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I am trying to follow a derivation in solution chemistry. I'm ok with the derivation to (i.e. the starting point): $$x_1d\ln f_1=-x_2d\ln f_2$$

Now, the next step presented is: $$x_1\left({\partial \ln f_1 \over \partial x_1}\right)_{T,p}=x_2\left({\partial \ln f_2 \over \partial x_2}\right)_{T,p}$$

It probably is important to know (this equation is the Gibbs-Duhem equation for a binary mixture) that $x_2=1-x_1$, but otherwise supposedly eqn 2 can be derived from eqn 1.

This seems like it must be really simple, but I am having difficulty following this step. Since $x_2=1-x_1$, I thought that would lead to a negative sign negating the original negative sign in the 1st equation. Not so. I tried partial differentiating the eqn 1 wrt $x_1$ and end up with a mess including partial derivatives of differentials i.e. ${\partial (d\ln f_1) \over \partial x_1}$ which I'm not sure is even defined. I know I must be missing something simple. It almost seems you can write down eqn2 "by inspection" but I am wanting some rigorous understanding of why this is so.

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up vote 3 down vote accepted

You're right in suspecting that the partial derivative of a differential doesn't make sense. You can think of the differential as standing for a change in the quantity in general, whereas a derivative (total or partial) specifies a rate of change relative to some other change.

In the present case,

$$x_1\,\mathrm d\ln f_1=-x_2\,\mathrm d\ln f_2$$

gives a relationship between the changes in $\ln f_1$ and $\ln f_2$, and this relationship holds in particular if the change is caused by a change in $x_1$ while holding $T$ and $p$ constant; thus

$$x_1\left(\frac{\partial\ln f_1}{\partial x_1}\right)_{T,p}=-x_2\left(\frac{\partial\ln f_2}{\partial x_1}\right)_{T,p}\;.$$

Now, again as you suspected, you can use $x_2=1-x_1$ to replace $\partial/\partial x_1$ by $-\partial/\partial x_2$:

$$x_1\left(\frac{\partial\ln f_1}{\partial x_1}\right)_{T,p}=x_2\left(\frac{\partial\ln f_2}{\partial x_2}\right)_{T,p}\;.$$

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