thermalQ

notess.pdf

Home >Engineering homework help >Mechanical Engineering homework help >thermalQ

MSE 708: THERMODYNAMICS OF MATERIALS Page 118

SOLUTION THERMODYNAMICS Solutions are multicomponent, single-phase (homogeneous) thermodynamic systems. Solutions may be in the solid, liquid or gaseous state. The mixing process to form a solution can be viewed as

pure components at T, P solution at T, P

So the change in any thermodynamic quantity B for the mixing process is given by

ΔBmix = B(solution) – B(pure components)

The energy functions U, H, F and G do not have absolute values in thermodynamics. To calculate ΔBmix for these functions, we must define a reference state (standard state in other books). The reference state is simply the pure components at the same T and P as the final solution. Having a common reference state allows changes in state functions like ΔG to be compared between different phases of the same composition (e.g. liq and solid, α and β, etc.). A superscript " º " is used to represent the reference state. If value of B per mole for component k in its reference state nk = number of moles of component k then where Bº = total value of B for all the pure components in the reference state. We already know that the contribution of each component to B(solution) is given by the partial molar B values for each component:

Bk o =

Bo = Bk

k ∑ nk

B(solution) = Bk

k ∑ nk

ΔBmix = Bk

k ∑ nk − Bk

k ∑ nk = Bk −Bk

o( ) k ∑ nk

so for ΔBmix: ΔBmix = B(solution) – B(pure components)

