Chemical potential question

1. Sep 14, 2015

Coffee_

The chemical potential is defined as $µ=\frac{\partial E}{\partial N}$ while keeping entropy and volume constant. Well how to keep the volume constant while adding particles is obvious but what about the entropy? Obviously adding a particle at a certain energy $E_{p}$ changes the total energy of the system and $N+1$ changes the combinatorics to calculate the entropy.

How does one keep entropy constant while adding particles?

2. Sep 14, 2015

Bystander

Don't get too hung up on physical interpretations of partial derivatives that imply "processes" that are unrealizable in the lab or on the bench --- for this specific case, T changes.

3. Sep 14, 2015

Coffee_

Would it be correct to define µ as follows then?

''I add a particle with energy $E_p$ to the system without changing the volume. The entropy S changes. The increase in internal energy is obviously $E_p$ and $dN=1$. This means that $E_{p}=TdS+\mu$. We can define $\mu=E_p-TdS$ - that is: when adding a particle of a certain energy to a system , the chemical potential of that system is the difference between this particles energy and the product of T by $\Delta S$? ''

4. Sep 14, 2015

Bystander

Not in the original context of "constant entropy."

5. Sep 14, 2015

Coffee_

I think you misunderstood my initial question as something I saw as very difficult to do in the lab. Or maybe not, but to be sure:

I see not how it's possible to add a particle without changing the entropy THEORETICALLY. Adding a particles changes the total energy and changes the combinatorics of the system that are used to calcualte the entropy. In general entropy WILL change by adding a particle if what I just wrote is correct. This is not some limitation of the lab, it's the theory of statistical mechanics I'm speaking of.

So if the answer to the above is ''entropy in general changes'' when adding a particle, then see my second comment as my attempt to make sense of it.

6. Sep 14, 2015

Bystander

Yes. And, there are some partial derivatives that are --- less easily interpreted than others --- those at constant entropy being one particular case; take them at face value --- in a universe where S can be held constant, partial of E with respect to N at constant V, S is μ, which is identical to μ derived from partial of A (Helmholtz) at constant V, T (accessible), G (or F, Gibbs) at P, T, or H at P, S.

7. Sep 14, 2015

Coffee_

I see, nevermind then you understood me correctly.

Well so we both agree, adding a particle DOES change entropy usually.

So adding a particle that has an energy 'e' while keeping the volume constant must be according to 'dU=TdS + µdN' where dN=1 right?

dU = e = TdS + µ

Rewriting , $µ=e-TdS$

This dS we can actually calculate if we know the system well enough.

Is this a correct way to view/define the chemical potential then? As adding a particle of a certain energy, then counting what dS is and just taking the difference?