L23 - "Chemical Potential and Free Energy expressed as...

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“Chemical Potential and Free Energy expressed as partition functions” MQ: Sections 23.5, 26.8, 26.9 Prof. GianluigiVeglia
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So far we used the “macroscopic” approach to thermodynamics to define our standard chemical potential. For an ideal gas, we obtained: Can we calculate μ o directly from the molecular properties? o o P P RT T P T ln ) ( ) , ( + = µ µ
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Another fundamental equation of the statistical mechanics is: This expression can be derived from that of the internal energy and the entropy Considering that the Helmholtz function is: V N B T Q T k E U , 2 ln = = Q k T Q T k S B V N B ln ln , + = Q T k T Q T k T Q T k TS U A B V N B V N B ln ln ln , 2 , 2 = =
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If we differentiate A with respect to the number of the molecules in the system: The last term can be expressed in terms of moles (since N and n differ from each other by the Avogadro’s constant): dN N A PdV SdT dN N A dV V A dT T A dA V T V T T N V N , , , , + = + + = dn n A dN N A V T V T , , =
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Therefore: The last term of this expression is another way to write the chemical potential. In fact, it we write: dn n A PdV SdT dA V T , + = PV A TS PV U G PV U H TS H G + = + = + = = PV A G + = PdV VdP dn n A PdV SdT PV d dA dG V T + + + = + = , ) ( dn n A VdP SdT PV d dA dG V T , ) ( + + = + =
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If we compare this expression With the total derivative of G=G(T,P,n) We can deduce that This means that we can use either G or A to determine the chemical potential as long as we keep the natural variable of each one fixed when we take the partial derivative with respect to n. dn n A VdP SdT PV d dA dG V T , ) ( + + = + = dn VdP SdT dn n G dP P G dT T G dG T P N T N P µ + + = + + = , , , V T P T n A n G , , = =
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If we take and substitute into We obtain: Where we used Q T k A B ln = V T P T n A n G , , = = µ R N k nN N A B A = =
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This note was uploaded on 10/30/2011 for the course CHEM 3501 taught by Professor Blank during the Fall '08 term at Minnesota.

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L23 - "Chemical Potential and Free Energy expressed as...

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