Plotting system of equations for a phase diagram

Currently I am struggling to graph a phase diagram similar looking to the attached photo. I have two equations, which depend on two variants of X (one for each line on the graph, respectively called x_1 and x_2. The formulas look like so

Equation 1: ln(x_1/x_2)=2070(1/T-1/1726)

Equation 2: ln((1-x_1)/(1-x_2))=1580(1/T-1/1357)

It should be possible to solve for x_1 and x_2 then overlay two graphs, one being x_1 on the x-axis and T on the y-axis, and the other with x_2 on the x-axis and T on the y-axis. The bounds for the graph are 0<x<1 and 1357<T<1726. Thank you so much for your help!

[David Goodmanson]

Hi Eric,

In the first equation, the maximum possible value of the rhs is at T = 1/1357: 10*(1/1357-1/1726) = .0016

so log(x1) - log(x2) = .0016 or less. That's a really small amount, so x1 and x2 will always be very near each other in value and the two curves will almost be overlays. Qualitatively that is a lot different than the plot above. Are you positive about the 10 and 9.6?

[Eric Yoshida]

Oh you are right! I misread my own work. The values have been updated, it should be a factor of 2070 for the first equation and 1580 for the second equation. Sorry about that!

Answer 1

[David Goodmanson]

Hi Eric,

let    T1 = 1357;    T2 = 1726;   T = T1:T2;

if you exponentiate both sides of each equation, you get

x1/x2 = f
(1-x1)/(1-x2) = g
where
f = exp(2070*(1./T- 1/T2));
g = exp(1580*(1./T -1/T1));

Solve for x1 and x2 in terms of f and g, then plot(x1,T,x2,T).

f and g are vectors, so calculating x1 and x2 requires element-by-element operators such as ./ .

[Eric Yoshida]

Thank you so much!