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Elastic Balloon Filled with Ideal Gas

  1. Nov 10, 2011 #1
    This is the equation I'm talking about:
    PV = nRT

    So say I had a bit of Ideal Gas in a completely insulated elastic balloon, and I raised a temperature just a little bit.
    PV = nR(2T)

    Because this is an elastic balloon, would the volume also double?
    P(2V) = nR(2T)

    But what if the balloon is really tight? Would the pressure increase because the volume couldn't expand as much? But to what extent? How much would the balloon expand?
    (?P)(?V) = nR(2T)
    There has to be a sort of limit, depending on the stretchiness of the balloon, right?
     
  2. jcsd
  3. Nov 11, 2011 #2

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    For a balloon that stretches ideally according to Hooke's law, I believe it resists with a constant pressure, meaning you would see the volume doubling.

    That is, because the surface increases with the square of the radius, the stretching force also increases with the square of the radius.
    But the resulting pressure, which is the quotient of the two, remains constant.
     
    Last edited: Nov 11, 2011
  4. Nov 11, 2011 #3
    Say the volume of the balloon increased by √2, then the pressure inside of the balloon would have to increase by √2.

    Look at it this way because the balloon is a closed system (xP)(yV) = nR(zT)

    x · y = z

    if you double z, then x and y have to change so that their product also doubles.
     
    Last edited: Nov 11, 2011
  5. Nov 11, 2011 #4
    You have one equation, PV = nRT - or, the version I think with, P = NT - and three variables: P, V or N, and T. Specifying a change to one does not specify how the other two change, you need another equation. Something about the properties of the container (Hooke's law spring), something about heat flow, whatever - but without more information you cannot say what happens to V if you double T.
     
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