How slow is a reversible adiabatic expansion of an ideal gas? A truly reversible thermodynamic process needs to be infinitesimally displaced from equilibrium at all times and therefore takes infinite time to complete.  However, if I execute the process slowly, I should be able to get close to reversibility.  My question is, "What determines when something is slow?"
For definiteness, let's take an insulating cylindrical piston with cross-sectional area $A$ and original length $L_0$.  There is an ideal gas inside with $n$ molecules of gas with mass-per-molecule $m$.  The temperature is $T_0$, and the adiabatic index is $\gamma$.
I plan to expand the piston adiabatically to length $L_1$, taking a time $t$ to do so.  If I take $t$ to be long enough, the process will be nearly-reversible.  However, $t$ being long does not mean "one minute" or "one year".  It means $t >> \tau$ for some 
$$\tau = f(A, L_0, L, n, m, T, k_B, \gamma)$$
What is $\tau$?
From purely dimensional considerations, I guess the relationship is something like 
$$\tau = \sqrt{\frac{mLL_0}{k_bT}}f(n, L/L_0, A/L^2, \gamma)$$,
but I don't have a strong physical explanation.
Edit
A meaningful answer should let me do the following:  I take a certain example piston and try expanding it a few times, putting it in a box so I can measure the heat released to the environment.  I calculate the entropy change in the universe for the expansions.  After doing several expansions, each slower than the last, I finally get $\Delta S$ for the universe down to a number that I think is sufficiently small.  Next, I plan to repeat the experiment, but with a new piston that has different dimensions, different initial temperature, etc.  Based on my results for the previous piston, how can I figure out how long I should take to expand the new one to achieve a similar degree of reversibility on the first try?
For reference, the pressure is
$$P = \frac{nk_BT}{V}$$
and the speed of sound is
$$v = \sqrt{\frac{\gamma k_b T}{m}}$$,
and I'm happy to have an answer in terms of these or other derived quantities.  Formulas for entropy and thermodynamic potentials can be found in the Wikipedia article.
 A: I am a student so please point out in gory detail anything I did wrong.
For a process to be quasistatic, the time scales of evolving the system should be larger than the relaxation time. Relaxation time is the time needed for the system to return to equilibrium. 
We have an adiabatic process, so equilibrium must be preserved at each point, that is to say
(Working within the validity of the kinetic theory for ideal gases and ignoring friction)
$(A L(t))^\gamma P(t)  = (A L_0)^\gamma P(t_0)$ 
Momentum gained by the piston: 
$\Delta p =  2 m v_x$
A molecule would impact the piston every 
$\delta t = \frac{2(L_0+ \delta x) }{v_x}$
The force exerted on the piston is $F =\frac{\Delta p}{\delta t} = \frac{m v_x^2}{L_0+\delta x}$  Pressure would be $P = \frac{P}{A}$ and for $N$ such molecules
$P = \frac{N m <v_x>^2}{A (L_0+\delta x)} = \frac{N m <v>^2}{3A (L_0+\delta x)}$ 
So at the instant $t=t'$ where the piston has been displaced by $\delta x$, we have 
$(A L(t))^\gamma P(t) = \frac{N m <v>^2}{3A^{1-\gamma}} (L_0+\delta x)^{\gamma -1}$
Expanding in series
$ = \frac{N m <v>^2 L_0^{\gamma-1}}{3A^{1-\gamma}} (1 + \frac{(\gamma-1) \delta x}{L_0}+ O(\delta x^2) )  $
Substituing $\frac{\delta x}{L_0} = \frac{\delta t v_x}{2 L_0} -1$
$(A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + (\gamma-1) (\frac{\delta t v_x}{2 L_0} -1))$
If we want our process to be reversibly adiabitic atleast to first order, we must have from above 
$\delta t = \frac{2 L_0}{<v_x>}$
Now, this is time until collision for the starting case. Investigating second orders
$ (A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + \frac{(\gamma-1) \delta x}{L_0}+ \frac{1}{2} (\gamma-1)(\gamma-2) (\frac{\delta x}{L_0})^2   +O(\delta x^3) )  $
Looking at just the series terms 
$ 1 + (\gamma-1)\frac{\delta x}{L_0} (1 +\frac{1}{2} (\gamma -2) \frac{\delta x}{L_0}) \approx 1$
This would be true for 
$\delta t = \frac{4 L_0}{<v_x>} (\frac{1}{2-\gamma} -1)$
Now, this is the "time until next collision" for a gas molecule hitting the piston. To maintain reversibility, at least to second order, the piston should be moved from $L_0$ to $L_0 + \delta x$ in time $\tau = \delta t$ so that the system variables follow the adiabatic curve. 
The $<v_x>$ can be calculated from the maxwell distribution
A: This is not a direct answer to the question but rather a slightly different perspective on this adiabatic expansion. I am not sure how correct it is. 
