Chemical potential on a solid and its vapor pressure

[Physics_Student_2018]

I have been reading the book "Nanostructures and Nanomaterials" by G. Cao and Y. Yang, and was intrigued by the following passage in page 33:
"Assuming the vapor of solid phase obeys the ideal gas law, for the flat surface one can easily arrive at:
μ_v − μ_∞ = −kTlnP_∞, where μ_v is the chemical potential of a vapor atom, μ_∞, the chemical potential of an atom on the flat surface, k, the Boltzmann constant, P_∞, the equilibrium vapor pressure of flat solid surface, and T, temperature."

My first impression is that the two chemical potentials should be equal for the solid and its vapor pressure at equilibrium. Could somebody please explain to me how this formula is derived?

Thank you.

 

[mjc123]

My impression is that by μ_v they mean μ_v⁰, the chemical potential of vapour in the standard state (P=1). Then if the vapour at P_∞ is in equilibrium with the solid:
μ_v = μ_v⁰ + kTlnP_∞ = μ_∞
and hence μ_v⁰ - μ_∞ = -kTlnP_∞
If μ_v = μ_∞ then P_∞ = 1 and the solid sublimes.

 

[Physics_Student_2018]

My impression is that by μ_v they mean μ_v⁰, the chemical potential of vapour in the standard state (P=1). Then if the vapour at P_∞ is in equilibrium with the solid:
μ_v = μ_v⁰ + kTlnP_∞ = μ_∞
and hence μ_v⁰ - μ_∞ = -kTlnP_∞
If μ_v = μ_∞ then P_∞ = 1 and the solid sublimes.

Thanks for your reply. I see that you agree with me that we should have μ_v = μ_∞ at equilibrium. In the last line, did you mean "μ_v⁰ = μ_∞ then..."? I am still not understanding the physics behind it.

 

[mjc123]

Yes, I must have done. μ_v = μ_∞ always at equilibrium; at the sublimation point μ_v = μ_∞ = μ_v⁰.