I think there is a weird logical loop I keep running into with the second law of thermodynamics. One sequence is

Clausius Equality: $\oint \frac{\delta Q}{T}=0$ for reversible processes. This means there exists some state function $S$ such that $\Delta S(P) = \int_P \frac{\delta Q}{T} $ for reversible $P$. From this, you can define it up to a constant over the whole space.

Then ... Second Law of Thermodynamics: $\Delta S \geq 0 $ for all physically realizable processes. However, I read that the Clausius equality is actually a result of the second law of thermodynamics so you can't actually work in this order.

Alternatively, you could say that the second law simultaneously proposes the existence of $S$ as a state function, its relation to heat and temperature for reversible processes, and its natural increase over time. However, this feels like a little too much to me and makes me think the Clausius equality is basically trivial (which is fine).

I just want to know the correct logical progression for defining entropy and making statements about it. Is it defined through the second law or does the second law merely make a statement about something that was created elsewhere? If so, how do we create it?

Edit: I was thinking about this and thought maybe a potentially nice order is starting with the definition that $\Delta S = \int \frac{\delta Q}{T}$ letting it be a process function for now. Then declaring the second law, using it to prove Clausius's equality, and, from there, proving its a state function.

My concern with all of this also however is that the proper definition of thermodynamic temperature also comes from entropy so, in that case, I have no idea how this progression goes.

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    $\begingroup$ There are many equivalent statements of the second law. The increase in entropy is convenient for many calculations in physics in general, but in formally constructing thermodynamics it is normal to start with Kelvin or Clausius statements, neither of which mention entropy explicitly. $\endgroup$ Nov 11, 2018 at 1:31
  • $\begingroup$ Aren't you asking for the details about the exact historical sequence of the development of the 2nd law of thermodynamics? $\endgroup$ Nov 11, 2018 at 2:59
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    $\begingroup$ No, not the historical development. If you are building up thermo from postulates, in which order would you go? $\endgroup$ Nov 11, 2018 at 9:38

3 Answers 3


OK. It would go something like this.

  1. For any two thermodynamic equilibrium states of a closed system, there are an infinite number of process paths that can be used to take the system from its initial thermodynamic equilibrium state to its final thermodynamic equilibrium state. A subset of this infinite number of process paths is the (also) infinite number of process paths that we currently refer to as reversible paths.

  2. Back in the day, geniuses like Clausius studied the various reversible- and irreversible process paths for many different systems and empirically noted something very strange: If they evaluated the integral of $dq/T_B$ (where dq is the differential heat transferred to the system at its boundary with its surroundings over time interval dt and $T_B$ is the temperature at the boundary where the heat transfer is occurring) for all the various paths, the value of this integral is not totally arbitrary. There is a maximum value of the integral that is never exceeded. This maximum value is found to be obtained for all the reversible process paths between the two end states. All the irreversible paths between the two end states give lower values for the integral.

  3. Since the maximum value of the integral depends only on the two thermodynamic equilibrium end states, it must be a function only of state. They called this function the entropy S. And, in terms of the entropy S, the mathematical embodiment of their empirical observations is the Clausius inequality: $$\Delta S\geq\int{\frac{dq}{T_B}}$$This constitutes the mathematical statement of the 2nd Law of Thermodynamics.

  • $\begingroup$ Yes, I understand such a statement is true from an empirical standpoint. My question is how to form these statements through postulates i.e. assuming temperature is defined as something that is equal during thermal equilibrium and heat is the difference in internal energy and work, what is the minimum number of postulates/definitions to define S, show that it is 0 for reversible, >0 for irreversible, and =dQ/T for reversible. $\endgroup$ Nov 11, 2018 at 13:33
  • $\begingroup$ Sorry. I did my best to lay out this Calculus of Variations development in a logical way. Apparently this approach doesn't resonate with you. Maybe some other member can express it differently. $\endgroup$ Nov 11, 2018 at 13:40

I just want to know the correct logical progression for defining entropy and making statements about it.

