# Phase Diagram Of Germanium Silicon Solution Biology Essay

Published:

The approach for the problem is the following. For every given value of temperature T in terms of ideal solution model equations for the Gibbs free energies of the liquid and solid phases will be derived. Then we search for the common tangent (which means balance between chemical potentials of the components in different phases or equilibrium state of the system) to the energy curves to find xL (T) and xS (T). Finally, varying the temperature in the range from TfA to TfB we can plot the phase diagram of Germanium-Silicon Solution.

Let - Gibbs free energy of pure component i in phase j counted from the reference state of the i-component. We have two components of the solution: Ge and Si. Let us mark them as A - Ge and B - Si (to have more general equations).

In terms of ideal solution model for two mixed components A and B the equation for the Gibbs free energy of some given phase (mark it as n-phase) is the next [1]:

### Professional

#### Essay Writers

using our Essay Writing Service!

(1.1)

,where is mole fraction of component A in the phase n and is mole fraction of component B in the phase n. - Gibbs free energy of pure component A in phase n with regard to reference state for the component A. - similarly.

The system being analyzed is binary with components A and B, so in the given phase xA + xB=1

Supposing the temperature T is given. To calculate the energies, reference states of the components need to be defined. It is useful to define reference states like this:

1. For the Ge (A - component): pure liquid Ge (A) at Tref = T

2. For the Si (B - component): pure liquid Si (B) at Tref = T

Then the energies of pure components will be defined as (counted from the reference states):

(1.2)

(1.3)

(1.4)

(1.5)

,where is Gibbs free energy of pure component A in the liquid phase with regard to the reference state of component A and is Gibbs free energy of pure component B in the liquid phase with regard to the reference state of component B. Analogously and for the solid phase could be defined.

Considering x is mole fraction of B-component (Si) and using equations (1.1) and (1.2-5) Gibbs free energies for the liquid and solid phases could be defined:

(1.6)

(1.7)

Then we are searching for the common tangent to and graph to define xL (T) and xS (T) at temperature T. The illustration of the common tangent and xL and xS points at some T is given in the Fig.1.

Analytical equations for the xL (T) and xS (T) could be derived from (1.6), (1.7):

(1.8)

(1.9)

Finally, using equations (1.8) and (1.9) for liquidus and solidus curves the phase diagram could be plotted. Phase diagram for the Ge-Si solution is shown in the Fig.2.

C:\Documents\Labs\Materials Modelling\part1\PhaseDiagrams\report\21.png

Fig.1. Illustration of GS (red) and GL (blue) Gibbs free energy curves at some temperature between TfA and TfB. Common tangent and xL and xS points are shown.

C:\Documents\Labs\Materials Modelling\part1\PhaseDiagrams\report\51.PNG

Fig.2. Phase diagram of Ge-Si binary solution in assumption of ideal solution model. Liquid, Solid and two-phase areas are marked.

## Question 2: Phase Diagram of Nickel-Aluminum Binary System.

Binary system of Nickel and Aluminum is being explored in assumption of regular solution model. γ/(γ+γ') phase border is built and some extra parameters and features of the model and of the solution are researched.

## Given data:

1. Binary system of Ni and Al is given.

2. Regular solution model of mixing the components is assumed.

3. For the pure elements in γ - phase:

(2.1)

(2.2)

4. Regular solution parameter for the γ - phase is:

(2.3)

, where and are the mole fractions of Al and Ni.

5. The Gibbs free energy of the γ' - phase:

(2.4)

## Problem:

The task is to explore γ/(γ+γ') border in terms of regular solution model as a function of temperature using given Gibbs free energy functions and considering energy of mixing.

## Solution:

### Comprehensive

#### Writing Services

Plagiarism-free
Always on Time

Marked to Standard

The approach to the problem is the same like it was in the first task, but we also need to take into account energy of mixing - use regular solution model instead if ideal one. Like it was in the first task we will mark the components as A-component - Al and B-component - Ni. From the [1]:

(2.5)

(2.6)

Then Gibbs free energy of some given phase (it is marked as n-phase like it was in the task 1) is:

(2.7)

## a)

Let T is given temperature. Gibbs free energies for pure elements in the γ - phase and Gibbs free energy of the γ' - phase are already given. Which means that some reference states of the components have been already defined. But for our convenience we will redefine the reference states:

1. For the Al (A - component): pure γ - phase Al (A) at

2. For the Ni (B - component): pure γ - phase Ni (B) at

Then the energies of pure components in γ - phase will be equal to zero because they are in their reference states. Also let us mark and. If so, the Gibbs free energies of the phases (from the equation (2.7) and (2.4) using new reference states) will be:

(2.8)

(2.9)

It can be seen from the equations above that Gibbs free energy of the γ' - phase is independent from the molar fraction of Ni. So, to find the γ / (γ+γ') border we need to search for the line that is the tangent line for the curve and it goes through the point for the given value of T. The illustration of such line, the point and curve is given in the Fig.3. The important thing is that we are looking for the x - border value in the range from 0.75 to 1.0.

C:\Documents\Labs\Materials Modelling\part1\PhaseDiagrams\report\61.png

Fig.3. Illustration of Gγ (blue curve) and Gγ' (red point) Gibbs free energy for some temperature between 900K and 1500K. solution is shown in the fig.

For a given value of temperature T will be a solution of the equation:

(2.10)

For the given functions we cannot find the analytical solution, so we do it numerically using Mathematica. The range of temperatures is 900K to 1500K and the step is 1K. The solution for the γ / (γ+γ') border is shown in the Fig.4. According to the initial data γ'- phase is a line x=0.75.

C:\Documents\Labs\Materials Modelling\part1\PhaseDiagrams\report\71.png

Fig.4. Phase diagram of Ni-Al binary solution in assumption of regular solution model. γ, γ' and γ+ γ' phases are shown.

## b)

In the Fig.5 comparison between published phase diagram and modeled one is shown. Phases are marked and phase borders are illustrated by red and blue curves (modeled lines). Range from 900K to 1500K is marked with green lines. As it could be seen from the figure, model gives the best results in the range explored (900K to 1500K). The maximum difference between modeled border and published one in terms of composition occurs at T=1500K (if consider range from 900K to 1500K). The difference is:

(2.11)

(2.12)

So, the maximum error (mistake) of the model in the range from 900K to 1500K is 2%.

C:\Documents\Labs\Materials Modelling\part1\PhaseDiagrams\report\91.png

Fig.5. Comparison between published phase diagram and modeled one. Modeled borders of the phases are shown with red and blue lines. The temperature range from 900K to 1500K illustrated with green lines.

## c)

The approach is based on Gibbs free energy calculation for every value of temperature T. In the a) we chose reference states of the components to make Gibbs free energy of pure components equal to zero. Now let's say that for a given value of temperature T the values of Gibbs free energies of the components are:

(2.13)

(2.14)

Then from the equations (2.4) and (2.7):

(2.15)

(2.16)

, where are members which are independent from H and K.

Then substitution of (2.15) and (2.16) into the (2.10) equation gives:

### This Essay is

#### a Student's Work

This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.

(2.17)

Is independent from H and K.

Then, the choice of reference energies is irrelevant to the problem; the solution for the border curve of the phase is calculated from (2.17) which are not depending on values of reference energies.