Chemical potential and fixed number of particles

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SUMMARY

The discussion centers on the chemical potential of a system with a fixed number of particles, emphasizing its physical significance. It is established that the chemical potential μ can be derived from the free energy concerning the number of particles, even when N is fixed. The example of an ideal gas illustrates this concept using Maxwell-Boltzmann statistics, leading to the expression μ = kBT ln(N/Z1), where Z1 is the partition function for a single particle. The relationship between the total particle number N, the partition function, and quantum concentration is also highlighted.

PREREQUISITES
  • Understanding of chemical potential and its definition as the partial derivative of free energy
  • Familiarity with the grand canonical ensemble and its formulation
  • Knowledge of Maxwell-Boltzmann statistics and its application to ideal gases
  • Basic concepts of partition functions in statistical mechanics
NEXT STEPS
  • Explore the derivation of chemical potential in the context of the grand canonical ensemble
  • Learn about Fermi-Dirac and Bose-Einstein statistics and their implications for chemical potential
  • Investigate the role of quantum concentration in statistical mechanics
  • Study the applications of partition functions in various thermodynamic systems
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Physicists, chemists, and students in statistical mechanics who are interested in the thermodynamic properties of systems with fixed particle numbers and the implications for chemical potential.

cryptist
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Can we talk about the chemical potential of a system with fixed number of particles? Is this physically meaningful? Why/why not?

P.s: I know that chemical potential is the partial derivative of free energy with respect to number of particles. But in the formulation of grand canonical ensemble, we write N=Ʃf (for example f would be fermi-dirac distribution function) and can't we fix N in this formula, and solve chemical potential μ by changing energy ε?
 
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cryptist said:
Can we talk about the chemical potential of a system with fixed number of particles? Is this physically meaningful? Why/why not?

Yes, for the same reason you can talk about the temperature of a system with a fixed entropy or the pressure of a system with a fixed volume.
 
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You can indeed solve for the chemical potential \mu(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_{MB}(\epsilon) = e^{-(\frac{\epsilon-\mu}{k_{B}T})}.

The number of particles N can be expressed as

N=\sum_{j}g_{j}f_{MB}(\epsilon_{j})
where g_{j} is the number of states at energy level \epsilon_{j}

The partition function for a single particle Z_{1} is

Z_{1} = \sum_{j}g_{j} e^{-(\frac{\epsilon_{j}}{k_{B}T})}=\eta_{q}V : \eta_{q}=(\frac{m k_{B}T}{2 \pi \hbar^{2}})^{\frac{3}{2}}

Here \eta_{q} is a characteristic quantum concentration (you can see it has dimensions of \frac{N}{V}). It's a large concentration indicating when quantum effects (bose/Fermi statistics) must be taken into account. For ideal gases, the real concentration \eta=\frac{N}{V} is much less than \eta_{q}.

We can relate the total particle number N to the partition function Z_{1}, giving us

N=Z_{1}e^{-\frac{\mu}{k_{B}T}}

We can solve for \mu to find

\mu =k_{B}T \ln (\frac{N}{Z_{1}})

Then, substituting out expression for Z_{1}, we arrive at the final result

\mu =k_{B}T \ln (\frac{\eta}{\eta_{q}}) :\eta=\frac{N}{V}.

It's a total non-sequitur, but it seemed worth explaining since I had already written lecture notes on the subject.
 
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