Show that a partial molar property is an intensive property

AI Thread Summary
The discussion focuses on demonstrating that a partial molar property is an intensive property through mathematical derivations involving derivatives of energy with respect to component amounts. Participants clarify the notation and index labeling in their equations to avoid confusion, emphasizing the importance of careful differentiation. The correct expression for the derivative of energy is established, leading to the conclusion that the partial molar property can be shown to be intensive. A reference to a relevant textbook is made for further insights on the topic. The conversation concludes with a participant noting that expressing the derivative of component amounts will finalize the proof.
mcas
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Homework Statement
Show that a partial molar property is an intensive property
Relevant Equations
Intensive property ##I##: ##\sum_{i=1}^{\alpha} (\frac{\partial I}{\partial n_i}n_i)=0##
Extensive property ##E##: ##\sum_{i=1}^{\alpha} (\frac{\partial E}{\partial n_i}n_i)=E##
Partial molar property for an extensive property ##E##: ##E^{(p)}_i=(\frac{\partial E}{\partial n_i})##
I started by taking a derivative:
$$E = \sum_{i=1}^{\alpha} (E_i^{(p)} n_i) \ \ \ | \cdot \frac{\partial}{\partial n_i}$$
$$\frac{\partial E}{\partial n_i}=\sum_{i=1}^{\alpha} [\frac{\partial E_i^{(p)}}{\partial n_i}n_i + E_i^{(p)} \frac{\partial n_i}{\partial n_i}]$$
$$\frac{\partial E}{\partial n_i}=\sum_{i=1}^{\alpha} [\frac{\partial E_i^{(p)}}{\partial n_i}n_i + E_i^{(p)}]$$
$$\frac{\partial E}{\partial n_i}=\sum_{i=1}^{\alpha} [\frac{\partial E_i^{(p)}}{\partial n_i}n_i + (\frac{\partial E}{\partial n_i})]$$
$$\frac{\partial E}{\partial n_i} - E=\sum_{i=1}^{\alpha} [\frac{\partial E_i^{(p)}}{\partial n_i}n_i ]$$
I'm not sure what to do know. I also have a question regarding theleft hand side -- does ##E## defined as ##E =\sum_{i=1}^{\alpha} (\frac{\partial E}{\partial n_i}n_i)## depend on ##n_i##? Or is ##\frac{\partial E}{\partial n_i}## equal to 0?
 
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Be careful with your indices. If
E = Σif(ni)
it does not follow that
dE/dni = Σidf(ni)/dni
You want to use a different index label and say
dE/dnj = Σidf(ni)/dnj for a given j.
 
mjc123 said:
Be careful with your indices. If
E = Σif(ni)
it does not follow that
dE/dni = Σidf(ni)/dni
You want to use a different index label and say
dE/dnj = Σidf(ni)/dnj for a given j.
Thank you! That looks much better now.
Now I get
$$\frac{\partial E}{\partial n_j}=\sum_{i=1}^{\alpha} [\frac{\partial E_j^{(p)}}{\partial n_i}n_i ]$$
And I'm still confused as to what ##\frac{\partial E}{\partial n_j}## is and how to deal with this term.
 
There is a very interesting derivation of this in Chapter 11, Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics.
 
mcas said:
Now I get
$$\frac {\partial E} {\partial n_j} = \sum_{i=1}^α[\frac {\partial E_j^{(p)}} {\partial n_i}ni]$$
No, that should be
$$\frac {\partial E} {\partial n_j} = \sum_{i=1}^α[\frac {\partial E_i^{(p)}} {\partial n_j}ni + E_i^{(p)}\frac {\partial n_i} {\partial n_j}]$$
I don't know if you're on the right lines with the proof, I'm just correcting your notation.
 
mjc123 said:
No, that should be
$$\frac {\partial E} {\partial n_j} = \sum_{i=1}^α[\frac {\partial E_i^{(p)}} {\partial n_j}ni + E_i^{(p)}\frac {\partial n_i} {\partial n_j}]$$
I don't know if you're on the right lines with the proof, I'm just correcting your notation.
Thank you! Now I just need to express ##\frac {\partial n_i} {\partial n_j}## as ##\delta_{ij}## and it's proven.
 
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