# A thermodynamic transformation that can be represented by a continuous quasistatic path in its state space may still be irreversible. Why?

A thermodynamic transformation that has a path (in its state space) that lies on the surface of its equation of state (e.g., $PV=NkT$) is always reversible (right?). However, if the path is a continuous quasistatic path in state space but not on the surface of equation of state, it is considered irreversible.

Why is this so? In this case the gas still has a uniform thermodynamic variables (because all the intermediate states are points in the state space). Why can it not be reversed exactly along the same path?

• "However, if the path is a continuous quasistatic path in state space but not on the surface of equation of state, it is considered irreversible." How would you make the system to go into state where $P,V,T$ are defined but do not obey the equation of state? – Ján Lalinský Feb 2 '14 at 23:16

Let's look at your first statement:

A thermodynamic transformation that has a path (in its state space) that lies on the surface of its equation of state (e.g., $PV=NkT$) is always reversible

I don't think this is right, but there may be some implicit qualifiers in your statement that I'm missing.

Here's an example to illustrate why. Consider a system consisting of a thermally insulated container with a partition in the middle dividing it into compartments $A$ and $B$. Suppose that we fill each side of the container with an equal number of molecules of a certain monatomic ideal gas. Let the temperature of the gas in compartment $A$ be $T_A$ and the temperature of the gas in compartment $B$ be $T_B$. Suppose further that the partition is designed to very very slowly allow heat transfer between the compartments.

If $T_B>T_A$, then heat will irreversibly transfer from compartment $B$ to compartment $A$ until the gases come into thermal equilibrium. During this whole process, each compartment has a well-defined value for each of its state variables since the whole process was slow ("infinitely" slow in an idealized setup). However, the process was not reversible because heat spontaneously flowed from a hotter body to a colder body. You can also show that the sum of the entropies of the subsystems $A$ and $B$ increased during the process to convince yourself of this.

So we have found a process for which the thermodynamic evolution of each subsystem can be described by a continuous curve in its state space, but the process is irreversible.

Addendum - 2017-07-02 After more thought, although I originally conceded to Valter Moretti in the comments that the composite system $A + B$ should not be considered in equilibrium during the process because each subsystem has a different temperature, I no longer believe that this means the process cannot be considered a quasistatic process for the composite system. I currently believe that as long as the process is sufficiently slow for the energy $U$, volume $V$, and particle number $N$ to be well-defined for each subsystem $A$ and $B$ during the process, then it can be considered quasistatic for the composite system. If we consider very slow heat transfer ("infinitely slow" in order to approach the ideal), then at all points in time, the each subsystem will be very close to equilibrium in such a way that the physical process is well-described by a continuous curve in the composite thermodynamic state space with coordinates $(U_A, V_A, N_A, U_B, V_B, N_B)$ lying on the surface satisfying the equations of state, but because it will not lie on the intersection of this surface with the hyperplane of constant total entropy, the process is irreversible.
• @pppqqq Yes that's rather vague I agree. I think I originally had in mind some sort of a membrane that can be designed to allow for the heat to flow between two gases as slowly as one wishes so that despite the finite temperature difference, the process is quasi-static. After any finite amount of time, there would be a net flux of heat, but the process would be sufficiently slow that one could still use $\delta Q = T dS$ to determine the entropy change the each subsystem on either side of the membrane. – joshphysics Feb 2 '14 at 20:23
• @V.Moretti Good points. I agree with everything you're saying. I think I should be more careful about differentiating between systems and subsystems. In particular, I suppose that my claim really is that there exists a system $\Sigma$ and an irreversible process $P$ on $\Sigma$ during which $\Sigma$ is out of equilibrium, yet for which certain of it's subsystems $\Sigma_1$ and $\Sigma_2$ undergo quasistatic processes $P_1$ and $P_2$. Perhaps this is not actually an answer to the original question; I'm not entirely sure now that I look back at it. – joshphysics Feb 2 '14 at 20:35