Physics Asked on February 11, 2021
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?
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.
Answered by user115350 on February 11, 2021
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.
Answered by warlock on February 11, 2021
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