From the second law of thermodynamics:

The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems always evolve toward thermodynamic equilibrium, a state with maximum entropy.

Now I understand why the entropy can't decrease, but I fail to understand why the entropy tends to increase as the system reach the thermodynamic equilibrium. Since an isolated system can't exchange work and heat with the external environment, and the entropy of a system is the difference of heat divided for the temperature, since the total heat of a system will always be the same for it doesn't receive heat from the external environment, it's natural for me to think that difference of entropy for an isolated system is always zero. Could someone explain me why I am wrong?

PS: There are many questions with a similar title, but they're not asking the same thing.

  • $\begingroup$ Because the entropy is not "the difference of heat divided for the temperature" and in general $dS≠\delta Q/T$. The equality $dS = \delta Q/T$ is only true for reversible exchanges of heat. For an isolated system $dS=d_{\rm{i}}S$, where $d_{\rm{i}}S≥0$ is the variation of entropy due to the internal processes, and $d_{\rm{i}}S>0$ if some of those processes are irreversible. $\endgroup$
    – The Quark
    Jun 26, 2023 at 3:10
  • $\begingroup$ Also, there is no such thing as "the total heat of a system". The "total energy", yes, but "heat" is only a form of exchange of energy. $\endgroup$
    – The Quark
    Jun 30, 2023 at 15:07
  • $\begingroup$ Closely related (Liouville theorem VS entropy increase) physics.stackexchange.com/q/761468/226902 and physics.stackexchange.com/q/306055/226902 $\endgroup$
    – Quillo
    Oct 10, 2023 at 11:43

5 Answers 5


Take a room and an ice cube as an example. Let's say that the room is the isolated system. The ice will melt and the total entropy inside the room will increase. This may seem like a special case, but it's not. All what I'm really saying is that the room as whole is not at equilibrium meaning that the system is exchanging heat, etc. inside itself increasing entropy. That means that the subsystems of the whole system are increasing their entropy by exchanging heat with each other and since entropy is extensive the system as whole is increasing entropy. The cube and the room will exchange, at any infinitesimal moment, heat $Q$, so the cube will gain entropy $\frac{Q}{T_1}$, where $T_1$ is the temperature of the cube because it gained heat $Q$, and the room will lose entropy $\frac{Q}{T_2}$, where $T_2$ is the temperature of the room because it lost heat $Q$. Since $\frac{1}{T_1}>\frac{1}{T_2}$ the total change in entropy will be positive. This exchange will continue until the temperatures are equal meaning that we have reached equilibrium. If the system is at equilibrium it already has maximum entropy.  

  • $\begingroup$ Ok I thought to have understood this: but then how can the entropy not decrease? In the case of an ice cube, it gains heat and the system loses heat to give it to the cube. The difference of heat is negative for the system, so why is the entropy greater than zero in this case? $\endgroup$ Jun 15, 2014 at 15:08
  • 2
    $\begingroup$ The key lies in the fact that the room and the ice cube are at different temperatures (the whole system is not at equilibrium otherwise it would have the same temp everywhere). Therefore, $\Delta S= Q(\frac{1}{T_1}-\frac{1}{T_2})$, where $T_1$ is room temp and $T_2$ is the ice cube's temp. If it's in equilibrium then $T_1=T_2$ then entropy is not increasing because it is maximum already. $\endgroup$
    – Bubble
    Jun 15, 2014 at 15:16
  • 6
    $\begingroup$ @RamyAlZuhouri, heat is always transferred from the hotter to the cooler subsystem making the entropy change to be always positive. $\endgroup$
    – Bubble
    Jun 15, 2014 at 15:51
  • 3
    $\begingroup$ @RamyAlZuhouri: if the ice cube melts, the ice cube gains entropy, and the room loses entropy. The key point is that the ice cube gains more entropy than the room loses, so the net entropy of the room/cube system increases. $\endgroup$ Jun 16, 2014 at 16:19
  • 1
    $\begingroup$ What if there is no subsystem ? $\endgroup$
    – user262060
    Jul 23, 2020 at 9:56

For completeness, an information theoretical answer is needed. Entropy is, after all, defined for arbitrary physical states and does not require a notion of thermal equilibrium, temperature, etc. We need to use the general definition of entropy, which is the amount of information that you lack about the exact physical state of the system given its macroscopic specification.

If you knew everything that is to know about the system then the entropy would be zero and it would remain equal to zero at all times. In reality, you will only know a few parameters of the system and there is then ahuge amount of information that you don't know. Now, this still does not explain why the entropy should increase, because the time evolution of an isolated system is unitary (there is a one to one map between final and initial states). So, naively, you would expect that the entropy should remain constant. To see why this is not (necessarily) the case, let's focus on the free expansion experiment caried out inside a perfectly isolated box. In this thought experiment we make the rather unrealsitic assumption that there is no quantum decoherence, so that we don't smuggle in extra randomness from the environment, forcing us to address the problem instead of hiding it.

