Advanced Problems textbook.(adsbygoogle = window.adsbygoogle || []).push({});

(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:

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}##? 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:

[tex](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. [/tex]

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:

[tex](2) \ \ \ \ \ \ \bar{G}_1'-\bar{G}_2'=\bar{G}_1''-\bar{G}_2''=\frac{\bar{G}''-\bar{G}'}{x_1''-x_1'}[/tex]

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:

[tex](3) \ \ \ \ \ \bar{G}_1'=\bar{G}_1'' \ \ \ \ and \bar{G}_2'=\bar{G}_2''[/tex]

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:

[tex](4)\ \ \ \ \ \ C'=\frac{x_1''-x_1}{x_1''-x_1'}, C''=\frac{x_1-x_1'}{x_1''-x_1'},[/tex]

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:

$$(5)\ \ \ \ \ \ C'\bar{G}'+C''\bar{G}''<\bar{G}(x).$$

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# A Question about Gibbs free energy from Kubos' Thermodynamics

Tags:

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**