See Ján Lalinský's commment on Entropy $dQ=TdS$ and Work $dW = -pdV$ conditions?

My question are:

  1. Is there a way to prove that $dQ=Tds$ is a quasistatic process?

  2. What's the funcitonal description for quasistatic process?

  • 1
    $\begingroup$ Quasistatic is not enough, it also has to reversible. The example of @Jan_Lalinsky is misleading because he ignores the irreversible entropy production in the heat-conductor connecting the two masses of different temperatures. It is true that the higher temperature body loses $q=T_{high}dS$ internal energy and entropy ("heat") but the lower temperature body gains more than that, it also gets the excess entropy generated in the conductor, i.e., $s_{irrev}=q(1/T_{low}-1/T_{high})$. This process is always quasistatic but unless the latter is negligible and neglected it is irreversible. $\endgroup$
    – hyportnex
    Commented Sep 26, 2018 at 14:50
  • $\begingroup$ @hyportnex I think quasi static and internally reversible(no friction losses) is enough for the equation to hold.The external reversibility (isothermal heat transfer)is not necessarily needed.Else we will not be able to apply these equations to any other cycle other than Carnot cycle. $\endgroup$
    – Mohan
    Commented Sep 26, 2018 at 18:54

2 Answers 2


Since it is necessary to first have a functional understanding of a quasi-static process I will attempt to answer question 2 first.

A quasi-static process is one that is carried out such that disequilibrium is minimized at each step in the process. Such a process is also a reversible process if it does not involve friction. Such disequilibria typically involve pressure and temperature differentials.

For example, heat transfer requires a temperature difference between two things. A temperature difference is an example of disequilibrium and it is that disequilibrium which drives the heat transfer process. To reduce the disequilibrium you need to reduce the temperature difference. That of course slows down the rate of heat transfer. In order to eliminate the disequilibrium, the temperatures would have to be equal. But then you would not have heat transfer. This demonstrates that in real life all processes are necessarily irreversible. We can only approach reversible processes in the limit.

Now regarding question 1.

Let's first rearrange the equation

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

This equation does not necessarily mean a quasi-static (reversible) process is involved unless it is accompanied by an explicit statement that $dQ$ is a reversible transfer of heat. That stipulation is explicit in the following commonly used definition of $dS$:


Where $dQ_{rev}$ means a process that involves a reversible (quasi-static) transfer of heat. Since entropy is a property of the system, the difference in entropy of the system between two equilibrium states does not depend on the process being quasi-static. However the total entropy change, system + surroundings, does depend on whether or not the process is quasi-static (reversible). The total will $=0$ if the process is quasi-static (reversible) and will be $>0$ if not quasi-static (irreversible). So I think the relevant “proof” you would want to know is how this is true.

The following is not necessarily a “proof” but rather an example of why a process needs to be quasi-static (reversible) in order for the total entropy change (system + surroundings) to be, or approach, zero.

Consider a system H (a hot body) and its surroundings C (a cold body). Further consider both H and C to be thermal reservoirs, that is, they are so massive relative to the desired heat transfer, that heat transfer between them doesn’t change their temperatures. The temperature of body H is $T_H$ and the temperature of body C is $T_C$. We bring the bodies together and desire to transfer heat $Q$ from H to C. Since the temperature of either does not change, the heat transfer occurs isothermally. Let’s look at the entropy changes:

For Body A (System): $$\Delta S_A = -Q/T_H$$ (which is a drop in entropy)

For Body B (Surroundings): $$\Delta S_C = +Q/T_C$$ (which is a rise in entropy)

The total entropy change $$\Delta S_A + \Delta S_C = -Q/T_H + Q/T_C$$

Then for any $$T_H > T_C$$ $$ Q/T_C - Q/T_H> 0$$

In order for the total entropy change to approach zero, the process must be carried out quasi-statically, meaning the temperature difference must approach zero. This of course, results in the heat transfer rate approaching zero and the time it takes to transfer Q infinitely long. This is why in order to have quasi-static (reversible) processes those processes must be carried out very slowly

Hope this helps

  • $\begingroup$ But I have learnt that quasi static process and reversible process are a bit different.If you take an isobaric expansion(piston made to expand slowly) then the process is not reversible but quasi static.You can plot the states in a graph and obtain a curve.But this doesnot satisfy the reversible condition because, during expansion heat should be transferred to the system by a hotter body and during contraction, heat should be rejected to a cooler body(cooler than our system).At the end after bringing back to initial state, the net change in entropy of the universe will be positive. $\endgroup$
    – Mohan
    Commented Sep 25, 2018 at 17:05
  • $\begingroup$ en.m.wikipedia.org/wiki/Quasistatic_process ,even in this wikipedia link, it is explained that quasi static process is a bit different from reversible process.And I think quasi static doesnt necessarily mean isothermal heat addition but reversible does.Please explain. $\endgroup$
    – Mohan
    Commented Sep 25, 2018 at 17:11
  • $\begingroup$ A quasi-static process occurs slowly enough that the system can remain in internal equilibrium. Any reversible process is necessarily a quasi-static process. A quasi-static process that is not reversible is one that involves friction, such as a quasi-static compression or expansion against a piston that is not frictionless. In my experience the terms are used interchangeably with the assumption that friction is not involved. I will edit my answer to reflect this. Hope this helps. $\endgroup$
    – Bob D
    Commented Sep 25, 2018 at 17:16
  • 1
    $\begingroup$ Neither am I an "expert". You didn't say anything wrong. On the contrary, you made me clarify that quasi-static is reversible only if doesn't involve friction (or other aspects that increase overall entropy, though I only know of friction). Thanks. $\endgroup$
    – Bob D
    Commented Sep 25, 2018 at 18:44
  • 1
    $\begingroup$ @silverbackgorilla Yes, your reasoning is correct. Take a look at my answer to the link below, in particular the figure that demonstrates that the more slowly the process is carried out, the nearer a reversible process occurs. Hope it helps. physics.stackexchange.com/questions/530018/… $\endgroup$
    – Bob D
    Commented Sep 29, 2021 at 23:04

I think, if the process is not quasi static, the properties will not be the same through out the system(i.e. the properties will change from point to point within the system).If we discuss about the pressure of the entire system or temperature of the entire system,entropy change of the system, etc.. means we are discussing about a quasi static process.This is why, quasi static processes are said to be made up of equilibrium states.These states can be plotted as points in a graph(PV or TS graphs).Since the terms T and dS in your equation are the properties of the entire system , you are describing a quasi static process.


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