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According to the inequality of Clausius, $$S\ge \dfrac {q_{rev}}T$$ , where $S$ is the entropy of the system, $q$ the heat absorbed by the system during a reversible process and $T$ is the temperature in kelvin.

Now, how can we calculate the entropy change during an irreversible process? I am confused about this since a long time.

This pdf made it clear to me that we cannot define the entropy change for an irreversible process as $\dfrac {q_{rev}}T$ because the entropy of a system consists of "produced entropy" (which is zero in a reversible process but not in an irreversible process) and entropy due to exchange.

So I searched more about the actual calculation, becuase I could easily calculate the change for an isothermal process but not for an adiabatic process.

I found this NASA article which clearly defines entropy as simply heat change (and not reversible heat change) upon temperature i.e. $$\Delta S = \dfrac{\Delta q}{T}$$

I was quite satisfied with this definition until I ran across this on MIT's website which made me confused again. According to MP $6..9$ on the site, for an irreversible process "we need to define a reversible process between the two states in order to calculate the entropy". I don't get what this means. How can we define a reversible process between two states connected by an irreversible process? Wouldn't it lead to erroneous results?

Tl;dr: Is entropy $\dfrac{q_{rev}}{T}$ or is it $\dfrac{q}{T}$? How can we calculate the entropy change for an adiabatic irreversible expansion?

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THE RECIPE

  1. Apply the First Law of Thermodynamics to the irreversible process to determine the final thermodynamic equilibrium state of the system

  2. Totally forget about the actual irreversible process (entirely), and focus instead exclusively on the initial and final thermodynamic equilibrium states. This is the most important step.

  3. Devise a reversible path between the same two thermodynamic equilibrium states (end points). This reversible path does not have to bear any resemblance whatsoever to the actual irreversible process path. For example, even if the actual irreversible process is adiabatic, the reversible path you devise does not have to be adiabatic. You can even separate various parts of the system from one another, and subject each of them to a different reversible path, as long as they all end up in their correct final states. Plus, there are an infinite number of reversible process paths that can take you from the initial state to the final state, and they will all give exactly the same value for the change in entropy. So try to devise a path that is simple to work with (i.e., for which it is easy to apply step 4).

  4. For the selected reversible path, evaluate the integral of dq/T from the initial state to the final state, where dq is the incremental amount of heat added to the system along the sequence of changes comprising the reversible path. This will be your change of entropy S. That is, $$ΔS=∫\frac{dq_{rev}}{T}$$ where the subscript rev refers to the reversible path.

Reference https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

I also wanted to mention a little-emphasized (but important) fact regarding the Clausius inequality: In applying the Clausius inequality to an irreversible process, it is important to use the temperature at the interface $T_I$ with the surroundings (where the heat transfer dq is occurring) as the T in the integral. That is $$\Delta S\geq\int{\frac{dq}{T_I}}$$where the temperature at the interface might be that of a constant temperature reservoir (for example), and, for an irreversible path, dq is the actual heat flow during the irreversible process.

ADDENDUM:

The Clausius Inequality represents a mathematical statement of the 2nd Law of thermodynamics that, as Clausius showed, is fully equivalent to various "word statements" of the 2nd Law (such as the Kelvin-Planck statement and the Clausius statement). The Clausius Inequality says that, if you look at the infinite number of possible process paths between the initial and final thermodynamic equilibrium states of a closed system (i.e., no mass entering or leaving, but heat flow and work permitted), each processes path is characterized by a different "time history" of the heat flows dq(t) and temperatures $T_I(t)$ at the boundary between the system and its surroundings.

Clausius discovered something astonishing: He found that, for any given pair of end states, if you calculate the value of the integral of $\frac{dq(t)}{T_I(t)}$ over the infinite set of possible process paths, the range of values you obtain for the integral is not infinite. There is an upper bound to the value calculated for the integral over all the possible process paths. Since this maximum value depends only on the two ends states, it must be a function of state. Clausius called this function the entropy S.

He then made another astonishing discovery. He found that, for the infinite set of process paths between the two equilibrium ends states, there is a closed subset of these paths all of whose members give exactly the maximum value for the integral of $\frac{dq(t)}{T_I(t)}$. This subset of process paths is what we describe today as reversible process paths. So, all reversible process paths give the maximum value for the integral (i.e., $\Delta S$), and all irreversible process paths between the same initial-and final thermodynamics equilibrium states give less than the maximum value.

This is the motivation for doing step 3. It is the only way we know of for determining the change in entropy between two thermodynamic equilibrium states of a closed system.

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  • $\begingroup$ Yep. And a perfectly good reason to say "sod this for a game of soldiers" and look instead for a statistical model where ever possible. $\endgroup$ – dmckee Dec 29 '17 at 3:34
  • $\begingroup$ @dmckee Are you saying that this straightforward procedure would be too difficult for you personally to apply? $\endgroup$ – Chet Miller Dec 29 '17 at 3:47
  • $\begingroup$ Please explain the reason for doing the 3rd step. $\endgroup$ – Abcd Dec 29 '17 at 9:22
  • $\begingroup$ For an explanation of the reason, see the ADDENUM in my answer. $\endgroup$ – Chet Miller Dec 29 '17 at 13:53
  • $\begingroup$ Is there a systematic way to model $\Delta S - \int dq/T_i$, the positive part that irreversible processes "misses"? Is the von neumann/gibbs/... entropy always the left hand side ($\Delta S$) also for irreversible processes? $\endgroup$ – Emil Dec 29 '17 at 14:05
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The pdf linked in the question has a good explanation, and it is correct that $$\operatorname{d}S_{\mathrm{transfer}} = \frac{\operatorname{\text{đ}}Q_{\mathrm{rev}}}{T},$$ where I have included the much needed differential symbols (exact on the left, inexact on the right) because this formula only applies when the heat and entropy transfers are small. The NASA article linked was just leaving the reversibility to be understood because it's based on an already cluttered slide.

The key thing to understand is that in a reversible process the entropy of the universe is constant. Thus you're describing the flow of existing entropy from one place to another. When the process is irreversible the entropy of the entire universe increases, and the only way to figure out by how much is to isolate the part of the universe that changed (the entropy of the unchanged part, clearly, has to be constant), find a reversible process that takes it back to its original state, and then keep track of how much entropy you have to push out into the rest of the universe during that process.

Clausius's inequality is a way to modify the equality that works for only reversible entropy flows to cover the case when the process isn't reversible and thus generates new entropy in the system in question.

At least, that's what you have to do if you're limited to working with the thermodynamic definition of entropy. It can be simpler to use the Gibbs entropy or Boltzmann's $$S=k\ln W,$$ though there can be sticky issues there in clearly defining "probability" and what is "allowed".

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