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Assume a closed and adiabatic (isolated) system (no energy or mass transfer at the boundaries) then is it possible for the average temperature to change (increase or decrease)? Corollary: 1. If yes how? Give example of a process. 2. If not. Assume a system that encloses a system that converts mass into energy (Ex. nuclear reaction). The result of the nuclear reaction is producing energy inside the closed system and decreasing the mass inside the closed system. I expect that the resulting increase of energy represents a increase in the internal energy and thus an average temperature increase. Then what is the process that offsets that to maintain the average temperature constant?

I will add an example which we were talking about. Assume a thermodynamic boundary - system - which does not exchange energy or mass with the surrounding (isolated). That closure includes a nuclear plant and the necessary energy sources to prime and sustain the nuclear reaction. All this system is isolated and has an average temperature T before initiating the reaction. The reaction is initiated - without any input of energy from outside - and a mass to energy conversion takes place. At this point the average temperature inside this system is going up, or stays the same? The argument for the temperature to go up would be that the conversion of mass to energy produces some sensitive heat which increased the average temperature of the system. But is that violating the first law - as at the boundary no energy exchange took place? If the average temperature stays the same then this heat production should be offset by something to keep the average the same. What is that?

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  • $\begingroup$ I did not get your last line? $\endgroup$ – Anubhav Goel Oct 13 '16 at 16:21
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    $\begingroup$ I mean, you're assuming that the temperature is equal to the average kinetic energy of the molecules in the gas, which is fine as long as you have an ideal gas. But as soon as you have potential energy (like that locked in nuclear bonds), you no longer have an ideal gas, and the temperature is no longer just a measure of average kinetic energy. $\endgroup$ – march Oct 13 '16 at 16:44
  • $\begingroup$ @AnubhavGoel I would say your question is correct and I can elaborate on it but that would make sense if the answer is yes. Is that the answer? Thanks! $\endgroup$ – Mihai Oct 13 '16 at 17:34
  • $\begingroup$ @march It is an hypothetical question but per my understanding ideal gas means perfect particles which have perfect collisions but they can undergo mass to energy conversions. To answer the questions I am not necessarily assuming ideal gas although in my opinion that would be a perfectly good model for a first step. $\endgroup$ – Mihai Oct 13 '16 at 17:38
  • $\begingroup$ Yeah, but if you have mass to energy conversions, the definition of temperature as the average kinetic energy doesn't work, because there are degrees of freedom other than just translational, vibrational, or rotational ones in the problem. $\endgroup$ – march Oct 13 '16 at 17:40
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First of all, let's go with the assumption that you are defining the temperature as the average kinetic energy per particle. This is the case for ideal gases where the constituent particles have no internal structure and don't interact with each other, and in this case, the temperature is a direct measure of the average energy in the system.

If the system is closed and isolated, then the total energy of the system is conserved. In the case where we allow nuclear decay, some mass is converted into a different form of energy, and in the end this increases the kinetic energy of the "ideal" gas in the system, which means that the temperature increases, even though the total mass-energy is the same.

The reason there's a disconnect here is that on one hand, we want the temperature to be a measure of the average energy of the system, but on the other hand, temperature-as-average-kinetic energy only works as this measure if we have an ideal gas where the constituents don't have any internal structure.

The upshot is that if you really want to do thermodynamics where nuclear decay goes on in the system, you have two options:

  • Redefine your temperature so that it takes into account the nuclear degrees of freedom.

  • Use the ideal gas temperature, and model the nuclear decay process as heating. That is, model your system as not isolated, and model the nuclear decay process as heating from external sources. The problem here is that you might also need to introduce chemical potentials if the nuclear decay is one in which there is transmutation, i.e. where you are changing the numbers of constituent particles.

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The argument for the temperature to go up would be that the conversion of mass to energy produces some sensitive heat which increased the average temperature of the system.

Correct. You're converting energy in one form to energy in another form. A simpler example would be burning wood inside the system.

But is that violating the first law - as at the boundary no energy exchange took place?

There's no violation. Energy is converted from one form (chemical or nuclear potential energy) to another form (thermal energy).

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Especially in an adiabatic process, you have to say with the fact that there is no heat exchange, where as if you have a mass to energy conversion, energy is given off as radiation which if not contained (well somehow, I personally don't think you can contain gamma rays because eventually the heat will be lost by the boundary by conduction) will simply break the definition of the process as adiabatic. So, if the energy is contained then it must be given to the gas (somehow).

On the other hand if you want to increase the temperature to change, then, the gas will have to expand or compress (against external pressure), thereby losing (& gaining) internal energy which (in case of ideal gas) is primarily temperature. Process remains adiabatic.

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  • $\begingroup$ All that is correct but I am also talking about an isolated system - no energy or mass exchange with the surrounding. Per my understanding your answers do not account for this. $\endgroup$ – Mihai Oct 13 '16 at 17:44
  • $\begingroup$ An isolated system doesn't allow the energy to leak out. So it has nowhere to go unless in the form of internal energy. We are assuming an ideal gas here. If it were some hypothetical radioactive molecule (gaseous), then we could conclude about its vibrational energy increasing an all that, but it is all essentially an increase in the internal energy of the gas relatable to the kinetic energy too $\endgroup$ – Pranshu Malik Oct 13 '16 at 17:48
  • $\begingroup$ I think I have given a description of this in the text, is it something else you're trying to point out? I am sorry I couldn't get your context clearly. $\endgroup$ – Pranshu Malik Oct 13 '16 at 17:49

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