dΔBmix = d

k ∑ Bk −Bk

o( )nk⎡⎣⎢ ⎤⎦⎥= Bkdnk +nkdBk −Bkodnk −nkdBko⎡⎣ ⎤⎦ k ∑

MSE 708: THERMODYNAMICS OF MATERIALS Page 119

is referred to as the partial molar B of solution for component k. (loosely referred to as the "partial molar B" by DeHoff). The "Δ" quantity must be used when B = U, H, F and G, because partial molar absolute values of these quantities cannot be measured. Only changes in these quantities from the reference state can be measured. Note that the two quantities are different: Absolute values can be obtained for . The quantities can also be measured and defined, but are not equal to the former quantities.

~~~~~~~~~~~~~~~~~~~~ The variation of ΔBmix with composition is given by:

The second summation is zero by the Gibbs-Duhem equation. The third summation also equals 0 because the values of Bk in the reference state do not change when the composition of the solution is changed.

dΔBmix = Bk −Bk

o( )dnk k ∑ = ΔBk

k ∑ dnkso:

Collecting the two terms which were equal to zero above:

This is another form of the Gibbs-Duhem equation, applied to the mixing process.

ΔBk

ΔBk ≠ Bk !!

Vk and Sk ΔVk and ΔSk

0 = nkdBk − nkdBk

o⎡ ⎣

⎤ ⎦ k

∑ = dΔBk k ∑ nk

nkdΔBkk k ∑ = 0

ΔBmix = ΔBk

k ∑ nk where ΔBk = Bk −Bk

oor:

dΔBmix = (Bk −Bk

o )dnk + nkdBk − nkdBk o

k ∑

MSE 708: THERMODYNAMICS OF MATERIALS Page 120

OBTAINING PARTIAL MOLAR PROPERTIES FROM TOTAL PROPERTIES For a 2-component system, we have the following two equations for ΔBmix:

ΔBmix = ΔBknk

k ∑ = ΔB1n1 +ΔB2n2

dΔBmix = ΔBkdnk

k ∑ = ΔB1dn1 +ΔB2dn2

If we now put ΔBmix on a per mole of solution basis (i.e. divide by nT = n1 + n2 + .... + nk), the mole numbers are converted to mole fractions, Xk:

ΔBmix = ΔB1X1 +ΔB2X2

dΔBmix = ΔB1dX1 +ΔB2dX2

(ΔBmix is now intensive)

(ΔBmix is extensive)

The mole fractions must sum to 1: X1 + X2 = 1

and so: dX1 + dX2 = 0 Combining the four preceding equations, it can be shown that (DeHoff section 8.2.1):

ΔB1 = ΔBmix +(1− X1)

dΔBmix dX1

ΔB2 = ΔBmix +(1− X2 )

dΔBmix dX2

Thus, measurement of ΔBmix as a function of X1 or X2 allows calculation of the partial molar quantities for both components. The derivatives are usually calculated by differentiating a suitable function which has been fit to the data. Note: The derivatives in these last two expressions are strictly partial derivatives at constant T & P. Since T & P are assumed constant for the mixing process, they are shown as total derivations here (and in DeHoff and many other textbooks).

(derivatives are actually partial derivatives at constant T, P)

MSE 708: THERMODYNAMICS OF MATERIALS Page 121

The inverted parabola is a typical curve shape for ΔHmix or ΔSmix. The partial molar properties at a composition represented by point P are illustrated. The intercepts of a tangent line drawn at P with the X2 = 0 and X2 = 1 axes yields the partial molar quantities. The derivative dΔBmix/dX2 is obtained from the slope of the tangent line = BC/PB. This method is not as accurate as differentiation of a fitted equation, but it is a good way to visualize how partial molar quantities vary with composition:

Δy1 = ΔBmix at X2 o

Δy2 = (Δx)(slope) = (1− X2

o ) dΔBmix

dX2

ΔB2 = Δy1 +Δy2

The partial molar quantities can also be obtained graphically (DeHoff Fig. 8.1).

for this curve shape, as X2 ⇑, ΔB1 ⇑ and ΔB2 ⇓

MSE 708: THERMODYNAMICS OF MATERIALS Page 122

PARTIAL MOLAR PROPERTIES OF ONE COMPONENT FROM THE OTHER The alternative form of the Gibbs-Duhem equation given earlier can be used to obtain the partial molar properties of the second component of a binary solution, if the properties of the first component are known as a function of composition.

nkdΔBkk

k ∑ = 0alternative Gibbs-Duhem eqn:

for a 2-component solution: dividing by nT: solve for

n1dΔB1 +n2dΔB2 = 0

X1dΔB1 + X2dΔB2 = 0

dΔB1 = −

X2 X1

dΔB2 or dΔB1 = − X2 X1

dΔB2 dX2

dX2 dΔB1

dΔB10

ΔB1∫ = − X2 X1

dΔB2 dX2

dX2 0

X2 ∫integrating:

When X2 = 0, ΔB1 = 0 since B1 → B1 o in this limit. So the integral on the left becomes:

This result is known as the Gibbs-Duhem integration. It yields the value of at composition X2. It requires data for vs. X2 over the full range of compositions. The derivative is also evaluated from this data (typically by differentiating an equation from a curve fit). Applying this equation to the partial molar Gibbs free energy, µ, it becomes:

ΔB2 ΔB1

ΔB1 = −

X2 X1

dΔB2 dX2

dX2 0

X2 ∫

Δµ1 = −

X2 X1

dΔµ2 dX2

dX2 0

X2 ∫

dΔB2 dX2

= slope of ΔB2 vs. X2

MSE 708: THERMODYNAMICS OF MATERIALS Page 123

The Gibbs-Duhem integration can be extended to solutions having more than 2 components. This leads to a strategy for obtaining the partial molar properties of all the components in the solution, as well as the total thermodynamic properties for the solution. 1. Starting information ! Δµk for component k vs. Xk at a given T and P. 2. From Gibbs-Duhem integration ! Δµk vs. Xk for all the other components at T and P (see Chemical Thermodynamics of Materials by C.H. Lupis)

3. From ΔGmix = ΔGk

k ∑ Xk = Δµk

k ∑ Xk → ΔGmix at T and P

4. If Δµk is also obtained for one component as a function of T and P (and composition), it is possible to compute all the other partial molar thermodynamic quantities through their relationships with µ and its derivatives with T and P. See Table 8.2 in DeHoff. 5. Using the results of (4) in ! all thermodynamic quantities for the solution process can be obtained.

~~~~~~~~~~~~~~~~~~~

(ΔSk , ΔFk , ΔHk , ΔUk )

ΔBmix = ΔBk

k ∑ Xk

MSE 708: THERMODYNAMICS OF MATERIALS Page 124

THE FUGACITY AND ACTIVITY CONCEPTS These quantities were initially defined to treat non-ideality in gases at higher pressures. The Gibbs free energy change accompanying a change in pressure of a gas at constant temperature is given by:

dG = – SdT + VdP = VdP If the gas is ideal ! V = RT/P and integration yields (V and G are per mole):

ΔG = RTln

P2 P1

If the gas is not ideal, then the dependence of P on V is not given by V = RT/P. To preserve the simplicity of the above equation, G.N. Lewis introduced a quantity called the fugacity, f, which represents the tendency of a substance to escape from the state it is currently in (e.g. liq ! vapor, vapor ! liq, etc.). f is defined through: G = RT lnf + B(T) where B is a constant depending on T and the substance. We can also write, for the reference state of the substance:

Gº = RT lnf º + B(T) so for an isothermal process, the free energy difference between the reference state and the current state is

ΔG = G− Go = RTln

f o

The activity, a, is now defined as which leads to:

ΔG = G – Gº = RTlna or G = Gº + RTlna

In the reference state: G = Gº ! RTlna = 0 ! a = 1 So the activity is a measure of the distance between the current state of the substance and the reference state.

