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Thermodynamics - PV Diagram of an Ideal Gas

  1. Feb 5, 2014 #1
    Thermodynamics -- PV Diagram of an Ideal Gas

    1. The problem statement, all variables and given/known data

    Right, I've got a PV diagram attached to make this easier to explain. The substance is an ideal gas.

    I need to calculate the heat absorbed along the different paths ACB, ADB and AB. The first two are easy. The red lines in the diagram are isotherms.

    2. Relevant equations

    PV = NKT for an ideal gas.

    QAB = ΔU + ΔW = ΔU + PΔV, U is internal energy, U = U(T) = CVT

    For a monatomic ideal gas, CV = 3/2

    3. The attempt at a solution

    The change in internal energy is obviously just 3/2NK(4T1 - T1) = 9/2NKT1

    The work done is P1(2V1 - V1) = NKT1

    Thus the heat absorbed on the path AB is 11/2NKT1.

    My problem is the solution manual says it's 6NKT1.

    It also says that ΔU + ΔW = 3/2(NK)(2T1) + 3/2P1V1 = 6NKT1 which obviously isn't true so I'm questioning it's integrity. It uses the fact that QAB = 6NKT1 in the proceeding problems as well so it's a little confusing.

    Thanks for any help.

    Attached Files:

  2. jcsd
  3. Feb 5, 2014 #2
    The answer book is correct. For path AB, the average pressure is 3/2P1, and the volume change is V1.

  4. Feb 6, 2014 #3

    Andrew Mason

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    Chestermiller is right. The area under the graph from A-B is (P1+P2)(V2-V1)/2 = 3P1V1/2.

    However, if the book says ΔU + ΔW = 3/2(NK)(2T1) + 3/2P1V1 , that is not correct. ΔU = 3/2(NK)(3T1) as you have noted.

  5. Feb 6, 2014 #4
    Thank you!
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