Clausius' theorem states that $$\oint\dfrac{\delta Q}{T}\leq 0,$$ $=$ for reversible cycles and $<$ for irreversible ones.

For a cycle with two reversible paths connecting points $a$ and $b$, $$\oint\dfrac{\delta Q}{T}= 0\implies \int_{a,\text{ path 1}}^b\dfrac{\delta Q_{\text{rev}}}{T}+\int_{b,\text{ path 2}}^a\dfrac{\delta Q}{T}=0\implies \int_{a,\text{ path 1}}^b\dfrac{\delta Q_{\text{rev}}}{T}=\int_{a,\text{ path 2}}^b\dfrac{\delta Q_{\text{rev}}}{T}$$ Thus, $\Delta S_{\text{rev}}=S(b)-S(a)\equiv\displaystyle{\int_a^b}\dfrac{\delta Q_{\text{rev}}}{T}$ is a state function.

Now, from what I understand, for a cycle with reversible and irreversible paths connecting points $a$ and $b$, $$ \begin{aligned} \oint\dfrac{\delta Q}{T}< 0&\implies \int_{a,\text{ rev path}}^b\dfrac{\delta Q_{\text{rev}}}{T}+\int_{b,\text{ irrev path}}^a\dfrac{\delta Q}{T}<0\\ &\implies \int_{a,\text{ rev path}}^b\dfrac{\delta Q_{\text{rev}}}{T}=S(b)-S(a)>\int_{a,\text{ irrev path}}^b\dfrac{\delta Q}{T} \end{aligned}$$

So my question is why when we want to calculate the entropy for an irreversible process (path), we argue that since entropy is a function of state, $\Delta S_{\text{irrev}}=\Delta S_{\text{rev}}$. Namely, that you can calculate it as though it were any other reversible process connecting the same points.

  • 1
    $\begingroup$ In classical thermodynamics, that is the only way we know of to determine the entropy change for a system between two thermodynamic equilibrium states. The real question is why is that so (I.e., why is the integral of dq/T a function of state)? $\endgroup$ Commented Jan 8 at 12:27
  • $\begingroup$ Just to clarify what it is you are asking, are you asking why entropy is a state property ?(per Chet Miller's comment) $\endgroup$
    – Bob D
    Commented Jan 9 at 14:28
  • $\begingroup$ Not really, what troubles me is the meaning of the last expression $S(b)-S(a)>\int_{\text{irrev}}\delta Q/T$ $\endgroup$
    – Conreu
    Commented Jan 9 at 15:45
  • $\begingroup$ @JoanS.GuillametF. Why does it trouble you and what do you think it means? $\endgroup$
    – Bob D
    Commented Jan 10 at 22:12
  • $\begingroup$ does this help? physics.stackexchange.com/q/189996/226902 $\endgroup$
    – Quillo
    Commented Jan 13 at 13:02

2 Answers 2


If you know the equilibrium states $a,b$, entropy change when going from $a$ to $b$ is definite and does not depend on details of the process, or whether it is reversible or not.

A reversible process is special because it allows us to define value of entropy at all accessible states, including $a,b$, via the formula

$$ S(x) = S(0)+\int_{0,reversibly}^x\frac{dQ}{T}. $$ The reference state 0 can by anything, including the state $a$. If you calculated this integral for $dQ$ and $T$ describing an irreversible process, the value obtained would not be guaranteed to be the correct value of change of entropy (that change is usually greater during an irreversible process).

To find the change of entropy $\Delta S$ due to an irreversible process that gets from $a$ to $b$, we can imagine a reversible process that brings $a$ to $b$, and only then we can use the above formula. Because while the fact of whether the process is reversible or not is immaterial for the value of change of entropy, it matters for applicability of the formula above. The integral of $dQ/T$ is a correct way to calculate change of entropy only for a reversible process. We can treat $a$ as the reference state, and then the change of entropy can be expressed as $$ \Delta S = S(b) - S(a) = \int_{a,reversibly}^b \frac{dQ}{T}. $$


what troubles me is the meaning of the last expression $𝑆(𝑏)βˆ’π‘†(π‘Ž)>\int _{irrev} 𝛿𝑄/𝑇$

The last expression can be deduced from the fact that the change in entropy between two states potentially consists of two components: (1) The entropy entering/exiting the system due to heat and (2) The entropy generated in the system due to any irreversibility associated with that heat. That is:

$$\Delta S_{ab}= S_{b}-S_{a}=\int_a^b\frac{\delta Q}{T_B}+\sigma_{ab}\tag{1}$$

Where $\sigma_{ab}$ = the entropy generated in the system due to irreversibility and $T_B$ is the temperature at the boundary (entropy point) between the system and the surroundings.

