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I Thermodynamic Process

  1. Jan 19, 2017 #1
    What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

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    Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p1, p2, V1, and V2 is given by the following.

    1) WBC=p2(VC-VB)
    2) Isothermal from A → B so p1V1 = p2VB so VB=(p1V1)/p2
    3) Similarly VC=(p1V2)/p2
    4) WBC= p2((p1V2)/p2-(p1V1)/p2)
    5) Factor out p1/p2 and simplify the you get;
    WBC=p1(V2-V1)​
     
  2. jcsd
  3. Jan 19, 2017 #2

    Charles Link

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    The isothermal processes have no change in internal energy. for an ideal gas. Much heat must be added to the system in the isobaric expansion for two reasons: 1)work is being done as it expands 2) It is going to a higher temperature state. In the isobaric compression, much heat must be removed from the system for two reasons 1) The system is going to a lower temperature 2) Work is being done on the system. The energy from this work must also be removed by removing heat. I do think your calculations are correct. Thermodynamic problems can sometimes be tricky, but I believe you analyzed it correctly.
     
  4. Jan 20, 2017 #3

    Charles Link

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    It should also be noted that you can readily compute ## \Delta Q=\Delta W ## for each of the two isotherms, since each has ## \Delta U=0 ##. ## \\ ## Meanwhile ## \Delta U ## for each of the isobars will be ## \Delta U=\frac{3}{2}nR \Delta T ## for an ideal monatomic gas, and ## \Delta U=\frac{5}{2}nR \Delta T ## for an ideal diatomic gas (where ## \Delta T=\Delta (PV)/(nR) ##). You can thereby readily compute ## \Delta Q ## for each of the isobars, if you know the type of gas. ## \\ ## I think I have this completely correct, but it's always good to have someone double-check the solution. @Chestermiller Did we get everything correct?
     
    Last edited: Jan 20, 2017
  5. Jan 20, 2017 #4
    Since this is a thermodynamic cycle with the final state being identical to the initial state would the total change in internal energy be equal to zero?
     
  6. Jan 20, 2017 #5

    Charles Link

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    Yes. And if you will notice, for the two isotherms ## \Delta U=0 ## and for the transitions on the two isobars, ## \Delta U_1=-\Delta U_2 ##. ## \\ ## Editing... perhaps we haven't proved yet that ## \Delta (PV) ## is equal and opposite for these two cases...but that is obvious because on one isotherm, ## PV=nRT_1 ## and on the other isotherm ## PV=nRT_2 ##.Thereby ## \Delta (PV)=nR \Delta T ## and ## \Delta U=\frac{3}{2} nR \Delta T ## or ## \Delta U=\frac{5}{2} nR \Delta T ##.
     
  7. Jan 21, 2017 #6
    Yes. So....

    And yes, since this a thermodynamic cycle, the change in internal energy for the working fluid has to be zero.
     
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