Thermodynamics behind a vacuum

I have a perfectly insulated container with $N$ molecules of air at an initial temperature $T_{0}$ and pressure $P_{0}$. Imagine that I now have a set of thermodynamic 'tweezers' that allows me to remove individual molecules from the container (i.e. i'm trying to mimic a vacuum pump type of device). How do I work out how my pressure and temperature will change as I remove molecules from the container?

Attempt:

$dU = dW + dQ + \mu dN$

Am I allowed to say that $dW=0$ since $dW = - P dV$ and since $dV=0$, then $dW=0$?

Also, can I say that $dQ = 0$ since it's perfectly isolated? I would initially think no because as I remove molecules from the container, I'm also removing some heat as I do this, correct? Or is this completely accounted from in the thermodynamic chemical potential/Fermi energy $\mu$ - i.e. the energy added to system for every one molecule added (if that's the correct definition).

I'm just a bit rusty on my thermodynamics. I feel like I'm also not given proper consideration to the fact that the processes are irreversible.

Any help/guidance is appreciated.

BTW if $dW=0$ and $dQ=0$, then the solution I get is something like:

$\Delta T = \frac{-\Delta N (T - (\mu/1.5k_{B}) )}{N + \Delta N}$

after integrating, etc...

this basically says as I remove molecules, my temperature of the system will go up. But I thought creating a vacuum should make the temperature go now (like in space)

• You might want to read the answers to Temperature in space. Pressure and temperature are not intrinsically related to one another (though they are coupled in many system), you can have any combination of pressure and temperature that you desire in most system (Fermi degenerate system are special). Feb 21 '16 at 19:46