Monday, October 29, 2007
PS4 notes
There might be some confusion about the difference between partial molar properties in mixtures and intensive properties in pure components.
A partial molar something is defined by the expression given at the top problem 5, and it involves a derivative with respect to the number of molecules of a component. It is not the total quantity divided by the number of molecules in the mixture.
On the other hand, an intensive property is the quantity divided by the number of molecules, for a single component system.
A partial molar something is defined by the expression given at the top problem 5, and it involves a derivative with respect to the number of molecules of a component. It is not the total quantity divided by the number of molecules in the mixture.
On the other hand, an intensive property is the quantity divided by the number of molecules, for a single component system.
Wednesday, October 24, 2007
Maxwell relations: be careful
You'll be tempted to start swapping around derivatives to get Maxwell relations, but remember not all derivatives can be transformed using this approach. For example,
(dS/dT)_(V,N) cannot be transformed using a Maxwell relation. It is not equal to (dP/dV)_(T,N), if that is what you were thinking.
Recall:
(dS/dT) = (d2A/dT2)
(dP/dV) = (d2A/dV2)
Therefore, these are not the same second derivative of a free energy.
If it had instead been (dS/dV)_(T,N) then we could have said that equals (dP/dT)_(V,N).
(dS/dT)_(V,N) cannot be transformed using a Maxwell relation. It is not equal to (dP/dV)_(T,N), if that is what you were thinking.
Recall:
(dS/dT) = (d2A/dT2)
(dP/dV) = (d2A/dV2)
Therefore, these are not the same second derivative of a free energy.
If it had instead been (dS/dV)_(T,N) then we could have said that equals (dP/dT)_(V,N).
Tuesday, October 23, 2007
Hint on PS3 problem 5
The relationship we derived in class between Cp and Cv may be useful in both parts (b) and (c).
Hint on PS3, problem 4
In converting from A(T,V,N) to E(S,V,N), it may simplify your life to use the following relationship:
-E/T^2 = d(A/T)/dT (constant V and N)
You can use this equation to first get E(T,V,N) and then you can make the appropriate substitution to get E(S,V,N).
Be careful of using Mathematica to do all of the manipulations here. Sometimes it doesn't simplify to the best answer.
-E/T^2 = d(A/T)/dT (constant V and N)
You can use this equation to first get E(T,V,N) and then you can make the appropriate substitution to get E(S,V,N).
Be careful of using Mathematica to do all of the manipulations here. Sometimes it doesn't simplify to the best answer.
Monday, October 22, 2007
Calculus manipulations handout correction
The definition of constant-pressure heat capacity should involve a derivative at constant P, not constant V.
PS3
A couple of notes:
For problem 8.1 in MDF, do two things: (1) express Z in terms of the Gibbs free energy in integrated form (e.g., Z = G + ...). (2) write Z in differential form (e.g., Z = x dy + ...)
For problem 3, don't forget that the chemical potential is also the Gibbs free energy per particle, for a single component system.
For problem 8.1 in MDF, do two things: (1) express Z in terms of the Gibbs free energy in integrated form (e.g., Z = G + ...). (2) write Z in differential form (e.g., Z = x dy + ...)
For problem 3, don't forget that the chemical potential is also the Gibbs free energy per particle, for a single component system.
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