a =

f f o

MSE 708: THERMODYNAMICS OF MATERIALS Page 125

We can now express ΔG for an isothermal process in terms of the activities:

ΔG = G2 – G1 = (Gº + RTlna2) – (Gº + RTlna1)

ΔG = RTln

a2 a1

Note the similarity between this equation and the earlier one based on P. The simplicity of this expression for ΔG is preserved, but the equation can now be applied to non-ideal systems. Fugacities are primarily used to describe non-ideal behavior in gases and are not of much use for solutions. The activity is a more general concept which can be applied to solutions in the solid, liquid and gaseous states. For solutions, the activity is defined through:

where ak is given by:

pk = vapor pressure of component k in solution at T

= vapor pressure of pure k at T This equation is the basis for experimental measurement of activities, which can then be used to calculate ΔGmix and other thermodynamic quantities. For a pure substance, µ = G/n. So the above expression for µk is very similar to the earlier expression for gases:

G = Gº + RTlna since G is on a per mole basis in this equation The above equation for µk can also be written:

µk = µk o + RTlnak

pk o

ak =

pk pk

Δµk = ΔGk = µk −µk o = RTlnak

This relates activity to the partial molar Gibbs free energy of solution for component k.

MSE 708: THERMODYNAMICS OF MATERIALS Page 126

The activity of component k in a solution is related to its concentration (mole fraction) through the activity coefficient, γ:

ak = γkXk

When γk = 1, ak = Xk, and the behavior of component k in the solution is determined completely by its concentration or mole fraction. If γk > 1, component k acts as if there is more k in solution than Xk (i.e. k is "more active"). If γk < 1, the opposite occurs; k is "less active". Rewriting the above equation for Δµk: Δµk = RTlnγk Xk

Δµk = ΔGk = RTlnγk + RTln Xk

excess ideal

The second term represents the for an ideal solution (more later). So the term containing the activity coefficient represents the deviation from ideal behavior, or the "excess" amount:

ΔGk (excess) = RTlnγk

Other partial molar properties can also be expressed in terms of activity coefficients. These are summarized in DeHoff Table 8.5. The Gibbs-Duhem equation can also be recast in terms of activity coefficients. For a binary solution, the Gibbs-Duhem equation for the chemical potential is:

X1dΔµ1 + X2dΔµ2 = 0

Differentiating the previous equation for Δµk (T = constant):

dΔµk = RT dlnγk + dln Xk( )

ΔGk

Substituting into the G-D equation:

X1RT dlnγ1 + dln X1( )+ X2RT dlnγ2 + dln X2( ) = 0

MSE 708: THERMODYNAMICS OF MATERIALS Page 127

X1dln X1 + X2dln X2 = X1

dX1 X1

+ X2 dX2 X2

= dX1 + dX2 = 0note that: so this leaves: X1dlnγ1 + X2dlnγ2 = 0

Starting with this equation, the Gibbs-Duhem integration can also be cast in terms of γ:

lnγ1 = − X2 X10

X2 ∫

dlnγ2 dX2

dX2

Substituting γk = ak/Xk into the previous two equations puts them in terms of the activity:

