# Are isobaric, isochoric, and isothermal processes quasistatic by definition?

Let us start from definitions, as they are given in many sources.

A quasistatic process is an idealized processes that happen out so slowly that a system go through a sequence of states arbitrarily close to equilibrium.

An isobaric process is a process, in which the pressure remains constant.

An isochoric process is a process, in which the volume remains constant.

An isothermal process is a process, in which the temperature remains constant.

Now, I could imagine how an isochoric process could not be quasistatic, but what about isobaric and isothermal processes?

For instance, for the following specific example (from Count Iblis answer):

Suppose we put a mixture of hydrogen gas and oxygen gas in a conducting cylinder that is kept at constant volume and in thermal contact with a heat bath. There is also a device in the cylinder that will produce a spark, igniting the gas.

I indeed object (as Count Iblis expected) that the process of ignition is isothermal: when we ignite the mixture chemical reaction takes place, the huge amount of heat is liberated, and besides, a complicated movement of the gas mixture occurs. Thus, it does not make a lot of sense to speak about keeping the temperature of the system constant, since the temperature is different even in different regions of the system. This should not surprise us, since right after the ignition and for some time then the system is clearly not in equilibrium, and thus, the notion of the temperature of the system is not even defined.

So, if we adopt this definition then, an isothermal process imply a quasistatic one, since it does not make sense to talk about constant temperature in a process that is not quasistatic, since temperature is not even well-defined in intermediate non-equilibrium states of such process.

In other words, the presence of the heat bath is not enough for the process described above to be isothermal: the process should also be performed slowly enough so that all heat liberated can be almost instantaneously transferred to the surroundings. Only in such a case the temperature of the system is indeed constant during a process. Clearly, the system should be in thermal equilibrium with its surrounding all the way, and thus, I think that for the process to be isothermal, it should be quasistatic.

Am I right?

P.S. Oh, how I feel like I see where I went wrong. An isothermal process indeed must happen at such a slow rate that the thermal equilibrium is maintained, however, thermal equilibrium is not the whole story. As we know systems in thermodynamic equilibrium are always in thermal equilibrium, but the converse is not always true. So, since only thermal equilibrium is required for an isothermal process, it is not necessarily quasistatic. A quasistatic process is always arbitrarily close to thermodynamic equilibrium, while isothermal is only in thermal equilibrium.

• It is incorrect to say in the P.S. that "since only thermal equilibrium is required for an isothermal process, it is not necessarily quasistatic". Thermal equilibrium of the intermediate states is a necessary but insufficient condition for an isothermal processes. An isothermal process also requires temperature to be continuously defined. Temperature is an equilibrium property of thermodynamic systems (zeroth law). To proceed through a continuous series of thermodynamic equilibrium states is the definition of a quasistatic process. Gabriel Landi’s answer should be accepted as correct. – tjd Sep 30 '19 at 1:32

No, not necessarily. The definition you gave for a quasi-static process is correct, and it clearly states "close to equilibrium", meaning that all three $P, V, T$ parameters should undergo infinitesimal changes, for the process to be considered quasi-static.

Whereas in either isobaric, isochoric and isothermal processes, each time only one of the 3 quantities are held constant, and neither of them will dictate whether the remaining quantities should evolve infinitesimally.

Of course it becomes very simple to understand when you just consider a set of examples where iso-processes occur far from equilibrium:

In Otto cycle: for example after the adiabatic compression, the follow-up combustion is considered isochoric wherein the pressure P shoots up, cannot be quasi-static.

Stirling engine: cycle: an isothermal expansion followed by an isochoric heat-removal, then an isothermal compression followed by an isochoric heat-addition, and all these transformations that the gas undergoes are non-quasistatic, meaning the energy changes of the system either due to heat exchange or work, do not take place in infinitesimal amounts.

EDIT: Further explanations:

Quasi-static processes:

