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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.

sbs95
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3 Answers3

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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.

eranreches
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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.

FGSUZ
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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.

Alex Alex
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