Show that a partial molar property is an intensive property

[mcas]

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?

 

[mjc123]

Be careful with your indices. If
E = Σ_if(n_i)
it does not follow that
dE/dn_i = Σ_idf(n_i)/dn_i
You want to use a different index label and say
dE/dn_j = Σ_idf(n_i)/dn_j for a given j.

 

[mcas]

Be careful with your indices. If
E = Σ_if(n_i)
it does not follow that
dE/dn_i = Σ_idf(n_i)/dn_i
You want to use a different index label and say
dE/dn_j = Σ_idf(n_i)/dn_j 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.

 

[Chestermiller]

There is a very interesting derivation of this in Chapter 11, Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics.

 

[mjc123]

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.

 

[mcas]

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.