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Thermodynamics - Heat Engine Cycle

  1. Feb 3, 2014 #1
    1. The problem statement, all variables and given/known data
    n = 5 moles, monatomic, ideal gas
    p1 = 7280 Pa; v1 = 2 m3
    p2 = 7280; v2 = 6
    p3 = 6500; v3 = 6

    What is the change in heat, from 2 to 3, and from 3 to 1?

    2. Relevant equations
    PV = nRT
    U = (3/2)nRT
    Q = du + dw

    3. The attempt at a solution

    PV = nRT
    (7280)(2) = (5)(8.32)T1

    T1 = 350 (correct)
    T2= 1050 (correct)
    T3 = 937.5 (correct)

    U = (3/2)nRT
    U1 = (3/2)(5)(8.32)(350) = 21840
    U2 = 65520
    U3 = 57796.88

    W1 = 29120
    W2 = 0
    W3 = -27560
    (all correct)

    Q = du + dw
    ΔQ1 to 2 = (65520-21840) + 29120 = 72800 (correct)
    ΔQ2 to 3 = (57796.88-65520) + 0 = -7723.12 (correct: -7020)
    ΔQ3 to 1 (21840-57796.88) - 27560 = -63516.88 (correct: -64220)

    This is driving me nuts. I can get the first heat quantity right, but not the next two. I've triple checked my work (though that doesn't totally eliminate the possibility of an arithmetic mistake). Where am I going wrong?
     
    Last edited: Feb 3, 2014
  2. jcsd
  3. Feb 3, 2014 #2

    BvU

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    Is Q = du + dw or is dQ = du + dw ?

    How come I get T3 = 312.5 from pV = nRT ? Any chance p3 = 19500 ?
    Any (other) typos or omissions (adiabatic, isothermal?) in the problem statement ?

    You say W1, do you mean W##_{1\rightarrow2}## ? There I get 29120 for the work pΔV the gas is doing.
    If so, why is W##_{2\rightarrow3}## = 0 ?
    If so, why is W##_{3\rightarrow1}## = -27560 ? The volume does not change ?!
     
    Last edited: Feb 3, 2014
  4. Feb 3, 2014 #3
    The first one is correct. The heat added or subtracted is equal to the change in internal energy and work.
    I'm unsure how to work the problem to find dQ; in ΔQ1→2 I equated it to Q1→2 because I assumed at "step 0" there was no heat energy.

    There are no omissions. V3 should actual be 6 instead of 2. I fixed it. Sorry about that.

    Right.
    W##_{2\rightarrow3}## = 0 because the volume does not change.
    W##_{3\rightarrow1}## = -27560 because I made a mistake in the original post; V3 should be 6 instead of 2.
     
  5. Feb 3, 2014 #4

    SteamKing

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    What happened to the units in your calculations?
     
  6. Feb 3, 2014 #5
    I added the units to the original problem. There was no unit conversion involved, so I didn't want to clutter my work.
     
  7. Feb 3, 2014 #6

    Andrew Mason

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    Your calculation of U3 is a bit low. I get 58500 J.

    You can use ΔU = 3nRΔT/2

    ΔU from 2-3 is (3)(5)(8.32)(112.5)/(2) = 7020.

    AM
     
  8. Feb 3, 2014 #7
    Ah, that's it. Thank you so much.
     
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