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
Specific heat capacity is defined for a substance not for a process. Specific heat capacity depends on the structure of the substance but, it ($C_P$ for example) can be measured by the formula $$C_P=\left(\large{\frac {\partial h}{\partial T}}\right)_P=\left(\large{\frac {\partial q}{\partial T}}\right)_P$$
$C_P$ isn't a function of heat ($q$) or kind of process (isobaric, isochoric, isothermal, etc.)
As an obvious example, we can measure the volume of a container by measuring the volume of water that can fill it. But, all of us know that volume of a container depends on its dimensions not on volume of the water (If we don't have any water, doesn't container have a quantity called volume?!!:-)
Best Answer
An adiabatic (zero heat exchanged with the surroundings) process is reversible if the process is slow enough that the system remains in equilibrium throughout the process.
Mathematically, the characteristic relaxation time has to be smaller than some characteristic time of action of the applied process: $\tau_{relax} << \tau_{process}$.
The system then equilibriates faster than the process is able to change it.
This $\tau_{process}$ can be linked with the speed $v$ of a changing system size $\Delta x$ (e.g. a moving piston) as $\tau_{process} \approx \frac{\Delta x}{v}$.
In your case this is violated due to the rapid expansion, thus rendering the process irreversible.