I have formula $PV=nRT$. As temperature rises in hot air balloon why does pressure inside hot air balloon and outside are the same? I have thought about it for long time and I think due to air pressure changes as height changes is one of the reason but it still is not satisfying. My book says "Hot air balloons are open at the bottom, so the pressure $P$ inside the balloon is equal to atmospheric pressure outside." But I think pressure inside should be larger since the force of hot air balloon fights with balloon cover and atmospheric pressure. To me only when balloon is not inflated the pressure inside is same as outside. Is there good explanation for it?


1 Answer 1


The balloon is open on the bottom.1 If the pressure inside the balloon is larger than the outside air pressure gas will move from the inside to the outside. When the pressure inside is lower, gas will move into the balloon through the opening. So, at equilibrium, the pressure inside equals the pressure outside.

What changes as you run the heaters is $T_{inside}$ and $n_{inside}$. The volume is (approximately) fixed (hot air balloon shells don’t stretch, much). The pressure equalized with the outside pressure via fluid movement through the opening.

You might visualize the non-equilibrium process as: heat is added, pressure goes up, fluid flows out, pressure is equalized again, but you can’t merely apply equilibrium thermodynamics to that process.

So why do hot air balloon have structure if not from the gas pushing the envelope out? It’s actually the mechanical structure based on how the gondola hangs off of the envelope containing the hot air. For the classic tear drop shape balloon, the top, slightly curved part is a cap that the hot gas pushes upwards against due to buoyancy. Ropes hang down from it and are under tension due to the upward buoyancy and the weight of the gondola (and the balloon material itself). The tapering part of the balloon is just tailored to match how the ropes will hang during flight. The fabric in that part of the balloon is not under any stress.

  1. For simplicity, let’s ignore the parachute valve at the top of the balloon that is used to vent the hot air in order to descend.
  • $\begingroup$ Plus the big (albeit controllable) flap on the top as well. $\endgroup$
    – Jon Custer
    Commented Oct 14, 2021 at 13:33
  • $\begingroup$ since my background is only classical mechanics I don't understand this one "but you can’t merely apply equilibrium thermodynamics to that process." Can you explain this in simple words? $\endgroup$
    – banned
    Commented Oct 14, 2021 at 14:43
  • $\begingroup$ @BannedUser $PV = nRT$ is the equation of state for a gas that is in equilibrium, i.e. not changing; really almost all of basic thermodynamics is equilibrium. So, for systems that are not in equilibrium there can be problems with even defining thermodynamic quantities like pressure or temperature for those systems as they change. In practice, as long as things aren't too far from equilibrium, one can still use relationships derived at equilibrium to get approximate results, but non-equilibrium theromodynamics is a really complicated problem, so the outline there is qualitative at best. $\endgroup$
    – Dave
    Commented Oct 14, 2021 at 14:55
  • $\begingroup$ @Dave thanks for excellent explaination. $\endgroup$
    – banned
    Commented Oct 14, 2021 at 15:17

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