There isn't such a thing as "the correct way to introduce thermodynamic entropy". There is the historical way, and there are other, later accounts, some more mathematical (integrating factors of differential forms, see Caratheodory), some more pedagogical (some mix of historical path and modern hindsight, see physical chemistry textbooks).

Is it defined through the second law or does the second law merely make a statement about something that was created elsewhere? If so, how do we create it?

The second law has different, not entirely equivalent formulations. Some of them rely on the concept of entropy, some of them do not. However, neither formulation by itself can tell you what entropy is. You need to have a separate definition for it in any case.

The historical way ( and the easiest, in my opinion) is to use 2nd law as formulated by Clausius (it is impossible for heat to go from hotter to colder spontaneously, no mention of entropy) to derive Clausius' inequality:

$$ \oint \frac{dQ}{T_r} \leq 0 $$

where $T_r$ is temperature of reservoirs used in the heat transfers and the circle in the integral means that the integration is to be done for a process that starts and ends in the same thermodynamic state. Clausius showed that for any reversible process, the integral is zero so the equality takes place. Also in this case, $T$ could be put in the integral instead of $T_r$.

So far, no mention of entropy. But we can use this result to define function of state, called entropy, as

$$ S(X) = S(X_0) + \int_{X_0}^{X}\frac{dQ}{T}. $$

So, 2nd law, through Clausius equality, definitely motivated introduction of entropy and is in fact necessary for this definition of entropy to make sense.

  • $\begingroup$ So does one use the second law to prove Clausius equality? $\endgroup$ Nov 12, 2018 at 3:32
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    $\begingroup$ In general, yes, because without it there is no justification for why inverse of ideal gas temperature should be integrating factor for every general thermodynamic system. For simple systems with only two degrees of freedom, such as single species gas in a container with variable volume, 2nd law is not necessary, because in 2D space every differential form has integrating factor, and thus entropy exists irrespective of 2nd law of thermodynamics. $\endgroup$ Nov 12, 2018 at 18:25

Traditionally, entropy is first defined thermodynamically as $\Delta S=\int \frac{dQ_{rev}}{T}$ up to a constant, which is possible via Clausius' equality. However, the proof for Clausius' inequality relies on Kelvin's statement of the second law of thermodynamics, which does not presuppose the existence of entropy.

Specifically, let $E$ be any cycle, reversible or irreversible, and let $E$ draw heat $đ Q_i$ at temperatures $T_i.$ By some process, $E$ can then produce work $\Delta W$ per cycle, with $$\Delta W=\sum_{\text{i, cycle}}đ Q_i$$ by energy conservation. Now imagine that the heats $đQ_i$ are provided each by Carnot engines $C_i$ depositing $đQ_i$ at each reservoir $T_i$ and drawing heat from a shared reservoir $T,$ producing work $đW_i$ in the process. One has, by the Carnot efficiency condition, \begin{align*} \frac{đQ_i+đW_i}{T}&=\frac{đQ_i}{T_i} \\ đW_i&=đQ_i \left(\frac{T}{T_i}-1\right). \end{align*} Since $E$ doesn't deposit heat to a cold reservoir, we appear to have violated Kelvin's statement of the second law of thermodynamics - unless the net work generated per cycle is nonpositive: \begin{align*} \Delta W+\sum_{i} đW_i \leq 0 \\ \sum_i đQ_i + đQ_i \left(\frac{T}{T_i}-1\right) \leq 0 \\ T\sum_i \frac{đQ_i}{T_i} \leq 0, \end{align*} and dividing through $T$ recovers the Clausius inequality. For the equality case, just note that if $E$ is reversible, all the heat transfers may be reversed, by definition. Then the direction of the inequality flips, requiring $\sum_i \frac{đQ_i}{T_i}=0.$ It's worth noting that you could have circumvented all of this by first defining entropy statistically using $S=k_B \log \Omega.$ The beauty of this formula is that it immediately reveals that entropy is a state function, and that it stays constant in reversible adiabatic processes (the phase volume doesn't change).


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