# Enropy irreversible Process

1. Apr 1, 2013

### Abigale

Hi,
I regard a irreversible cycle process.

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

12: isothermal compression (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

2. Apr 1, 2013

### 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 $\delta Q\leq0$.

3. Apr 18, 2013

### Rap

/Th

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.

Last edited: Apr 18, 2013