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Pressure inside a balloon

  1. Jun 11, 2016 #1
    here is something confusing me about air pressure inside balloons. I got three questions:

    Assume I am sitting at sea level, with a balloon. I fill it to 1/3 its capacity with air.
    Then I place an absolute pressure sensor at its opening.
    1- What pressure value do I read ?
    Do I get the 1013 mbar which is the atmospheric pressure at sea level ? Because I read that the pressure inside the balloon must equal the outside pressure.

    Now, I fill it to 2/3 its capacity with air.
    2- What do I read now ? Is it double the pressure I got at question 1 or is it still the same ?
    Remember, the outside atmospheric pressure did not change. It is still 1013 mbar. Also note the balloon has doubled its air mass.

    Now, with 2/3 its capacity still filled with air, I move the balloon up to an altitude where the atmospheric pressure is half that at sea level.
    3- What pressure value do I read there ?
  2. jcsd
  3. Jun 11, 2016 #2


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    Sort of depends on what you mean by 'capacity'. It's really a maximum volume before bursting.

    The balloon's volume will contain a certain mass of air. The pressure will be a direct function of the balloon's volume at any altitude.
    So, when you fill it from 1/3 to 2/3 capacity, you are doubling the mass, but not doubling the volume. Depending on how large the balloon gets, that will determine the pressure.
  4. Jun 12, 2016 #3

    jack action

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    A balloon is an elastic. To deform an elastic, you must provide a force. If the force inside is greater than the one outside, the area extends. Since we're talking about force and area, we can also talk about pressure. If a balloon has deformed, then the pressure inside has to be greater than the one outside. The greater the deformation, the greater the pressure differential (Pin - Pout). This takes care of your points 1 & 2.

    For point 3, what happens is that you are reducing the outside pressure which has the same effect as increasing the inside pressure, i.e. increasing the deformation of the balloon. The interesting thing is that the inside of the balloon is a closed system so, as the balloon increases in volume, the inside pressure will also decrease. The equilibrium will be reached faster, so the increase in size will be smaller than one could have initially thought.
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