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A quasi-static process is often defined as a process "that occurs infinitely slowly such that equilibrium holds at all times."(Harvard, Matthew Schwartz, statistical mechanics Spring 2019). My question is a simple but possibly subtle one which I haven't seen mentioned anywhere.

Simply put, does the system need to maintain equilibrium with the surroundings at all times during the process in order for the process to be classified as a quasi-static process or is the lesser requirement of simply having the system maintain internal equilibrium with itself good enough to classify the process as quasi-static?

As an example, suppose we have a fixed volume system immersed in a heat bath (the surroundings). The system is at temperature $T_{sys}$ while the heat bath is at fixed $T_{bath}$ with $T_{sys}<<T_{bath}$. The walls/boundary of the system are virtually adiabatic but not totally (i.e they posses a very low thermal conductivity) and so heat can and will seep into the system across a large temperature gradient but this will happen very slowly (perhaps even infinitely slowly). After a very long time, the systems temperature will equal the heat baths temperature. Does this count as a quasi-static process? Throughout the process, the system had a well-defined internal equilibrium however it was never in equilibrium with the surroundings and so I am not sure whether it counts as quasi-static or not.

Any help on this issue would be most appreciated!

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2 Answers 2

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In branches of physics other than thermodynamics, the word "adiabatic" is used as synonym of "infinitesimally slow." This creates some confusion.

In equilibrium thermodynamics, as you correctly intuit, you need to be more specific. Infinitesimally slow processes are introduced as a useful concept in order that the equation of state be valid throughout. In this way, you can calculate work, specific heats, etc. by using calculus. As an example of why you need to distinguish "adiabatic" from "quasi static", consider the slow expansion of an ideal gas, in which you wish to keep it at constant pressure (calculation or measurement of $C_p$). You would obviously have to supply some heat so that your gas doesn't lose pressure. But you would have to do it very slowly, so that pressure itself, as well as temperature, are well defined. This would be in sharp contrast with the adiabatic expansion (also infinitesimally slow, but with no heat supplied, and thereby suffering a decrease in pressure, as well as in temperature).

Why do you need to distinguish "quasi static" at all? Because sometimes you want to discuss changes that occur almost instantaneously. The system immediately readjusts to a new equilibrium condition, so to speak, so that it jumps over all possible intermediate equilibrium conditions.

In a quasi-static process, on the contrary, the system immediately re-adjusts to an intermediate equilibrium condition at every infinitesimal step.

I hope that was helpful.

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    $\begingroup$ Thanks for the response! I still don't understand whether my example is a quasi-static process or not though? In a quasi-static process, we must have a well-defined equilibrium state at all times during the process. Does this mean that the system must simply be in internal equilibrium at all times during the process or does the system need to be in equilibrium with its surroundings at all times during the process or does it have to be both of these? $\endgroup$ Commented Mar 24, 2021 at 9:27
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    $\begingroup$ In this case you can manage to make the process quasi-static by guaranteeing that the membrane that separates the system from the surroundings is a very poor heat conductor, even though the system is far from equilibrium with the surroundings. The heat transfer can in principle be made as slow as you wish. Something in the way of a Dewar flask: en.wikipedia.org/wiki/Cryogenic_storage_dewar $\endgroup$
    – joigus
    Commented Mar 24, 2021 at 9:32
  • $\begingroup$ Thanks again! Okay so I definitely want to agree with you because your explanation would make more sense to me. That is, for a process to be quasistatic, the system must simply maintain internal equilibrium at all points in time during the process and need not be in equilibrium with the surroundings. Yet Matthew Schwartz from his harvard lectures states that "Or if we take two gases at different temperatures and put them in contact, as heat flows from one to another, equilibrium does not hold, so this heat transfer is not quasistatic." I think what he may mean though is that if you take the... $\endgroup$ Commented Mar 24, 2021 at 11:43
  • $\begingroup$ system to be both gases, then it would not be quasistatic. But if the system is only one of the gasses and they are separated by a boundary with a very poor conductivity, then technically this heat transfer could still be quasistatic. $\endgroup$ Commented Mar 24, 2021 at 11:45
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    $\begingroup$ The example of bringing into contact two gases at different temp's is IMO, a very good one of non-reversible (and thereby non-quatistatic) process. As the system quickly equilibrates, it's neither here nor there thermodynamically speaking. There would be mechanics going on (waves, and such). And mechanics must be "dead", for thermodynamics to proceed. $\endgroup$
    – joigus
    Commented Mar 24, 2021 at 13:05
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I would consider the process you described as quasi static. In fact, in that particular process, both the system and the surroundings experience reversible changes. However, for the low thermal conductivity medium in-between, the process is not reversible, and entropy is generated within this medium. This generated entropy is transferred to the system. So the increase in entropy of the system is greater than the decrease in entropy of the surroundings, and the net result is an increase in entropy of the "universe."

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  • $\begingroup$ Thanks Chet! Okay so as long as the system remains in internal equilibrium throughout the process, the process is a quasi-static one? The system need not maintain equilibrium with the surroundings during the process in order for the process to be called quasi-static but rather the system simply needs to be at a well-defined equilibrium state (in internal equilibrium so to speak) at each point in time during the process so that all state functions and properties are always well defined during the process? $\endgroup$ Commented Mar 24, 2021 at 12:40
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    $\begingroup$ This is a subtle, but extremely important point. $\endgroup$
    – joigus
    Commented Mar 24, 2021 at 13:01
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    $\begingroup$ I would say your assessment is correct. The system passes through a continuous sequence of nearly equilibrium states. $\endgroup$ Commented Mar 24, 2021 at 14:02

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