So, let us assume that the piston moves(in direction of $x$-axis) infinitely slowly with velocity $\vec{v}_p$. Let a molecule flies to the piston at a velocity $\vec{v}_k$. With respect to the piston, its velocity will be $\vec{v}_{k_{rel}}=\vec{v_k}-\vec{v_p}$. Normal component(relative piston) of relative velocity is $(v_{k_{rel}})_x=v_{kx}-v_p$. Let's denote by $\vec{v'}_{k_{rel}}$ the velocity of the molecule with respect to the piston, after reflection. Tangential component of relative velocity as a result of the reflection does not change, and the normal changes sign.  $$(v'_{k_{rel}})_x=-(v_{k_{rel}})_x=-v_{kx}+v_p$$ Let's denote by $v'_k$ the velocity of the molecule relative to the fixed cylinder walls after reflection. Its normal component is $v'_{kx}=(v'_{k_{rel}})_x+v_p=-v_{kx}+2v_p$ and the tangential component is the same as that of the velocity $\vec{v}_k$. As a result of the reflection from the piston, the
kinetic energy of the molecule is incremented:  
$$\frac{1}{2}m(-v_{kx}+2v_p)^2-\frac{1}{2}mv_{kx}^2=-2mv_pv_{kx}+2mv_p^2$$ Let's denote by $n_k$ the number of molecules per unit volume whose velocities are approximately equal to $\vec{v}_k$. Number of hits these molecules on the piston during the time $dt$ is equal to $z_k=An_k(v_{kx}-v_p)dt$ where A is area of the piston. As a result, the kinetic energy of molecules in this group in time dt will increase:  
$$(-2mv_pv_{kx}+2mv_p^2)An_k(v_{kx}-v_p)dt=-2mn_k(v_{kx}^2-v_p^2)dV$$ where $dV=Av_pdt$ is an increase of the volume of gas for the same time.  
The increment of the kinetic energy of the whole gas:$$dE_{kin}=dU=-dV\sum_{v_{kx}>0}2mn_kv_{kx}^2+2dVv_p^2\sum_{v_{kx}>0}mn_k$$ Here $U$ is the internal energy of an ideal gas. The summation is only for those groups of molecules, which move in the direction of the piston. When summarizing of all groups of molecules, moving like a piston, and from him, then the sum should be divided in half. In this case: $$dU=-dV\sum mn_kv_{kx}^2+dVv_p^2\sum mn_k$$ But by definition, the first sum is the pressure of the gas $P$ and the second sum is simply the density of the gas $\rho$. Thus we get a differential equation:  
$$dU+PdV=\rho v_p^2dV$$ Internal energy of an ideal gas may be expressed as follows: $$U=\frac{f}{2}PV$$ where $f$ is a number of degrees of freedom of molecule. Using the fact that the adiabatic index is $\gamma=\frac{f+2}{f}$, the differential equation can be rewritten:  
$$\frac{dP}{dV}+\gamma\frac{P}{V}=(\gamma-1)\frac{\rho}{V}v_p^2$$ If $v_p=0$ then we get from it the adiabatic equation: $PV^{\gamma}=const$  
Because $\rho$ depends on the pressure and temperature it is impossible to integrate the differential equation directly. But for small shifts of the piston we can assume that the density is approximately constant, i think.
A: The good answer to your question was indeed a condition on the velocity of the piston much lower than the average molecule velocity. To understand why, you need to study kinetics and fluid theories. From Boltzmann's equation one can deduce the fluid equations that give rise to classical thermodynamics. The passage from the kinetics scale to the fluid scale is only valid if macroscopic time scale and macroscopic gradient lengths are much greater than the microscopic relaxation time and the particle mean free path.
A: The funny thing is that the answer to your particular question is not even "one minute" or "one year". Expansion/contraction of gases is effectively reversible for the regimes where hydrodynamic description is valid, that is gas motion is governed by Navier-Stokes equations.
The easiest way to see this is to remember how is the formula for the speed of sound you mention is derived:
$$v = \sqrt{\frac{\gamma k_b T}{m}}$$
You assume air to contract/expand adiabatically and assume no dissipation, that is the work done by pressure goes entirely to the internal energy and vice versa. And you come to the acoustic wave equation. So expansion/contraction of the gas is reversible to the extent the sound propagation is described by the usual wave equation.
The primary reason for the phenomenon is that for gases second viscosity is zero in the assumptions of kinetic theory. Actual value is above zero, but it is neglected in practice. In fluid dynamics second viscosity $\zeta$ measures entropy production due to expansions/contractions:
$$\sigma_\zeta = \frac{\zeta}{T} (\operatorname{div} \boldsymbol v)^2$$
Here $\boldsymbol v$ is the fluid velocity. So it's not the expansion that causes irreversibility for the gas in piston. There are two other sources of entropy in a fluid flow. The first one is heat conduction:
$$\sigma_{\lambda} = \frac{\lambda}{T^2} (\operatorname{grad T})^2$$
If you have no temperature gradients, it is zero.
The other one is arises from shear viscosity:
$$\sigma_\mu =\frac{2 \mu}{T} \left[\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{1}{3} \frac{\partial v_k}{\partial x_k} \right]^2 $$
The expression above is written in Cartesian coordinates, repeating indexes mean summation, $a_{ij}^2 = a_{ij} a_{ij}$.
I think it is possible to construct a piston where no shear stresses near the wall will arise, so the entropy production would be zero.
To answer your question, we may assume gas expansion in piston reversible if the entropy produced is small compared to the overall entropy
$$\Delta S = \int_0^t \int_V \sigma(\boldsymbol r, t') \; d \boldsymbol r dt'  \ll S$$
this is the condition for $t$.
Once again, the explanation above is true whenever hydrodynamic description is valid, if you have shock waves, continuum description is not applicable for part of the region.

Let's assume $\zeta$ be non zero. Than entropy produced would be
$$\Delta S = \frac{\zeta}{T} \left[\frac{|L_1 -L_0| / t}{L} \right]^2 t A L \sim \frac{1}{t}$$
So making expansion slow really reduces entropy produced.
