I understand the archimedes principle as it relates to water, but am having trouble reconciling a thesis related to a hot air balloon. If theoretically, I could increase the temperature of air contained in a fixed pocket/container, would the increased pressure exerted upward lift the container? The archimedes principle would argue against the prospect of lifting the container due to the fact that the container remains the same size and therefore displaces the same amount of air outside. However, I find it difficult to understand why increased upward pressure on the container would not lift the container. Can you explain?

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    $\begingroup$ This question appears to be off-topic because it is about basic physics and better served on physics.SE. $\endgroup$ Oct 3, 2014 at 13:26
  • $\begingroup$ I am wanting to design an aircraft using elements of buoyancy rather than lift, and am thinking about some basic fundamental questions...so I guess the question is basic, but it is related to a possible aircraft. $\endgroup$
    – user3736
    Oct 3, 2014 at 13:30
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    $\begingroup$ @user3736 It is related to aircraft, certainly, but I think what ratchet freak is trying to say is that it is more related to the physics stack, and you are likely to get a better answer there. $\endgroup$
    – Jae Carr
    Oct 3, 2014 at 13:46
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    $\begingroup$ The problem with that is that Peter gave a great and understandable answer, leading me to tend to disagree with the hold. The answer given fits the question asked, and is about why a hot air balloon works. $\endgroup$
    – CGCampbell
    Oct 3, 2014 at 22:08

2 Answers 2


I case of an hot air balloon, the volume of the balloon remains unchanged when the air is heated. But because the heated air expands, part of it escapes from the balloon since it is not closed (the bottom is open). Therefore the weight of the air inside the balloon is reduced. The hot air in the balloon has a lower density that the cool air around it. Therefore the balloon lifts.

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Would a closed balloon of unstretchable fabric be used, then the pressure in the balloon would rise but the volume of the balloon would remain unchanged. Therefore the density of the air inside the balloon and the buoyancy would remain the same and the balloon would not lift off.

  • $\begingroup$ I get it now thank you...it is not about upward pressure on the balloon it is about the downward pressure causing some amount of air to escape and create lower density in the balloon which is displaced by the higher density outside, forcing the lower density upward. So, if you could close the balloon and create a vacuum, you could create similar buoyancy of the balloon...actually likely more buoyancy due to the fact of no air vs the less dense air in the conventional balloon. If this is the case, what are the equations that would govern the amount of weight you could lift? $\endgroup$
    – user3736
    Oct 4, 2014 at 16:13
  • $\begingroup$ @user3736: Since pressure does not change, this is the isobaric heating of an ideal gas (air is close enough to be considered ideal). The basic law is called Gay-Lussac's law and postulates that the volume of a gas increases with its temperature. If you lift the temperature from 288°K to 338°K, the volume will increase by 17.36%, or 17.36% of the gas will escape from the envelope. $\endgroup$ Oct 4, 2014 at 18:17
  • $\begingroup$ If instead we close the envelope and instead of heating the gas we remove the air, it seems clear that the envelope will be buoyant and a function of the relationship between amount of displaced air and the weight of the envelope. What are the equations and values of air density governing these relationships? $\endgroup$
    – user3736
    Oct 5, 2014 at 14:20
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    $\begingroup$ @user3736 Only if the envelope maintains the same volume when the air is removed. If you remove air from a closed balloon, it will just become a smaller balloon without becoming buoyant. $\endgroup$
    – DeltaLima
    Oct 5, 2014 at 18:56
  • $\begingroup$ Assuming the volume remains the same or there is a fixed volume, what volume can support what amount of weight, given that the density inside is zero (a vacuum) and the density outside is our atmosphere at sea level. $\endgroup$
    – user3736
    Oct 6, 2014 at 1:00

When you increase the pressure in a closed volume, the increase pressure will act on all parts of the surface. Since both the upward-facing and the downward-facing area is the same, those pressures cancel each other.

If you want to reduce the area on the lower end of the volume, you need to cut a hole in it. Now the upper area is bigger by the area of the hole, and the volume will be forced upwards by a force which is the product of the pressure difference to outside pressure and the area difference, which is the area of the hole. This kind of lift, however, is not buoyancy, but rocket thrust. Unfortunately, as soon as you open that hole, gas will flow out and the pressure difference will vanish, unless you pump more gas into the volume.

  • $\begingroup$ So a hot air balloon behaves similarly, only the air leakage is minimal because the pressure differential between the inner and outer air is minimal, but because the expansive area of the balloon, it works on another principle 'the archimedes principle'...is this the correct way to understand this? $\endgroup$
    – user3736
    Oct 3, 2014 at 16:51
  • $\begingroup$ @user3736: A ballon works like a submarine: It displaces air of a mass which is equal to its own mass. You may call it the Archimedes principle, but the normal expression is buoyancy. To compensate for the mass of the envelope and payload, the gas filling the ballon must be less dense than the air outside. This can be achieved by heating the air in the ballon or by using a less dense gas like helium. $\endgroup$ Oct 3, 2014 at 18:47
  • $\begingroup$ This is how I thought it worked, but from what i understand from others is that you do not get less density simply by heating the balloon, rather you must use the increased pressure from heating the gas to allow gas to escape and cause less density inside in order to create buoyancy. $\endgroup$
    – user3736
    Oct 4, 2014 at 16:31
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    $\begingroup$ @user3736: Or you allow for the envelope to expand. Yes, this understanding is correct and my description was sloppy, leaving out the escaping part. Since the hot air ballon envelope is open at the bottom, it will always be in a pressure equilibrium with outside, and heating will expand the air, so that some has to escape. The reduction of the air mass inside the envelope by the escaping air is creating the lift which is needed to counteract the mass of the envelope and the payload/basket (whatever hangs under the ballon). $\endgroup$ Oct 4, 2014 at 18:08

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