Question

Problem 5.58. In this problem you will model the mixing energy of a mixture in a relatively simple way, in order to relate the existence of a solubility gap to molecular behaviour. Consider a mixture of A and B molecules that is ideal in every way but one: The potential energy due to the interaction of neighbouring molecules depends upon whether the molecules are like or unlike. Let n be the average number of nearest neighbours of any given molecule (perhaps 6 or 8 or 10). Let n be the average potential energy associated with the interaction between neighbouring molecules that are the same (4-A or B-B), and let uAB be the potential energy associated with the interaction of a neighbouring unlike pair (4-B). There are no interactions beyond the range of the nearest neighbours; the values of ${\mu }_{o}and{\mu }_{AB}$are independent of the amounts of A and B; and the entropy of mixing is the same as for an ideal solution.

(a) Show that when the system is unmixed, the total potential energy due to neighbor-neighbor interactions is $\frac{1}{2}Nn{u}_0$. (Hint: Be sure to count each neighbouring pair only once.)

(b) Find a formula for the total potential energy when the system is mixed, in terms of x, the fraction of B.

(c) Subtract the results of parts (a) and (b) to obtain the change in energy upon mixing. Simplify the result as much as possible; you should obtain an expression proportional to x(1-x). Sketch this function vs. x, for both possible signs of $u_{AB}-u_0$.

(d) Show that the slope of the mixing energy function is finite at both end- points, unlike the slope of the mixing entropy function.

(e) For the case $u_{AB}>u_0$, plot a graph of the Gibbs free energy of this system

vs. x at several temperatures. Discuss the implications.

(f) Find an expression for the maximum temperature at which this system has

a solubility gap.

(g) Make a very rough estimate of $u_{AB}-u_0$for a liquid mixture that has a

solubility gap below 100°C.

(h) Use a computer to plot the phase diagram (T vs. x) for this system.

Short answer

Therefore,

$$\begin{gathered} Thefunctionisfiniteatbothendpoints. \\ WecanseefromthegraphthatGibbsfreeenergyisfallingwithtemperatureri\sin g. \end{gathered}$$

Step-by-step solution

Given information and explanation

An ideal mixture of A and B molecules is given to us:

n= the average number of nearest neighbours

$u_o$=average potential energy associated with the interaction between neighbouring molecules that are the same

u_AB= the potential energy associated with the interaction of a neighbouring unlike pair

A) If there are N molecules and we count every interaction twice, the total potential energy due to all neighbor-neighbor interactions is given by:

$U=\frac{1}{2}·N·n·u_0$

B)Total potential energy when the mixture is mixed

$$\begin{gathered} U_{mix}=\frac{1}{2}·N·n·\left(\left(x·u_0+(1-x)·u_{AB}\right)·x+\left(x·u_{AB}+(1-x)·u_0\right)(1-x)\right) \\ =\frac{1}{2}·N·n·\left(u_0·\left(x^2+(1-x{)}^2\right)+u_{AB}·(x·(1-x)+x·(1-x\left)\right)\right) \\ =\frac{1}{2}·N·n·\left(u_0·\left(1-2x+2x^2\right)+u_{AB}·2x·(1-x)\right) \end{gathered}$$

C)Change in energy upon mixing

$$\begin{gathered} \Delta U=U_{mix}-U \\ =\frac{1}{2}·N·n·\left(u_0·\left(1-2x+2x^2\right)+u_{AB}·2x·(1-x)\right)-\frac{1}{2}·N·n·u_0 \\ =N·n·x(1-x)\left(u_{AB}-u_0\right) \end{gathered}$$

Explanation

Case1: $u_{AB}-u_0>0$

Case 2: $u_{AB}-u_0<0$

Calculations

D)The mixing energy function's slope is:

$$\begin{gathered} \frac{d\Delta U}{dx}=\frac{d}{dx}\left(N·n·x(1-x)\left(u_{AB}-u_0\right)\right) \\ =N·n·\left(u_{AB}-u_0\right)(1-2x) \end{gathered}$$

If x=0

localid="1646988551444" $\frac{d\Delta U}{dx}=N·n·\left(u_{AB}-u_0\right)$

If x=1

localid="1646988555728" $\frac{d\Delta U}{dx}=-N·n·\left(u_{AB}-u_0\right)$

Hence the function has end points.

E)To plot the graph, we can use a variety of Gibbs free energy values and temperatures. Allow the temperatures to be as follows:

localid="1646988561564" $$\begin{gathered} T=50K \\ T=100K \\ T=150K \\ T=200K \end{gathered}$$

Explanation

The Gibbs free energy function is:

$$\begin{gathered} G=(1-x)G_A+x{G}_B+\Delta U-\Delta S_{mix}·T \\ =(1-x)G_A+x·G_B+N·n·x(1-x)\left(u_{AB}-u_0\right)-NkT·\left(x\ln \right(x)+(1-x\left)\ln \right(1-x\left)\right) \end{gathered}$$

Deriving it with x and equating to zero,

$$\begin{gathered} \frac{dG}{dx}=\frac{d}{dx}\left((1-x)G_A+x·G_B+N·n·x(1-x)\left(u_{AB}-u_0\right)-NkT·\left(x\ln \right(x)+(1-x\left)\ln \right(1-x\left)\right)\right) \\ =-G_A+G_B+N·n·\left(u_{AB}-u_0\right)(1-2x)-NkT\left(\ln \right(x)-\ln (1-x\left)\right)=0 \end{gathered}$$

Finding the value of x

$$\begin{gathered} 1-2x=0\Rightarrow x=\frac{1}{2} \\ \ln \left(x\right)-\ln (1-x)=0\Rightarrow x=\frac{1}{2} \end{gathered}$$

By doing double derivation, we get

$$\begin{gathered} \frac{d^{2}G}{d{x}^2}=\frac{d}{dx}\left(-G_A+G_B+N·n·\left(u_{AB}-u_0\right)(1-2x)-NkT\left(\ln \right(x)-\ln (1-x\left)\right)\right) \\ =-2N·n·\left(u_{AB}-u_0\right)-NkT·\left(\frac{1}{x-x^2}\right) \end{gathered}$$

Second derivation for maximum point

$$\begin{gathered} \frac{d^{2}G}{d{x}^2}\left(x=\frac{1}{2}\right)=-2N·n·\left(u_{AB}-u_0\right)-NkT·\left(\frac{1}{\frac{1}{2}-\frac{1}{4}}\right) \\ =-2N·n·\left(u_{AB}-u_0\right)-4NkT \end{gathered}$$

Now let's set it to zero and determine the maximum temperature expression:

$$\begin{gathered} 0=-2N·n·\left(u_{AB}-u_0\right)-4NkT \\ T=\frac{n\left(u_{AB}-u_0\right)}{2k} \end{gathered}$$

Explanation

Using these values in temperature term

$$\begin{gathered} n=4 \\ {u}_{AB}-u_0={10}^{-21}\mathrm{J} \\ \mathrm{k}=1.38·{10}^{-23}\frac{\mathrm{J}}{\mathrm{K}} \\ \mathrm{T}=\frac{\mathrm{n}\left({\mathrm{u}}_{\mathrm{AB}}-{\mathrm{u}}_0\right)}{2\mathrm{k}}=\frac{4·{10}^{-21}}{2·1.38·{10}^{-23}}=144.93\mathrm{K} \end{gathered}$$

At T = 144.93 K, a liquid mixture has a solubility gap.

H)Graph for T vs x: