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From statistical mechanics one cab derive the change of entropy of the mixing of two ideal gases with the result

$$\Delta S_\text{mix} = - n R \Sigma_{i} x_i \ln(x_i)$$

where $n$ is the total amount of mols of gas, $R$ is the gas constant and $x_i$ are the mol ratios of the various gases.

My main question has to do with the interpretation of this equation:

Would it be correct to say this equation determines the required change of entropy to completely mix the separate gases?

If this is true then it should further lead to a way to determine how much energy in some interval of time is required to mix gases at a particular temperature. Is space (volume) also a constraint? How do you formulate the energy required to mix gases given these constraints?

It's interesting to point out that for ideal gases the mol ratio is equal to the volume ration (Avogadro, I believe).

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  • $\begingroup$ The equation you gave is strictly valid only for mixing ideal gases. Also, for ideal gases, the "heat of mixing" (enthalpy change, internal energy change, and temperature change) are all zero. $\endgroup$ – Chet Miller May 26 at 11:45
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Your formula for the entropy is simply a reformulation of the Gibbs entropy. The Gibbs entropy is given by $$ S = -k_B \sum_i p_i \ln p_i,$$ where $p_i$ is the probability of finding a particle in state $i$. Now if only one state is accessible then $p_i = 1$, therefore $S=0$. Now consider that two states with the same energy are accessible. If we are in equilibrium then both states will be as likely. Therefore $$ p_1 = \frac{1}{\Omega_1}, p_2 = \frac{1}{\Omega_2}$$ where $\Omega_1$ and $\Omega_2$ are the number of accessible states of state 1 and state 2 respectively. The Gibbs entropy becomes $$ S = -k_B (\frac{1}{\Omega_1} \ln(\frac{1}{\Omega_1}) + \frac{1}{\Omega_2} \ln(\frac{1}{\Omega_2})).$$ This are all reasoning within the framework of the micro-conanical ensemble. The next step is then to introduce accessible states of different energy, where the lowest energy will be more likely. If you are interested you can find more information in the excellent lectures of David Tong (http://www.damtp.cam.ac.uk/user/tong/statphys.html).

Now to answer your question this has nothing to do with mixing. Mixing would involve taking two gasses in equilibrium which are described by statistical mechanics pull them out of equilibrium by mixing, then let them eventually reach a new equilibrium. What you need to describe this is kinetic theory, which you can view as the out of equilibrium extension of statistical physics. David Tong has also an excellent lecture on kinetic theory, however it is best to have some knowledge of statistical physics before going to kinetic theory.

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