So, let's assume that before the free expansion the gas can be in one of N states, and we don't know which of the N states the gas actually is in. The entropy is proportional to Log(N) which is prioportional to the number of bits you need to specify the number N. But this N does not come out of thin air, it is the number of different physical states that we cannot tell apart from what we observe. Then after the gas has expanded there are only N possible final states possible. However, there are a larger number of states that will have the same macroscopic properties as those N states. This is because the total number of physical states has increased enormously. While the gas cannot actually be in any of these additional states, the macroscopic properties of the gas would be similar. So, given only the macroscopic properties of the gas after the free expansion there are now a larger number of exact physical states compatible with it, therefore the entropy will have increased.

  • $\begingroup$ "If you knew everything that is to know about the system then the entropy would be zero ...": entropy is not a measure of ignorance, but rather it is a measure of possible configurations of the system that results in the same "macro" state, where the definition of what is macro depends on what you want to understand about the system. $\endgroup$
    – Our
    Mar 19, 2020 at 12:59

While Bubble gave a nice example, let me try to explain this with "Clausius inequality". (You can read this on several sources, I like the explanation from Atkins' Physical Chemistry)

Let's start with the statement: $$ |\delta w_{rev}| \geq |\delta w| \\ $$ Furthermore, for energy leaving the system as work, we can write $$ \rightarrow \delta w - \delta w_{rev} \geq 0 $$ where $\delta w_{rev}$ is the reversible work. The first law states $$ du = \delta q + \delta w = \delta q_{rev} + \delta w_{rev} $$ since the internal energy $u$ is a state function, all paths between two states (reversible or irreversible) lead to the same change in $u$. Let's use the second equation in the first law: $$ \delta w - \delta w_{rev} = \delta q_{rev} - \delta q \geq 0 $$ and therefore $$ \frac{\delta q_{rev}}{T} \geq \frac{\delta q}{T} $$ We know that the change in entropy is: $$ ds = \frac{\delta q_{rev}}{T} $$ We can use the latter equation to state: $$ ds \geq \frac{\delta q}{T} $$ There are alternative expressions for the latter equation. We can introduce a "entropy prodcution" term ($\sigma$). $$ ds = \frac{\delta q_{rev}}{T} + \delta \sigma, ~~\delta \sigma \geq 0 $$ This production accounts for all irreversible changes taking place in our system. For an isolated system, where $\delta q = 0$, it follows: $$ ds \geq 0 \,. $$

  • $\begingroup$ How you have written the latter step . And can you tell me where you find this article in atkins $\endgroup$
    – Koolman
    Nov 26, 2016 at 12:46
  • $\begingroup$ See Atkins' Physical Chemistry (9th edition) on page 102ff. $\endgroup$
    – g.b.
    Nov 29, 2016 at 8:27
  • $\begingroup$ To get the last expression, set heat (delta q) to zero since the system is isolated. All that remains is entropy production which is always larger or equal to zero. $\endgroup$
    – g.b.
    Nov 29, 2016 at 8:31
  • $\begingroup$ What do you mean by ff in 102ff $\endgroup$
    – Koolman
    Nov 29, 2016 at 12:14
  • 1
    $\begingroup$ I mean page 102 and the following. $\endgroup$
    – g.b.
    Dec 4, 2016 at 19:29

We know that $ds_{\rm (universe)}$is equal to $ds_{\rm(system)} + ds_{\rm (surroundings)}$,and for an isolated system $ds_{\rm (surroundings)} = 0$ because $dq_{\rm (reversible)} = 0$; therefore, for an isolated system, $ds_{\rm (universe)}$ is equal to $ds_{\rm (system)}$.

Now, we know that the spontaneity criteria for any process is $ds_{\rm (universe)} > 0$, or if not, at least should be $0$ for equilibrium.

Therefore, $ds_{\rm (system)} \geq 0$.


This may directly answer your question.

Note that the requirement in the law of entropy is for the heat of the initial and final states of the system to be the same, not that no heat can be exchanged at all with the outside world in the path between them.

Now notice that entropy is defined as $\displaystyle dS=\frac{dQ_{rev}}{T}$, where $\displaystyle dQ_{rev}$ is the reversible heat exchanged with the outside. We must specify a reversible path between the initial and final states of the system to calculate this quantity. If the heat is reversible in this path, it can be exchanged back to the system, and so in general $\displaystyle dQ_{rev}$ is not zero for each step in the reversible path.

But hold on, doesn't the total $\displaystyle dQ_{rev}$ need to still be zero for the system to be isolated, as you said in the beginning? Yes, but that doesn't meant that the integral of $\displaystyle dS=\frac{dQ_{rev}}{T}$ needs to be zero. This means that the total entropy can still sum to a non zero value even though the total heat can't.

Just for completeness, note that $\displaystyle dQ_{irrev}$ needs to be zero at all points along any path, unlike $\displaystyle dQ_{rev}$, because it is heat that cannot be recovered from going between the initial and final states and therefore doesn't fulfil our requirement for an isolated system.


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