Questions And Answers 6u1s6b

QUESTIONS 1. Explain the meaning of the following concepts, by presenting some figures when required. a. The degree of freedom (f) and Gibbs phase rule (5). b. Intensive and extensive properties of thermodynamic functions(5). c. Chemical potential (5). d. Compressibility (5).

Answer 1. a. Degree of freedom: the number of variables that can be changed independently without disrupting the equilibrium of the system, i.e. the variables when changed a new face is not introduced or an existing face is not dispelled. Gibbs Phase rule: Gibbs phase rule states that if a system containing C components and P phases is in equilibrium the number of degrees of freedom F is given by P+F= C+2 A degree of freedom is an intensive variable such as T, P. X A, XB . . . that can be varied independently while still maintaining equilibrium. If pressure is maintained constant one degree of freedom is lost and the phase rule becomes P+F= C+1

b. (i)The intensive properties are those which are independent of the size of the system such as T and P, (ii) The extensive properties are directly proportional to the quantity of material in the system, e.g. P, E, H, S and G.

c. Chemical potential: If a small quantity of element A, dnA, mol, is added to a large amount of a phase at constant temperature and pressure, the size of the system will increase by dnA, and therefore the total free energy of the system will also increase by a small amount dG'. If dnA is small enough dG' will be proportional to the amount of A added. Thus we can write dG ' =μ A dn A

(T, P, nB constant)

(1.28)

The proportionality constant μA, is called the partial molar free energy of A or alternatively the chemical potential of A in the phase which depends on the composition of the phase. When equation 1.28 is rewritten it can be seen that a definition of the chemical potential of A, μ A , is μ A ¿( ∂G ' / ∂ n A )T ,P ,n

B

where G' corresponds to the Gibbs free energy of the whole system.

d. Compressibility: At constant T0C the free energy of a phase increases with pressure such that

=V , ( ∂G ∂P)

which

T

shows that the slope of the free energy surface on isothermal planes equals the molal volume of the corresponding phase. The curvature of the free energy surface on isothermal planes (T=const.), which is the second derivation of the free energy, directly relates to the compressibility of the corresponding phases according to the equation

∂2G ∂V = =−βV 2 ∂P T ∂P T

( ) ( ) where

β is called compressibility, the change in volume with pressure and expressed as

2.

Sketch T-P phase diagram for pure iron and explain why the slope of a®g phase transformation line is negative while g®liquid transformation line is positive. Explain this question by driving Clausius – Clapeyron equation.

Answer 2. The lines of phase transformation can be calculated according to the Clasius-Clapeyron equation as expressed;

dT TΔV = dP ΔH Transformation from α to γ with increasing temperature at a constant pressure results in a decrease in volume since γ has a close packed structure. Change in enthalpy with increasing temperature causes an increase in enthalpy of the system. Therefore the slope

dT /dP<0

is negative. Since the transformation from γ to liquid involves increase in both volume and enthalpy of the system, the slope of (γ to liquid) transformation will be positive. 3.

a. Draw a schematic G-T-P diagram ( G=f ( P , T ) ) of one component system and show that the slopes of the free energy surface surfaces in isobaric and isothermal planes are equal to the molal entropies, ∂G ∂G =−S , and volumes , =V , of the respective phases. ∂T P ∂P T b. Compare the positions of the transition temperature, T tr with melting temperature of a pure substance for the temperature dependent allotropy to occur.

( )

( )

Answer 3. a. The partial differentiation of the equation G=f ( P , T ) is ∂G ∂G dG= dP+ dT ∂P T ∂T P .

( ) ( )

This equation can be compared with

dG=V dP−SdT

equation, because each represents the change in molal free energy of the system. Therefore it can be easily seen that the following equation hold:

=V ( ∂G ∂P ) T

and

=−S ( ∂G ∂T ) P

b.

For temperature dependent allotropy of a pure substance to occur melting temperature of the substance should be greater than the transition temperature. Otherwise melting would occur before allotropic transformation takes place. This is indicated in part (a) of the following figure.

4.

a. Molar Free energy (XA + XB = 1) of a solid solution consisting of element A and B is given in of the corresponding potential energies, mA and mB as follows: G = mAXA + mBXB J / mol .....................................................….. (1) and also given in of molar free energies of pure elements follows:

G A and GB as

G= X A G A +X B G B + RT ( X A . ln X A + X B . ln X B )

........................……… (2) By using equation (1) and (2) prove that chemical potential of elements A and B can be calculated as shown below: mA = G A + RT . ln X A

mB = GB + RT . ln X B b. Draw the molar free energy curve of this solid solution as a function of XB at two different temperatures T1 and T2 (T1< T2) and interpret the change in curvature of these free energy curves.

