The equation

$$dS = \dfrac{dQ}{T}$$

is said do hold only on reversible processes. Indeed this is almost always emphasized by writing

$$dS = \dfrac{dQ_{\mathrm{rev}}}{T},$$

to be clear that this is during one reversible process.

Now there are some irreversible processes on which this is used. For instance, if one mass $m$ of a substance melts at temperature $T_0$ and if it has latent heat of fusion $q_L$ then it is usually computed that

$$\Delta S = \dfrac{mq_L}{T_0}.$$

Another example is when heat enters a system at constant temperature. In that case if the heat is $Q$ we have

$$\Delta S = \dfrac{Q}{T}.$$

All these processes are clearly irreversible. It is intuitively clear, but more than that we have $\Delta S > 0$ in all of them.

Still we are finding $\Delta S$ using

$$\Delta S = \int \dfrac{dQ}{T},$$

and it is obviously that this integral is being carried along irreversible processes in the examples I gave.

In that case, whey can we use this formula to find the change in entropy if the processes are irreversible?

  • 3
    $\begingroup$ "All these processes are clearly irreversible. " Who says? $\endgroup$ Nov 1, 2016 at 0:23
  • $\begingroup$ Well, by the computations we see that $\Delta S > 0$, which is the same as saying that the processes are irreversible right? Furthermore, I admit can't see how can a melting process be naturally reversed, that's what I meant by "these processes are clearly irreversible". $\endgroup$
    – Gold
    Nov 1, 2016 at 0:32
  • 2
    $\begingroup$ $\Delta S>0$ for the system does not mean that a process is irreversible. Melting can be reversed by putting the system in contact with a reservoir that is at a slightly lower temperature, rather than a slightly higher temperature (and waiting long enough). A reversible path is one in which the system passes through a continuous sequence of thermodynamic equilibrium states. Did they not cover this in your thermodynamics course? $\endgroup$ Nov 1, 2016 at 1:39
  • $\begingroup$ Related physics.stackexchange.com/q/677501/226902 $\endgroup$
    – Quillo
    Mar 14, 2023 at 0:45

2 Answers 2


Entropy $S$ is a state function.

It only depends on the final and initial states, and not on how the state was reached.

The Clausius' Inequality states

$$\oint \frac{đQ_\textrm{system}}{T_\textrm{source}}\leqq 0\,.\tag{I }$$

When the cycle is reversible, $T_\textrm{source}~=~ T_\textrm{system}$ and equality of $\rm(I)$ applies i.e.,

$$\oint \frac{đQ_\textrm{system}}{T_\textrm{system}} = 0\tag{I.i}$$

It is a matter of few steps from $\rm(I.i)$ to show that the integral $\displaystyle\int \dfrac{đQ}{T}$ takes the same value for two different reversible paths.

So, we can define $S(\rm A) = \displaystyle\int_{\rm O_\textrm{reference state}}^{\rm A}~ \dfrac{đQ_\textrm{rev}}{T_\textrm{system}}$ i.e., for a reversible transformation.

Now, consider two paths equilibrium states $\rm A$ and $\rm B$ such that $\mathsf I$ connecting $\rm A$and $\rm B$ is arbitrary path (reversible or irreversible) and the second path $\mathsf I^\prime$ connecting $\rm B$ and $\rm A$ is reversible.

So, using $\rm(I),$ we get

$$\begin{align} \oint_{\mathrm A \mathsf I \mathrm B \mathsf{^\prime}\mathrm A } \frac{đQ}{T} & \leqq 0 \\ \implies~~~~~~~~~~~~~~~ \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathsf{I}} + \left (\int_{\mathrm B}^{\mathrm A} \frac{đQ}{T}\right)_{\mathsf{I^\prime}} & \leqq 0\\\implies \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathsf{I}} - \left[\int_{\mathrm O}^{\mathrm B}\frac{đQ}{T}-\int_{\mathrm O}^{\mathrm A} \frac{đQ}{T}\right]&\leqq 0~~~~~~~~~~~~~~~~~~~~~(\mathsf I^\prime~\textrm{is reversible})\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ S(\mathrm B)- S(\mathrm A)&\geqq \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathsf{I}}\tag{II}\end{align}$$

From $\rm (II)$

$$S(\mathrm B)-S(\mathrm A)~=~\left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathsf{I}}~~~\textrm{iff}~~~\mathsf I ~\textrm{is a reversible transformation}$$ that is, $$S(\mathrm B)-S(\mathrm A)~=~\int_{\mathrm A}^{\mathrm B} \frac{đQ_\textrm{rev}}{T}\tag{III.i}$$

When $\sf{ I}$ is irreversible, then from $\rm{(II)}$

$$S(\mathrm B)-S(\mathrm A)\gt \int_{\mathrm A}^{\mathrm B} \frac{đQ_\textrm{irrev}}{T}$$

More precisely, using the first law $$\mathrm dS= \dfrac{\textrm{đ} q_\textrm{irrev}}{T}+ \dfrac{\left [ \textrm{đ} w_\textrm{rev}-\textrm{đ} w_\textrm{irrev}\right]}{T}\;.\tag{III.ii}$$

Both $\rm{(III.i)}$ and $\rm{(III.ii)}$ would yield the same entropy change as after-all entropy $S$ is a state function.

But which of $\rm{(III.i)}$ and $\rm{(III.ii)}$ would one use to compute the change in entropy?



$\bullet$ Thermodynamics by Enrico Fermi.

  • 1
    $\begingroup$ What is "$đQ$" supposed to be? $\endgroup$
    – Ryan Unger
    Nov 12, 2016 at 3:59
  • 1
    $\begingroup$ It means that $đQ$ is not an exact differential, @0celo7. $\endgroup$
    – user36790
    Nov 12, 2016 at 4:18

When a system at constant temperature $T$ receives (loses) heat $Q$, the system's gain (loss) of entropy $\textit{due to this particular operation}$ of heating addition (removal) may indeed be written as $\Delta S_{heating/cooling}=\frac{Q}{T}$. Such a situation occurs during phase change. However this is only one mechanism by which a system's entropy may change; entropy of a system can also change due to flux of mass which carries entropy; dissipative processes such as friction generate entropy all on its own.

To show that a process is reversible (irreversible), you must show that entropy of system+surroundings remains the same (increases). Just from knowledge of $\Delta S_{heating/cooling}$ pertaining to the system you cannot conclude anything.


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