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Homework Help: Heat Transfer problem

  1. Sep 14, 2006 #1
    I've been working on the following problem for several hours but am not sure how to approach part (c).

    Consider the insulated vessel shown here, with compartment A of volume 0.03 m3, which is empty, and separated by an insulating membrane from compartment B of volume 0.01 m3, which contains 0.15 kg of air at 25°C. The air is stirred by a fan until the membrane ruptures. The membrane is designed to rupture at a pressure of 2 MPa.

    http://img375.imageshack.us/img375/6605/37jf9.jpg [Broken]

    a) What is the temperature when the membrane ruptures?
    b) Calculate the work done by the fan.
    c) Find the pressure and temperature of the air after the membrane ruptures and the air reaches equilibrium state.

    For part (c), I'm not sure which equation to apply. I'm attempting to get the temperature T first and then insert that into the gas law equation to find the pressure. Neglecting kinetic, potential and other energies, the law of conversation of energy would say that

    [tex](mc \Delta_T)_{before rupture} = (mc \Delta_T)_{after rupture}[/tex].

    Am I on the right track? If so, I'm confused about which values of Tfinal and Tinitial to apply to delta T on both sides. Please help me out.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 15, 2006 #2


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    c) The air does no work in expanding into the additional chamber since there is no air in it. The container is also isolated. This means that the energy lost is zero and the temperature will therefore stay the same. This is contradictory to everyday experience since a gas normally cools when it expands, but in this case the expansion occurs at no cost. The pV product will therefore be constant.
    Last edited: Sep 15, 2006
  4. Sep 15, 2006 #3
    Yeah, I messed that up and got a weird number because I kept thinking that any gas that expands must cool. Mathematically,

    Energy of the gas at equilibrium = \Delta_U = mc\Delta_T

    Since there is no change in energy, [itex]\Delta_U=0[/itex] and therefore

    [tex]mc\Delta_T = 0[/tex]

    Since the mass m and specific heat of the gas cannot be 0,

    [tex]\Delta_T = 0[/tex] and therefore [itex]T_{before} = T_{after}[/itex]

    Only realized that after the solution was given out. :biggrin:
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