# Why does rapidness come in the examples for irreversible processes?

One of the well known examples for reversible vs irreversible process is the thought experiment of having a million stones over a piston that encloses a gas chamber and removing them one by one slowly. This slowness in removing the pebbles vs removing all of them together is cited as a difference in the way reversible vs irreversible process works. But this is misleading.

There is no time factor involved in definition of reversible processes, even mathematically. This is all about driving force isn't it?

also, what bugs me is even if you remove one small pebble a time from the piston so that heat released is infinitesimally small, all the way from point A to point B, can't the sum of all such infinitesimal heat losses be significant? even if the individual steps produced very less heat(due to friction), the fact that you have a million such steps, makes the cumulative heat loss still significant. so how is doing it slowly makes a difference? I need a better explanation to understand this

• Your discomfort is understandable; to paraphrase an old saying, you can't be a little bit irreversible. Once energy spontaneously moves down a gradient (in temperature, voltage, pressure, whatever), entropy is generated and reversibility is lost. The examples of slow processes are intended merely to approximate what reversibility would look like if we could observe it in the real world, which we can't. – Chemomechanics May 2 '17 at 15:05
• THanks, yes, this lack of ability to visualize is what makes it difficult for me to grasp the concept of reversibility. I am not a fan of mathematical explanations, more a fan of simple thought experiments that can intuitively explain the concepts. That pebbles on a piston experiment does a great job of almost explaining it(thanks to Andre who posted it in another question) but still leaves some questions for me – daraj May 4 '17 at 5:01
• This answer of mine to another question might be helpful: physics.stackexchange.com/questions/211782/… – Steeven May 25 '17 at 21:36

What you need to do is to remove an (infinite) number of (infinitesimally) small pebbles making sure that after each removal of a small pebble the system is allowed to reach an equilibrium state with the surroundings before the next pebble is removed.
The time factor is there as you have to wait for the system to reach an equilibrium state before removing the next pebble.

In practice what is needed is the system responding to any changes in much less time than the time scale of the changes.

• when you are "allowing system to reach equilibrium" you are in disequilibrium, even if the irreversibility is really small. tHat's what bugs me – daraj May 2 '17 at 8:55
• also in a reversible process, useful work derived is maximum. why is this so? what is "useless" work then? is useful work, work done on the surroundings by the system (in the form of PV expansion, heat transfer etc.)? – daraj May 2 '17 at 8:59
• going back to the piston/pebble example, if the vessel was frictionless, then even if I removed 1000 pebbles at the same time can I still get useful work via expansion? if not, what causes entropy increase? gas cooling? – daraj May 2 '17 at 9:01
• @daraj The problem is the transient states. By having infinitely small changes over long periods of time, you have infinitely small transient states which are able to reach equilibrium. As you start increasing the magnitude and duration of the changes, the transient effects become more important, affecting the results and reversibility. – JMac May 2 '17 at 11:11
• Jmac, so why do transient effects lead to entropy loss, irreversibility? – daraj May 4 '17 at 4:55

I would like to state, $$\Delta G=\Delta H-T\Delta S$$ Where $\Delta G$ is the spontaneity measure by the heat and $\Delta S$ the Entropy Change of the System $\Delta H$ is the heat of the system Here we consider the case if the process is Irreversible Which means it is spontaneous.So by which $$\Delta H-T\Delta S \lt 0$$ $$\Rightarrow\Delta H \lt T\Delta S$$ $$\Rightarrow\frac{\Delta H}{T} \lt \Delta S$$ Hence entropy Is Depending upon the heat loss When ($\Delta H \lt 0$) The process is said to be Irreversible as the(There is a increase in entropy)So when($\Delta H \gt 0$) The process has highly increased its entropy so we can say this process Remains Irreversible.

Which proves that every Spontaneous or quick process is Irreversible.Hence rapidness is also one of the Irreversible process.

Now For time Factor. $\Delta H$ is the heat transfer of this process and hence when the heat is transferred slowly the $\Delta H \lt\lt 0$. Which is already taken above. Now the pebble case the $\Delta H \lt\lt 0$ As where i consider the heat transfer to be equal to the heat because the time is larger
$\Delta Q t=\Delta H$ .Here $\delta Q$ is very small and the t is greater (process occurring slowly).So which agrees with the point of Above proof for irreversibly. Moreover if $\delta Q$ is small then the temprature change would be negligible.And hence $\frac{\Delta H}{T}$ is large so $\Delta S$ would be larger.

• Rishi, you said "ΔH=δQ/t" which is a bit confusing as dimensionally it is not correct. Where is the time unit in enthalpy definition? I think it is more to do with magnitude of driving force than time. If driving force is very small, then naturally time taken will be very large. – daraj May 4 '17 at 4:57

In a very rapid deformation, a gas behaves very differently from when it is deformed slowly. The force per unit area that the gas exerts at the piston face depends not only of the gas volume but also on the rate of change of volume. That's where time comes in.

Moreover, the gas pressure is not even constant within the cylinder; it depends on spatial location. Even at the piston face, the force per unit area is not given by the ideal gas law. The force per unit area on the piston face is equal to the local ideal gas value $\rho RT$ (where $\rho$ is the local density), plus term related to viscous stresses in the gas (that are proportional to the local rate of gas deformation).

So, in summary, because a gas behaves differently in rapid deformations, the work that it does on the piston is also very different.

This is a very smart question! Yes, reversibility does not have any intrinsic reference to time, but no, reversible processes are slow in practice. Let's talk about why.

We'll just start with the pebbles on a piston. What realistically happens if you pull the pebbles away too fast? Well, it's like making the pebble suddenly vanish: afterwards, the piston jumps up a tiny bit. When this happens, the piston immediately starts vibrating like a damped harmonic oscillator, and eventually the damping provided by the air and the walls of the piston leads to an energy transfer which increases entropy overall.

To minimize the vibration, you have to extract all of that would-be vibrational energy as work when you remove the pebble; this requires the forces to be applied slowly so that the piston is moving arbitrarily slowly when the pebble loses contact, so that the piston is left at rest at its new equilibrium point. In fact it should be moving arbitrarily slowly throughout the process to minimize friction losses as well.

Now what's the general principle here? The general principle is that we're reducing a source of pressure pushing down on the piston $P_0 \mapsto P_1$, and the resulting pressure gradient $P_1 - P_0$ between the gas and our pressing, causes the piston to move and the object to change volume with some $dV/dt.$ The gas loses energy at some rate $-P_0~dV/dt$ but we're harvesting energy at some rate $P_1~dV/dt$, so to recoup as much energy as possible we want $P_0 = P_1,$ or $P_1 - P_0 \to 0.$ That's what we want for reversibility.

Well if the pressure gradient is really driving the volume change then we'd expect something like $\frac{dV}{dt} = -\alpha~(P_1 - P_0),$ so reducing then reducing this difference means changing the volume over a very long time interval as $P_1 - P_0 \to 0.$

But the great thing about this is that it is such a generic argument. You might know that isothermal processes are reversible. What does this really mean? It means that you touch two objects at the same temperatures together, and let them exchange their thermal energy. "B.S.", you should be calling: "if they have the same temperature they shouldn't be able to trade internal energy." Very true. The above argument tells us that actually what we mean is the limit of a process where both objects have similar temperatures, but slightly different, so that they trade energy very very slowly. It has to be slow because $\frac{dE}{dt} \propto T_1 - T_0.$