloon is closed, then yes, both volume and pressure will increase when the gas inside is heated. Let's look at two simpler cases first.

If the gas were completely free to expand against ambient pressure (say, inside of a container sealed with a freely moving piston, with no friction), then the heated gas would expand until it created as much force per area (gas pressure) as the force per area acting on it (ambient pressure), so that the forces cancel out and the piston stops moving. Here, a temperature increase in the gas would translate solely to a volume increase.

If the gas were confined in a perfectly rigid box, then an increase in gas temperature would cause the molecules inside to bump harder against the inner surfaces, but to no avail, as the walls do not budge and the box stays exactly the same size. Here, a temperature increase in the gas would translate solely to a pressure increase.

In a balloon, the gas is free to expand, but not completely free. In other words, it's a situation somewhere between the two described above. The balloon skin is an elastic which pulls itself, creating a force vector pointing towards the interior of the balloon, and the more the skin is stretched, the stronger the force becomes. Now the gas inside the balloon has to create enough pressure to compensate for both the ambient pressure, and the elastic force trying to pull the balloon skin inwards. This means that after heating, the gas inside a balloon will expand since the balloon is not perfectly rigid, but the equilibrium pressure of the gas inside the balloon will be higher than before because the balloon is pressing against a more tightly stretched balloon.

This should be provable experimentally without much difficulty. Take a rubber balloon, preferably one with as thin of a membrane as you can get, and open its mouth to the atmosphere. Then, clamp it shut without blowing any air inside (in reality, you might have to blow some air in to unstick the rubber walls, but then let all the excess air out). The air inside will have a pressure exactly equal to ambient pressure, because the elastic is not being stretched. Now cover the balloon completely in warm water for a few minutes, and it should inflate slightly. Finally, remove the balloon from the water and quickly perforate it with a sharp object. You might be able to hear a small pop, and feel a rush of air. Both are an indication that pressure inside the heated balloon was higher than ambient pressure.


  1. Here is a very related Physics.SE question - Why is the pressure inside a soap bubble higher than outside? which deals with bubbles rather than balloons. The diagram and equations are applicable in both cases.

  2. Evidently, taking into account the elasticity of the balloon requires a more subtle treatment of the problem. Take a look at Floris' interesting input in the comments. It seems that even an idealized balloon starts off acting as a quasi-rigid container. After reaching a maximum pressure, the balloon starts expanding, and from there the walls get weaker as expansion continues. This means that for a range of temperatures higher than some critical value, the pressure of the gas actually decreases as the temperature increases. The interior pressure of the balloon will still always be greater than ambient pressure, though.

| improve this answer | |
  • 2
    $\begingroup$ It is worth pointing out that the elasticity of the balloon material usually varies (a lot) with temperature: as the material becomes more compliant (lower modulus) the balloon will expand more and the pressure could actually drop. Note also that an expanding balloon has lower pressure as the volume increases. See for example this answer on Physics.SE $\endgroup$ – Floris May 25 '15 at 2:06
  • $\begingroup$ @Floris Very interesting, thank you for the link! I'm trying to wrap my head around this. Even assuming the elasticity doesn't vary with temperature, does that mean that the unstretched balloon with gas at ambient pressure, when heated, initially goes through a quick increase in pressure with little volume change, reaches a pressure maximum, and from there the pressure decreases as the balloon expands? $\endgroup$ – Nicolau Saker Neto May 25 '15 at 2:29
  • $\begingroup$ @NicolauSakerNeto - yes, I think that is correct when you assume perfect elasticity. The thinner balloon is less able to support high pressures. $\endgroup$ – Floris May 25 '15 at 2:36
  • $\begingroup$ This answer provides a great treatment of buoyant gas balloons. Aeronautics fans: A hot air balloon is open at the bottom - the gas is completely free to expand and escape. Pressure and volume stays constant, density and weight decreases. $\endgroup$ – Jirka Hanika May 25 '15 at 12:16

You may recall the ideal gas law: $$PV = nRT.$$

Here, $P$ is pressure, $V$ is volume, $n$ is the amount of gas present (in moles), $R$ is the ideal gas constant, and $T$ is temperature.

In an enclosed system, with no gas flowing in or out, $n$ is constant (as is also, obviously, $R$). We can rearrange the equation above to pull all the constant terms to one side, like this: $$\frac{PV}{T} = nR = \text{const}.$$

Thus, we can see that, with $nR$ staying constant, the left-hand side of the equation, $\frac{PV}{T}$, must also stay constant. Thus, the three quantities $P$, $V$ and $\frac1T$ are inversely related to each other: if you increase one of them, at least one of the other two must decrease so that their product remains constant.

Having noted this, it should be obvious what's happening with your balloon. By heating it, you're increasing the temperature $T$ (and thus decreasing $\frac1T$); to keep $\frac{PV}{T}$ constant, at least one of $P$ and $V$ must therefore also increase.

As it happens, in a rubber balloon, both the pressure and the volume will typically increase, since any increase in pressure will cause the balloon skin to expand, increasing the volume (and reducing the pressure) until the inward force exerted by the stretched balloon skin is balanced by the pressure of the gas inside it.

With a mylar balloon, however, things would be different, since mylar is not particularly elastic. In that case, starting with a partially inflated balloon, the pressure would remain approximately constant (and equal to ambient pressure) up to the point where the mylar skin was stretched taut; at that point, the volume would stop increasing, and the pressure would start to build up instead (potentially up to the point where the balloon would burst).

| improve this answer | |

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