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Inflating balloon

  1. Dec 8, 2004 #1
    let's say a rubber balloon is not streched (but almost) when its volume is 1L. what force (or pressure) does the balloon exert on a gas inside it when its volume is stretched by xL? I'm measuring this experimentally, but I'd like to know a theoretical way of calculating it.
     
  2. jcsd
  3. Dec 8, 2004 #2
    The ideal gas law states that:
    [tex]PV=nRT[/tex]
    where [itex]n[/itex] is amount of gas in moles, [itex]T[/itex] is the kelvin temperature, and [itex]R[/itex] is the universal gas constant. Define initial pressure as [itex]P_{1}[/itex], and establish a relationship between volume and pressure when amount of gas and temperature are held constant.

    Note: think about how to control these variables effectively when you conduct the experiment for best results.
     
  4. Dec 9, 2004 #3
    yes, that is not what I was asking. Maybe I misexpressed myself. Let me reformulate. P-inside=P-outside + F-balloon/area. I'm looking to predict what the force exerted by the balloon on the gas will be when the fabric is stretched by a certain amount. I know how to measure it experimentaly (as you said, we can use the gas law to do this), but I'm looking for a theoretical answer: is the force exerted by the fabric proportional to the area it's been stretched by?
     
  5. Dec 11, 2004 #4
    come on, don't tell me nobody knows
     
  6. Dec 12, 2004 #5
    I could have a go using Hooke's law wich states that the force the rubber exerts is proportinal to the elongation: [itex]F=C \Delta x[/itex]. This constant C you can ofcourse easily measure, and at the same time check if your baloon indeed obeys Hooke's law!


    Let's also make the assumption the balloon is approximately spherical with a radius [itex]r[/itex] when it is unstretched. and r' if you inflate it a little bit. Now if you draw a small circle on it with an angle [itex]\delta \theta[/itex] from the center of the balloon to the side of the circle the circumference is [itex]r \delta \theta[/itex]. The elongation is in this case the extension of the circumference:

    [tex]\Delta x = 2 \pi (r'-r)[/tex]

    And the force is thus:

    [itex]F=C 2 \pi (r'-r)[/itex]

    If you would like to calculate the pressure you only take the part of this force in the radial direction. Wich amounts to multipying by [itex]\delta \theta[/itex]. And ofcourse pressure is the force per unit area and you have to divide by [itex]\pi (r \delta \theta)^2[/itex]:

    [tex]p(r')=2C\frac{r'-r}{r^2}[/tex]
     
  7. Dec 12, 2004 #6

    rcgldr

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