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
 A: 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.
A: 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.
A: 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.
A: 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 $\frac{dE}{dt} \propto T_1 - T_0.$
