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A reversible process can be represented on a $T-S$ plane, and the area under the curve is the heat exchanged by the system.

On $P-V$ plane a irreversible process is conventionally represented with a dashed line, since the curve cannot be drawn, as intermediate states do not have defined thermodynamical variables.

Does the same hold for $T-S$ plane? Is a process represented (conventionally) with dashed line there? Are the reasons of the impossibility of representation the same of $P-V$ plane?

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  • $\begingroup$ The state of a simple compressible system, can be defined by two independent, intensive properties. If $T$ and $S$ are known for a simple compressible system, then its state will be determined. This means if you know $T$ and $S$ at each moment, you can find $P$ and $v$ at that moment and this means that you can draw $P-v$ diagram and vise versa. $\endgroup$
    – lucas
    Commented Jun 26, 2016 at 21:04
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    $\begingroup$ In an irreversible process, the temperature and pressure are not uniform throughout the system. So what value of the temperature or pressure do you use on the plot.? $\endgroup$
    – Chet Miller
    Commented Jun 26, 2016 at 21:53
  • $\begingroup$ @ChesterMiller. This might be a matter of different disciplines using different terminology, but I use the term quasi-static (or quasi-equilibrium) for a process in which a system goes through a sequence of equilibrium states. For such a process, $p$, $V$, etc. are all well-defined and hence the process can be plotted in the $p$-$V$ diagram. Such a process does not have to be reversible, however, because it could for example be a slow expansion of a gas where there is friction between the piston and the cylinder, which would not be reversible. $\endgroup$
    – march
    Commented Jun 27, 2016 at 4:00
  • $\begingroup$ (@ChesterMiller. However, your comment is how I would have answered had the OP used the term quasi-static, so I'm pretty sure we're on the same page here.) $\endgroup$
    – march
    Commented Jun 27, 2016 at 4:02
  • $\begingroup$ @March. Thanks. Friction is a very interesting situation. If we include the piston and the interface as part of our system, then there is definitely dissipation of mechanical energy (irreversibility) occurring within our system. It is somewhat analogous to viscous dissipation, but viscous dissipation increases as the square of the velocity gradients. On the other hand, frictional dissipation occurs even if the piston movement is very slow. Still, you are correct in saying that, in the case of friction, the p-V diagram can still be plotted even though the process is irreversible. $\endgroup$
    – Chet Miller
    Commented Jun 27, 2016 at 12:36

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It is for the same reason that you need to use a dashed line for irreversible process which is higher than reversible process. This is because we know it is higher but don't know how high. If the dashed line is lower, we will conclude this is not feasible. This is the first thing what we want to know from the diagram.

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It is even worse for the $T$—$S$ diagram. An irreversible process goes through non-equilibrium states. When the volume $V$ and entropy $S$ are defined for all states of the gas — equilibrium and non-equilibrium, the pressure $P$ and temperature $T$ are not. And if the pressure can be additionally defined in the case of a non-equilibrium state — for example, as some average pressure, then temperature is defined only for an equilibrium state in principle.

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