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Thermodynamics of Helium using a P-V Diagram

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Helium gas is initially at a pressure of 16 atm, a volume of 1L, and a temperature of 600K. It is expanded isothermally until its volume is 4L and then compressed at constant pressure until its volume and temperature are such that an adiabatic compression will return the gas to its original state. (a) Sketch this process on the P-V diagram. (b) Find the volume and temperature after the isobaric compression. (c) Find the work done during each cycle. (d) Find the efficiency of the cycle.



    2. Relevant equations
    (P1V1) / T1 = (P2V2) / T2

    PV[itex]\gamma[/itex] = constant

    P1V1[itex]\gamma[/itex] = P2V2[itex]\gamma[/itex]

    [itex]\gamma[/itex] ≈ 1.67

    3. The attempt at a solution

    At this time i have only attempted a and b. I believe i have drawn the P-V diagram correctly, but do not know how to draw one here, so i will describe what i have drawn.

    a) Point A is at a pressure of 16atm and a volume of 1L. There is an isothermal curve between point A and B in the direction from A to B. Point B is at a pressure of P and volume of 4L. There is an isobaric line from Point B at a higher volume to Point C where there is a lower volume. Point C is at a pressure of P(same as point B due to isobaric process) and a volume between 1L and 4L. There is an adiabatic curve from Point C to Point A.

    b) I began by finding the pressure at Point B by using (P1V1) / T1 = (P2V2) / T2. From this i found that the pressure at point B was 4atm, making this the pressure at Point C as well. From there ive tried to use P1V1[itex]\gamma[/itex] = P2V2[itex]\gamma[/itex] in order to try and find the volume at point C, but havent been able to find much. My understanding of thermodynamics is still shaky at this point so any help would be welcome.
     
  2. jcsd
  3. Sep 16, 2012 #2

    Simon Bridge

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    You have the relation for B→C and another for C→A and you know the complete states for A and B. Therefore you have two unknowns and two simultaneous equations. What's the problem?
     
  4. Sep 16, 2012 #3
    When i use the relation B→C and solve for one of the unknown i find that Tc = 150Vc. I cant seem to perform the math correctly for the process from C→A using P1V1γ = P2V2γ. Thanks for the reply as well Simon.
     
    Last edited: Sep 16, 2012
  5. Sep 16, 2012 #4

    Simon Bridge

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    Check:

    State A: ##P_A##, ##V_A##, ##T_A## ... all known quantities.

    State B: ##P_B##, ##V_B##, ##T_B=T_A## ... all known quantities

    State C: ##P_C=P_B## ... known: need V and T.

    From ##B \rightarrow C## you get ##V_C=\frac{V_B}{T_A}V_C## ... (1)

    From ##C\rightarrow A## you get ##P_BV_C^\gamma = P_AV_A^\gamma## ... (2)

    in equation 1 the unknowns are ##V_C## and ##T_C##.
    in equation 2 the only unknown is ##V_C##

    make ##V_C## the subject in equation (2) and evaluate. Then substitute into (1) to find ##T_C##
     
  6. Sep 16, 2012 #5
    Okay i finally figured out why i couldn't solve for Vc. Just a simple calculation error. Thanks again for all the help
     
  7. Sep 16, 2012 #6

    Simon Bridge

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    No worries.
     
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