[kgwgk]

Is it easy to see how the number of microstates changes when you increase the temperature - everything else being equal?

How would you say that it changes then?

I'd say that the number of compatible microstates doesn't change. The probability of each microstate does change though.

[Jensson]

Your statement doesn't make sense, temperature is defined in terms of entropy changes, you can't calculate temperature without first calculating entropy changes.

[kgwgk]

Have you heard of thermometers? 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.

How do you calculate the number of microstates for the sample of gas before and after? How do you think these numbers are related? You said it was easy!

[Jensson]

Thermometers is a way to measure temperature, it isn't its theoretical definition. Temperature is defined as energy required per change in entropy. There is no other reasonable way to define temperature, since at its core it measures which way energy flows when two macro systems are connected. Temperature tends to go up as you add energy to things, but not always, for example temperature doesn't go up when you melt ice, it starts and ends at 0 degrees C.

> How do you calculate the number of microstates for the sample of gas before and after? How do you think these numbers are related?

If you added energy to the gas by heating it, lets say you doubled the energy, then you now have twice as many energy packets to distribute between the particles. 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. You can calculate the change in states yourself, it is just discrete normal probability theory. Note that the base rate isn't interesting, you care about the change of the logarithm of number of states.

[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