Question about Gibbs free energy from Kubos' Thermodynamics

[MathematicalPhysicist]

Advanced Problems textbook.
(It's on pages 212-213).
I'll post the question and following it the solution in the book:
The question:

The Gibbs free energy per mole of a two component system is given by:
$\bar{G}(x)=x_1\bar{G}_1+x_2\bar{G}_2$.
Here $x_1$ and $x_2$ are the molar ratios of the components 1 and , and $\bar{G}_1$ and $\bar{G}_2$ are chemical potentials per mole, respectively.
Show that, if the plot of $\bar{G}(x)$ ($x=x_1$ or $x_2$) as a function of $x_1$ has a common tangent $Q'Q''$, the states between $Q'Q''$ separate into two phases, $Q'$ and $Q''$.

The Solution:

Set $x=x_1$. From the Gibbs-Duhem relation, $x_1\partial \bar{G}_1/\partial x_1 + x_2 \partial \bar{G}_2/\partial x_1 = 0$, we have:

(1) \ \ \ \ \ \ \frac{\partial \bar{G}}{\partial x_1}= \bar{G}_1+(dx_2/dx_1)\bar{G}_2+x_1 \partial \bar{G}_1/\partial x_1+x_2\partial \bar{G}_2/\partial x_1 = \bar{G}_1-\bar{G}_2.

If the values of the functions at $x_1'$ and $x_1''$ are denoted by ' and '', the condition that a common tangent $Q'Q''$ can be drawn is:

(2) \ \ \ \ \ \ \bar{G}_1'-\bar{G}_2'=\bar{G}_1''-\bar{G}_2''=\frac{\bar{G}''-\bar{G}'}{x_1''-x_1'}
Using $\bar{G}= x_1\bar{G}_1+x_2\bar{G}_2$, we get from $(2)$:
$$\bar{G}'_1-\bar{G}'_2 = \bar{G}''_1-\bar{G}''_2=\frac{x_1''\bar{G}_1''+x_2''\bar{G}_2''-x_1'\bar{G}_1'-x_2'\bar{G}_2'}{x_1''-x_1'}$$

from which we get:
$$\frac{x_1'\bar{G}'_1-x_1'\bar{G}_2'}{x_1'}=\frac{-x_1''\bar{G}_1''+x_1''\bar{G}_2''}{-x_1''} = \frac{x_1''\bar{G}_1''+x_2''\bar{G}_2''-x_1'\bar{G}'_1-x_2'\bar{G}'_2}{x_1''-x_1'}$$
$$= \frac{\bar{G}_2''-\bar{G}_2'}{0},$$
and
$$\frac{-x_2'\bar{G}_1'+x_2'\bar{G}_2'}{-x_2'}=\frac{x_2''\bar{G}''_1-x_2''\bar{G}_2''}{x_2''}=\frac{x_1''\bar{G}''_1+x_2''\bar{G}_2''-x_1'\bar{G}_1'-x_2'\bar{G}_2'}{x_1''-x_1'}$$
$$=\frac{\bar{G}_1''-\bar{G}_1'}{0}.$$

This means that:
(3) \ \ \ \ \ \bar{G}_1'=\bar{G}_1'' \ \ \ \ and \bar{G}_2'=\bar{G}_2''

which is equivalent to (2). (If (3) is valid, (2) is obviously valid.) Eq. (3) is the condition that two phases with different molar concentrations (denoted by single and double primes) can coexist.
Thus a system that has a density $x_1$ between $Q'$ and $Q''$ separates into two phases with the ratio of:

(4)\ \ \ \ \ \ C'=\frac{x_1''-x_1}{x_1''-x_1'}, C''=\frac{x_1-x_1'}{x_1''-x_1'},

and its free energy (point Q in figure 4.5) is lower than $\bar{G}(x)$ (point R in fig. 4.5).
In other words:
$$(5)\ \ \ \ \ \ C'\bar{G}'+C''\bar{G}''<\bar{G}(x).$$

What I don't understand is how did they arrive at the identities with $\frac{\bar{G}_1''-\bar{G}_1'}{0}$ and $\frac{\bar{G}_2''-\bar{G}_2'}{0}$?

I am perplexed as to what is going on with this problem really,does anyone know?

 

[Chestermiller]

Are you able to sketch of the graph that depicts how the molar average G is varying with x in this situation?

 

[MathematicalPhysicist]

@Chestermiller the figure is given in figure 4.5 in the book.

I still don't see how did they derive those identities, did they derive it from the figure or algebraically?

 

[Chestermiller]

@Chestermiller the figure is given in figure 4.5 in the book.

I still don't see how did they derive those identities, did they derive it from the figure or algebraically?

I haven't gone through the derivation yet, but, from your knowledge, it should be clear to you that the partial molar free energies of each of the two species at the two tangential points must be equal, correct?

