In VASP, how is the chemical potential of elements calculated?

I would like to calculate the chemical potential of elements having different environmental condition (rich or poor) using VASP. How is this accomplished?

The chemical potential in VASP or any other computational chemistry software is calculated based on the theory of chemical potential relation to Helmholtz energy functional as:

$$\mu_{i} = \Bigg(\frac{\partial F(V,T,N_{1},N_{2},...,N_{i}...,N_{n})}{\partial N_{i}}\Bigg)_{V,T,N_{j \neq i}}$$

You can approximate the above equation up to first order as:

$$\mu_{i} \simeq F(V,T,N_{i}+1,N_{j \neq i}) - F(V,T,N_{i},N_{j \neq i})$$

Here we omitted the terms of $$\mathcal{O}\Big( \frac{\partial^{2} F}{\partial N_{i}^{2}} \Big)$$.

I call $$F(V,T,N_{i}+1,N_{j \neq i}) - F(V,T,N_{i},N_{j \neq i}) = \Delta F_{N_{i} \rightarrow N_{i}+1}$$

So, you have:

$$\mu_{i} = \Delta F_{N_{i} \rightarrow N_{i}+1} = \mu_{\text{XC}}^{i}+\mu_{\text{ideal}}^{i}$$

Where $$\mu^{i}_{\text{ideal}}$$ is the chemical potential of the ideal gas and $$\mu^{i}_{\text{XC}}$$ is the chemical potential of exchange-correlation. The ideal gas chemical potential is trivial:

$$\mu^{i}_{\text{ideal}} = -k_{B}T\ln{\Bigg( \frac{V}{\Lambda^{3} (N_{i}+1)}\Bigg)}$$

Where $$\Lambda$$ is the Broglie wavelength: $$\Lambda = \sqrt{\frac{2\pi\hbar^{2}}{mk_{B}T}}$$.

The exchange-correlation chemical potential is calculated as:

$$\mu^{i}_{\text{XC}} = -k_{B}T \ln{\Bigg ( \frac{1}{V} \frac{\int \exp{\Big(-\frac{U(\mathbf{r}^{N_{i}+1})}{k_{B}T}\Big)} d^{3}\mathbf{r}^{N_{i}+1}}{\int \exp{\Big(-\frac{U(\mathbf{r}^{N_{i}})}{k_{B}T}\Big)} d^{3}\mathbf{r}^{N_{i}}} \Bigg )}$$

Or in ensemble form:

$$\mu^{i}_{\text{XC}} = -k_{B}T \ln{\Bigg ( \frac{1}{V} \Bigg \langle \int \exp{\Big( -\frac{\Delta U_{N_{i} \rightarrow N_{i}+1}}{k_{B}T} \Big)} d^{3}\mathbf{r}^{N_{i}+1} \Bigg \rangle \Bigg )}$$