2
$\begingroup$

To calculate the entropy change of the universe when a closed system undergoes an irreversible process, one must choose a reversible process and calculate the following integral

$$\Delta S=\int \frac{\delta Q_\mathrm{rev}}{T}$$

However, it contradicts the fact that during reversible processes, $\Delta S=0$. Am I misunderstanding something?

$\endgroup$
4
  • $\begingroup$ On a general note: thermodynamics doesn't say anything about the entropy of the universe. It only says that the entropy of something else (what Bob calls "surroundings" in his answer) must increase. If you are trying to apply the definition of entropy to "the universe", then you are running into a contradiction because there is nothing else. The universe is not "surrounded" by anything by definition. Philosophically you can try a workaround by saying "visible universe", but since the universe is not even energy conserving, even the first law of TD falls away, so why not the second? $\endgroup$ Apr 23, 2023 at 18:23
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/questions/743208/… $\endgroup$
    – Themis
    Apr 23, 2023 at 20:24
  • $\begingroup$ @Themis This is different. I understand that entropy is a state function but when one uses a reversible process to calculate the entropy change of the universe, doesn't it contradict the fact reversible processes should have no change of entropy in the universe? This is where I am confused. $\endgroup$
    – Jimmy Yang
    Apr 23, 2023 at 20:39
  • $\begingroup$ Would you at least be able to determine the entropy change of the system for the irreversible process path? $\endgroup$ Apr 23, 2023 at 21:36

1 Answer 1

6
$\begingroup$

How it is possible for entropy of universe to increase during reversible process?

It's not.

Your equation is for the change in entropy of the system only. The change in entropy of the system for a reversible process can be zero, less than zero, or greater than zero depending on the process. The change in entropy of the universe is the sum of the change in entropy of the system and the surroundings.

For a reversible process

$$\Delta S_{univ}=\Delta S_{system}+\Delta S_{surroundings}=0$$

or

$$\Delta S_{system}=-\Delta S_{surroundings}$$

So if the change in entropy of the system is greater than zero for a reversible process, the change in entropy of the surroundings will be less than zero by the same amount for a total change of zero. For example, the change in entropy of the system for a reversible isothermal expansion is $+Q/T$ whereas the change for the surroundings is $-Q/T$ for a total entropy change of zero.

I feel uncomfortable by the fact one can determine the entropy change of an irreversible process by selecting a reversible path. If a path is reversible, there should be no entropy change in universe.

If the actual path is reversible, then there will be no change in the entropy of the universe. If the actual path is irreversible, then there will be an increase in entropy of the universe, where the change in entropy of the universe is the sum of the changes in entropy of the system plus the surroundings. But the change in entropy of the system will be the same for a reversible and irreversible path between the same two equilibrium states since entropy is a state function.

Selecting a reversible path does not "replace" the irreversible process. Its only purpose is to enable calculation of the entropy change of the system using the definition of entropy change

$$\Delta S_{sys}=\int_1^2 \frac{\delta Q_{rev}}{T}\tag{1}$$

An example is calculating the entropy change for the free adiabatic expansion of an ideal gas (Joule expansion). The process is irreversible with the initial and final temperatures being the same, $T_{2}=T_{1}$, and where $P_{2}V_{2}=P_{1}V_{1}$. Although any reversible path can be used, the logical choice is a reversible isothermal expansion. The change in entropy of the system using eq (1) above is then

$$\Delta S_{sys}=R\ln\frac{V_2}{V_1}$$

Since the actual irreversible process is adiabatic, $\Delta S_{surr}=0$ making $\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}= R\ln\frac{V_2}{V_1}$ or $\Delta S_{univ}\gt 0$. The increase in entropy is due to entropy generated within the system.

If the actual process was a reversible isothermal expansion, where heat transfer can occur between the system and surroundings (in contrast to an adiabatic expansion where heat transfer cannot occur), then the change in entropy of the surroundings will be $\Delta S_{surr}= -R\ln\frac{V_2}{V_1}$ and therefore $\Delta S_{univ}=0$.

Hope this helps.

$\endgroup$
19
  • $\begingroup$ For more context, I was trying to understand this article about calculating entropy change in irreversible processes. My problem is that I do not understand why it is possible to use reversible processes to calculate the entropy change of the universe. $\endgroup$
    – Jimmy Yang
    Apr 23, 2023 at 18:29
  • $\begingroup$ What, in particular, do you not understand about the article? note that they use the reversible process to calculate the change in entropy of the system, not the universe. $\endgroup$
    – Bob D
    Apr 23, 2023 at 18:37
  • $\begingroup$ Hmm, it seems like one can calculate the entropy change of the universe by assuming that the universe is in contact with some external thermal reservoir. By using a reversible process, the total entropy change of universe+reservoir is zero? $\endgroup$
    – Jimmy Yang
    Apr 23, 2023 at 18:48
  • $\begingroup$ I’m not sure I follow you. The “universe” is by definition everything. It makes no sense to say the universe is in contact with something else (external thermal reservoir) $\endgroup$
    – Bob D
    Apr 23, 2023 at 19:05
  • 2
    $\begingroup$ @JimmyYang If the ACTUAL path is reversible, then there will be no change in the entropy of the universe. If the ACTUAL path is irreversible, then there will be an increase in entropy of the universe, where the change in entropy of the universe is the sum of the changes in entropy of the system plus the surroundings. But the change in entropy of the SYSTEM will be the same for a reversible and irreversible path between the same two equilibrium states. It's getting late here (US east coast) but tomorrow I will update my answer to include some concrete examples to demonstrate this. $\endgroup$
    – Bob D
    Apr 25, 2023 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.