Since entropy is a state function we have, for the system,

$$\Delta S_{irrev}=\Delta S_{rev}=\int_a^b\frac{\delta Q_{rev}}{T}=\int_a^b\frac{\delta Q}{T_B}+\sigma_{ab}\tag{2}$$

It therefore follows from eqs (1) and (2) that

$$\int_a^b\frac{\delta Q_{rev}}{T}\gt\int_a^b\frac{\delta Q_{irrev}}{T_B}\tag{3}$$

by the amount of entropy generated, $\sigma_{ab}$, due to irreversible heat.


The following is an example to demonstrate that the Clausius inequality applicable to an irreversible process, applies when the entropy exiting the system exceeds the entropy entering the system by an amount equal to the entropy generated by irreversible heat. Refer to Figures 1 and 2.

The cycle consists of (1) an isothermal compression from state $a$ to $b$, (2) isochoric (constant volume) heat extraction from state $b$ to $c$ and (3) an isobaric (constant pressure) expansion from state $c$ to $a$ to complete the cycle. In both figures the path from $a$ to $b$ is a reversible isothermal compression. The system under consideration is an ideal gas in an cylinder fitted with a massless, frictionless piston. Example data for temperature, pressure and volume for states $a$, $b$, and $c$ are given. A reversible and irreversible cycle is considered.

CASE A: Reversible Cycle

In FIG 1 the ichoric + isobaric path is reversible. For each process in this path to be reversible the system is exposed to an infinite series of thermal reservoirs so that the temperature $T$ of the system is always in equilibrium with the surroundings, i.e., $T_{b}=T$ (the temperature at the boundary equals the temperature of the system).

Applying Clausius's theorem to the cycle we have

$$\oint \frac{\delta Q}{T}=\int_{Ta}^{Tb}\frac {\delta Q}{T}+\int_{Tb}^{Tc}\frac {\delta Q}{T}+\int_{Tc}^{Ta}\frac {\delta Q}{T}\tag{4}$$

For the reversible isothermal compression $T$=constant and $Q=W=\int_{V1}^{V2}PdV$, therefore

$$\int_{Ta}^{Tb}\frac {\delta Q}{T}=nR\ln\frac{V_b}{V_a}=-0.69 nR\tag{5}$$

For the reversible isochoric heat extraction:

$$\int_{Tb}^{Tc}\frac {\delta Q}{T}=nC_{V}\int_{Tb}^{Tc}\frac{dT}{T}=nC_{V}\ln\frac{T_c}{T_b}=-0.69nC_{V}\tag{6}$$

For the reversible isobaric expansion:

$$\int_{Tc}^{Ta}\frac {\delta Q}{T}=nC_{P}\int_{Tc}^{Tb}\frac{dT}{T}=nC_{P}\ln\frac{T_a}{T_c}=+0.69nC_{P}\tag{7}$$

Substituting the results of equations (5), (6), and (7) into equation (4)

$$\oint \frac{\delta Q}{T}=-0.69nR-0.69nC_{V}+0.69nC_{P}=-0.69nR+0.69n(C_{P}-C_{V})\tag{8}$$

Then, given $C_{P}-C_{V}=R\tag{9}$

$$\oint \frac{\delta Q}{T}=0\tag{10}$$

as expected for a reversible cycle.

CASE B: Irreversible Cycle

See FIG 2. In this case for the isochoric and isobaric processes the system is subjected to single fixed temperature reservoirs equal to the final temperature of each process. As a result the temperature at the boundary, $T_B$, is not the system temperature which varies spatially in the system. The finite temperature difference between the system and the temperature at the boundary (reservoir temperature) results in irreversible heat.

