Why doesn't reversible adiabatic expansion generate heat? In analogy to a compressed gas allowed to reversibly expand and do work, I've been thinking about the behavior of a compressed spring that is very slowly and incrementally relieved of its pressure.  At each incremental removal of a delta mass, the spring will expand to a new equilibrium length by expanding a delta x.  However, the spring has also acquired a tiny bit of momentum and I'm curious about whether the spring always overshoots the equilibrium, perhaps by (delta x)?  If this process is repeated but infinitely slowly, do those overshoots add up (integrate) to a value > 0?  If the integrated overshoot value is greater than 0, then to get the spring to the final resting equilibrium after an infinite time, heat dissipation would be needed to dampen the overshoot.  
However, a gas can undergo reversible adiabatic expansion as described in pchem books, which suggests a well-defined final state, and yet by definition, heat exchange cannot occur.  So does this disprove the idea of adding infinitesimally small overshoots?
 A: I like this question! We rarely think about these processes so explicitly, but they can teach us a bit about what's going on microscopically. I'll consider a very concrete model and solve it.
Let's divide the expansion into $N$ pressure increments, tuned so that ideally, each one would move the piston by $L/N$. Let these increments occur at time $t_i$, and assume no dissipation whatsoever. Let $x(t)$ be the 'overshoot amplitude' of the piston. Roughly speaking, the equation of motion is
$$\ddot{x} + \omega^2 x = \sum_{i=1}^N \frac{L}{N}\, \delta(t - t_i)$$
where we get a 'bump' driving force every time we do one of the increments. The final result is
$$x(t) \sim \frac{L}{N} \sum_{i=1}^N \sin(\omega (t - t_i)) \, \theta(t - t_i)$$
and your question is about what the typical amplitude $A$ of $x(t)$ is after all $N$ increments in the limit $N \to \infty$. Let's consider several cases.
Case one: fast expansion. Suppose all of the $t_i$ are within a time interval smaller than about $1/\omega$. Then all the terms are roughly in phase, so the final amplitude is $A \sim L$, ruining the adiabatic approximation. We knew this already, because adiabatic processes must be slow compared to the internal dynamics of the gas. 
Case two: random expansion. Let's say the $t_i$ increments are random and large. Then the phases will be approximately evenly distributed, so the amplitude will almost certainly be about $A \sim (L/N) \sqrt{N} = L/\sqrt{N}$, which vanishes as $N \to \infty$. Thus the overshoots cancel themselves out. (Here, 'almost certainly' means 'ignoring exponentially unlikely events', as we always do in thermodynamics. The random phases could coincidentally align in the same sense that an egg could uncook itself.)
Case three: synchronized expansion. Suppose the $t_i$ increments are large, but synchronized in phase. Then the final amplitude is $A \sim L$, again ruining the adiabatic approximation. This seems unrealistic, but it is not only realistic but ubiquitous -- this is a model of heat transfer!
Think about how a gas picks up heat from a hot wall. On the microscopic level, the wall itself is jiggling, in such a way that when a gas molecule hits the wall, it bounces off with a higher speed than it had before. Our model is exactly the same, but since we've crudely ignored dissipation the gas only has one degree of freedom, $x(t)$. So this case is a model of heating up the gas. It doesn't happen in an adiabatic process, because by definition such processes exchange no heat.

Summary: the two ways for the overshoots to not cancel out are going too fast, and allowing heat transfer. A real adiabatic process has neither of these features by definition, so the overshoots will cancel out, producing no dissipation in the limit of smooth expansion $N \to \infty$.
