My book says

A process is said to be reversible if it is carried out infinitesimally slowly so that in each step thermodynamic equilibrium of the system remains unchanged & any infinitesimal change in the condition can reverse the process to restore the initial state of the system & universe . Here the driving force is infinitesimally greater than opposing force.

I have some confusions with this definition.

  1. In the definition,they have mentioned 'step of the process'. Is 'step' at an instant or for a period of time? If it is for instant,how can I explain thermal equilibrium and mechanical equilibrium (which are to be maintained for reversibility)?

  2. Suppose the process connects two states of a gas in a cylinder connected with piston headed by lead shots. The initial and final pressures of gas can be given as $P_i$ & $P_f$ . When we remove some lead shots, the gas does work to go to balance the force by moving the piston so that the pressure of the piston and that of gas become equal . How can I then explain the existence of mechanical equilibrium during the process? (It is after the process the pressure becomes equal.)

  3. In the last,they have mentioned about driving force and opposing force. What are they? Why the former is greater than the later? I know it is big but it will be a great help if anyone explain me the three questions (or confusions) . The definition is very intricate. Help.


3 Answers 3


This is a very interesting, important and at the same time subtle matter which is useful not only in thermodynamics, but other areas as well, and not everyone quite understands it.

First off, two statements that you'd better just memorize for now:

  1. In a reversible process, equilibrium is never left. Yes, that is a paradox. That's why all real processes are irreversible. However, a real, irreversible process can approximate a reversible one.
  2. A reversible process does not waste work: no other process can do more work going from state A to B of a substance than a reversible one. That also mean that all irreversible processes are wasting work, that is, when carrying out an irreversible process, you're losing the opportunity to have even more work done.

Equilibrium, Reversibility and Driving vs Opposing Forces

Let me try to give you an informal but intuitive notion of what reversibility has to do with equilbrium and balance between driving and opposing forces.

Equilibrium is achieved when a driving force in some direction is countered by an opposing force in the opposite direction. The concept of force here can be expanded to include whatever wants to steer any situation to a particular direction. Let's work with the example you gave:

Gas in cylinder with pressure $p_i$ under piston of area $A$ over which lie one thousand lead shots each weighing $w = \frac{p_iA}{1000}$.

The gas wants to expand (driving force), but the thousand shots exactly oppose it (opposing force). All is well and in equilibium here; nothing moves or changes.

Now imagine that you removed one shot. For a brief moment, the force in the piston due to the gas pressure would be marginally bigger than that due to the shots' weight, such that the piston would accelerate upwards a little, thus expanding the gas. This expansion would lead to a small decrease in pressure, which would eventually equal the current weight (99,9% of the original). In other words, for a brief period of time, the driving force trumps the opposing force. Then we have equilibrium again.

If your goal is, say, $p_f = \frac{1}{2}{p_i}$, you can keep doing that until only 500 lead shots are left. Each removal is one step of your process, and note that every step is a small period of disequilibrium between two equilibrium states. We can say that our example process is an approximation of a process that never leaves equilibrium (in this case, mechanical equilibrium!). Also note that our process did some work!

On the other hand, what would happen if, instead of patiently removing one lead shot at a time, we abruptly took away 500 shots? Well, suddenly the driving force (gas pressure on piston) would be twice as big as the opposing force. The piston would crazily accelerate upwards and the gas would expand rapidly. After quite some time, of course the same state would be reached: $p_f = \frac{1}{2} p_i$, $v_f = 2 v_i$. Note however, that the work done is smaller! In this case, only 500 shots were elevated to the final height, while, in the former, the same 500 shots were elevated PLUS 500 others were left along the way, at each height of each step of the process.

The former process (slow) is a better approximation of a reversible process than the latter (fast). Note that, when something is at an equilibrium, a very small push is all you need to tip the state in the direction you want, minimizing time spent in disequilibrium. You don't need to roundhouse-kick it to the desired direction. The more the driving force surpasses the opposing force, the greater the irreversibilities of the process. However, it does happen faster that way.

