Question

Chemical potential at fixed number of particles

Answer 1

You can indeed solve for the chemical potential μ ( N , V , T ) μ(N,V,T) in the way you suggest. Consider the ideal gas as an example. Here, the occupancy comes from the classical maxwell-Boltzmann statistics f M B ( ϵ ) = e − ( ϵ − μ k B T ) fMB(ϵ)=e−(ϵ−μkBT). The number of particles N can be expressed as N = ∑ j g j f M B ( ϵ j ) N=∑jgjfMB(ϵj) where g j gj is the number of states at energy level ϵ j ϵj The partition function for a single particle Z 1 Z1 is Z 1 = ∑ j g j e − ( ϵ j k B T ) = η q V : η q = ( m k B T 2 π ℏ 2 ) 3 2 Z1=∑jgje−(ϵjkBT)=ηqV:ηq=(mkBT2πℏ2)32 Here η q ηq is a characteristic quantum concentration (you can see it has dimensions of N V NV). It's a large concentration indicating when quantum effects (bose/Fermi statistics) must be taken into account. For ideal gases, the real concentration η = N V η=NV is much less than η q ηq. We can relate the total particle number N N to the partition function Z 1 Z1, giving us N = Z 1 e − μ k B T N=Z1e−μkBT We can solve for μ μ to find μ = k B T ln ( N Z 1 ) μ=kBTln⁡(NZ1) Then, substituting out expression for Z 1 Z1, we arrive at the final result μ = k B T ln ( η η q ) : η = N V μ=kBTln⁡(ηηq):η=NV. It's a total non-sequitur, but it seemed worth explaining since I had already written lecture notes on the subject.

Reference https://www.physicsforums.com/threads/chemical-potential-and-fixed-number-of-particles.706641/