What is the relationship between entropy and the irreversibility of a process?

[Abigale]

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 Δs_{\text{gas}} > 0 and Δs_{\text{heat reservoir}} >0.

So the Δs_{\text{gas+reservoir}} 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

 

[kevinferreira]

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 \Delta S_{heat~reservoir}>0? 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 \delta Q\leq0.

 

[Rap]

/Th

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 \Delta S_{heat~reservoir}>0? 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 \delta Q\leq0.

If the reservoir is at a higher temperature (T_{hot}) than the system being heated (T_{cold}) then entropy \delta Q/T_{hot} is lost by the reservoir $$\Delta S_{heat~reservoir}=-\delta Q/T_{hot}$$ and entropy \delta Q/T_{cold} 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.