Because the isobaric process is irreversible, the pressure of the gas also varies spatially, thus the pressure is the external pressure and not the pressure of the gas, except at the equilibrium states $c$ and $a$.

We now apply Clausius' theorem to the irreversible cycle:

For the reversible isothermal compression the entropy change is once again

$$\int_{Ta}^{Tb}\frac {\delta Q}{T}=-0.69 nR\tag{11}$$

For the irreversible isochoric heat extraction:

$$\int_{Tb}^{Tc}\frac{\delta Q}{T}=\frac{nC_{V}(T_{c}-T_{b})}{T_{c}}=-nC_V\tag{12}$$

It is instructive to relate the results of equation (12) for the irreversible isochoric process and equation (6) for the reversible isochoric process, together with the general relationship between reversible and irreversible entropy change in equation (2), in order to quantify the entropy generated in the irreversible process.

$$-0.69 nC_{V}=-nC_{V}+\sigma_{bc}\tag{13}$$ $$\sigma_{bc}=0.31 nC_{V}\tag{14}$$ $$-nC_{V}=-0.69nC_{V}-0.31nC_{V}\tag{15}$$

So from equation (15) we see that the entropy transferred to the surroundings consists of two components: that of the reversible heat transfer (first term on the right) plus the entropy generated due to the irreversible heat transfer (second term on the right).

For the irreversible isobaric expansion:

$$\int_{Tc}^{Ta}\frac {\delta Q}{T}=\frac{Q_{ca}}{T_{a}}\tag{16}$$

Here it would appear we can't use $Q=C_{P}\Delta T$ because $C_P$ is the specific heat at constant gas pressure, whereas the gas pressure is not constant as it varies spatially during the expansion (only the external pressure is constant). So we turn to the first law which applies to both reversible and irreversible processes:

$$Q_{ca}=\Delta U_{ca}+W_{ca}\tag{17}$$



With the interesting result that not only the work done, but the heat added for the irreversible isobaric process is the same as that for the reversible isobaric process, i.e., $Q=nC_{P}\Delta T$ for both the reversible and irreversible processes.

We now substitute for $Q_{ca}$ from eq (29) into eq (16)

$$\int_{Tc}^{Ta}\frac {\delta Q}{T}=\frac{Q_{ca}}{T_{a}}=0.5nC_{P}\tag{20}$$

Once again it is instructive to relate the results of equation (20) for the irreversible isobaric process and equation (7) for the reversible isobaric process, with the general relationship between the reversible and irreversible heat transfer of equation (2), to identify the entropy generated in the irreversible isobaric process.

$$+0.69nC_{P}=0.5nC_{P}+\sigma_{ca}\tag{21}$$ $$\sigma_{ca}=0.19nC_{P}\tag{22}$$ $$0.5nC_{P}=0.69nC_{P}-0.19nC_{P}\tag{23}$$

So from equation (23) we see that the entropy transferred to the system for the irreversible isobaric process is less than that for the reversible isobaric process by an amount equal to the entropy generated by the irreversible process, which is retained by the surroundings.

We now apply Clausius' theorem to the irreversible cycle:

$$\oint\frac{\delta Q}{T}=-0.69nR-nC_{V}-0.31nC_{V}+0.69nC_{P}\tag{24}$$

Splitting the last two terms into their reversible and generated entropy components from the right sides of equations (15) and (23) we have

$$\oint\frac{\delta Q}{T}=-0.69nR-0.69nC_{V}-0.31nC_{V}+0.69nC_{P}-0.19nC_{P}$$ $$=-0.69nR+0.69n(C_{P}-C_{V})-0.31nC_{V}-0.19nC_{P}$$ $$=-0.69nR+0.69nR-0.31nC_{V}-0.19nC_{P}$$ $$=-0.31nC_{V}-0.19nC_{P}\lt 0$$ $$=-(\sigma_{bc}+\sigma_{ca})\lt 0\tag{25}$$

The result being less than zero by an amount equal to the sum of the entropies generated in the irreversible isochoric and isobaric path.

Hope this helps.

enter image description here

enter image description here


Your Answer

By clicking β€œPost 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.