I'm currently looking at Principles of Chemistry by Atkins, and I have a few questions involving entropy. These questions probably arise from my poor understanding of reversible/irreversible processes.

  1. Why are reversible processes called reversible? Is it necessarily true that reversible processes can be reverted spontaneously, or is this a common misconception that people have?
  2. So my chemistry textbook gives the definition of entropy as $$\Delta S = \frac{q_{rev}}{T}$$ where $q_{rev}$ is heat transferred reversibly. However, I get lost at how entropy could increase during a reversible process.
  3. Why is the reversibility condition necessary for the definition of entropy?
  • 2
    $\begingroup$ Just me with a trivial comment again, but I'll say that entropy is more correctly defined with infinitesimals: $\mathrm{d}S = \mathrm{d}q_{\text{rev}}/T$. The equation in OP's question is obtained via integration of this, and is therefore only valid if the process is isothermal, i.e. the temperature of the system is constant throughout the process. $\endgroup$
    – orthocresol
    Jul 14, 2015 at 11:41

3 Answers 3

  1. They're called "reversible" because the processes can run backwards without additional input of energy. There aren't really any truly reversible processes in the universe, though (some come very close, but the construct of "reversible" is an ideal, not a reality).

  2. Entropy of a system can increase just fine during a reversible process: expanding a gas "slowly enough" is reversible ("slowly enough" means in an infinite number of infinitesimal steps), but the entropy of the gas increases as it expands as it has more microstates to go into. The entropy that remains constant in a reversible process is that of the system plus the surroundings (in other words, the universe).

  3. The process does not need to be reversible to have entropy. However, we measure entropy based on the heat of the reversible process because entropy is a state function (it depends on the beginning and end points of your process, nothing more) while heat is a path function (it depends on the path you take to get there).

  • 1
    $\begingroup$ Regarding item 2, if the process is adiabatic and reversible, the entropy of the system does not change. If the process is isothermal and reverisble, the entropy of the system increases. $\endgroup$ Jul 16, 2015 at 3:43

Second law of thermodynamics tells that:

Energy tends to disperse from being localized if it is not hindered from doing so.

Why does a hot cup of tea become cool? Because the higher vibrational energy of the hot mug & tea was localized only to the tea-mug system. When it is brought in the surroundings, the localized energy disperses to the surroundings. This is what second law of thermodynamics says about.

Let me give you another example of energy dispersal approach to entropy.

Suppose there is a gas enclosed by a stop-cock in of the bulbs of a two-bulb container. When you open the stop-cock, the gas expands & covers the two bulbs. Why did it happen so? It happened because as you open the stop-cock(removed the hinderance), the localized energy of the gas molecules spread over the whole volume to disperse its energy over a larger volume.

Change in entropy measures how much energy has been dispersed at a specific temperature. Entropy is a state property & how it changes doesn't matter at all. However it is far easy to take the course of the change as a reversible process.

Entropy of the universe(system + surroundings) always increases or doesn't change but never decreases.

For example, take a hot body at temperature $T$. Now, bring in another body which is at a temperature $T - dT$. Since, the difference is infinitesimal, the process of transfer of heat can be considered as reversible. Let $dQ$ heat energy be transferred from the hotter body to the colder body. Decrease in entropy of the hotter body after the heat transfer is $\dfrac{-dQ}{T}$ & increase in entropy of the colder body is $\dfrac{dQ}{T - dT}$. By simple inequality, we can show that $$\dfrac{dQ}{T - dT} > \dfrac{dQ}{T} \implies -\dfrac{dQ}{T - dT} < -\dfrac{dQ}{T} \implies \dfrac{dQ}{T - dT} - \dfrac{dQ}{T} > 0 \implies dS > 0$$.

In classical thermodynamics, there is no definition of entropy actually; what is concerned here is actually the change of entropy.

However, in statistical Mechanics, entropy can be defined as the number of microstates the system has at a particular state.

What is a microstate? each particle in a system collides & now & then they redistribute the energy among themselves without changing the total energy of the system. Microstate is one of the arrangements of the particle's energy for a particular state. It describes how each molecule is distributed at various energy levels at a certain instant. ( As you are reading Atkins, you can get the taste of this approach soon; he has described it beautifully.) It is not that the particles are at various microstates at a single time; it rather describes at a certain instant how the particles are distributed among various energy levels(Total energy is always same; it is just redistributed now & then due to the random collisions of the particles.)

More the microstates for a given state(temperature, pressure, internal energy of the system), more is the entropy at that state. Why? Because here the total energy has more ways to get dispersed to a certain microstate. Suppose, you have two systems $A$ & $B$ having the same energy $U$. However, $A$ has more microstates than $B$. So, the energy $U$ in $A$ owing to its having larger number of microstates has more ways to redistribute among the particles at a certain instant than in case of $B$. So, the energy $U$ disperses more in $A$( as it has more ways to distribute its internal energy) than in $B$ where the energy $U$ disperses lesser (since it has lesser number of ways to redistribute its energy) & hence its energy is localized. Thus entropy of $A$ is more in $A$ than in $B$ as in the former, energy isn't localized & disperses to more microstates than in $B$ .

  • $\begingroup$ @thkim1011: Has it helped you somehow? Have you read my answer? Please comment. $\endgroup$
    – user5764
    Jul 14, 2015 at 18:57
  • $\begingroup$ uhh.. I guess it was informative. But most of my misunderstanding was in question 2. I didn't consider the fact that the surrounding lost entropy which results in no change in the entropy of the universe. Thanks for the response though. I appreciate it. $\endgroup$ Jul 16, 2015 at 5:09
  • $\begingroup$ @thkim1011: Do follow this site for the understanding of second law of thermodynamics & entropy. $\endgroup$
    – user5764
    Jul 16, 2015 at 5:40

If you'd like to get a better understanding of entropy and the second law of thermodynamics so that you have what you need to solve homework problems, see the following link:


  • $\begingroup$ I generally don't like the Physics Forum means of question & answering contrary to the SE. However, the link you have offered is really greatly intuitive . Lucidly written about how Clausius comprehended the concept of entropy; the page is excellent. An awesome job, sir(hope that you have written). However, for understanding , I do follow Frank Lambert's approach of energy dispersal which is highly conceivable unlike the term disorder. +1 for providing the link, sir. $\endgroup$
    – user5764
    Jul 16, 2015 at 4:39
  • $\begingroup$ However, sir, if you don't mind, can you please at least write the gist, what your link provides, in the answer because in case your link gets broken, then also the future users may have a view to your excellent intuitive work ? It is better to provide the info a link supplies in the answer as sometimes the link may not work:) $\endgroup$
    – user5764
    Jul 16, 2015 at 4:44
  • $\begingroup$ Glad you liked it and hope it will help with your understanding and with your ability to do your homework problems. The reason I wrote the Physics Forums Insight article is so that I wouldn't have to always paste it into answers to Physics Forums thread questions. That's why I don't want to give the gist again here. It is one of the most accessed Insight articles to date, and is sure to survive over time. $\endgroup$ Jul 16, 2015 at 13:31
  • $\begingroup$ It needs to get survive! This is one of the best articles I have ever read on classical thermodynamics. Most texts do not at all clarify the things. I got the intuition first from Frank Lambert's energy-dispersal approach; that was greatly intuitive . Now after reading your article, I wish I could get that earlier!! Really an amazing page:) $\endgroup$
    – user5764
    Jul 16, 2015 at 13:57
  • $\begingroup$ You're making me blush. How would you summarize the gist of the article? $\endgroup$ Jul 17, 2015 at 0:57

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