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Unformatted text preview: 4.12 Gibbs Energies of Mixtures AIM: Introduce the concept of chemical potential. BACKGROUND: We are interested in developing a framework for free energies when the composition of a substance changes (chemical reactions, mixtures of particles or open systems). In this case, the state functions are also dependent on the amount of the individual molecular species in the system. For example, the Gibbs energy of the system would be written as G = G ( T , P , n 1 , n 2 , n 3 , . . . ) , (4.147) where n i is the number of moles of species i . The total differential of this expression is d G = G T P , n 1 , n 2,... d T + G P T , n 1 , n 2,... d P + G n 1 T , P , n 2 , n 3 ,... d n 1 + G n 2 T , P , n 1 , n 3 ,... d n 2 + . (4.148) We would like an expression for d G that does not have the partial derivatives in it. This is easy for red and blue terms. If the concentration of all 447 species is constant, then d n i = for all i , and this reduces to d G = G T P , n 1 , n 2,... d T + G P T , n 1 , n 2,... d P = S d T + V d P , (4.149) where we have applied Eqs. 4.141 and 4.142. For the green terms we now define a new thermodynamic quantity i = G n i T , P , n j 6 = n i (4.150) called the chemical potential of species i . Then we can write Eq. 4.148 as d G = S d T + V d P + i i d n i . (4.151) NOTE: The units of chemical potential are Energy mol . QUESTION: How can we intuitively understand chemical potential? ANSWER: The Gibbs energy of a system will change by n i i if we add n i mol of substance i to the system at constant concentration (this is an infinitesimal change, therefore the concentration of particles does not change). 448 EXAMPLE: Lets perform an thought experiment to allow us to better understand this concept. We have a system in which: we are at constant T and P the composition is constant; i.e. the ratio of all n i n j is constant Now we start from a very small system (all n s i very small, we assume 0 and G s = ) and add particles to system in the correct...
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 Winter '09
 scotty
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