# Chemical potential on a solid and its vapor pressure

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• Physics_Student_2018

#### 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:
μvμ = −kTlnP, where μv 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.

My impression is that by μv they mean μv0, the chemical potential of vapour in the standard state (P=1). Then if the vapour at P is in equilibrium with the solid:
μv = μv0 + kTlnP = μ
and hence μv0 - μ = -kTlnP
If μv = μ then P = 1 and the solid sublimes.

My impression is that by μv they mean μv0, the chemical potential of vapour in the standard state (P=1). Then if the vapour at P is in equilibrium with the solid:
μv = μv0 + kTlnP = μ
and hence μv0 - μ = -kTlnP
If μv = μ then P = 1 and the solid sublimes.

Thanks for your reply. I see that you agree with me that we should have μv = μ at equilibrium. In the last line, did you mean "μv0 = μ then..."? I am still not understanding the physics behind it.

Yes, I must have done. μv = μ always at equilibrium; at the sublimation point μv = μ = μv0.