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

This preview shows pages 1–8. Sign up to view the full content.

“Chemical Potential and Free Energy expressed as partition functions” MQ: Sections 23.5, 26.8, 26.9 Prof. GianluigiVeglia

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 , , =
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 , ) ( + + = + =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 , , = =
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 = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 21

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

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online