• Along a quasi-static path all intermediate states are equilibrium states;
• if a system progressing along a quasi-static path is “isolated” from its environment, then the values of all properties will remain constant and equal to those just before the isolation.
• Quasi-static processes occur at finite rates but not so rapidly that the system is not able to adjust on a molecular level, in order to reach equilibrium. (perturbation relaxation times $>$ system's relaxation time to reach equilibrium, necessary condition for quasi-static) There would not be in general gradients of any intensive properties such as pressure, density, temperature, etc.
• No, I don think so. As I mentioned, even to meaningfully speak about pressure, the system must be in equilibrium. And besides, I could not tell for all the examples you provided, but one with Carnot cycle is wrong: isothermal expansion/compression in Carnot cycle are reversible, and thus, quasistatic. That is for sure. – Wildcat Aug 3 '14 at 18:34
• my question is essentially the following: what is the meaning of the phrase that some quantity "is held constant" for a non-quasistatic process, during which this quantity is not even defined? After reading reply by Count Iblis, I have a feeling that the phrase that some quantity "is held constant" should be taken literally only for quasistatic processes, while for non-quasistatic it simply means that final and initial values of the quantity are equal to each other. – Wildcat Aug 4 '14 at 7:12
• Two things: First, a non-quasi-static process is not necessarily an iso-process (e.g. isothermal etc), so be careful when you say "some quantity held constant for a non-quasi-static process". Second: Quasi-static doesn't mean final-initial values are equal either, it simply means that if there's any change of any kind in the system, it has occur so slowly, by infinitesimal changes so that the system always feels basically at equilibrium. – Ellie Aug 4 '14 at 7:48
• Why do you find "keeping a quantity at constant" so difficult? For isothermal processes, the system is coupled with a thermal bath, for isobaric ones the gas is subject to a mobile piston at all times (an example), and for isochoric just consider the case where the surrounding volume of the gas does not change. This is what one means by "held constant". – Ellie Aug 4 '14 at 7:52
• Yes with that in mind, try to realize now that although the heat exchange has to be happening slowly for the process to remain isothermal, it does not necessarily imply that the other quantities such as P,V have cannot (or should not) have a gradient either, they may very well do, in which case the slowness of the heat exchange does not make the transition quasi-static. If the gradients of all three parameters (or more) are non-existent, then we can talk about quasi-static processes... – Ellie Aug 4 '14 at 8:13

Most textbooks will still use these terms for processes that are not quasi-static. Let's consider a few examples.

• Isochoric process Suppose we do a free expansion of a gas constrained to a part of a cylinder into the entire volume of the cylinder. The fact that the volume of the entire cylinder is constant tells you that no work is done on the system. People sometimes say that it is because the gas is expanding into a zero pressure vacuum that explains why the gas performs no work, but this is not the correct explanation. If the gas would bump into the boundary of a cylinder which could move and e.g. lift a weight upward, then it would perform work.

Here you could say that the volume that the gas actually occupied did increase. But the usefulness of thermodynamics is that you can define the system in any way you like by choosing your system boundaries and external parameters in the way that suits you. So, I am free to define the system as the entire cylinder. The point of doing that is that you can immediately see that the internal energy will remain constant. You can also consider reaching the final state using a quasi-static method. In this case, you need to let the gas perform work as it expands, so the actual volume the gas expands becomes an additional external parameter needed to describe the system as the gas slowly expands. The work done must be extracted and put back into the system as heat.

• Isothermal + isochoric process Suppose we put a mixture of hydrogen gas and oxygen gas in a conducting cylinder that is kept at constant volume and in thermal contact with a heat bath. There is also a device in the cylinder that will produce a spark, igniting the gas.

Of course, you can object to this process being isothermal, a lot of heat will obviously be produced by the chemical reactions. However, this is a non equilibrium process and we are free to only consider the initial and final equilibrium states where temperature is rigorously defined. What then matters is that the system is coupled to a heat bath which will always remain in internal thermal equilibrium at constant temperature no matter what the system does.

What matters in all these processes is that the relevant constraints always apply in thermal equilibrium. The reason why equilibrium thermodynamics is so useful in practice is precisely because you can compute the outcome of non-equilibrium processes with it. The constraints therefore only need to be defined for the initial and final equilibrium states. Quasi-static processes are useful to do computations by exploiting the path independence of thermodynamic state variables, so if thermodynamics itself doesn't care if you cheat with the constraints during the process, why would we?

• So, if I understand you correctly, when we say that, for instance, temperature is kept constant during a process, the meaning of this phrase is actually different for quasistatic and non-quasistatic processes. For a quasistatic process we indeed mean that $T$ has the same value all the time during the process, while for a non-quasistatic process we simply mean that $T_f = T_i$, or, $\Delta T = T_f - T_i = 0$. Am I right? – Wildcat Aug 4 '14 at 7:07
• Yes, at least this is what the authors of quite a few textbooks will say. Some may avoid saying "temperature is constant" and say that "system is in contact with a heat bath" and avoid using the word "isothermal". You have to note here that if a process isn't quasistatic, it no longer has a rigorously defined temperature anyway. So, if you use the term ""isothermal" for a real process where the thermodynamic state changes then because in practice it can only be approximately quasistatic, you are already making use of the definition that I mentioned. – Count Iblis Aug 4 '14 at 18:18

I may be one year late, but Wildcat's question is correct. Processes where an extensive quantity is kept constant can always be defined, irrespectively of whether they are quasi-static or not. But to keep an intensive quantity constant, the process must be quasi-static since intensive quantities are only defined in equilibrium. Hence, there is no such thing as a non-equilibrium isothermal process.

Some people define an "isothermal" process as being such that the bath is kept at a constant temperature. But that is not the most usual definition. Usually, "isothermal" means the system is kept at a constant temperature and for this to be accomplished, the process must be quasi-static, otherwise temperature will not be constant or homogenous.