X1dlna1 + X2dlna2 = 0

lna1 = −

X2 X10

X2 ∫

dlna2 dX2

dX2

γk is obtained from measurements of ak and Xk.

~~~~~~~~~~~~~~~~~~~~~~~~

G-D eqn. in terms of activity coefficients

MSE 708: THERMODYNAMICS OF MATERIALS Page 128

THERMODYNAMIC APPROACHES FOR DESCRIBING SOLUTIONS So far we have only identified general relationships between the various thermodynamic quantities for solutions. To make these equations useful, we must relate them to measurable quantities. This requires application of a model or certain assumptions regarding the behavior of the solution. 1. IDEAL SOLUTIONS Assumes that there are no interactions between the components of the solution, as in an ideal gas. In this case, the thermodynamic properties of the solution should be determined only by the concentrations (mole fractions) of the components.

i.e. γk = 1 and ak = γkXk = Xk (ideal solution) Using the earlier expression for Δµk:

Δµk = ΔGk = RTlnak = RTln Xk

So for ΔGmix: ΔGmix = ΔGk

k ∑ Xk = RT Xk ln Xk

k ∑

from ΔS = − ∂ΔG ∂T

% &

( ) P,nj

⇒ ΔSmix = −R Xk k ∑ ln Xk

from ΔH = ΔG + TΔS ! ΔHmix = RT Xk ln Xk

k ∑ −RT Xk ln Xk

k ∑ = 0

from ΔU = ΔH – PΔV ! ΔUmix = 0 − 0 = 0

from ΔV = − ∂ΔG ∂P

% &

( ) T,nj

⇒ ΔVmix = 0

from ΔF = ΔG – PΔV ! ΔFmix = RT Xk ln Xk

k ∑

These relationships are summarized in DeHoff Table 8.3, along with partial molar quantities.

MSE 708: THERMODYNAMICS OF MATERIALS Page 129

For a binary solution, the variation of the non-zero quantities with composition and temperature are shown in DeHoff, Fig. 8.3.

In the ideal solution model, everything is determined by the composition. The results do not depend in any way on the chemical identity of the components! Thus, while the ideal model is a useful starting point for thermodynamic description of solutions, it is not very useful without additional information specific to the components being mixed. 2. NON-IDEAL (REAL) SOLUTIONS Already discussed in terms of "ideal" and "excess" quantities, e.g.:

Δµk = ΔGk = RTlnγk + RTln Xk

excess ideal

To use this approach, the activity coefficients of one component must be measured experimentally as a function of composition (X). The activity coefficients of the other components can be calculated from the Gibbs-Duhem integration. Other partial molar quantities are given in DeHoff, Table 8.5. From these, the ΔBmix values can be obtained from For non-ideal solutions, ΔUmix, ΔHmix and ΔVmix ≠ 0. This is a very general approach but requires experimental data.

ΔBmix = ΔBk

k ∑ Xk

MSE 708: THERMODYNAMICS OF MATERIALS Page 130

3. DILUTE SOLUTIONS These systems are described by Raoult's law and Henry's law. A liquid of pure component 1 in a closed container at a constant temperature will develop an equilibrium vapor pressure above its surface. This vapor pressure represents a dynamic equilbrium between the rate of evaporation from the liquid surface and condensation onto the surface. Now add a few atoms or molecules of component 2. The mole fraction of 1 is now X1 < 1. If 2 is uniformly distributed throughout the solution, the vapor pressure of 1 will be reduced by a factor of X1:

p1 = X1p1 o

p2 = X2p2 o

Since a1 =

p1 p1

o ⇒ a1 = X1

Either expression describes Raoult's law. Since a1 also equals X1 for an ideal solution, ideal solutions correspond to Raoultian behavior. Starting with pure component 2 and adding a few atoms of 1 leads to a similar expression:

Raoult's law

The total vapor pressure of a binary solution which follows Raoult's law is the sum of the partial pressures of the two components (p1 + p2) :

0 X2 1

vapor pressure

p1 o

p2 o

p1 + p2

p1 p2

MSE 708: THERMODYNAMICS OF MATERIALS Page 131

When X1 increases to the point where the component 1 atoms begin interacting, Henry's law no longer holds. Thus, this is also a dilute solution limiting law. If 1-2 interactions are stronger than 1-1 or 2-2 ! Henry's law line lies below Raoult's law. If 1-2 interactions are weaker than 1-1 or 2-2 ! Henry's law line lies above Raoult's law. Illustrated in the figures below. The relationship between Raoult's law, Henry's law and the experimentally measured activity is also illustrated.

The solution on the left is said to exhibit positive deviations from Raoult's law while the one on the right exhibits negative deviations from Raoult's law. Examples: Positive deviation: Thallium-Tin at 352 ºC

Negative deviation: Iron-Nickel at 1600 ºC

k1 or γ1 o = Henry's law constant

1-2 interactions stronger 1-2 interactions weaker

X1 X1

p1 o

dilute in X1

dilute in X2

a1 or p1

Raoult's law holds when the 1-2 interactions are identical with the 1-1 and 2-2 interactions; i.e. solute atoms behave identically to solvent atoms. If the 1-2 interactions have a different strength than the 1-1 or 2-2 interactions, the rate of evaporation of 1 changes, leading to:

p1 = k1X1 or a1 = γ1 oX1 Henry's law

k1 = p1 oγ1

MSE 708: THERMODYNAMICS OF MATERIALS Page 132

4. REGULAR SOLUTIONS - have two characteristics. a. ΔSmix = ΔSmix (ideal), i.e. no excess entropy of mixing. ! b. ΔHmix ≠ 0. ΔHmix depends on composition. The form of this dependence varies with different regular solution models. In general:

ΔSmix xs = 0

ΔGmix xs = ΔHmix

xs − TΔSmix xs

ΔGmix xs = ΔHmix

xs = ΔHmix

ΔGk = RTlnγk + RTln Xk

excess ideal

so ΔGk xs = RTlnγk = ΔHk

This allows the activity coefficient to be related to the partial molar heat of mixing for each component:

γk = e ΔHk /RT

Once the activity coefficients are known for each component, all the partial molar properties for the solution components can be calculated (DeHoff Table 8.5). From these, the ΔBmix quantities for the solution can be obtained. The simplest regular solution model uses the following expression for ΔHmix with one adjustable parameter, ao (binary solution):

ΔHmix = aoX1X2 = ΔGmix xs ao ≠ activity

The form of equation selected for ΔHmix must allow ΔHmix to pass through zero when X1 or X2 = 0. This is the simplest form which meets this criterion. Inserting this into ΔGmix:

Recall that the excess partial molar free energy is related to the activity coefficient:

ΔHmix = ΔHmix xs +ΔHmix

ideal since ΔHmix ideal = 0, ΔHmix = ΔHmix

ΔGmix = aoX1X2 +RT X1ln X1 + X2 ln X2( )

(excess) (ideal)

ΔGmix = ΔGmix xs + ΔGmix

ideal

Since ΔSmix xs = 0 ⇒

ΔHmix can also be related to ΔGmix(excess):

MSE 708: THERMODYNAMICS OF MATERIALS Page 133

For this model, the partial molar heats of mixing become (DeHoff example 8.1):

ΔH1 = aoX2 2

ΔH2 = aoX1 2 (binary solution)

so the activity coefficients are: γ1 = e aoX2

2 /RT

γ2 = e aoX1

2 /RT

in the dilute solution limit, X2 ! 1 as X1 ! 0, and vice-versa. In this limit, the activity coefficients become the Henry's law constants:

γ1 o = eao /RT γ2

o = eao /RT

The regular solution model predicts that the activity coefficients are equal at both ends of the composition range (i.e. for dilute solutions). This is a shortcoming of this simple model and a result of the mathematical form of ΔHmix. This implies that the properties of component 1 surrounded by component 2 are the same as the properties of component 2 surrounded by component 1. May be applicable when 1 and 2 are very similar (e.g. Cu & Ni). Why does γk become the Henry's law constant as Xk ! 0?

γ1 o and γ2

where γ1

o = lim X1→0

γ1 γ 2

o = lim X2→0

γ 2

The sign of ao determines the nature of the departure from ideal behavior (or from Raoult's law):

positive ao ! positive deviation from ideal behavior negative ao ! negative deviation from ideal behavior

Recall that activity ak = γkXk. So γk is the secant slope of a plot of ak vs. Xk at each point. As Xk ! 0, the slope of this secant line becomes the slope of ak vs. Xk at Xk = 0. This is also the Henry's law constant, as shown in the figure on page 131 of these notes. If the ordinate is p1, Henry's law constant = k1. If the ordinate is a1, Henry's law constant = .

0 X1 1

slope = γ1 o

slope = γ1

γ1 o

MSE 708: THERMODYNAMICS OF MATERIALS Page 134

ΔHmix and ΔSmix do not depend on T. At any given composition, ΔGmix vs. T is linear (not shown). This comes from ΔG = ΔH – TΔS. ΔH and ΔS are functions of composition only, not temperature. So at fixed composition, ΔG varies linearly with T. For positive departures from ideal behavior (ao > 0), the ΔGmix curves develop two minima and a maximum at lower temperatures. This leads to a miscibility gap in the solution which can be observed on phase diagrams. Refinements to the simple regular solution model: -- more complicated dependence of ΔH on composition with additional adjustable parameters -- add temperature dependence to the adjustable parameters, which makes ΔHmix dependent on T.

