Consider a piston head in a arbitrarily long cylinder. One side of the cylinder is sealed and pressurized, the other side has no pressure. How would you model the acceleration or velocity of the piston head with respect to time?
I began with the gas law $PV = P_0V_0$. P and V are instantaneous values wheras $P_0$ and $V_0$ are initial values, thus constants. I used the relations of $P = \frac{F}{A}$ and $F = ma$ to rewrite it in terms of acceleration.
$$a = \frac{P_0V_0A}{Vm} \rightarrow A = \frac{aVm}{P_0V_0}$$
Solving for $A$, area of the piston head, will come in handy in a bit. To model the changing volume I came up with this expression:
$$dV = A\ ds$$
$s$ is the displacement the piston head. Then I substituted the value of $A$ in the top equation into the lower one. (Note that you can use this same approach with $dt$ instead, but using $A$ was more straight forward. Both approaches gave me the same result in the end.)
$$dV = \frac{Vm}{P_0V_0}a\ ds$$
By definition, $a\ ds$ equals $v\ dv$ and with some other algebra, the equation can be rearranged to:
$$ \frac{dV}{V} = \frac{m}{P_0V_0}v\ dv$$
And finally integrate:
$$\int_{V_0}^{V}\frac{dV}{V} = \int_{0}^{v} \frac{m}{P_0V_0}v\ dv$$
$$\ln{V} - ln{V_0} = \frac{m}{2P_0V_0}v^2 - 0$$
$$ v = \sqrt{\frac{2P_0V_0}{m}\ln{\frac{V}{V_0}}}$$
I found a fallacy with this equation. At initial conditions, the acceleration is infinite. My intuition tells me the velocity should be asymptotic, but its logarithmic, and that the acceleration should be an exponential decay, but its inverse. What did I do wrong?