1
$\begingroup$

In the usual formalism of thermodynamics, the internal energy of system is said to be extensive in a sense that if we construct n identical copies of same system, the energy of the composite system will be n times of the energy of the subsystem.

The above statement is related to multiple copies of “identical system”.

But in some derivation of the equilibrium condition, even two subsystem is not identical, we still assume the additivity of internal energy.

$$E_a+E_b = const.$$

Does this make sense?

If two system when brought into contact (removing the wall) start reaction violently, the energy should be different from the sum of subsystems before removing the wall?

$\endgroup$
  • $\begingroup$ Why do you feel that the total energy is not constant? $\endgroup$ – Chet Miller Jul 27 '19 at 12:40
  • $\begingroup$ I think it is a const but not the sum of energies of two subsystem $\endgroup$ – Z. Sun Jul 27 '19 at 13:03
  • $\begingroup$ If it is constant, how can it not be equal to the sum of the initial energies of the two subsystems? $\endgroup$ – Chet Miller Jul 27 '19 at 13:54
  • $\begingroup$ I was thinking that there will be some "new" interaction which only exist after we move the wall. For example, there are protons in left chamber, electrons in right chamber, and the wall screens the electric force. $\endgroup$ – Z. Sun Jul 27 '19 at 14:08
  • 1
    $\begingroup$ Even so, that doesn't change the total energy. $\endgroup$ – Chet Miller Jul 27 '19 at 14:32
1
$\begingroup$

Extensiveness is always an approximation, to a certain extent. If you bring two systems $A$ and $B$ into contact, there will always be some interaction energy $E_{int}$ such that the total energy $E_{tot}$ will be $$E_{tot} = E_A + E_B + E_{int}$$

The approximation we make is then that $E_{int}$ is negligible compared to $E_A+E_B$. This can be physically justified if we assume that the number of particles participating in that interaction between systems at any given time is much, much smaller than the total number of particles in the system, which would be the case if that interaction were happening at some kind of physical boundary.

As far as chemical reactions go, it's worth remembering that the internal energy also includes the binding energies of molecules. As long as the system is isolated from its environment, the change in internal energy will be zero - it's possible, however, that there will be a trade-off between chemical energy and kinetic energy as the system heats up as the (exothermic) chemical reaction occurs.

$\endgroup$
1
$\begingroup$

I would think it is true if the two subsystems when brought together form an isolated system. That is, a system which cannot exchange mass or any form of energy (heat or work) with its surroundings.

But classically when we combine two system, there still will be some potential energy term arising from the interaction of two subsystem. If such term exists, the energy is not additive right ? Thank you!

When you refer to $E$ I assume you mean the internal energy of the system. The internal energy of a system is the sum of its internal kinetic and potential energies, that is, the kinetic and potential energies at the atomic/molecular level. So a chemical reaction can result in both a change in internal kinetic energy (as reflected by temperature change) as well as potential energy (as reflected in making or breaking chemical bonds). The sum of both is conserved if the combined system is isolated, as defined above.

Hope this helps.

$\endgroup$
  • $\begingroup$ Thank you very much. But if we think this quantum mechanically. H(A tensor B) = H(A) + H(B) + H(A,B interaction) != H(A)+H(B) If we use this argument the energy is not additive ? $\endgroup$ – Z. Sun Jul 27 '19 at 12:26
  • $\begingroup$ Sorry not up on QM. My answer based on thermodynamics, the tag given to your question $\endgroup$ – Bob D Jul 27 '19 at 12:33
  • $\begingroup$ But classically when we combine two system, there still will be some potential energy term arising from the interaction of two subsystem. If such term exists, the energy is not additive right ? Thank you! $\endgroup$ – Z. Sun Jul 27 '19 at 12:37
  • $\begingroup$ @Z.Sun I've revised my answer to answer your follow up question. Hope it helps. $\endgroup$ – Bob D Jul 27 '19 at 13:04
  • $\begingroup$ Thank you very much! $\endgroup$ – Z. Sun Jul 27 '19 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.