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From Zemansky and Dittman's Heat and Thermodynamics (6th ed.), p. 214:"A throttling process is obviously an irreversible one, since the gas passes through nonequilibrium states on its way from the initial equilibrium state to its final equilibrium state. These nonequilibrium states cannot be described by thermodynamic coordinates..."

The authors then compute the work done: $W=-\int_0^{V_f} P_f dV-\int_{V_i}^0P_idV$. In what sense can the nonequilibrium states not be described by thermodynamic coordinates ? We fix the pressure to be $P_i$ on one side of the porous plug and $P_f$ on the other, so the thermodynamic coordinates are known, and the work done can be evaluated.

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  • $\begingroup$ Can you provide more context? A throttling process does not normally involve work. It is one where the initial and final enthalpy is the same. $\endgroup$
    – Bob D
    Jan 20, 2021 at 20:38
  • $\begingroup$ Initial and final enthalpies are shown to be equal by computing the work done and adding to the internal energy, via the first law of thermodynamics $\endgroup$
    – Frost
    Jan 21, 2021 at 0:18

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As the passage you cite states, the initial (i.e. pre-throttle) and final (i.e. post-throttle) states are equilibrium states. Therefore, you have no difficulty in describing them in equilibrium thermodynamics language, for example by the pressures $P_{i}$ and $P_f$. They are true states.

The difference between nonequilibrium and equilibrium isn't necessarily that state variables cannot be used (for example, you could talk about a variable like pressure in a local sense, $P(x)$ with $x$ along the throttle). It's rather that the name state variable is a misnomer, because they do not describe a thermodynamic state.

The equilibrium state is - loosely - defined to be the state that you just end up in if you fix certain state variables long enough and don't fondle with the system otherwise. By that, equilibrium is defined as a state after an infinite amount of time passes, and with that, it cannot change over time and one wouldn't even notice a reversal in time. Non-equilibrium states, on the other hand, are states the system passes through before reaching equilibrium. If you will, the two regimes are separated by a system-dependent equilibration time scale $\tau_{eq}$, and equilibrium happens at $t \gg \tau_{eq}$, while non-equilibrium is observable at $t < \tau_{eq}$. These states depend on time, some initial conditions and the properties of the environment (while the only role of the environment in an equilibrium system is that it demands certain values for state variables). This additional dependence makes it much more difficult (and often enough impossible) to define clear-cut relations between different "state" variables.

As an example, think about a hot cup of tea in a large environment of fixed lower temeprature. The true equilibrium state is the one where the cup of tea has the environment's temperature. Only in that state will you have well-defined relations between temperature and density that do not depend on the way the cup was heated up or whether or not you stirred it during cooling. While the cooling takes place, the relations between temperature, density and pressure may depend on the history of the cooling process and other environmental factors (which would for example determine the evaporation rate or the heat exchange etc.).

It is true that in this cup-of-tea example, you may, depending on the circumstances, be able to assume that the heat transport within the water that is required for the whole thing to reach equilibrium is faster than the rate at which heat is exchanged with the environment. Then you can basically say that the water passes through different equilibrium states, because the water sample at the current temperature doesn't differ much from water that equilibrated to exactly this temperature. But this is only valid under these assumption, which can be too strcit and wouldn't apply if you stirred the water or had a much less humid environment etc.

The throttle works much the same way. Within the throttle, your fluid will suffer turbulences and energy loss. These can't be described by equilibrium variables, and even local equilibrium approximations become problematic when there is a lot of eddies going through the system that transport energies to smaller and smaller time scales.

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    $\begingroup$ Thank you, that clarifies the matter $\endgroup$
    – Frost
    Jan 21, 2021 at 0:34

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