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Ideal gas law mechanical work problem

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    4g70SY0.jpg

    Calculate the gas' mechanical work on its environment for the thermodynamic process path IF.

    Solution says: 505 J


    2. Relevant equations

    W=[tex]\int_i^f p,dv[/tex]

    i = Vi, initial volume
    f = Vf, final volume


    3. The attempt at a solution

    My approach is to find a function for p(v) between 2 and 4 liters.

    [tex] p(v) = 709275 - \frac{3}{2} * 101325 V (Pa) [/tex]

    And then integrate it from 2 to 4 liters, which is 0.002m3 to 0.004m3

    Problem is, I get: rKpGaeu.jpg

    Which is way off. I should get 505 J..

    I'm stumped. What am I doing wrong?
     
  2. jcsd
  3. Apr 17, 2013 #2

    rude man

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    where did you get your 709275 number?

    Anyway, I make it about 306 J.
     
  4. Apr 17, 2013 #3
    Oh, I should have stated that. I extended the IF line to where it intersects with the Y-axis, which is 7 atm.

    7 * 101325 Pa = 709275 Pa.

    You got 306 J... it should be 505, unless the document is wrong.
     
  5. Apr 17, 2013 #4

    rude man

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    The first time I just calculated area = 1/2 bh and got 306.

    This time I did the integration and got ~ 518. And I found your mistake - in your 2nd term under the integral you divided by 2 instead of 0.002.

    Still don't know why 1/2 bh didn't work! :frown:
     
  6. Apr 18, 2013 #5
    Thank you for your help. Gonna ask the professor about it, he made the task after all.

    The second term is -(3*2) because that is the gradient of the curve. That should be correct.
     
  7. Apr 18, 2013 #6

    rude man

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    It's not correct. The second term should be -(3/2) * 101,325/0.001 V.
    I wouldn't go to your professor 'till we agree on this.
     
  8. Apr 18, 2013 #7
    My God, that seemed to do the trick.

    DcgSep2.jpg

    This seems right to me.

    I gather since I converted from atm to Pa, and integrated using m3 instead of liters, I also should have taken that into account for the gradient.

    Is that assessment correct?
     
  9. Apr 18, 2013 #8

    rude man

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    That assessment is fully correct! :smile:
     
  10. Apr 18, 2013 #9
    Thank you very much!
     
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