Chemical+Potential - x = kT ln x x eq ´ µ ¸ ¹ So...

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Chemical Potential For the case of B solutes being dissolved into pure A, the chemical potential of the B atoms in solution relative to the pure B state would be ± μ = ²± G ² n B = 1 N ²± G ² x = ²± g ² x . From the relations on page 67 of the notes we have ± = ²± g ² x = ± g f + kT ln x 1 ³ x . Since for small concentrations the equilibrium solute fraction is x eq ± exp ² ³ g f kT ´ µ · ¸ ¹ , we can express the solute chemical potential as ± = ²± g ² x = ³ kT ln x eq + kT ln
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Unformatted text preview: x = kT ln x x eq ´ µ ¶ · ¸ ¹ . So when the defect solid is in equilibrium with pure B, there is no change in chemical potential in going from pure B to the solution. That occurs only when x = x eq . The chemical potential for x < x eq is obviously negative, meaning that the free energy decreases when a B atom is removed from pure B and placed into the solution....
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This note was uploaded on 12/01/2011 for the course MS&E 206 taught by Professor Nix during the Spring '08 term at Stanford.

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