A few days ago I re-visited my undergraduate thermodynamics notes and (as you may have seen from the amount of questions I have been posting) I have encountered many difficulties that I overlooked at that time. And I think most of my problems come from how unclear is the concept of quasi-static process in books. Many textbooks define a quasi-static process as:

A process that has its macroscopic variables well defined and in the system it is at each instant of time in a state infinitesimally close to the state of equilibrium.

The first part of that definition is logical. But the second part, related to thermodynamical equilibrium confuses me. What is clear is that if it is stated that a certain process is quasistatic, then the pressure of the system is infinitesimally similar to the external pressure, which allows us to easily calculate the volumetric work.

My concise doubt comes from the relationship of the external temperature with the internal temperature. We know that for a system to be in true equilibrium with the outside, both pressure and temperature must be the same (without considering material equilibrium).

If this is the case, T system should be infinitesimally equal to T of the environment which would completely eliminate the generation of entropy by transfer between finite temperatures.

Now, when one studies theoretical cycles, it is generally stated that they are internally reversible ( quasi-static internally and without internal friction) but irreversible at last because the unstopabble heat transfer between finite temperatures in many steps.

I see a contradiction here: As I understand it, many define a quasistatic process as:

  • "An extremely slow process where internally the P and the T are equal, homogeneous". (Completely agree)
  • "The system at each instant of time in a state infinitesimally close to the state of equilibrium". Dont' get it since many problems considers Pexternal=Psystem but T System different to T surroundigs.

To me, it should be called partially quasistatic process if Tsys does not equals to Tsurroundings.


  • $\begingroup$ I will try to word a full answer later, but quasi-static does not mean Text=Tinternal. It means if you make a change, the change happens so slowly as to be continuous. This is convenient if you want to calculate the change by integration. For instance the work done is $\int PdV$ and in a quasi static process P is continuous and thus, a differentiable and integrateable function. Assuming ideal gas you can replace P with $\frac{nRT}{V}$ and integrate wrt V. Quasi-static does not mean that the temperature inside the system is equal to the temperature of the surroundings during a change. $\endgroup$ – Charlie Crown Feb 15 '19 at 17:48
  • 1
    $\begingroup$ As they explained to me (which does not mean that it is correct) the volumetric work always, no matter the nature of the process is equal to Pext.dV. And the pressure of the system is considered equal to the external pressure in a quasi-static process, "since it is so close to equilibrium". Now, * casually * temperature issue is always omitted at this point. $\endgroup$ – Zincman Feb 15 '19 at 17:56
  • $\begingroup$ pressure is not equal inside and outside. If that was the case, nothing would move. $\endgroup$ – Charlie Crown Feb 15 '19 at 17:58
  • $\begingroup$ quasi-static is not a measure of the difference between inside and outside for anything. It is a statement about how changes occur. Changes only occur when there is a driving force. There is no driving force when things are equal. $\endgroup$ – Charlie Crown Feb 15 '19 at 17:59
  • $\begingroup$ Ok, no doubt about it. Let's say that quasi-static only refers to the evolution is "slow" and "continuous", easily integrable. Now, I understand that the first law of thermo is related to energy transferred to the system by external agent, so Volume Work =Pext.dV . Is not obvious the relation of Pext to Psys. And in many many places, with no further explanation that is considered quasiestatic they do Psys=Pext. :( $\endgroup$ – Zincman Feb 15 '19 at 18:28

In my judgment, when they are talking about irreversible cycles, they are referring to idealized cases in which (a) the process is internally reversible with respect to the working fluid but (b) irreversible with regard to the region outside the working fluid where there is a finite thermal resistance (e.g., a cylinder wall with finite thermal conductivity) between an (external) ideal constant-temperature reservoir and the working fluid. So the rate of heat transfer to the working fluid is very low (quasi static), and all the entropy generation takes place within the (higher) thermal resistance wall. At least, this is how the irreversibility is often modeled.

To illustrate all this, see the following thread from Physics Stack Exchange: https://physics.stackexchange.com/questions/443527/entropy-generation-in-an-endoreversible-engine

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.