~~~~~~~~~~~~~~~~~~~~~~

Variation of thermodynamic mixing properties with composition for this model shown in DeHoff Figure 8.5. ao = 12,500 J/mol, max T = 1200 K.

MSE 708: THERMODYNAMICS OF MATERIALS Page 135

5. QUASICHEMICAL MODEL Allows for interactions between the molecules or atoms of a solution. Only nearest- neighbor interactions are considered. For an A-B solution, there are 3 types of interactions ! A–A, B–B and A–B. Each type of interaction is assigned an interaction energy: eAA, eBB, eAB These interaction energies can be thought of as the value of the energy at the minimum of the energy-distance curve between two atoms undergoing some type of bonding:

Energy

0 x

• eaa

The total internal energy of the solution = the sum of all the nearest-neighbor interactions between all the atoms or molecules in the solution. This requires knowledge of the number of A–A, B–B and A–B bonds in the solution ("P" terms).

Usoln = PAAeAA + PBBeBB + PABeAB

It can be shown that (DeHoff sec. 8.7.3):

PAA = 0.5(XANoz – PAB) PBB = 0.5(XBNoz – PAB) where No = total number of atoms or molecules in the solution

z = coordination number (number of nearest neighbors) substitution into U yields, after simplification:

Usoln = PAB eAB −.5(eAA + eBB )

"# $%+.5Noz XAeAA + XBeBB "# $%

MSE 708: THERMODYNAMICS OF MATERIALS Page 136

We desire an expression for ΔUmix. To get this, we imagine formation of a solid solution of A and B from pure solid A and pure solid B. All 3 solids are assumed to have the same crystal structure.

