4
$\begingroup$

I have just begun reading Huang's Statistical Mechanics textbook and am confused by his definition of a quasi-static process. In his definition, he states that a quasi-static process is one in which "the external condition changes so slowly that at any moment the system is approximately in equilibrium". In my mind, this brings to mind that such a process can be represented by a series of points on the state space, representing equilibrium states, thus defining a curve. I would imagine that this curve be continuous, else there may be a non-equilibrium state involved in the discontinuity.

However, there seems to be some conflict of this interpretation with his definition of a reversible process as he states that only those can be represented as a continuous path on the state space surface.

Is my interpretation of his definition of quasi-static too restrictive? If only reversible processes can be represented as a continuous curve on the state space surface, how, if they can be, are non-reversible quasi-static processes represented on the state space?

$\endgroup$
2

2 Answers 2

6
$\begingroup$

The problem with irreversible transformations, even if quasi-static, is that one needs to add some non-equilibrium quantities to the set of variables describing the system's state. In this sense, we cannot represent an irreversible process by using only state variables.

Indeed, if there is some internal entropy production, this gives an additional contribution to the change of entropy related to the heat fluxes. From a more mathematical point of view, entropy production would make a function of the thermodynamic state a multi-valued object. In this sense, it cannot be represented as a unique curve on the state space.

$\endgroup$
3
$\begingroup$

Quasistatic processes encompass reversible processes. The inclusion is strict, because there are some irreversible quasistatic transformations.

A typical example of a quasi static process would be a system being slowly heated up by a thermostat. Say your system is an ideal solid whose state depends only on its temperature $T$. For simplicity, assume that the internal energy is: $$ U = CT $$ with $C$ the heat capacity and entropy is therefore: $$ S = C\ln T $$

Say it starts at temperature $T_0$ and the thermostat is at temperature $T_e$. The process is merely temperature equilibration. The transformation is quasistatic if the transformation is slow enough for $T$ to be well defined throughout the transformation. However, no matter whether it is irreversible, the transformation is fundamentally irreversible. Quantitatively, you have a positive creation of entropy: $$ \begin{align} S_c &= \Delta S-\frac{Q}{T_e} \\ &= C\left[\ln\left(\frac{T_e}{T_0}\right)-1+\frac{T_0}{T_e}\right] \\ &\geq 0 \end{align} $$ (equal to zero iff $T_e=T_0$ by strict convexity of the logarithm). Even if you break down the process in smaller increments so that the sub-processes are almost reversible, the total process is still irreversible (the irreversibility "adds up" to the same amount).

Hope this helps.

Answer to comments

The usual argument of second order in created entropy does not apply in my example. Indeed, I am assuming that the thermostat's temperature is constant, whereas in the usual argument, the thermostat's temperature is slowly raised from the initial temperature of the system to the final desired temperature.

Mathematically, $\Delta S$ is the same since it is a state function. Similarly, $Q = \Delta U$ since there is no work and I can use the first principle. Thus, under these assumptions, $S_c$ is actually path independent. This is why dividing the transformation into smaller steps will not change the final total created energy. If you do the calculation explicitly, the increments are actually of first order, not second order, so everything is consistent.

$\endgroup$
6
  • $\begingroup$ see, some more calculational details physics.stackexchange.com/questions/317690/… $\endgroup$
    – hyportnex
    Jun 29, 2023 at 10:17
  • $\begingroup$ I'm not sure your conclusion is correct ? It is easy to prove that the entropy created is second order with respect to the temperature difference. As a result, the entropy created does indeed tend towards 0 within the limits of a series of infinitesimal transformations. $\endgroup$ Jun 29, 2023 at 11:44
  • $\begingroup$ @VincentFraticelli why would LPZ's conclusion for pure thermal exchange be incorrect if it is indeed the case, as you have stated, that "it is easy to prove"? I for one have never seen a real proof for the general quasi-static case that does not assume the conclusion. It is not enough to prove that entropy production is 2nd order relative to temperature difference but that it is second order relative to its production by all other quasi-static changes. $\endgroup$
    – hyportnex
    Jun 29, 2023 at 12:10
  • $\begingroup$ @LPZ Thanks for the additional comment ! With these assumptions, we would have a temperature discontinuity on the surface? In a more realistic model, we would have a strong gradient near the surface and the transformation would not be quasistatic? The entropy creation would be localized near the surface. $\endgroup$ Jun 29, 2023 at 14:20
  • $\begingroup$ Yes, this isn't unusual for simplified model. The situation could be modelled as a solid of infinite thermal conductivity (for the temperature to be homogeneous so that the evolution is quasistatic) and Newton's law of cooling at the boundary (Robin boundary condition). Indeed, the creation of entropy is at the boundary as it should. $\endgroup$
    – LPZ
    Jun 29, 2023 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.