The change in entropy during any reversible process between 2 states, state 1 and state 2, is given as:

$${\Delta S}_{rev} = \int_1^2 \delta Q / T $$

and if the process between state 1 and 2 is irreversible then the change in entropy is given as:

$${\Delta S}_{irrev} = \int_1^2 \delta Q / T + S_{gen} $$

How can entropy, S, be a state function if it has different values at states 1 and 2 depending on if the process is reversible or irreversible i.e: for the entropy to be a state function shouldn't the change in entropy between the 2 states be the same no matter if the process is reversible or irreversible?

Note: For reversible process $S_2$ would be: $$ S_2 = S_1 + \Delta S_{rev} $$ & for irreversible process $S_2$ would be: $$ S_2 = S_1 + \Delta S_{irrev} $$

So then $S_2$ for the reversible process will be different than $S_2$ for the irreversible process, which shouldn't be the case if entropy is a state function and a property of the system.


4 Answers 4


I think the way you write the two conditions may be misleading. $$ \begin{align} {\Delta S}_{rev} &= \int_1^2 \delta Q / T \\ {\Delta S}_{irrev} &= \int_1^2 \delta Q / T + S_{gen}. \end{align} $$ The starting point to clarify the issue is the Clausius theorem stating that $$ \oint \frac{\delta Q}{T}\leq 0 $$ in every thermodynamic cycle, where the equality holds for the particular case of a reversible cycle.

Such a property of reversible cycles allows introducing a state function, the entropy, through $$ \Delta S = S(2)-S(1)= \int_1^2 \delta Q_{rev} / T. $$ That means that there is only one difference of entropy, not dependent on the process (since it is a state function) whose definition hinges on the particular case of a reversible transformation. But, this is just the scaffold for defining (and assigning a value of) such a function to every thermodynamic state. Once you have the value of the function at each thermodynamic state, you remove the scaffold, but the function is there.

Therefore, there is nothing like a $\Delta S_{rev}$ different from $\Delta S_{irrev}$. The correct way of rewriting the two equations you wrote is the following: $$ \begin{align} {\Delta S} &= S(2)-S(1)= \int_1^2 \delta Q_{rev} / T = \int_1^2 \delta Q_{irr} / T + S_{gen}. \end{align} $$ where both terms in the right-hand side of the last equality depend on the specific, irreversible process and, as a consequence of Clausius' theorem $S_{gen}>0$.

  • $\begingroup$ Exactly! And so then $dQ_{rev}/T$ is greater than $dQ_{irrev}/T$ ? $\endgroup$
    Dec 19, 2021 at 18:09
  • 1
    $\begingroup$ @GRANZER sure. Again a consequence of Clausius' theorem. $\endgroup$
    – GiorgioP
    Dec 19, 2021 at 18:52

Your error is in assuming that $\int_1^2 \frac{\delta Q}{T}$ is the same for the two processes.

As an example, consider an ideal gas whose initial state is given by $(V_0,T_0)$ and whose final state is given by $(2V_0,T_0)$. One way to achieve this is to reversibly heat the gas and allow it to expand in such a way that the temperature remains constant. Another approach would be to heat the gas at constant volume until its pressure and temperature reach $(p',T')$, and then allow it to expand adiabatically until its pressure returns to its initial value and its volume doubles. Lastly, we could separate two halves of an evacuated container of volume $2V_0$ with a membrane, fill one side with the gas at temperature $T_0$, and then suddenly remove the membrane and allow the gas to freely expand into the vacuum.

In the first case, from the first law we have that $$\delta Q = p \mathrm dV \implies \boxed{\Delta S = \int_{V_0}^{2V_0} \frac{p}{T}\mathrm dV = \int_{V_0}^{2V_0} \frac{nR}{V} \mathrm dV = nR\ln(2)}$$

In the second example, all of the heat is provided in the first stage, in which $$Q = \frac{3}{2} nR (T'-T_0) = \frac{3V_0}{2}(p'-p_0)$$ The work done in the second stage is $$\int_{V_0}^{2V_0} p \mathrm dV= p' V_0^{5/3} \int_{V_0}^{2V_0} V^{-5/3} \mathrm dV =\frac{3}{2}p' V_0 (1-2^{-2/3})$$ For these quantities to be equal to one another (which would return the temperature back to its initial value) we must have that $p' = 2^{2/3} p_0 \iff T' = 2^{2/3} T_0$. As a result, the entropy increase during this process (which again, occurs entirely during the first phase) is given by $$\boxed{\Delta S = \int_{T_0}^{2^{2/3}T_0} \frac{3}{2}n R \frac{dT}{T} = nR\ln(2)}$$

We see that the entropy change is the same for the two reversible processes. The third case is not a reversible process and so we cannot compute the entropy as $\int \delta Q/T$, but we do observe that $\delta Q = 0$. As a result, we see that $$\Delta S = \int_1^2 \frac{\delta Q_{rev}}{T} > \int_1^2 \frac{\delta Q_{irrev}}{T}$$

Put differently, given any two states $1$ and $2$, the entropy difference can be computed by finding any reversible process which takes $1$ to $2$ and calculating $\int_1^2 \delta Q/T$ for that process. If the process in question is irreversible, then computing $\int \delta Q/T$ will yield a result which is strictly less than $\Delta S$.

  • $\begingroup$ The two cases seem reversed to me. Entropy change is zero for a reversible process and it is larger for an irreversible process. If entropy were to increase for a reversible process you would end up with an increased entropy after performing a process and its reverse so you wouldn't end up in the same state. $\endgroup$ Dec 20, 2021 at 9:37
  • $\begingroup$ @AccidentalTaylorExpansion Entropy change is zero for a reversible cycle. Note that inequality is not comparing the change in entropy in a reversible process to the change in entropy in an irreversible process; given a starting and ending state, the entropy change is non-negotiable. Rather, it is comparing the entropy change to $\int \delta Q/T$, and noting that for irreversible processes, the latter is less than the former. The fact that no heat is absorbed during free expansion but the entropy obviously increases is a good mnemonic to remember this. $\endgroup$
    – J. Murray
    Dec 20, 2021 at 11:34

A reversible path between the two equilibrium end states 1 and 2 does not have to bear any resemblance whatsoever to the irreversible path between the same two end states, aside from matching at the two ends. So the histories of $\delta Q$ and/or T along a reversible path are not going to be the same as those along the irreversible path. The entropy generated for the irreversible path is going to be the difference between the integral of $\delta Q/T$ along a reversible path and the integral of $\delta Q/T$ for the irreversible path. I should also mention that the T in the equation should really be the value of temperature at the boundary between the system and surroundings, $T_B$, through which the heat $\delta Q$ flows. For the case of a reversible path, this is the same as the system temperature T, but for an irreversible path it is usually not the same.


Being a state function does mean that is defined at the equilibrium state which in this case are 1 and 2. It does not mean that it is defined during the process between 1 and 2 (exp for the irreversible case), and it does not mean either that the function is continuous across the process between the two states (the continuous case is the case of quasi-static i.e. always at equilibrium transformations).

Maybe it can get more clear if you consider the general case where the final state $2_{irr}$ is different than the state $2_{rev}$ therefore you immediately see that given the two processes in the middle are different there is no good reason for the two end entropies to be same.

The special case where the final entropy is the same is essentially when $S_{gen}=0$, i.e. when the process in the middle is reversible, for the process in this case there can be still multiple paths or options, or "detailed reversible processes" to arrive to the same final state.


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