A (solid) + B (solid) ! AB (solid) The reference state is therefore pure solid A and B, which are mixed in the same proportions as the final solid solution (i.e. same mole fractions). The internal energy of the reference state is:

ΔUmix = Usoln − XAUA

o + XBUB o#

$ % &

where UA o and UB

o = internal energies per mole of pure A and B. These are

UA o = .5NozeAA UB

o = .5NozeBB

substituting for these quantities and Usoln and simplifying:

ΔUmix = PAB eAB −.5(eAA + eBB )#$ %& ΔHmix

since ΔU ≅ ΔH for solids and liquids at constant P. So ΔHmix depends only on the number of A–B bonds in the solution and an energy term related to the difference between the A–B bond energy and the average energy of A–A and B–B bonds. This is a general result as it does not depend on the arrangement of A and B in the solid solution (note the absence of z). The primary limitation of the quasichemical theory is that it considers only nearest-neighbor interactions, so the "e" quantities are independent of composition. The A and B atoms in a solid solution may be distributed among the lattice sites in a random, clustered or ordered way. PAB will be different for each of these cases.

(factor of .5 to avoid double-counting)

Uref = XAUA o + XBUB

so ΔUmix is

MSE 708: THERMODYNAMICS OF MATERIALS Page 137

Consider random mixing of A and B: The probability that a given site is occupied by A is XA, and the probability it is occupied by B is XB. The probability that two adjacent sites are occupied by A is XA2 and the probability they are occupied by B is XB2. The probability that two adjacent sites are occupied by one A and one B is

XAXB + XAXB = 2XAXB = fraction of A–B bonds because there are two ways to populate two sites with an A and a B. The total number of bonds in the solid solution is .5Noz. So the total number of A–B bonds is

PAB = .5Noz(2XAXB) = NozXAXB

Inserting this into the ΔHmix equation yields the following result for random mixing:

It will be shown later that the entropy of mixing for a random solution of A and B is the same as for ideal solutions:

ΔSmix

ran = ΔSmix id = −R XA ln XA + XB ln XB( )

ΔHmix = NozXAXB eAB −.5(eAA + eBB )#$ %&

Compare with the result from the regular solution model: ΔHmix = aoXAXB

apparently

So the regular solution model corresponds to the quasichemical model in which the A and B atoms are randomly mixed. However, the constant ao has now been given physical meaning: it depends on the energies of the 3 types of bonds in the system.

Using this entropy of mixing with the previous equation for the enthalpy of mixing leads to a free energy of mixing which is also identical to the regular solution model, except for the extended interpretation of ao:

ΔGmix

ran = ΔHmix ran − TΔSmix

ran = aoXAXB +RT XA ln XA + XB ln XB( )

ao = Noz eAB −.5(eAA + eBB )

"# $%

MSE 708: THERMODYNAMICS OF MATERIALS Page 138

Recall that the sign of ao determines whether the solution exhibits positive or negative departure from ideal solution behavior or Raoult's Law. We can now interpret this in terms of the interactions between atoms A and B.

positive departure: ao > 0 ! eAB > .5(eAA + eBB) negative departure: ao < 0 ! eAB < .5(eAA + eBB) ideal solution: ao = 0 ! eAB = .5(eAA + eBB) (ΔHmix = 0)

