[Eternal Student (OP)]

Hi.

    Just had a minor thought about temperature which seemed a bit odd, so I would like to have it checked by others.

Background:   
   Temperature, T, is a measure of the average kinetic energy  < 1/2 mv² >   of some particles in a substance.
For an ideal gas we can assume the usual relationship:
      The average kinetic energy of one particle =  (3/2) K.T        [with K = Boltzman constant,  T = temperature in kelvin]

The problem:
   Now suppose you have an ideal gas in an insulated container so that no energy gets in or out.   So that's a big box full of little particles like peas and they have some temperature T, that is to say the peas are moving around.   The exact kinetic energy of a given pea varies but on average we have the usual relationship:   The average k.e. = 3/2 kT.
    The problem is that you didn't buy high quality peas, you got the "two-for-one" bargain peas.   If you bounce them around then they just break in half.   Now, the new half-sized peas are what you have bouncing around in the box.  Those are the new particles you have, each identical and for simplicity exactly half the size of the original particles BUT there are, of course, twice as many of them.

   Now the question is - does the temperature T change at the moment when the peas (particles) split into two?

This is my reasoning:
   Energy is conserved, the container is insulated.
   Initial total kinetic energy = E and this remains the final total k.e.  of all the particles at the end.
   So  the average k.e. of a particle at the end (when there were 2n particles instead of just n) is   E/2n,    while it was  E/n  initially.
   Now just re-arrange the usual formula     (average k.e. of one particle) = (3/2)kT     using E/n    and  E/2n  for the initial and final averge energy of a particle.

   So ,  initially we have   T_i =  (2/3k) .(E/n)
    At the end we have,   T_f  = (2/3k) . (E/2n)    =     T_i  / 2

The temperature must fall to half it's initial value when the particles split into two.   (Have a think about it, seems a bit odd to me - but it keeps coming up any way that I look at it).

- - - - - - - - - -

Some other things to consider:
      Since each of the final particles is precisely half the mass of the initial particles,  we have that the average speed of a particle does not change.   
[Briefly,     (2n).  (1/2) m_f V_f ²     =    E      =      (n).  (1/2) m_i V_i ²,
substitue  m_f = final particle mass =  (1/2) m_i     to obtain   V²  is unchanged and hence  |V| is unchanged  ]
    So you can imagine that two halves of a pea started off just by holding hands to make one large pea.  In that state the gas had a temperature T_hot.  If the two half-peas just decide to let go of each others hand, then it's not as if they need to abruptly part ways with some significant relative velocity like a Uranium nucleus undergoing fission.   Just drifting along at much the same velocity will change the temperature to T_cold  just because the total number of individual particles has changed.

- - - - - - - - - -
Why it matters:
   Well, I just read a school chemistry book that was talking about releasing energy when N₂ is formed from atomic Nitrogen.   It said an increase in temperature is evident in the reaction   N + N --> N₂. 
   The number of particles present has halved so you'd expect the temperature to double even if no energy was released during bond formation.
   Furthermore, going back to Uranium and run-away nuclear fission reactions, once everything is hot enough to be gaseous or genuinely atomised, the effect of doubling the number of particles on each fission helps to prevent any further rapid increases in temperature.   That's quite nice, although it's already hot enough to be a bit of a problem but you've got to be grateful for small things.

   I'd be grateful if someone could check this reasoning.

Best Wishes. 
     

[paul cotter]

Hi ES, i'm shooting in the dark here as physical chemistry was one of my weakest disciplines(among many!): I suggest you analyse in terms of enthalpy, I would expect the temperature to go down and the pressure to increase. I now patiently await denunciation.  SCRAP that, no change in pdv, but I still think pressure will increase.

[Bored chemist]

Imagine just one particle.
It falls apart into two but, thanks to momentum and energy conservation, the two parts must carry on along almost identical paths (and their CoG must follow exactly the same path).

So, each one has half the mass, but the same speed.
So it has half the  energy.

(and the overall energy is the same- obviously).

The temperature of a gas is a measure of the average energy per particle.
And there are now twice as many particles with only half as much energy each and that, in principle, means that the temperature has fallen by half.

I don't think it's a realistic scenario, but it's an interesting bit of thermodynamics.

[Bored chemist]

I'm now trying to think of a situation where the effect would occur.

I think it's impossible.
If the "pairs" were so weakly held together that you could ignore the "bond energy" between them, then they would fall apart as soon as they bumped into each other.
If the binding energy was high enough to avoid that, then you could no longer ignore it when the molecules split apart- they would have a bit less kinetic energy after the split.

[Eternal Student (OP)]

Thanks to all.    I'll reply later (probably).  I must get to a train station now.

Best Wishes.

[Petrochemicals]

Endo or exo thermic molecular bindings would have an effect, but half the mass at half the energy gives the same temperature?

[Eternal Student (OP)]

Hi.

   Well actually I'm not sure there is more that needs to be said.   Thanks for checking the reasoning, it seemed a little odd.

....but half the mass at half the energy gives the same temperature?

     It's numbers of particles that seems important.   When someone says "half the mass" it isn't clear if you had half the number of particles    OR   the same number but each was half the mass.   Similarly,  "half the energy" could be that the total kinetic energy is halved     OR    that each particle had half the average kinetic energy.

    Depending on which combinations of these things you take,  the temperature may change.   At least one combination shows no change:   Half the total kinetic energy  AND  half the number of particles    --->   That results in the same average k.e. per particle   and thus the same temperature.

Best Wishes.

[Petrochemicals]

Hi.

   Well actually I'm not sure there is more that needs to be said.   Thanks for checking the reasoning, it seemed a little odd.

....but half the mass at half the energy gives the same temperature?

     It's numbers of particles that seems important.   When someone says "half the mass" it isn't clear if you had half the number of particles    OR   the same number but each was half the mass.   Similarly,  "half the energy" could be that the total kinetic energy is halved     OR    that each particle had half the average kinetic energy.


Best Wishes.

Sorry about that. I was writing from an ideal stand point, half the mass any particles you like, similarly half the energy refers to an equal share by measurement of mass content. I suppose if you had dinitrogen being split into singular, if you had managed to seperate the atoms into singular they may well have a different specific heat capacity at the same energy quotent.