For a dense system we assume that the two types of molecules are the same size and that molecules of different types interact with each other in the same way as molecules of the same type (same forces, etc.). Such a system is called an ideal mixture.

The Question asks us to explain why, for an ideal mixture, the mixing entropy is given by
$$\Delta S_{mixing}=kln{N \choose N_A}$$,

And here's the given answer from the solution manual:

In the unmixed state, the system could have quite a bit of entropy due to molecular energies and (for fluids) configurations. When we allow the system to mix, assuming that the mixture is ideal, the only change is that molecules of different types can now switch places with each other ta random. Therefore, to compute the mixing entropy, we can ignore the initial entropy and pretend that the molecules are initially frozen in place. Upon mixing, molecules randomly switch places with each other but still occupy the same collection of $N$ fixed sites. The increase in multiplicity due to mixing, therefore, is the number of ways of assigning the two species of molecules to the $N$ sites, that is, the number of ways of choosing $N_A$ of the sites to be occupied by molecules of type $A$:
$$\Omega_{mixing} = {N \choose N_A}$$
The entropy of mixing is then k times the natural log of this expression:
$$\Delta S_{mixing}=kln{N \choose N_A} = kln \left(\frac{N!}{N_A!N_B!} \right)$$

Although there's an explanation for why we can ignore the initial entropy in the solution manual, I still don't quite get the reasoning behind it. Why the switching places of molecules lead to the ignoration of initial entropy? Can someone please shed some light on this! Thanks!


Can't say I care from the description there, so I'm gong to take a different tack entirely.

In equilibrium thermodynamics and thermostatistics, entropy is a state variable meaning exactly that you can determine it from the current state of the system without reference to the history. Once the mixing is complete the system is indistinguishable from one that was formed by combining two half-sized but pre-mixed systems, so we have to be able to compute the entropy of the final system without reference to the prior conditions.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.