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Reversible cyclic process

  1. Oct 8, 2006 #1
    An ideal gas undergoes a reversible, cycli process. First it expands isothermally from state A to state B. It is then compressed adiabatically to state C. Finally, it is cooled at constant volume to its original state, A.

    I have to calculate the change in entropy of the gas in each one of the three processes and show that there is no net change in the cyclic process.

    O.K. From A to B Delta S is nRln(V2/V1) since the process from A to B is isothermal. Delta S is 0 from B to C since that process occurs adiabatically.
    But I am having trouble with the Delta S for the process from C to A. So far I have Delta S is Cv ln(T1/T2) But I'm having trouble converting this to something similar to the Delta S for A to B

    Could someone please tell me if I'm on the right track with this problem, and possibly give me a hint
     
  2. jcsd
  3. Oct 8, 2006 #2

    Andrew Mason

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    You are right on the first two parts:

    [tex]\Delta S_{CA} = \int_C^A ds = \int_C^A dQ/T[/tex]

    Since: [itex]dQ = dU + PdV[/itex] and dV = 0, [itex]dQ = dU = nC_vdT[/itex]

    [tex]\Delta S_{CA} = \int_C^A nC_v\frac{dT}{T} = nC_v\ln\frac{T_A}{T_C}[/tex]

    So the total change in entropy is:

    [tex]\Delta S_{AA} = nR\ln\frac{V_B}{V_A} + nC_v\ln\frac{T_A}{T_C}[/tex]

    So the question is whether:

    [tex]nR\ln\frac{V_B}{V_A} = nC_v\ln\frac{T_C}{T_A}[/tex]

    AM
     
  4. Oct 9, 2006 #3
    I solved it.

    Thanks :smile:
     
  5. Oct 9, 2006 #4

    Andrew Mason

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    I trust that you used the fact that the adiabatic condition applies to the compression phase. So for the B-C phase:

    [tex]\left(\frac{V_C}{V_B}\right)^{1-\gamma} = \frac{T_C}{T_B}[/tex]



    AM
     
    Last edited: Oct 9, 2006
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