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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?)

iwab
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1 Answers1

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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.

Bob D
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  • Thanks, so my assumption at the end of my question in the (..) is correct? – iwab May 14 '22 at 02:21
  • Not sure what you mean by (...) – Bob D May 14 '22 at 02:28
  • (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?) – iwab May 14 '22 at 11:22
  • @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"> – Bob D May 14 '22 at 12:30
  • But imagine only one single molecule in a 1m^3 isolated system. – iwab May 14 '22 at 13:26
  • To quote from Rococo's answer on the following post: https://physics.stackexchange.com/questions/65690/can-a-single-molecule-have-a-temperature "For a single molecule that is in complete isolation, it is indeed generally not true (or at least not useful) to assign it a temperature" – Bob D May 14 '22 at 13:38
  • Ok, that threw me off track a bit, so to reshape my picture, imagine so many molecules in a large volume that there is a noticeable amount of time between the collisions between the molecules and the themometer, say 0.5s. In between, the material of the thermometer would have no time to give off energy again in the form of heat ("it cools down"), because there is nothing to transfer, right (because there is no molecule that would somehow have a lower momentum to take a little off the thermometer.)? – iwab May 14 '22 at 14:43
  • @iwab, it is not the case that a collision heats up and lack of collision cools down. Collisions can cool or heat. Reducing the collisions reduces the rate of transfer, not the total amount (or direction) of transfer. – BowlOfRed May 14 '22 at 17:44
  • @iwab Whether or not something warms up or cools down depends on the average direction of the transfer of kinetic energy due to multiple molecular collisions, not due to the transfer of kinetic energy due to a single molecular collision. On average, kinetic energy is transferred from the warmer object to the colder object per the second law. – Bob D May 14 '22 at 18:32
  • But at the molecular level, it is possible, in the case of a single collision, for a higher energy molecule in the cold object to transfer kinetic energy to a lower energy molecule in the warmer object (the kinetic energies of individual molecules can vary around the average per the Boltzmann distribution). That doesn’t mean the warmer object warms up and the colder object cools down. Because cooling or warmer is an average effect of many collisions and that average effect is in the direction of warm to cold, in the absence of work. – Bob D May 14 '22 at 18:32