I am confused about the difference between a spontaneous process and an irreversible process. Based on what I read so far, both processes increase universe's entropy. I never heard of any reversible spontaneous processes before. Is it safe to assume that they both mean the same thing? I can't find any rigorous definition of spontaneous processes.

Wikipedia seems to characterize spontaneous processes as any process that decreases a system's free energy. I learned that free energy change can either be irreversible or reversible (e.g. system doing reversible work on surroundings). I think a spontaneous process can be either reversible or irreversible as long as there is a change in a system's free energy?

Edit: Maybe a process is spontaneous if a system increases or decreases its entropy? This entropy change can be either reversible or irreversible.

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    $\begingroup$ The terms reversible and irreversible have different context with regard to chemical reactions than with regard to non-chemical processes. $\endgroup$ Commented Apr 28, 2023 at 10:22
  • $\begingroup$ @ChetMiller Yes, by irreversible I mean processes in which the entropy of the universe increases and reversible processes keeps the universe's entropy constant. $\endgroup$ Commented Apr 28, 2023 at 10:29
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    $\begingroup$ For such a system, I would regard spontaneous as the subset of irreversible processes over which you exert no control while they are occurring. $\endgroup$ Commented Apr 28, 2023 at 10:34
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    $\begingroup$ > "both processes increase universe's entropy." Don't say this. Nobody knows whether Universe has meaningful thermodynamic entropy and whether it increases or decreases (we can't do thermodynamic experiments on the Universe, and no heat exchange with the Universe has been ever observed). It's better to say those processes generate entropy. $\endgroup$ Commented Apr 28, 2023 at 15:28
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    $\begingroup$ A reversible reaction is one in which, if you start out only with reactants present, the reaction will proceed in the forward directions, and the equilibrium state will feature both reactants and products; and, if you start out only with products, the reaction will proceed in reverse and the equilibrium state will feature both reactants and products. On the other hand, a reversible process is one in which, if the initial state of a system is A and the final state is B, you can return both system and surroundings from state B to state A without affecting any other entity. $\endgroup$ Commented Jun 25, 2023 at 11:27

3 Answers 3


Spontaneous and irreversible is the same. Think of two systems at different temperature that come to thermal contact:

  • Spontaneous means we do not need to intervene to get the process going. Heat transfer will happen on its own.

  • Irreversible means that the entropy of the universe increases.

If a spontaneous process takes place under constant temperature and constant volume in a closed system, then the Helmholtz energy (free energy) of the system decreases. Mathematically this is equivalent to the entropy increase of the universe (system plus surroundings).

An example is a box divided into parts that contain a gas at different pressures. If the box is in a bath and the walls between the internal partitions are removed, then the equilibrium state where pressure is equalized throughout the box has lower Helmholtz energy than the initial state. Equivalently, the sum of the entropy of the system plus the entropy of the bath is higher than before.

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    $\begingroup$ Thank you. From your second paragraph, I assume you mean this equation $\Delta F=-T\Delta S_{univ}$. It implies $\Delta F\le0$ but I do not understand why it is true. I thought Helmholtz free energy is a state function. The value of $\Delta F$ should not depend on whether a path between two states is reversible or not? $\endgroup$ Commented Apr 28, 2023 at 11:24
  • $\begingroup$ Start with $\Delta S_\text{univ} = \Delta S + \Delta S_\text{bath} = \Delta S - Q/T$ and use the first law to write $Q = \Delta U$ since $V$ is constant. $\endgroup$
    – Themis
    Commented Apr 28, 2023 at 11:44

This answer is too long for a comment, and it is for your question following @Themis's answer below.

Denote by $T^0,S^0$ the reservoir's parameters and by $\delta W^0$ the externally controlled work, then for any process connecting two infinitesimally close equilibrium states $$dU=TdS+\delta W=T^0dS^0+\delta W^0 \tag{1}\label{1}$$ For an isothermal process $T=T^0$ and then $$T(dS-dS^0)=\delta W^0-\delta W=T\sigma \ge 0 \tag{2}\label{2}$$ where $$\sigma = dS-dS^0 \ge 0\tag{3}\label{3}$$ is the internally generated entropy that is never negative, and is zero only for a reversible process. Therefore $dF=d(U-TS)=dU-TdS-SdT=-SdT + \delta W$ that can be written for an isothermal process, $dT=0$, as $$ dF|_T = \delta W = - T\sigma +\delta W^0 \tag{4}\label{4}$$ which in the special case of no work done, $\delta W^0=0$, is just $$ dF|_{T,\delta W^0=0} = - T\sigma \le 0 \tag{5}\label{5}$$ with equality only in a reversible process.

The above considerations are for infinitesimal changes between two equilibrium states. One might think that we could integrate $\eqref{4}$ or $\eqref{5}$ and obtain a similar inequality for finite, not infinitesimal, changes. Unfortunately, that would not work in general because in the $\eqref{4}$ we have assumed that each infinitesimal step is bracketed with an equilibrium states, that is the finite path connecting one equilibrium state with another not close to it is almost reversible.

It is possible to generalize our result but we must assume that the internal energy of the system undergoing the irreversible process between two non-neighboring equilibrium states satisfies the Gibbs equation $$U=TS+W\tag{6}\label{6}$$ where $\delta W=\sum_{k=1}^K Y_k dX_k$ is the work function. In other words the internal energy is a 1st order homogeneous function of the entropy.

That being the case assume that the system is connected to a single thermal reservoir at temperature $T^0$ and to various work reservoirs with which it can interchange entropy $dS^0$ and work, resp. that is $\delta W^0 = \sum_{k=1}^N Y_k^0 dX_k^0$ with $dX_k=dX_k^0; k=1,2,..,K$ but $dS-dS^0 \ge 0$.