This leads to a conceptual difficulty in applying the regular solution model with its implicit assumption of random mixing. If the solution exhibits positive departure from ideal behavior, bonds between like atoms are more favored than bonds between unlike atoms [since eAB > .5(eAA + eBB)]. Thus, the atoms would be expected to arrange themselves so that more A–A and B–B bonds would be formed. This is no longer random mixing, as there would be a higher preference for a site next to an A (B) to contain another A (B) atom. A similar argument can be made for negative departures from ideal behavior. In this case, A–B bonds would be favored over A–A or B–B bonds. More sophisticated versions of the quasichemical model incorporate deviations from random mixing by introducing long- and short-range order parameters into the entropy of mixing term. The regular solution model still finds application at high temperatures where ΔGmix is dominated by the ΔSmix term.

~~~~~~~~~~~~~~~~~~~~~~~~~ 6. POLYMER SOLUTIONS Here we will consider only liquid polymer solutions in which a polymer is dissolved in a solvent. The development of the theory of polymer solutions was initially carried out by Flory and Huggins, so it is usually referred to as the Flory-Huggins theory. The F-H theory is essentially a modified version of the quasichemical model (or the regular solution model with random mixing). The primary difference for polymer solutions is in the entropy of mixing term.

MSE 708: THERMODYNAMICS OF MATERIALS Page 139

Polymer molecules are of course thousands of times larger than solvent molecules. Flory treated this by assuming that a solvent molecule was the same size (volume) as a monomer unit (repeat unit) of the polymer. The difference in the entropy of mixing for two small molecules vs. a small molecule and a polymer can be visualized by imagining that the molecules occupy sites on a lattice (this is the configurational entropy).

mixture of 2 small molecules entropy of mixing is high

mixture of a polymer with a small (solvent) molecule

entropy of mixing is low

The entropy of mixing is related to the number of arrangements for the system through S = k ln Ω. Clearly the number of possible arrangements is higher for 2 small molecules than for a polymer and a small molecule.

The entropy of mixing for two polymer molecules is very low. This is why blends of two different polymers are almost always immiscible. For miscibility, ΔGmix must be negative. For mixing of small molecules, the entropy term is large and negative, and this provides the driving force for mixing, even if the enthalpy term is positive. For mixing of two polymers, the entropy term is small and usually overwhelmed by the positive enthalpy term.

MSE 708: THERMODYNAMICS OF MATERIALS Page 140

As with the earlier models, the goal is to obtain an expression for

ΔGmix = ΔHmix – T ΔSmix We first consider the random mixing of two small molecules for comparison with the polymer/solvent case. Using the lattice model, we define:

n1 ! number of molecules of solvent n2 ! number of molecules of solute

(n1 + n2) ! total number of cells in the lattice For the ΔSmix term: ΔSmix = S(solution) – S(pure components) To compute the S(solution) from S = k ln Ω, we need the total number of ways to arrange n1 solvent and n2 solute molecules in (n1 + n2) cells. This is given by the same combinatorial equation we encountered earlier:

Ω =

(n1 +n2 )! n1!n2 !

To get S(pure components), we recognize that there is only 1 way to arrange pure solvent or solute molecules on the same lattice. Therefore,

S (pure components) = k ln(1) = 0. So ΔSmix = S(solution)

Applying Stirling's approximation to each term and simplifying yields:

ΔSmix = −k n1ln X1 + n2 ln X2

#$ %&

ΔSmix = k ln

(n1 + n2 )! n1!n2 !

= k ln(n1 + n2 )! – lnn1! – lnn1!⎡⎣ ⎤⎦

ΔSmix = −k n1ln X1 + n2 ln X2

#$ %& NA NA

= −R N1ln X1 + N2 ln X2 #$ %&

Multiplying and dividing the rhs by NA converts the n's to mole numbers N1 and N2:

Dividing both sides by (N1 + N2) converts the N's to mole fractions and puts ΔSmix on a per mole basis:

ΔSmix = −R X1ln X1 + X2 ln X2

#$ %& (ΔSmix is per mole)

MSE 708: THERMODYNAMICS OF MATERIALS Page 141

The same equation was obtained earlier for the entropy of mixing for ideal solutions. The statistical approach used here gives us the additional insight that formation of an ideal solution corresponds to random mixing of the pure components. For polymer solutions, determination of Ω and ΔSmix are more complicated. The additional issues are: -- each monomer unit must be constrained to occupy a cell that is adjacent to a cell already occupied by another monomer unit. -- the probability that an adjacent cell is already occupied by a monomer unit must be considered. -- the formation of the solution involves two steps for the polymer: disorientation of the polymer molecules from an ordered (crystalline) state, followed by mixing of the disoriented polymer with solvent molecules. The entropy of each step is not zero. -- for details, see Chap. XII in "Principles of Polymer Chemistry", P.J. Flory (1953). The net result for the entropy of mixing is:

ΔSmix = −k n1lnφ1 + n2 lnφ2

$% &'

or, multiplying by NA/NA: per mole basis: (divide by [N1 + N2])

ΔSmix = −R N1lnφ1 + N2 lnφ2

$% &'

ΔSmix = −R X1lnφ1 + X2 lnφ2

$% &'

Here ϕ1, ϕ2 = volume fractions and N1, N2 = moles of solvent and polymer. Note the similarity between this equation and the earlier one for random mixing in an ideal solution. The difference is replacement of mole fractions with volume fractions in the ln terms. In terms of the n's, the volume fractions are given by:

φ1 =

n1 n1 + xn2

φ2 =

xn2 n1 + xn2

n1 = number of solvent molecules n2 = number of polymer molecules x = number of monomer units per polymer molecule

Note that x really represents the ratio of the volume of a polymer molecule to the volume of a solvent molecule. It may or may not equal the number of monomer units in a polymer molecule (x). The very low value of ΔSmix for polymer solutions (compared to small molecules) comes from the small value of X2. For example, consider a lattice consisting of 1000 cells, half of which are occupied by one polymer molecule having x = 500:

n1 = 500 n2 = 1

If we now replace the polymer molecule having x = 500 with 500 individual small molecules (different from the solvent):

n1 = 500 n2 = 500

So the X2lnϕ2 term in ΔSmix is ~2 orders of magnitude smaller for polymer solutions compared to solutions of small molecules or atoms. The difference gets larger as x increases if the concentration of polymer remains the same.

