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Polytropic process

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A pistol-cylinder device contains 0.8kg of nitrogen initially at 100kPa and 27 degrees C. The nitrogen is compressed in a polytropic process during which PV^1.3 = constant until the volume is reduced by one half. Determine the work done and the heat transfer for this process.

    2. Relevant equations

    [tex]
    W = \int_{1}^{2} {P}dV
    [/tex]

    [tex]
    W_{b} = \Delta U
    [/tex]


    3. The attempt at a solution

    I already figured out the work done, but I can't figure out where the heat transfer comes from.
     
  2. jcsd
  3. Aug 24, 2010 #2

    Andrew Mason

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    You don't have to know where the heat transfer comes from. It is required in order to maintain the relation PV^1.3 = constant.

    To find the heat transfer, apply the ideal gas law to find the change in the change in internal energy (hint: you have to find the change in temperature. It is not equal to the work done). Apply the first law to determine [itex]\Delta Q[/itex].

    AM
     
  4. Aug 25, 2010 #3
    Why is the change in internal energy not equal to the work done?

    I'm also not sure how I can equate change in temperature to change in internal energy. The only relevant equation that we have covered so far is

    [tex]
    \Delta U = m C (T_{2} - T_{1})
    [/tex]

    But in all other cases, either volume or pressure has been constant. I don't know how to apply this for a system where nothing is constant.
     
    Last edited: Aug 25, 2010
  5. Aug 25, 2010 #4

    Andrew Mason

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    The change in internal energy is only equal to the work done when there is no heat flow (dQ = 0). That means the process is adiabatic, which means that:

    [tex]PV^\gamma = constant[/tex]

    But [itex]\gamma = 1.4[/itex] for a diatomic molecule, not 1.3. So we know that this is not adiabatic.

    That is correct where C is the gas heat capacity at constant volume. This does not mean the process has to be constant volume. It just tells you what the change in internal energy during the process is. If volume changes, there will be work done as well.

    Substitute nRT/V for P in [itex]PV^{1.3} = k[/itex] to work out the change T.

    AM
     
  6. Sep 1, 2010 #5
    I'm not quite understanding what I should conclude from this. There is heat transfer independent of the work of the piston? So,

    [tex]
    Q - W_b = \Delta U
    [/tex]

    Can Q be calculated assuming volume is constant, since the change in internal energy is already accounted for in the work term?
     
  7. Sep 1, 2010 #6

    Andrew Mason

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    Yes. If the cylinder is not perfectly insulated, heat can flow into or out of the cylinder. In this case it flows out of the cylinder.
    The internal energy change is NOT equal to the work done. This is only the case if the process is adiabatic, which this is not.

    In order to solve this problem, you have to find the change in temperature. You do this using the relation:

    [tex]PV^{1.3} = \text{constant}[/tex]

    You know the initial pressure and the ratio of initial to final volumes so you can easily find the final pressure and, from that, the final temperature.

    When you have found the final temperature, use [itex]\Delta U = nC_v\Delta T[/itex] to determine the change in internal energy. Add that to the figure you found for work done by the gas (make sure you get the sign right) and that equals [itex]\Delta Q[/itex].

    AM
     
    Last edited: Sep 1, 2010
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