The change in entropy for a reversible process is given as-

$\Delta$S = $\frac{q_{rev}}{T}$

Where qrev is the heat supplied isothermally and reversibly.

The change in entropy for a irreversible process is given as-

$\Delta$S = $\frac{q_{irrev}}{T}$

Where qrev is the heat supplied isothermally and irreversibly.

Could someone please explain these equations while giving the meaning of qrev and qirrev? I didn't really get a proper understanding of them.


2 Answers 2


For a reversible process that is not isothermal, you need to integrate between the initial and final states of the system: $$\Delta S=\int{\frac{dq_{rev}}{T}}$$where $dq_{rev}$ is the differential amount of heat transferred across the boundary of the system during the process and T is the temperature.

The change in entropy for a system experiencing an irreversible process is not $q_{irrev}/T$ or even $\int{\frac{dq_{irrev}}{T}}$. To get the entropy change for a system that experiences an irreversible process, you need to devise (i.e., dream up) an alternate reversible process path between the same initial and final states as the irreversible process, and evaluate $\int{\frac{dq_{rev}}{T}}$ for that reversible process path.

  • $\begingroup$ "For any reversible process carried out change in entropy (S2-S1)=Q(rev)/T." -Is there a derivation for this, or is it just the definition? $\endgroup$
    – AlphaRogue
    Commented Aug 4, 2021 at 13:07
  • $\begingroup$ Somehow, back in the day, Clausius recognized observationally that the integral of dq/T is the same for all possible reversible paths between the same two end points (there are an infinite number of reversible paths). That means that the quantity is a function of state (like internal energy or enthalpy). He called this function of state the entropy. $\endgroup$ Commented Aug 4, 2021 at 14:26

For any reversible process carried out change in entropy (S2-S1)=Q(rev)/T. If Temperature varies then you have to integrate the differential changes in entropy. Q(rev) is the amount of heat transferred to the gas in this "reversible" process. I hope there is no confusion till now. Next consider an irreversible process. Let initial state of the system be A and final state be B. Now the problem with irreversible process is that during such a process all parts the system is not in equilibrium with each other. In other words you will not be able to find out a particular equation representing the process. So in this case you can't find out work done by integration and hence you would not be able to apply first law of thermodynamics to find heat exchanged. So we exploit the intensive property of entropy. An intensive property is one which depends only on initial and final state of the system and does not depend on the path. So it may occur to you that entropy change in the irreversible process will be same as that in a HYPOTHETICAL REVERSIBLE PROCESS CONNECTING THE TWO STATES A AND B. Now in this reversible process you can easily find out the heat exchanged during an infinitely small step during which temperature can be assumed to be constant. Thus you can integrate and find total entropy change. This method is valid for any process not only for isothermal. In isothermal since temperature is constant you need not integrate and the problem becomes easy


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