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Simple entropy calculation

  1. Jul 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A Carnot engine operates on 1 Kg of CH4, which we shall consider an ideal gas. Take [tex]\gamma[/tex]=1.35. The ratio of max to min volume is 4 and the cycle efficency is 25%. Find the entropy increase during the isothermal expansion.
    [Classical and Statistical Thermodynamics, Ashley H. Carter, pg.104]

    2. Relevant equations

    dU = TdS-PdV

    3. The attempt at a solution

    Ideal gas, isothermal expansion,[tex]\Rightarrow[/tex] dU = 0

    [tex]\Rightarrow[/tex] dS=[tex]\frac{P}{T}[/tex]dV=nR[tex]\frac{dV}{V}[/tex]

    [tex]\Rightarrow[/tex] [tex]\Delta[/tex]S=R(62.5)ln(4)=720J/K


    The correct answer is 294J/K

    I know my problem has to be with the number of moles, probably accounting for the efficiency. working backwards I see there should be 25.5 moles of methane for n.
    How do I account for the efficiency and get this answer?
     
  2. jcsd
  3. Jul 24, 2009 #2

    Mapes

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    In a Carnot cycle, the efficiency is connected to the temperatures of various processes. Does this help? Have you sketched out the cycle?

    (Also, 1 kg methane is not 25.5 moles.)
     
  4. Jul 24, 2009 #3
    [tex]\eta=1-\frac{T_{1}}{T_{2}}[/tex]

    Is that what you mean? How does that help for an isotherm? I'm not sure what information to use regarding the Carnot cycle itself considering the question only asks about the isothermal expansion.

    For just the expansion with the ratio of volumes given my answer should be correct, there is some relevant information I am not seeing.


    I know, its 62.5 moles. I got 25.5 from solving
    [tex] nRln(\frac{V2}{V1})=294J/K[/tex]
    for n.
     
  5. Jul 24, 2009 #4

    Mapes

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    Sketch the cycle; are the volumes at the beginning and end of the isothermal process the maximum and minimum volumes in the cycle?
     
  6. Jul 27, 2009 #5
    Thank you Mapes.

    Of course those are not the max/min volumes. I was able to obtain the correct volume ratio buy utilizing [tex]PV^{\gamma}=constant[/tex] for an an adiabat,[tex] \eta=1-\frac{T_1}{t_2}[/tex], and the given information.
     
  7. Mar 7, 2011 #6
    I'm doing this question right now, and I am wondering how to get 294 J/K as an answer. I don't understand why the final volume doesn't relate to the max volume, and how to use PV^gamma to get the correct ratio.

    Thanks.
     
  8. Mar 13, 2011 #7
    Seem you guys forget that the initial volumn is not at the isothermal expansion, you have to trace it back to the start of adiabatic expansion then the isothermal expansion. (draw the diagram and you will see)

    Ratio Max/Min volume is the volume at the tip of adiabatic expansion.

    You'll need to use both PVg = PV/T = constant

    Good brush up exercise though.
     
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