Answer 4. a. G=μ A . X A +μ B . X B

… … G= X A G A + X B GB + RT ( X A ln X A + X B ln X B )

When rearranging equation (2)

… .…

… .…

....

(1) (2)

G= X A (G A +RT ln X A )+X B (G B RT ln X B ) And comparing this with Equation (1) yields the followings: μ A =G A + RT ln X A μB =G B +RT ln X

b. The free energy of the solution G will depend on G A and GB. As the temperature increases GA and GA decrease and the free energy curves assume a greater curvature. The decrease in GA and GB is due to the thermal entropy of both components and is given by the following equation,

( )

∂G =−S ∂T P

5.

Components A and B in a binary system are fully soluble in each other in liquid and solid states at all composition range. Assuming that the free energy of liquid is lower than that of solid solution, free energy curves of liquid and solid phases in the system at different temperatures are given in Fig. 3a, b, c, d, and e. By making use of these free energy curves; a. Draw the binary phase diagram in Fig 3f and indicate the face domains on the diagram. b. Apply the phase rule at temperature T2 between composition points b and c

f=c-p+1 ® f=2-2+1=1 6.

An ideal solid solution formed by elements A and B has molar free energy as expressed, G= X A G A + X B G B + RT ( X A ln X A + X B ln X B) . Show that if the schematic curve given below is properly plotted, this free energy curve must end asymptotically at the vertical axes of the pure components, i.e. tangential to the vertical axes of the diagrams.

7.

Cite the difference between the intensive and extensive properties of thermodynamic functions and give examples for each property. A.7 Intensive properties such as T and P are independent of the size of the system, whereas extensive properties like V, E, H and G have dependencies on the size of the system. 8.

Draw the relative position of the free energy curves for both liquid and solid phases as a function of temperature and explain why the rate of change in the free energy of liquid is faster than that of solids. A.8 The rate of decrease in G with increasing T corresponds to – S. The relative position of G curves of solid and liquid is given in Figure on the right (G=H-TS). Enthalpy of the liquid is higher than that of the solid at all temperature, therefore at all temperature GLGS since entropy for liquid being always higher than that of solid. G curves of liquid decreases more rapidly due to its higher increase in entropy with increasing temperature.

9.

Draw temperature – pressure diagram for pure iron and explain why the slope of the curve separating a phase from g phase is negative. Use Clausius- Clapeyron equation to explain it. A.9 Clausius- Clapeyron equation:

dP ΔH = dT eq T eq ΔV

( )

Which gives the change in dT required to maintain equilibrium between the two different phases (a g) if pressure is increased by dP. The change in ΔH is always positive with increasing temperature. However There will be a volume decrease when a is transformed to phase because molar volume

γ α of g is smaller than that of a V = V m−V m< 0 ¿ . Therefore the sign of ¿

slope of (a 10.

g) transformation is negative.

dP ¿ ) which is the ¿ ¿

The molar free energy-composition diagram for an ideal solution is given below. Show that the free energy curve, when properly drawn, must end asymptotically at the vertical axes of the pure component. And explain shortly why the molar free energy curve at higher temperature assumes a greater curvature. A.10 The molar free energy of an ideal solution is G1

ΔG mix

⏞ G=G 2 =X ⏞ A G A + X B G B + RT ( X A ln X A + X B ln X B )

GA and GB, the molar free energy of the components of the solution, decrease with increasing temperature. Therefore the molar free energy-composition diagram assumes a greater curvature. 11.

Components A and B in a binary system are fully soluble in each other in liquid and solid states at all composition range. Assuming that the free energy of liquid is lower than that of solid solution, free energy curves of liquid and solid phases in the system at different temperatures are given in Fig. 5a, b, c, d, and e. By making use of these free energy curves, draw the binary phase diagram in Fig 5f and indicate the face domains on the diagram. A.11

12.

A typical, hypothetical and simple binary phase diagram is as follows:

Where L stands for liquid, and A and B are the two components and α and  are two solid phases rich in A and B respectively. a. Draw the free energy curves of the existing phases at the given temperatures [T m(A), Tm(B), T1, T(Eutectic) and T2] as indicated on the phase diagram above. b. Indicate liquidus, solidus and solvus lines on the diagram depicting by the corresponding letters . A.12.a

A.12.b Liquidus: Tm(A)ypzk Tm(B) Solidus: Tm(A)xapqtlTm(B) Solvus: Amo and Bnq