What is the entropy generated by opening up a vacuum chamber of volume $V$ to the atmosphere? (let's say with some constant ambient pressure $P_{atm}$)
I was reading about Joule expansion, but it doesn't seem to apply as easily in this case since the environment always stays at a constant $P_{atm}$ and more so seems to have a rather "infinite" volume that isn't really much affected by expanding into some small additional space $V$ of a vacuum chamber. So, the $(V_f/V_i)$ or $(P_i/P_f)$ terms in the formula for free/isothermal expansion entropy there always just tend to 1 thus leading to zero entropy generation (per the formulas there). But, this is clearly an irreversible process, since gas wouldn't spontaneously exit out on its own and recreate the vacuum after filling up the chamber!
Also, in order to create a vacuum of volume $V$ in the first place — we'll need to perform $W = P_{atm}V$ amount of work (e.g, raising a piston originally at the bottom end of a cylinder). However, as the surroundings "fill up" a vacuum, the pressure inside $V$ starts to rise and soon it becomes harder and harder for the surroundings to "push" each additional unit of gas in (since it's no longer expanding against a pure vacuum $P_{internal} = 0$). So, does that mean some work is actually done by the environment against the vacuum chamber as well? (Maybe, even temporarily "heating up" some of the gas inside as well, adiabatically?)
On one hand, if filling up the chamber was a reversible process, we could probably "extract" $W=P_{atm}V$ amount of work back from the environment filling the chamber back up. But, in our case of "just opening a valve" to let air back in, the environment initially does no work expanding against a pure vacuum $P_{internal} = 0$, and only at the very end would have to try to push gas against a $P_{internal} = P_{atm}$ (when the chamber is essentially already filled up!). In essence, only doing what looks like $W = P_{atm}V/2$ onto the system instead…
Best Answer
This is a great twist on the Joule-expansion or free-expansion problem. We’re familiar with the result there that the average temperature is ultimately unchanged, as the total gas does no work on anything (it’s adjacent to a vacuum) and no heating occurs. As you note, entropy is generated because this dispersion is irreversible.
The notable twist here is that the gas being pushed into the box certainly has work done on it by the atmosphere. We conclude (and experimentally verify) that that gas heats up. (In balance, the atmosphere cools down, but this is negligible because the atmosphere is large.) Heat transfer then equalizes the temperature, producing the free-expansion outcome.
How much entropy is generated? Just replace the free expansion with an easier-to-analyze process that produces the same state. You’ve already done half of this—reversible collection of work through quasistatic expansion—so just tack on the step of irreversibly dumping that energy magnitude back in as thermal energy (by passing electrical work through a resistor, for example). Follow me?