# Proving that the free energy is extensive

If I have two system of an Ideal gas $A$ and $B$ each of these system has a partition function:

$Z_{A,B} = \left ( \frac{V_{A,B}}{\lambda_T} \right )^{N_{A,B}}$

Where:

$\lambda_T = \left ( \frac{m}{2\pi\beta \hbar } \right )^{\frac{1}{2}}$

The free energy is:

$F_{A,B} = -kT \ln \left ( Z_{A,B} \right ) = -kT N_{A,B}\ln \left (\frac{V_{A,B}}{\lambda_T}\right)$

For the free energy to be extensive the following must be true:

$F_{A} + F_B = F_{A+B} \Rightarrow Z_A \cdot Z_B = Z_{A+B}$

However:

$Z_A \cdot Z_B = \left (\frac{V_A}{\lambda_T} \right )^{N_A} \left ( \frac{V_B}{\lambda_T} \right)^{N_B}$

and:

$Z_{A+B} = \left ( \frac{V_{A+B}}{\lambda_T} \right )^{N_A + N_B}$

So, for $Z_A + Z_B = Z_{A+B}$ to be true $V_A^{N_A} V_{B}^{N_B} = \left(V_A + V_B \right)^{N_A + N_B}$ must be true as well but this isn't true for any system.

Since we cannot create energy by mixing two containers of Argon in the same pressure and temperature, something in my understanding is wrong. Where is my fault?

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Isn't the right spelling "Arragorn"? –  Georg May 22 '11 at 14:58
What you observe is Gibbs' paradox. The resolution comes about by postulating that the particles are indistinguishable (and thereby introducing a factor $1/N!$). Then the entropy and the free energy becomes extensive (in the thermodynamic limit $N,V\to\infty$, $N/V=\text{const}$).
No, undistinguishability of particles has nothing to do with that. The point is that the thermodynamical definition of entropy is given up to an arbitrary function of the number of particles (it is defined using a differential relation, with fixed number of particles). So the same is true when you define the entropy in stat. mech., since this is done by analogy with the thermodynamic one (they satisfy the thermodynamic relations). The only reason why you need an N! (or something like $N^N$) is that you want entropy to be extensive. This is an additional assumption, which does not always hold. –  Yvan Velenik Jun 5 '12 at 14:17