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I'm reading a thermodynamics textbook which states the following (translation from spanish):

The fundamental relation $E=E(S,T,n_1,...,n_r)$ is a first-order homogenous function of the extensive variables $S,V,n_1,...,n_r$. That is, for each value of $\lambda$ the following relation is satisfied: $$E(\lambda S,\lambda V,\lambda n_1,...,\lambda n_r)=\lambda E(S,V,n_1,...,n_r)$$

Is this true? Why? The author immediately proceeds to explore the consequences of this, but he doesn't account for the reason why it is the case in the first place.

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3 Answers 3

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That's exactly what an extensive quantity is: if you increase the system size times $\lambda$, i.e. $\lambda$ times the entropy, the volume and the particle numbers, then the new system has $\lambda$ times the energy of the old one. Just because it is 'composed' of $\lambda$ old systems.

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  • $\begingroup$ Why is the "system size" specifically? Is scaling it the same as scaling every possible extensive variable? Does that mean that there is no possible extensive quantity that is a higher order homogenous function, for example? Thank you! $\endgroup$
    – sbs95
    Commented Dec 17, 2017 at 0:42
  • $\begingroup$ @sbs95 If this function was homogeneous of higher order, say $2$ for example, then if you'd double the system - you'd quadruple the energy, which doesn't make sense since energy is additive. I am not sure I understand your other questions though. $\endgroup$
    – eranreches
    Commented Dec 17, 2017 at 0:49
  • $\begingroup$ The author of the book I'm reading says at one point that, for energy to be additive out of equilibrium, then both subsystems have to interact in some way, and they should be able to exchange energy. That would mean that the total energy wouldn't just be the sum of both energies, but you would also have to include some energy of interaction $\epsilon$. That is, unless you neglect that energy (which is a standard approximation, according to the author). $\endgroup$
    – sbs95
    Commented Dec 17, 2017 at 1:53
  • $\begingroup$ @sbs95 That's because there can be interactions that occur when you attach two small systems together, for instance. Anyway, keep in mind that you'll mostly deal with systems in equilibrium in first courses on thermodynamics, so the assumption of additive energy is fine. $\endgroup$
    – eranreches
    Commented Dec 17, 2017 at 12:17
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It is a direct consequence of the fact that Internal energy is additive respect the energy of its subsystems.

In other words, if you duplicate the volume, you can see the "new system" as the sum of two identical "initial systems", beause duplication implies also duplicating the number of particles, and other extensive quantities. Consequently, you will have

$U_{Total}=U_1+U_2=2 U_1 = \lambda U_1; \ \ \lambda=2$

This is works for every number you use to "re-escalate" your original system.

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  • $\begingroup$ About this, the author also says that the additivity of energy is only satisfied, out of equilibrium, if the system's subsystems have some energy of interaction at the surface of separation $\varepsilon$ that is negligible in comparison with both subsystem's energies. What does this imply for your answer? Thank you! $\endgroup$
    – sbs95
    Commented Dec 17, 2017 at 0:44
  • $\begingroup$ Sure, but I was talking about equilibrium. If you are in not-equiilbrium, then your comment is the correct one. It's just about common sense: you can use the same reasoning if the effect of frontiers is negligible. $\endgroup$
    – FGSUZ
    Commented Dec 17, 2017 at 14:07
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Take a look at Why internal energy $U(S, V, N)$ is a homogeneous function of $S$, $V$, $N$?. So for every extensive function $F(x_1,x_2,...)$ of extensive variables $x_1, x_2,...$, $F$ will be homogeneous function of $x_1, x_2, ...$. Internal energy is extensive function, entropy, volume and quantities are too, so internal energy is homogeneous function of entropy, volume and quantities.

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