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Thermodynamics piston–cylinder assembly Question

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data
    A piston–cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m3. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure–volume relationship is pV 5 constant. Assuming ideal gas behavior for the air, determine the mass of the air, in kg, and the work and heat transfer, each in kJ.


    2. Relevant equations
    Pv=RT, v=specific volume
    W=[itex]\int[/itex]PdV
    Q-W=ΔE


    3. The attempt at a solution
    Pv=RT
    (2 bar)(100 kPa/bar)v=(.286)(300 K)
    v=0.429 m3/kg

    v=V/m
    m=V/v
    m=2 m3/0.429 m3/kg
    m=4.66kg

    I feel like this may be incorrect.

    W=[itex]\int[/itex]PdV
    Would W=0 because PV is constant?

    I'm kind of lost at this point. If anyone could point me in the right direction, I would really appreciate it
     
  2. jcsd
  3. Nov 5, 2013 #2

    rude man

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    You're not necessarily dealing with 1 kg of air. So your formula should be
    pV =mRspecificT, V = actual volume, m= actual mass, p in Pascals. 1 Pascal ~ 1e-5 bar.

    Rspecific for air is 287 J kg-1 K-1 so where did you get 0.286? Are you using SI units thruout I hope?
     
  4. Nov 5, 2013 #3
    Technically, the formula you put is the same as mine, I just divided V by m to get a separate variable v. That formula shouldn't only apply to 1 kg of air. In the steps after, I substituted V/m in for v and solved for m. There isn't anything wrong with this; I guess it's just the longer way of doing it. As for units, I thought that I had them correct. Maybe the .286 is kJ/kgK? I'm not really sure
     
  5. Nov 5, 2013 #4

    rude man

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    Yes, that would account for your being off by a factor of 1000.

    You need to stick to SI units. The SI unit is the Joule, not kilojoule.
     
  6. Nov 5, 2013 #5
    Ok I'll make sure to watch units. The answer I got for mass wouldn't change, the decimal place will just move. Can anyone help with the work part of the problem? Would work be 0 because pV remains constant?
     
  7. Nov 5, 2013 #6

    rude man

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    Work = ∫pdV.
    pV0.5 = cosntant.

    Since you're told p changes, then so must V. And if V changes, work will be done either on or by the system.
     
  8. Nov 6, 2013 #7
    To expand on what rude man said, there is also a temperature change during the expansion. That is how the pressure and volume are controlled to satisfy pv0.5 = const. Use rude man's equations of the previous post to calculate the amount of work done. Tell us what you get.

    Chet
     
  9. Nov 6, 2013 #8
    I don't understand why there would be a temperature change in this problem. It says pV as a whole remains constant, but the values of p and V change. If pV is constant, mRT remains constant and, assuming there isn't a change in mass, the temperature remains the same. I finished the problem and got a mass=4.65kg, W=277.2kJ, and Q=277.2kJ. If T remains constant, Q must equal W because ΔE=0.
     
  10. Nov 6, 2013 #9

    rude man

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    It does not say pV remains constant. I think it says pV0.5 remains constant. So T changes.

    Your answers are correspondingly wrong.

    I have to admit that the statement as you posted it is questionable. Could you dig up the original for us?
     
    Last edited: Nov 6, 2013
  11. Nov 6, 2013 #10
    Yeah I just reread the question statement that I posted and it must have changed when I copy and pasted it from the pdf of the book. Instead of pV 5 constant, it says pV=constant. Sorry for all the confusion.
     
  12. Nov 6, 2013 #11

    rude man

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    OK, so temperature does remain constant.

    But work is still not zero.
     
  13. Nov 7, 2013 #12
    YEah I figured out how to solve for work and my answer was 277.2 kJ I believe.
     
  14. Nov 7, 2013 #13

    rude man

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    That answer is correct if you computed the mass right, which I did not check.
     
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