captdallas: “It is wrong because you are not solving the same problem Rob posed.”
We are looking at the same problem, and evaluating the end state after the “compressed gas” has cooled at the same temperature. I don’t think the gas remembers that it was warmer in the past, though I understand it’s part of the argument for the conclusion. Still, I am evaluating the end state: two containers of identical volume, same temperature, and pressures 1atm (“uncompressed”) and 2atm (“compressed”). I am saying that they have the same average kinetic energy per molecule but I now see that you really are agreeing with Rob.
“Temperature is a measure of the average energy of collisions. You can have lots of collisions with lower energy or fewer with more energy.”
Yes, this seems to agree with Rob. I am denying this. I am rather going with standard physics textbooks and saying that (kinetic) temperature is a measure of average kinetic energy per molecules. This is standard classical statistical and thermal physics, not QM.
“So you are looking at the average energy per container not the average energy per molecule.”
Do you mean the volumetric density of kinetic energy? Or the gross molecular flux of KE per surface area units? The latter, not the former, would relate rather more to the collisions with the surface, though it seems irrelevant to temperature, in my opinion. (It is relevant to rates of heat transfer).
“The 2g container has twice the molecules at half the energy per molecule.”
So you also are agreeing with Rob that both containers have the same total thermal energy content.
This would mean that the container with twice the molar amount of air (hence twice the mass), at the same temperature, though twice the pressure, has the same total internal energy content.
Are you really believing this to be true?
“Sometimes it is better to start simple, restate the actual problem and use the simplest approach before jumping into quantum physics.”
Agreed.