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Enropy irreversible Process

  1. Apr 1, 2013 #1
    Hi,
    I regard a irreversible cycle process.

    It is a cylinder filled with gas, which underlies the following processes:

    12: isothermal compression (reversible)
    23: adiabatic expansion (reversible)
    31: isochor heating (irreversible) [This happens by connecting the cylinder with a heat reservoir]

    I have callculated for the irreversible process 13, that [itex]Δs_{\text{gas}} > 0[/itex] and [itex]Δs_{\text{heat reservoir}} >0[/itex].

    So the [itex]Δs_{\text{gas+reservoir}}[/itex] for the full cycle should be >0.

    If i would have a completely reversible process, would the enropy change of the working gas be zero?
    (After the whole cycle?)

    THX
    Abby
     
  2. jcsd
  3. Apr 1, 2013 #2
    In a completely reversible cycle the total entropy change is zero, yes. But in this case you cannot consider some heat exchange with the heat reservoir (it is irreversible).

    Also, I have a question. How come [itex] \Delta S_{heat~reservoir}>0[/itex]? If the reservoir heats the gas, then by $$dS=\frac{\delta Q}{T}$$ this should give a negative or zero value (since we're talking of a reservoir, the heat exchange may be so small compared to the reservoir's capacity that one may just consider it zero). The reservoir heats the gas, thus [itex]\delta Q\leq0[/itex].
     
  4. Apr 18, 2013 #3

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    /Th

    If the reservoir is at a higher temperature [itex](T_{hot})[/itex] than the system being heated [itex](T_{cold})[/itex] then entropy [itex]\delta Q/T_{hot}[/itex] is lost by the reservoir $$\Delta S_{heat~reservoir}=-\delta Q/T_{hot}$$ and entropy [itex]\delta Q/T_{cold}[/itex] is gained by the system. $$\Delta S_{system}=\delta Q/T_{cold}$$ The total entropy is $$\Delta S_{total}=\delta Q/T_{cold}-\delta Q/T_{hot}$$ which you can see is positive. Since entropy is lost by the reservoir, the change in entropy of the reservoir is negative, not positive.
     
    Last edited: Apr 18, 2013
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