Gibbs Mixing Paradox

Gibbs Mixing Paradox

As mentioned in Partition Function and Density of States, we reach a paradox when mixing the many non-interacting classical particles. In classical statistical mechanics, the free energy as derived in (4) is

(7)\[A = -k_B T N ( \ln V - 3 \ln \lambda_T )\]

Imagine we have two systems, one has \(N_1\) particles and volume \(V_1\) while the other has \(N_2\) particles and volume \(V_2\). Now we mix the two systems. Our physics intuition would tell us that the free energy of this new system should be \(A = A_1 + A_2\) since they are non-interacting particles. However, the free energy shown in (7) tells us that

\[A_{\text{mixed}} = \cdots - k_B T \ln (V_1^{N_1} V_2^{N2})\]

which is different from the result we expect, i.e.,

\[A = \cdots - k_B T (N_1 + N_2) \ln (V_1 + V_2) = \cdots - k_B T \ln \left( (V_1 + V_2)^{N_1+N_2} \right)\]

That is, free energy becomes neither intensive nor extensive in our theory.

To make the free energy extensive, we could choose to divide volume by the particle number. Then a new term will appear in the epression for free energy, i.e., \(N\ln N\). On the other hand, we have \(\ln N! = N\ln N -N\) from Sterling approximation. In large systems, we can define the free energy in the following way

\[A = - k_B T N ( \ln(V/N!) - 3 \ln \lambda)\]

which is equivalent to

\[Z_N = \frac{Z_1^N}{N!}\]

This definition “solves” the Gibbs mixing paradox. The explanation of this modification requires quantum mechanics.

Warning

We can’t just pull out some results from statistical mechanics and apply them to a small system of a few particles. In stat mech we use a lot of approximations like Sterling approximation, many of which are only valid when particle number is huge.


Back to top

© 2019, Lei Ma | Created with Sphinx and . | On GitHub | Physics Notebook Datumorphism | Index | Page Source