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If I were to instantaneously remove the wall as drawn above, then there is no loss of molecular velocities --> T1 = T2.

But I have a problem imagining that there really is no change in temperature. If I put a thermometer in the system with the wall still inside, wouldn't there be significantly more molecules bumping the thermometer in a period of time than in the larger volume because it is more concentrated there? Or would I have to imagine a thermometer that somehow covers the entire system?

I imagine that as more time passes between two impacts of a molecule with the thermometer (which must be the case with larger volumes) the measured temperature is lower.

I know that the temperature must remain the same, as nothing changes in the mean kinetic energy, but I would like to have my error of reasoning found, please, can someone tell me what is wrong?

(I strongly suspect that my thought is wrong, that the thermometer "cools down" again in the time between impacts, that mercury actually has no possibility to transfer heat when there is nothing there (there is nothing to transfer between two impacts), in this case I have the following question: Suppose I had a body and would transfer a lot of energy to it in the form of heat, thereby its temperature rises strongly to T0, if I now instantaneously convey it into a perfect evacuated space, would T0 remain constant?)

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If I put a thermometer in the system with the wall still inside, wouldn't there be significantly more molecules bumping the thermometer in a period of time than in the larger volume because it is more concentrated there?

Yes that is correct, but what it means is the pressure with the wall still inside is greater than the pressure after the wall is removed because the number of collisions per unit time is greater with the wall in place due to the smaller volume.

But the thermometer doesn't measure the pressure exerted on it by the gas. It measures the average kinetic energy of the gas molecules bumping into it. That doesn't change simply because the volume increases allowing more room for the same number of molecules having the same average translational kinetic energy to move between collisions.

The assumptions are (1) the chamber in your diagram is rigid (cannot expand or contract doing work so $W=0$ (2) it is thermally insulated so there is no heat $Q$ transfer with the surroundings, so $Q=0$, and (3) the gas behaves like an ideal gas so that its change in internal energy $\Delta U$ depends only on temperature change $\Delta T$.

Then, from the first law, since $\Delta U=Q-W$, $\Delta U=0$ and $\Delta T=0$.

Hope this helps.

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  • $\begingroup$ Thanks, so my assumption at the end of my question in the (..) is correct? $\endgroup$
    – iwab
    May 14, 2022 at 2:21
  • $\begingroup$ Not sure what you mean by (...) $\endgroup$
    – Bob D
    May 14, 2022 at 2:28
  • $\begingroup$ (I strongly suspect that my thought is wrong, that the thermometer "cools down" again in the time between impacts, that mercury actually has no possibility to transfer heat when there is nothing there (there is nothing to transfer between two impacts), in this case I have the following question: Suppose I had a body and would transfer a lot of energy to it in the form of heat, thereby its temperature rises strongly to T0, if I now instantaneously convey it into a perfect evacuated space, would T0 remain constant?) $\endgroup$
    – iwab
    May 14, 2022 at 11:22
  • $\begingroup$ @iwab There are probably billions of gas molecule impacts with the thermometer per second. There is virtually no time between impacts for the thermometer to "cool down"> $\endgroup$
    – Bob D
    May 14, 2022 at 12:30
  • $\begingroup$ But imagine only one single molecule in a 1m^3 isolated system. $\endgroup$
    – iwab
    May 14, 2022 at 13:26

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