2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Are you confusing a change in temperature with an exchange of heat from an external source/sink? Adiabatic expansions and compressions can and do change the temperature of a gas. The requirement for no external heat source/sink is part of the adiabatic definition of reversibility and no entropy change. $\endgroup$ – honeste_vivere Mar 26 '18 at 13:24
  • $\begingroup$ My understanding is that the internal energy drops as the gas performs work on the piston, and this energy loss is due to collisional transfer of energy from the gas to the piston. In a real-world irreversible expansion, there is overshoot and rapid dampening of the piston from the transferred momentum, resulting in the piston equilibrating at its new reduced pressure position, and excess energy (diff btw the work of irrev vs rev) was lost as heat. I'm wondering if in the reversible case, is there still an infinitesimally-small overshoot for each reduction of weight on the piston? $\endgroup$ – lamplamp Mar 26 '18 at 16:07
  • $\begingroup$ I assume you are thinking about something similar to when you accelerate in a car that has balloons floating in it, the balloons will move toward the front of the car due to the pressure gradients in the air? Meaning, during the expansion, the rush of gas toward the newly opened end will carry slightly more gas toward the open end than is necessary to evenly fill the container? If so, I do believe that is how it works but the time scales of things like this would be on the order of microseconds at STP. $\endgroup$ – honeste_vivere Mar 27 '18 at 11:30
  • $\begingroup$ When a finite weight is removed from a compressed spring, it elongates and overshoots, going into an oscillation, until dampening establishes new length of compression = new reduced Force /spring constant. The difference in stored energy is lost as heat. In the limit that the reduction of weight approaches zero, the overshoot oscillation should approach zero. My question is if you integrate all of these tiny overshoots in a reversible infinitely-long decompression of a gas without heat exchange, does this integrate to zero, which would seem to be necessary since heat can't be lost. $\endgroup$ – lamplamp Mar 27 '18 at 16:30
  • $\begingroup$ Are you talking about a purely elastic spring with non-zero mass? $\endgroup$ – Chet Miller Feb 20 at 12:10
3
+50
$\begingroup$

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 $N \to \infty$.

$\endgroup$
  • $\begingroup$ Thanks for the explanation! I've never seen this issue discussed either, and I was thinking about the cooling that occurs in the cloud in a bottle experiment, and what that means in terms of molecular collisions. I studied pchem in Atkins text and reversibility was presented with a vague statement about the system always being in equilibrium, but that didn't offer much insight into the mechanics of energy transfer. Can you recommend a text that addresses issues like these in more detail? Thanks. $\endgroup$ – lamplamp Mar 29 '18 at 20:58
  • 1
    $\begingroup$ @lamplamp The intuitive thing here I just thought up, I haven't seen it anywhere. I think for the full math you'd want a course in kinetic theory, like this, which basically does the calculation I did but for all the molecules at once. Learning about the 'adiabatic theorem' in quantum mechanics also helps, because it's closely related to adiabaticity in statistical mechanics. For general conceptual stuff (like what entropy and heat are) I recommend anything written by E.T. Jaynes. $\endgroup$ – knzhou Mar 29 '18 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.