[kgwgk]

> This adds a lot more microstates that wasn't available before, and none of the old microstates are now possible since all old microstates had a total energy level half of what each new microstate has.

As I'm sure you know, the microstates of that sample of gas at some fixed temperature don't have all the same energy. For each temperature there will be a distribution of possible energies. If the temperatures are close enough there will be a large overlap between those distributions.

You cannot just count the microstates of the sample of gas. (You can count microstates of the gas plus reservoir system though.)

[Jensson]

> As I'm sure you know, the microstates of that sample of gas at some fixed temperature don't have all the same energy.

Depends if you do classical statistical physics or quantum. If you do classical they all have the same energy. If you do quantum you have to weight the states according to their probability densities, and the probabilities that the energy deviates are very small which is why classical works fine even when ignoring those.

But quantum statistical physics is way more complex, you should learn the classical statistical physics first before you try to discuss quantum statistical physics. Classical works a lot like the computer science version where you just count states, quantum doesn't.

[kgwgk]

> Depends if you do classical statistical physics or quantum. If you do classical they all have the same energy.

That's just completely wrong. With all due respect, you should (re?)learn the classical statistical physics :-)

https://en.wikipedia.org/wiki/Boltzmann_distribution

[Jensson]

That is for a subset of a larger system, not for a closed system. A closed system has fixed energy in classical.

So for example, if you have a gas as a closed system, takes a part of that, then you can measure that parts energy distribution by just accumulating the different micro states, yes. But if you view that part as a closed system, then if it has higher energy then it has higher temperature (if it is in the same phase) and thus energy will on average flow out of it to the other parts, if it didn't have higher temperature then it would be a stable system and that part would just have higher energy, which we know doesn't happen in for example gasses, (but it can happen with phase transitions, like an ice cube in water).

Edit: And it doesn't make sense to talk about temperature of subsets of systems. Temperature only deals with what happens when you connect two large systems, it is a macroscopic property, so when you calculate temperature you always deal with calculate that via entropy of closed systems.

So Boltzmann distribution happens when you have calculated the temperature for a macroscopic system as if it was closed, and then you start to calculate properties of some subsystem of that, that is how you get Boltzmann distribution. Not sure how you think these things were derived, have you tried reading a physics book on the subject and gone through how the formulas are derived?

Also, in these calculations you never ever mix two temperatures in a single system, as that isn't stable. The formulas only works for stable states. So if you have two different temperatures you have two different systems.

Edit Edit: I can't post more since I sometimes post stupid political stuff. Anyway, I answered your post already, the formulas we are talking about here doesn't deal with the case where systems of different temperatures interact. They only work for closed systems. If you heat a subset of a room to a higher temperature, then you now have a dynamic system where energy flows out of that subset, none of the formulas we have discussed here applies in that scenario. You can use them to calculate approximate properties of the systems by treating them as if they were closed though, still the Boltzmann distribution doesn't apply there for the entirety of those subsystems, since it assumes that everything surrounding it has the same temperature, which in your scenario the surroundings do not.

Anyway, all of these formulas are derived based on completely closed systems with no transfer of energy or particles. that is the source of entropy calculations, if you want to understand entropy you have to deal with such systems. Boltzmann's law is an example of a formula derived from entropy calculations, you can't use that to talk about entropy.

(Note, I learned these things in another language than English, so I might use the wrong words for some things, but I know how statistical physics, temperature and entropy works, are derived etc)

[kgwgk]

> A closed system has fixed energy in classical.

Wrong again!

https://en.wikipedia.org/wiki/Closed_system

"In thermodynamics, a closed system can exchange energy (as heat or work) but not matter, with its surroundings."

Maybe you meant "isolated" instead of "closed" - but then I don't know what we have been talking about. (When I say "we" I mean "you".)

> It doesn't make sense to count the microstates for a volume of gas at some pressure and temperature.

A volume of gas at some pressure and temperature is a closed system but not an isolated system.

> I can have a container with 1l of some gas at room temperature T1 and proceed to heat the room - and the container - to temperature T2.

The containers in that example are closed systems but not isolated systems.

> the microstates of that sample of gas at some fixed temperature don't have all the same energy

That sample of gas at some fixed temperature is a closed system but not an isolated system.

Hopefully you'll agree with everything I wrote with the understanding that we were not talking about isolated systems.

[ have _you_ tried reading a physics book? :-) ]