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Volume of a helium baloon inside a fridge

  1. Jul 10, 2014 #1
    If a helium-filled balloon initially at room temperature is placed in a freezer, will its volume increase, decrease, or remain the same?


    3. The attempt at a solution
    If you decrease the T of a vessel filled with gas, its molecules will have less kinetic energy, and the pressure inside the vessel will decrease.
    Since He is an ideal gas, PV=nRT. So, as both T and P decrease (I assume that in the same proportion), there is no reason for that the volume should decrease in order to keep PV/T constant.
    But it happens, why?
     
  2. jcsd
  3. Jul 10, 2014 #2
    I imagine that it can be explained by the fact that the balloon "walls" are flexible and then equilibrate both pressures. Therefore, when the balloon is inside and outside the fridge, it actually has the same pressure, which is the atmosphere pressure.
    Is this right?
     
  4. Jul 10, 2014 #3

    Doug Huffman

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    Gold Member

    Young's modulus of the balloon substance varies with temperature.
     
  5. Jul 10, 2014 #4

    jbriggs444

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    Science Advisor

    A typical helium balloon is constructed from Mylar. Mylar does not stretch appreciably under pressure. Standard practice is to inflate the balloon at room temperature and pressure until the material is just taut.

    [I believe that you are expected to assume this as background knowledge]

    What does this mean about the pressure inside of the balloon before it is placed in the freezer?

    What happens as the temperature of the helium inside the balloon decreases
     
  6. Jul 10, 2014 #5
    Irrespective of the balloon material, when the balloon is inflated, the gas pressure inside is higher than the gas pressure outside, and this requires the balloon membrane to be under tensile stress. If R is the radius of the balloon (assume a sphere), h is the thickness of the membrane, σ is the tensile stress in the stretched balloon membrane, and ΔP is the inside pressure minus the outside pressure, then
    [tex]σh=\frac{(ΔP)R}{2}[/tex]
    When the balloon is cooled, the inside pressure decreases, while the outside pressure remains the same. So ΔP decreases, and the stress and strain of the balloon membrane decreases. This means that the gas volume enclosed within the balloon membrane decreases. Eventually, if the pressure difference goes to zero, the stress goes to zero, and the balloon material will buckle. This buckling allows the balloon enclosed volume to decrease even further.

    Chet
     
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