Since for an irreversible process, $$dS_\mathrm{Surrounding} =-\frac{\text{dq}_\mathrm{irr,sys} }{T_{\text {surr }}}\tag{1}$$ where $\text{dq}_\mathrm{irr,sys}$ is heat exchange of system and $-dq_\mathrm{irr}$ is heat absorbed by the surrounding. Also $d u=d q+d w$. Now in an irreversible isothermal process involving only ideal gas. $$ \begin{aligned} & du=d w+d q \quad . \quad(d v=C v \cdot d T) \\ & -d w=d q \\ & P_{\text {ext }} \cdot d v=d q \end{aligned} $$ Now Substituting in (i) Now $T_{\text {syst }}=T_{\text {surr }}$ As the process is Isothermal . Clearly $dS_{\text {Surr}}=-dS_{\text {Sys}}$. Does that mean change in entropy of universe in irreversible isothermal process is always $0$?

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    $\begingroup$ Why call it irreversible if there is no change in entropy? $\endgroup$
    – Karsten
    Dec 28, 2022 at 2:15
  • $\begingroup$ Can you please elaborate what you mean ? $\endgroup$ Dec 28, 2022 at 4:29
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    $\begingroup$ Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of at least essential parts. Suitable formatting can be done according to formatting math/chem expressions/equations. $\endgroup$
    – Poutnik
    Dec 28, 2022 at 6:52
  • $\begingroup$ Hey thanks I’m new here I will keep that in mind next time and will edit this question shortly $\endgroup$ Dec 28, 2022 at 7:10
  • $\begingroup$ The amount of heat received by the system for the irreversible path is not the same as the amount of heat received by the system for the reversible path. $\endgroup$ Dec 28, 2022 at 13:30

1 Answer 1


For the irreversible case, we are talking about a situation in which the ideal surroundings reservoir is maintained at the same temperature as the initial temperature T of the system, such that the final equilibrium temperature of the system is also T. This is what we mean by an isothermal irreversible process, even though, within the bulk of the system during the process, the temperature may not be spatially uniform at T.

If, in determining the change in entropy of the system, you devise a reversible path between the same initial and final thermodynamic equilibrium states of the system as for the irreversible process, you will find that the final thermodynamic equilibrium state of the surroundings will not be th[e same as for the irreversible process. So this approach gives you the correct change in entropy for the system in the irreversible process, but it will not give you the correct value for the surroundings. Since the reservoir is ideal, the change in entropy of the surroundings for the irreversible process will be $$\Delta S_{surroundings, irreversible}=\frac{q_{surroundings, irreversible}}{T}=-\frac{q_{system, irreversible}}{T}$$ The change in entropy of the system for both the irreversible and reversible path will be $$\Delta S_{system,irreversible}=\frac{q_{system,reversible}}{T}$$ So, $$\Delta S_{total, irreversible}=\frac{(a_{system,reversible}-q_{system,irreversible})}{T}$$

  • $\begingroup$ Thanks a lot sir . So such proceses should always have $ds_{total} =0$ $\endgroup$ Dec 29, 2022 at 1:36
  • $\begingroup$ No. That's the opposite of what I said. What I said was that for an irreversible process, $$\Delta S_{Total}>0 $$Only for a reversible process is $$\Delta S_{Total}=0$$ $\endgroup$ Dec 29, 2022 at 11:31

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