Recently I have learnt Second Law of Thermodynamics that entropy for an isolated system is nondecreasing (such statement is true with a very high probability - I have heard that it can also decrease but chance of that is extremely small).

One of the most common exercises is to calculate entropy increase for an isolated system - room and ice cube, from some state A to state B where both bodies are in thermal equilibrium. What one can find is that even though total entropy of an isolated system increases (which is consistent with Second Law of Thermodynamics), entropy of one of the bodies decreases.

I understand that there is no fallacy here because obviously each of the bodies cannot be considered isolated systems for this process, but my question is this:

Why does Second Law of Thermodynamics holds only for isolated systems?

How can I intuitively understand this? Is it just that it is always consistent with experimental results therefore assumed to be true? Is there some fundamental principle underlying this problem?

  • 1
    $\begingroup$ "Why do we state that Second Law of Thermodynamics holds only for isolated systems?" because it only holds true for isolated systems. As well as we state any other law of physics under the conditions they actually hold true. $\endgroup$
    – gented
    Commented Nov 8, 2016 at 20:35
  • $\begingroup$ @GennaroTedesco, thank you for mentioning mistake in my question. I edited it. $\endgroup$ Commented Nov 9, 2016 at 17:58

4 Answers 4


With the advent of non-equilibrium thermodynamics, second law in suitably generalized form is made applicable to all systems, isolated or not. I am not an expert in this new field, but what I know comes from reading of $Modern ~Thermodynamics$ by Kondepudi and Prigogine. They divide total entropy change of a system into two parts: that due to flux of matter and heat, $\Delta S_{flux}$, and that generated internally (due to friction, mixing, heat transfer internally etc.), $\Delta S_{internal}$. They state second law as $\Delta S_{internal}\geq 0$ for any system. For an isolated system, there is neither flux of energy nor matter, so $\Delta S_{flux}=0$. So $\Delta S_{internal}\geq 0$ becomes identical to $\Delta S_{total}\geq 0$, which is the way second law is usually stated in reference to isolated systems.

Of course, second law has the status of a postulate in classical thermodynamics, so it cannot be justified theoretically. Its acceptance lies in the fact that its predictions are verified by experiments (within the realm of classical thermodynamics). However one may anticipate the existence of a function such as entropy on the grounds of an extremum principle. If for given conditions, system always seeks a particular state, then one may surmise that perhaps that final state corresponds to extremum of some function. For light traveling between two points, that function is the time of travel itself; for a particle in motion between two points under conservative force field, that function is the Lagrangian; for thermodynamic systems moving from one constrained state to another, that function is the entropy. See also $Thermodynamics$ by Callen.

  • $\begingroup$ Nice answer. Does a system "seeks" for a global or a local extremum? $\endgroup$ Commented Jun 8, 2020 at 14:47

Why do we state that Second Law of Thermodynamics holds only for isolated systems?

Because this is a theorem deduced by Clausius in 1865:

http://philsci-archive.pitt.edu/archive/00000313/ Jos Uffink, Bluff your Way in the Second Law of Thermodynamics, p. 37: "Hence we obtain: THE ENTROPY PRINCIPLE (Clausius' version) For every nicht umkehrbar [irreversible] process in an adiabatically isolated system which begins and ends in an equilibrium state, the entropy of the final state is greater than or equal to that of the initial state. For every umkehrbar [reversible] process in an adiabatical system, the entropy of the final state is equal to that of the initial state."

Clausius' deduction was based on three postulates:

Postulate 1 (implicit): The entropy is a state function.

Postulate 2: Clausius' inequality (formula 10 on p. 33 in Uffink's paper) is correct.

Postulate 3: Any irreversible process can be closed by a reversible process to become a cycle.

All the three postulates are totally unjustified - clever scientists are well aware of that:

Uffink, p.39: "A more important objection, it seems to me, is that Clausius bases his conclusion that the entropy increases in a nicht umkehrbar [irreversible] process on the assumption that such a process can be closed by an umkehrbar [reversible] process to become a cycle. This is essential for the definition of the entropy difference between the initial and final states. But the assumption is far from obvious for a system more complex than an ideal gas, or for states far from equilibrium, or for processes other than the simple exchange of heat and work. Thus, the generalisation to all transformations occurring in Nature is somewhat rash."


In my judgment, a more useful form of the 2nd law, applicable to a closed system (no mass entering or leaving) is $$\Delta S\geq \int{\frac{dQ}{T_B}}\tag{1}$$ where dQ is the heat flow occurring across part (or all) of the system boundary with its surroundings and $T_B$ is the temperature over the portion of the boundary through which the heat is flowing. The = sign applies to a reversible change for the system, and the > sign applies to an irreversible change. For an isolated system, this equations reduces, as required, to $$\Delta S\geq 0$$

The advantage of this form of the so-called Clausius inequality (Eqn. 1) is that, in applying the relationship, one does not need to consider the entropy changes that occur in the surroundings. The focus is totally on the system itself (in the same sense that we use when we apply the first law to the system).

Eqn. 1 is fully consistent with the various other empirical statements of the 2nd law, such as the Kelvin statement. All these are based on an overwhelming body of observational evidence over hundreds of years.


The second law is an inequality that becomes an exact equality at equilibrium. It can be written for any type of system, isolated or not, but the form of the inequality differs in each case:

  • Isolated system (constant $U$, $V$, $n_i$)

$$dS \geq \frac{dU}{T}+\frac{pdV}{T}-\sum_i\frac{\mu_i dn_i}{T} \Rightarrow (dS)_{U,V,n_i}\geq 0$$

  • Closed isothermal isochoric system (constant $T$, $V$, $n_i$)

$$ dA \leq -Tds - p dV + \sum_i \mu_i dn_i \Rightarrow (dA)_{T,V,n_i} \leq 0$$

  • Closed isothermal isobaric system (constant $T$, $P$, $n_i$)

$$dG \leq -S dT + V dP + \sum_i \mu_i dn_i \Rightarrow (dG)_{T,P,n_i}\leq 0$$

We can go on to generate more inequalities by fixing different sets of variables. All of them are equivalent statements of the second law.

To interpret such inequalities see my response to this question.

  • $\begingroup$ In the case of $dG$ must $n_i$ remain constant? For example during a chemical reaction $n_i$ varies but Gibbs free energy continues to decrease. $\endgroup$ Commented Jun 7, 2021 at 12:27
  • $\begingroup$ @ Antonios Sarikas When a reaction reaches equilibrium none of the ni change. This is precisely how we derive the equilibrium constant: we express all ni in terms of the extent of reaction, then we identify the extent that gives dni=0. $\endgroup$
    – Themis
    Commented Jun 8, 2021 at 13:36
  • $\begingroup$ At equilibrium $dG = 0$. But doesn't $n_i$ change until we reach equilibrium? So why we can put the subscript of constant $n_i$ at the differential? Constant $n_i$ applies only when the system reaches equilibrium ($n_i$ then no longer varies). $\endgroup$ Commented Jun 26, 2021 at 10:34

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