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
The number $\Omega$ of microstates accessible to the particle system is defined as :
$$
\Omega=\omega(E)\Delta
$$
where $\omega(E)$ is the density of state. To determine this guy, you need first to calculate the number $\chi(E)$ of µ-states corresponding to an energy inferior or equal to $E$.
Let us call the associated phase space region $\Sigma_{[0,E]}\equiv\lbrace\Gamma,0\leqslant\mathcal{H}(\Gamma)\leqslant E\rbrace$.
Then, it basically reads :
$$
\chi(E)=\frac{1}{h^{3N}}\int_{\Sigma_{[0,E]}}\mathrm{d}\Gamma \quad \text{with} \quad \mathrm{d}\Gamma=\prod^N_{i=1}\mathrm{d}\textbf{q}_i\,\mathrm{d}\textbf{p}_i
$$
$h^{3N}$ is here the element phase space volume corresponding to one µ-state.
It is quite straighforward to compute :
$$
\chi(E)=\frac{1}{h^{3N}}\left[\prod^N_{i=1}\int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}\textbf{q}_i\right]\left[\prod^N_{i=1}\int_{\Sigma_{[0,E]}}\mathrm{d}\textbf{p}_i\right]=\frac{1}{h^{3N}}\,V^N\times V_{3N}(\sqrt{2mE})
$$
where $V_{3N}(\text{r})$ is the volume of a $3N$ dimension hyperball with a $\text{r}$ radius, and $\Sigma_{[E,E+\Delta]}$ is the phase space region $\lbrace\Gamma,E\leqslant\mathcal{H}(\Gamma)\leqslant E+\Delta \rbrace$.
For a 3D gas, we have for a given particle $i$ :
$$
\int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}q_{i,x}\mathrm{d}q_{i,y}\mathrm{d}q_{i,z}=V\quad\text{volume of the gas}
$$
and for $N$ particles :
$$
\int_{\Sigma_{[0,E]}}\prod_{i=1}^N\mathrm{d}p_{i,x}\mathrm{d}p_{i,y}\mathrm{d}p_{i,z}=V_{3N}(\sqrt{2mE})
$$
by integrating over all possible direction of the total momentum $\sum_i\textbf{p}_i$ and all magnitudes below the energy shell $\lbrace\Gamma,\mathcal{H}(\Gamma)=E \rbrace$ so that :
$$
\left|\sum_i\textbf{p}_i\right|=\sqrt{\sum_i\textbf{p}_i^2}=\sqrt{2mE}\quad\text{with}\quad E=\frac{1}{2m}\sum_i\textbf{p}_i^2
$$
Since all direction are possible, the integration is performed over a $3N$-ball.
Then, it follows :
$$
\omega(E)=\frac{\mathrm{d}\chi}{\mathrm{d}E}(E)=\frac{1}{h^{3N}}\,V^N\times S_{3N}(\sqrt{2mE})
$$
Best Answer
It looks to come from the Sakur-Tetrode relation of the entropy, $$ S\sim\ln\left(P\rho^{-\gamma}\right)=\alpha\ln K $$ The constant of proportionality $\alpha$, in this case is the heat capacity at constant volume, $C_v$ (see also this answer of mine that briefly discusses the relation), which follows $C_v\propto1/(\gamma-1)=3/2$.