# Homework Help: Graphing the Gibbs free energy of mixed gases

1. Apr 8, 2017

### grandpa2390

1. The problem statement, all variables and given/known data
I am needing to graph the Gibbs free energy of mixed gases to determine the range when the gases will form an ideal mixture
The two gases have the same Gibbs free energy.

2. Relevant equations
$G = (1-X)G_A + XG_B$ for unmixed
$G = (1-X)G_A + (X)G_B + RT(x*ln(x) + (1-X)*ln(1-X))$ for mixed
$U = 6nTx_Ax_B$ x_A and x_B are molar fractions of the gases.

3. The attempt at a solution

so my attempt is that since the Gibbs of both gases equal the same, then $G_A = G_B$ so $G = G_{AB} + RT(x*ln(x) + (1-X)*ln(1-X))$

I am not sure how to graph that. I don't know what T is.
the second formula assumes that U does not change. But in this problem, it appears that U does. So I am thinking that my formula would be to add U in that equation:

$G = G_{AB} + 6nT(x)(1-x) + RT(x*ln(x) + (1-x)*ln(1-x))$

I don't know how to graph this because there are 3 independent variables still... ( T, x, and n) But I have to graph it somehow in order to estimate the range where the gases will form an good mixture.

Last edited: Apr 8, 2017
2. Apr 14, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Apr 14, 2017

### grandpa2390

after I turned in the assignment, I asked my professor about it. He gave me a few hints. I went home and did it, emailed it to him, and now I am praying I will get a bit of credit for the late addition...

For Posterity, the Gibbs free energy is equal so it is just a horizontal line. fill in any values for the constants (our independent variable is x) and graph the equation I posted and you get a curve that crosses the x axis a couple times in a wave form. the Gibbs line is the x-axis and everything below it is a point where the stuff will mix nicely.

I'm surprised nobody could help me with this. it is so simple...

anyways, thanks for the courtesy bump :)