Let the initial and final states be denoted by $\mathcal A$ and $\mathcal B$. Since the end states are assumed to be in equilibrium, both the initial and final temperature must equal to that of the reservoir, $T^0$.

Now calculate the change in the internal energy over the process $$\Delta U = U(\mathcal A)-U(\mathcal B)\\=(TS)|_{\mathcal A}+W(\mathcal A)-(TS)|_{\mathcal B} - W(\mathcal B)\\ =T^0(S({\mathcal A})-S({\mathcal B}))+W(\mathcal A) - W(\mathcal B) \\ =T^0 \Delta S + \Delta W \tag{7}\label{7}$$

By definition $F=U-TS$, $\Delta F=\Delta (U-TS)$ but at the endpoints $T=T^0$ is fixed, therefore we can write: $$\Delta U - T^0 \Delta S =\Delta (U - T^0 S) \tag{8}\label{8}.$$ $$\Delta F = \Delta W \tag{9}\label{9}.$$ By the conservation of energy $\Delta U = T^0 \Delta S^0 + \Delta W^0$ where $\Delta W^0$ is the "experimental", ie., external, work done by the environment on the system and $\Delta S^0$ is the entropy absorbed by the system from the thermal reservoir whose temperature is $T^0$ throughout. But from the 2nd law we know that $\Delta S \ge \Delta S^0$ therefore for some $\sigma = \Delta S-\Delta S^0 \ge 0$:

$$\Delta U - T^0 \Delta S = \Delta F = \Delta W \\ =T^0\Delta S^0 +\Delta W^0 -T^0 (\delta S^0 +\sigma) =\Delta W^0 +T^0\sigma$$ or $$\Delta F + T^0\sigma = \Delta W^0 \tag{10}\label{10}.$$

This $\eqref {10}$ is the general equation for the finite variation of the free energy. It shows that since the internal dissipation $T^0\sigma$ is never negative and is always positive for an irreversible process, it reduces the amount the free energy can be increased by the external, experimental, work. Or, if written in the "active mode", $\eqref{11}$ $$-\Delta W^0 + T^0\sigma = - \Delta F \tag{11}\label{11}$$ it shows how much the work done on the environment is reduced by the dissipation for a given change in the free energy.

Important: for this result to be true it is not needed that the system go through an isothermal process throughout. Instead, all is needed is that the beginning and end equlibrium states are at the same temperature and the reservoir with which thermal energy (entropy) is exchanged stay at the same temperature.

  • $\begingroup$ Thank you. So for infinitesimal changes in $F$, $dF\le0$ is true but for large changes $\Delta F$, it cannot depend on type of process, right? $\endgroup$ Commented Apr 28, 2023 at 16:06
  • $\begingroup$ yes, that is correct, $F$ is a state function in thermostatics. $\endgroup$
    – hyportnex
    Commented Apr 28, 2023 at 16:19
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    $\begingroup$ for your question on finite changes in the free energy I added a more detailed explanation in the answer above. $\endgroup$
    – hyportnex
    Commented Apr 29, 2023 at 19:15

To reply more briefly to your question in a comment to Themis' answer:

[$\Delta F = -T\Delta S_{\rm{univ}}$] implies $\Delta F ≤ 0$, but I do not understand why it is true. I thought Helmholtz free energy is a state function. The value of ΔF should not depend on whether a path between two states is reversible or not?

Even though $\Delta F$ depends only on the initial and final states and not on any specific path, the relation $\Delta F = -T\Delta S_{\rm{univ}}$ is valid only on certain paths: it applies when the process takes place at constant volume and temperature and when no other work than expansion work is involved.

In the more general case (see derivation below): $$\Delta F = W' -T\Delta S_{\rm{univ}}$$ where $W'$ is any work other than expansion work received (algebraically) by the system.


First law: $$dU=\delta W + \delta Q$$

Work received: $$\delta W = -P_{\rm{ext}}dV + \delta W'$$

where $-P_{\rm{ext}}dV$ is the expansion work and $\delta W'$ is any other non-expansion work.

Second law: $$dS = d_{\rm{e}}S + d_{\rm{i}}S$$ where:

  • $d_{\rm{e}}S$ is the variation of entropy due to exchanges with the surroundings (here, only heat is exchanged so $d_{\rm{e}}S=\delta Q/T$)
  • $d_{\rm{i}}S$ is the variation of entropy due to processes internal to the system and $d_{\rm{i}}S≥0$ (if those processes are irreversible then $d_{\rm{i}}S>0$).

Combining the above relations: $$dU=-P_{\rm{ext}}dV+\delta W'+T(dS-d_{\rm{i}}S)$$

Defining $F=U-TS$: $$dF=-P_{\rm{ext}}dV+\delta W'+SdT-Td_{\rm{i}}S$$

Assuming $V$ and $T$ constant: $$dF=\delta W'-Td_{\rm{i}}S$$ $$(\mathrm{assumption:}\ V\ \mathrm{and}\ T\ \mathrm{constant})$$

Here, $dF$ does not depend on the path, but both $\delta W'$ and $-Td_{\rm{i}}S$ do.

On all paths for which no other work than expansion work is involved: $$dF=-Td_{\rm{i}}S$$ $$(\mathrm{assumption:}\ V\ \mathrm{and}\ T\ \mathrm{constant,\ and}\ \delta W'=0)$$ And if $dF<0$, then $d_{\rm{i}}S=-(1/T)dF>0$ and all such paths are spontaneous and irreversible.


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