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Adiabatic Process in a heat engine

  1. Jul 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A heat engine uses a diatomic gas that follows the pV cycle shown in Figure.
    Part 2→3 is adiabat, part 3→1 is isotherm, V=1040 cm3, P=100 kPa, T1=212 K.

    Phys.jpg


    Determine the pressure at point 2.

    Determine the volume at point 2.

    Determine the temperature at point 2.

    Find Ws for process 1→2.


    Find Q for process 1→2.

    Tries 0/9
    Find ∆E for process 1→2.

    Tries 0/9
    Find Ws for process 2→3.

    Tries 0/9
    Find Q for process 2→3.
    0 J

    Find ∆E for process 2→3.

    Find Ws for process 3→1.

    Find Q for process 3→1.

    Tries 0/9
    Find ∆E for process 3→1.

    Tries 0/9
    What is the thermal efficiency of this heat engine? (in percent)
    2. Relevant equations



    3. The attempt at a solution

    phys3-1.jpg


    I know how to do the rest if I can find the temperature at point 2. I tried 2 different ways.

    1. I tried using the Ideal Gas law PV=nRT and solved for mols at pt 1. Then tried using that for point 2 and then used PV=nRT but that didn't work.

    2. Then I tried just simply using PV/T=PV/T and came up with the same answer and they are both wrong.

    What am I doing wrong?
     
  2. jcsd
  3. Jul 3, 2009 #2
    this would help out a lot!
     
  4. Jul 3, 2009 #3
    WHat is the equation i use to find the Work from point 2 to point 3?
     
  5. Jul 3, 2009 #4

    Redbelly98

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    It looks like there is a typo here. We know pv=constant for an isotherm, yet we have pv=2PV at 1, and pv=PV at 3. I suspect it should be p=0.5P at 3, judging from the scale of the p-axis.

    For the adiabat, a useful equation is
    pvγ = constant​

    (Look up γ[/SUP] for an ideal diatomic gas in your textbook, if you're not sure what it is.)
     
  6. Jul 6, 2009 #5
    I understand what that symbol means for monatomic its 5/3 and for diatomic its 7/5. But for the W for 2-->3 I tried using the following....

    W=nCv(delta T)
    n-mols Cv-constant volume Delta T- Change in temp.

    W=(.118 mol)(5/2)(8.31)(369.1-212)=138
     
  7. Jul 6, 2009 #6

    Redbelly98

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    I get a different number of moles than you did, calculated using p, v, and T at point 1. Perhaps you should reproduce that calculation.

    I agree with the 369 K temperature at point 3.
     
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