Chemical potential of ideal gas from canonical ensemble I am trying to derive an equation for the chemical potential of an ideal gas. I am following example 22.1 in Blundell & Blundell "Concepts in Thermal Physics".
In this example, an expression for $F$ is used which has been derived from the canonical ensemble
$$F = Nk_BT[\ln{n\lambda_{th}^3}-1].$$
This is then substituded into
$$\mu = \left(\frac{\partial F}{\partial N}\right)_{V,T}.$$
My question is: why is it valid to use an equation for $F$ derived from the canonical ensemble? Surely this equation is incorrect in this context, and will be missing various factors of $\mu N$, etc. This same process of stealing various results from the canonical ensemble is repeated throughout chapter 22 of Blundell & Blundell and I find it very confusing.
 A: While the other answer provided a detailed calculation of how to obtain chemical potential in microcanonical and canonical ensembles, it did not cover the grand canonical ensemble. However, using the grand canonical ensemble to calculate chemical potential is straightforward.
Starting with the grand canonical ensemble, we can obtain the chemical potential for an ideal monoatomic gas with N particles. The Hamiltonian for such a system can be written as
$$
H = \sum_i^N \frac{\mathbf{p}_i}{2m}
$$
The grand partition function is given by:
\begin{align}
\Xi(T, V, \mu) &= \sum_{N=0}^{\infty}  \int dq dp \frac{1}{N! h^{3N}} e^{-\beta (H - \mu N)} \\
&= \sum_{N=0}^{\infty} Z(T,V,N) e^{\beta \mu N} \\
&= \sum_{N=0}^{\infty} \frac{1}{N!} \left( \frac{V e^{\beta \mu}}{\lambda_{th}^3} \right)^N \\
&= \exp \left( \frac{V e^{\beta \mu}}{\lambda_{th}^3} \right)
\end{align}
where $Z(T, V, N)$ is the partition function that can be inferred from the other answer:
$$
Z(T, V, N) = \frac{1}{N!} \left( \frac{V }{\lambda_{th}^3} \right)^N 
$$
The grand canonical potential is then given by:
$$
J = -k_B T \ln \Xi = - \frac{k_B T V e^{\beta \mu}}{\lambda_{th}^3}
$$
Using the differential expression of $J$,
$$
dJ = -SdT - PdV - Nd\mu
$$
we can determine the average particle number:
$$
N = - \frac{\partial J}{\partial \mu} = \frac{V e^{\beta \mu}}{\lambda_{th}^3}
$$
Finally, the chemical potential can be obtained as:
$$
\mu = k_B T \ln \left( \frac{\lambda_{th}^3 N}{V} \right)
$$
As we can see, using the grand canonical ensemble to calculate chemical potential is simpler than the other ensembles. It is worth noting that the thermodynamic quantities obtained from any ensemble are equivalent.
