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I have to calculate the air pressure inside of an hot air balloon. After some searching I found out that I can use the ideal gas law: PV = nRT (from Wikipedia)

So to get the pressure in the balloon I would have to know n, which is the amount of gas in moles. For that amount I'm currently using the pressure (kg/m3) * the volume. So for the pressure I need the pressure which of course doesn't work.

Is there a way to calculate this? Right now I do know the normal air pressure (outside of the balloon) volume of the balloon etc. etc., but I can't know the weight of the amount of air in the balloon, because I use the pressure to calculate it. One thing to note is that it isn't an actual hot air balloon. It's just one made up from physics formulas in a program that calculates them every 0.1 seconds (for example) for me.

Is there some thing I'm missing?

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"So to get the pressure in the balloon I would have to know n" Well, you could try that, but the balloon is open to the surrounding air and can gain or lose moles as it warms and cools. – dmckee May 15 '13 at 16:51
Use the ideal gas law with the specific gas constant for air instead. $P = \rho R_{specific}T$ For air $R_{specific} = 287$ J/kg/K – OSE May 15 '13 at 17:14
I'm not the best on fluid dynamics and ideal gasses. But it seems to me that since the balloon is open to the outside air, the pressure in the balloon equals the pressure outside the balloon. Otherwise the air in would flow out or vice versa. I know the density in is lower, but I thought the pressure is equal – Jim May 15 '13 at 17:52
@OSE Isn't the case that the pressures inside the hot-air balloon and outside are the same at the open mouth at the bottom of the balloon? At the top of the balloon, the pressure on the inside must be greater than outside. That's where the lift comes from... – DJohnM May 16 '13 at 2:22
@User58220 You are right, I jumped at the answer a little too quickly. I believe the pressure inside the balloon should obey $dp/dy = \rho g$ with the pressure at the mouth equal to the atmospheric pressure. I'm going to remove my comments to hopefully prevent further confusion. – OSE May 16 '13 at 14:22

3 Answers 3

Yes, the pressure is atmospheric, normally hot air balloons are not even sealed at the bottom. If it was sealed then it could have pressure higher than atmospheric, balanced by the wall elastic forces.

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The pressure inside a hot air balloon must be slightly higher than the ambient pressure, otherwise the balloon would have no shape. – james large Sep 2 at 14:51

The pressure inside the balloon will be the same as that of atmosphere. Remember that the balloon is made of a rubber which is a flexible material and not a rigid body. The latter would not deform and the more air you introduce into the chamber will will increase the pressure inside the chamber. But a rubber balloon will keep on expanding (i.e. whenever the pressure inside the balloon is greater than ambiance) until it breaks. Basically the mass of air entrapped in a balloon will be $\frac{PV}{RT}$ where $P=P_{amb}$.

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"hot air balloon" means one of these: Not made of rubber. – james large Sep 2 at 14:49

It's confusing why any problem would ask you to find the pressure in a hot air balloon. For an elastic balloon, the gauge pressure comes from the elasticity of the balloon in a similar manner to Hooke's Law.

A hot air balloon gets lift from buoyancy, which comes from the density difference between the inside and the outside, which comes from the difference in temperature. You could consider the change in air pressure from the bottom to the top. This is a $\rho g h$ calculation. The pressure change over 50 vertical feet in normal air is around $194 Pa$. Then, if you imagine the hot air balloon is $110^{\circ} F$ with the outside air at $70^{\circ} F$, then you come up with about $180 Pa$ pressure drop from bottom to top inside the balloon. So the elevation pressure change is small compared to atmosphere pressure (of $100,000 Pa$) to begin with, and the difference that provides the lift is smaller still.

So it's not clear what pressure you're interested in to begin with, but this is a vague sense of the relative pressure scales involved.

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