We can also think of an example involving thermal equilibrium. Suppose you want to raise the temperature of an object from 0 to 100 degrees. If you just put it in contact with an object at 100 degrees, you'll have a driving force disproportionately bigger than the opposing force — that is, nature is DYING to heat that cold object. However, if you used one hundred objects, each at a different temperature, to incrementing our temperature 1 degree at a time, the process would be more gentle. We stay in equilibrium more. Our newfound intuition tells us that this process is closer to a reversible one.


We can sum this up answering your points:

  1. A step would be a period of time where the system moves toward its goal. In a reversible process, the driving forces (e.g. temperature or pressure differences) would be so small that each step would take forever. In a real process, generally, the "more reversible" it is, the longer each step will take.
  2. This is an idealization which can be approximated by having a series of "small disequilibriums" instead of a large pressure difference in each side of the piston (akin to a free expansion)
  3. These are abstract terms (i.e. may not be real forces). The driving force is that which leads the process in the direction you want. The opposing force is that which leads it in the reverse direction. Like pressure differences in a mechanical problem, or temperature differences in a heat transfer problem. The greater the difference between these forces, the farther from equilibrium the system will be, and the process will generally happen faster and more violently. This also means a less reversible process.

What the book is saying is that your system has to be in thermal equilibrium at every point in time during this process. A quasi-static process is slow with regards to the relaxation-time of your system.

If your system is in thermal equilibrium (constant N), that means it can be described by 2 variables - i.e. pressure P and volume V. A system that changes from state A to state B by means of a quasi-static process, can be described by a line on a P-V-Diagram. In case of a reversible process, you can 'go through' this line in both directions.

In order for a process to be reversible, it has to be quasi-static. Otherwise there'd be turbulence and temperature-fluctuations that would lead to irreversible heat-dissipation. Being quasi-static alone is not sufficient, though: there are also non-reversible quasi-static processes (temperature equalization over a bad conductor, Gay-Lussac experiment, etc.). For it to be reversible, the process has to be isentropic (ΔS = 0).

Regarding (2): Why would you need your system to be in mechanical equilibrium during the process? This is about thermal equilibrium only.

(3): Not sure, what exactly the author is trying to say here. I guess, he/she wants you to think of the system as being out-of-equilibrium by an infinitely-small amount? Perfectly reversible processes only exist in theory.

  • $\begingroup$ For it to be reversible, the process has to be isentropic: I disagree, $\Delta S_\text{system}$ can be nonzero as long as $\Delta S_\text{surroundings}$ is the opposite of it. $\endgroup$ Sep 23, 2014 at 21:35
  • $\begingroup$ Whut? o.O How is the total entropy change that the process is causing not zero if $ΔS_{system} = -ΔS_{surroundings}$? $\endgroup$
    – MuH4hA
    Sep 23, 2014 at 21:52
  • $\begingroup$ Let me try again: I understand what you said in your post as $\Delta S_\text{system} = 0$, but what I think is right is $\Delta S_\text{system} + \Delta S_\text{surroundings} = 0$. $\endgroup$ Sep 23, 2014 at 21:59
  • 2
    $\begingroup$ This is a non-issue. Yes, you are right, the process also needs to be adiabatic, to be isentropic - but nobody said anything about surroundings until you came along, so we'll just imagine a nice, perfectly isolated system. (I thought it would be clear, I meant the total change in entropy this process is causing - should've been more specific. sry). However, I don't think this is the part OP is having trouble with, so this discussion serves no purpose. $\endgroup$
    – MuH4hA
    Sep 23, 2014 at 22:48
  • $\begingroup$ The author also never mentioned entropy. There's really no need to that confusing oversimplification. $\endgroup$ Sep 23, 2014 at 22:55

generally a process is said to be reversible if it responds fastly to a change in state than the change is occurring;and the physical properties unaltered. the above definition's opposite is irreversible

  • $\begingroup$ I don't believe there is any need to address a question answered 4 years ago $\endgroup$ Dec 10, 2018 at 17:11

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