 

[MathematicalPhysicist]

I haven't gone through the derivation yet, but, from your knowledge, it should be clear to you that the partial molar free energies of each of the two species at the two tangential points must be equal, correct?

From what does it follow that:"the partial molar free energies of each of the two species at the two tangential points must be equal"?

 

[Chestermiller]

From what does it follow that:"the partial molar free energies of each of the two species at the two tangential points must be equal"?

$$\bar{G}=x_1\bar{G}_1+x_2\bar{G}_2$$
So, $$\bar{G}=x_1\bar{G}_1+(1-x_1)\bar{G}_2=\bar{G}_2+(\bar{G}_1-\bar{G}_2)x_1\tag{1}$$
If we evaluate the change in $\bar{G}$ between two adjacent points separated by a change of $dx_1$ along this curve, we obtain: $$d\bar{G}=(\bar{G}_1-\bar{G}_2)dx_1+x_1d\bar{G_1}+x_2d\bar{G_2}$$ But, from the Clausius-Duhem equation, the sum of the 2nd and 3rd terms on the right hand side of this equation are zero. Therefore, along this curve, the slope of the tangent line is $(\bar{G}_1-\bar{G}_2)$ at each point. So Eqn. 1 must also be the equation for the straight line tangent to the curve at that point (holding $\bar{G}_1$ and $\bar{G}_2$ constant along the line). And the intercepts of this line with x1= 0 and x1=1 must be the partial molar free energies of the substance at the point. So this is a graphical method of determining the partial molar free energies as a function of mole fraction, if the relationship between $\bar{G}$ and $x_1$ is known.

 

[MathematicalPhysicist]

Yes, I believe you have just repeated what was written in the book in the first few paragraphs, just notifying me that we have: $x_1+x_2=1$ which really eluded me, thanks for that.

But I still don't understand what comes after that, i.e. between: "from which we get:" and equation (3).

How did they get the equalities: $(\bar{G}_2''-\bar{G}'_2)/0$ and the second equality with 1 replacing 2?

I thought that because of the equality: $\frac{-x_1''\bar{G}_1''+x_1''\bar{G}_2''}{-x_1''} = \frac{x_1''\bar{G}_1''+x_2''\bar{G}_2'' - x_1'\bar{G}_1'-x_2'\bar{G}_2'}{x_1''-x_1'}$, after multiplying by $-x_1''(x_1''-x_1')$ the two hands of this equation I get:

$$x_1''^2\bar{G}_2''-x_1''^2\bar{G}_1''-x_1'x_1''\bar{G}_2''+x_1'x_1''\bar{G}_1''=-x_1''^2\bar{G}_1''-x_1''x_2''\bar{G}_2''+x_1''x_1'\bar{G}_1'+x_1''x_2'\bar{G}_2'$$
After rearranging and noticing that: $x_1'+x_2'=1=x_1''+x_2''$ I get
$$x_1'' \bar{G}_2''-x_1'x_1''\bar{G}_2''+x_1'x_1''\bar{G}_1''=x_1''x_1'\bar{G}_1'+x_1''x_2'\bar{G}_2'$$

What else should I do to derive these last two identities in my first OP in the solution's quote?

 

[MathematicalPhysicist]

Ah, wait a minute I can reduce the last RHS to $x_1''\bar{G}'$ and the LHS to: $x_1''x_2'\bar{G}_2''+x_1'x_1''\bar{G}_1''$.

But how to proceed from here?

 

[Chestermiller]

I'm not going to spend my time trying to figure out what they did because, in my view, it's unnecessarily complicated. Here is my simpler assessment:

If the straight line is tangent to the curve of $\bar{G}$ vs $x_1$ at two points, $x_1'$ and $x_1''$, then two alternate forms of the equation for this same straight line are $$\bar{G}=\bar{G}_2'+(\bar{G}_1'-\bar{G}_2')x_1'$$ and $$\bar{G}=\bar{G}_2''+(\bar{G}_1''-\bar{G}_2'')x_1''$$But, since it is the same straight line, the intercepts in these equations at $x_1=0$ and $x_1=1$ must be identical. Therefore,

$$\bar{G}_2'=\bar{G}_2''$$and $$\bar{G}_1'=\bar{G}_1''$$

These are the conditions for equilibrium of the two states at the tangent points.

 

[MathematicalPhysicist]

OK, thanks.

BTW, why do you call the relation Clausisus-Duhem and not Gibbs-Duhem?

Did he find it earlier than Gibbs?

 

[Chestermiller]

OK, thanks.

BTW, why do you call the relation Clausisus-Duhem and not Gibbs-Duhem?

Did he find it earlier than Gibbs?

Nope. Just terminology confusion on my part.