~~~~~~~~~~~~~~~~~~

MSE 708: THERMODYNAMICS OF MATERIALS Page 142

X2 =

n2 n1 +n2

= 500

500 + 500 = .5

X2 =

n2 n1 +n2

= 1

500 +1 = .002

We need an expression for n12. If z = coordination number of the lattice, (z – 2) = number of cells adjacent to each monomer unit, and (z – 2)x + 2 = number of cells adjacent to one polymer chain (each chain end has 1

additional adjacent cell). For large x ! (z – 2)x + 2 is approximated as zx Now we assume: ϕ1 = volume fraction of solvent, and can also be interpreted as = the probability that a cell adjacent to a monomer unit is occupied by a solvent molecule. so: n12 = (adjacent cells per chain, zx) x (# of chains, n2) x (probability an adjacent cell is occupied by solvent, ϕ1) n12 = zxn2ϕ1 = total number of 1-2 contacts in the entire solution

MSE 708: THERMODYNAMICS OF MATERIALS Page 143

Next we consider the ΔHmix term. As in the quasichemical model, this term contains the nearest-neighbor pairwise interaction energies exx, redefined as follows for polymer solutions:

e11 = interaction energy between solvent molecules e22 = interaction energy between monomer units e12 = interaction energy between a solvent molecule and a monomer unit

For the quasichemical model, we found that ΔHmix depends on the number of A-B bonds formed, which in this case corresponds to the number of monomer unit- solvent molecule contacts, n12:

ΔHmix  ΔUmix = n12 e12 −.5(e11 + e22 )#$ %&

MSE 708: THERMODYNAMICS OF MATERIALS Page 144

n1 + xn2 =

n1 φ1 =

xn2 φ2

Recall that:

rewriting:

so:

So the total number of 1-2 contacts can also be written as n12 = zn1ϕ2. Putting this into ΔHmix:

ΔHmix = zn1φ2 e12 −.5(e11 + e22 )$% &'

We now define χ = Flory-Huggins polymer-solvent interaction parameter:

χ =

z e12 −.5(e11 + e22 )#$ %& kT

χ is a dimensionless quantity which is determined by applying the F-H theory to experimental measurements. Incorporating χ into ΔHmix yields:

ΔHmix = kTχn1φ2

As before, this equation can be converted to mole numbers or mole fractions:

ΔHmix = χRTN1φ2 ΔHmix = χRTX1φ2 ! ΔHmix is per mole

Finally, the expressions for ΔSmix and ΔHmix are combined to obtain ΔGmix:

ΔGmix = ΔHmix − TΔSmix

ΔGmix = RT χN1φ2 +N1lnφ1 +N2 lnφ2

$% &' (ΔGmix in cal)

ΔGmix = RT χX1φ2 + X1lnφ1 + X2 lnφ2

$% &' (ΔGmix in cal/mole)

φ1 =

n1 n1 + xn2

and φ2 = xn2

n1 + xn2

xn2φ1 = n1φ2

Combining this eqn. with provides a relationship for a1:

MSE 708: THERMODYNAMICS OF MATERIALS Page 145

The chemical potential difference for the solvent can be obtained by differentiating ΔGmix (homework problem):

Δµ1 = RTlna1

Δµ1 =µ1 −µ1

o = RT lnφ1 +φ2 1− 1 x

% &

( )+ χφ2

2 +

, -

/ 0

a1 can be obtained experimentally by measuring the vapor pressure of the solvent as a function of polymer concentration (assuming the vapor behaves as an ideal gas):

a1 =

p1 p1

lna1 = lnφ1 +φ2 1−

1 x

$ %

' (+ χφ2

This is the usual route to testing solution theories via experiment. Measurement of a1 vs. concentration (ϕ1, ϕ2) yields values of χ vs. concentration. If F-H theory is correct, χ should be constant with ϕ2.

from Flory, Principles of Polymer Chemistry, Figure 111.

Seems to work well for rubber-benzene system (first system tested), but not for the others. The existence of a molecular dipole in either the polymer or solvent apparently results in specific interactions that aren't accounted for.

PDMS/benzene

PS/methyl ethyl ketone

PS/toluene

rubber/benzene

(= Φ2)

MSE 708: THERMODYNAMICS OF MATERIALS Page 146

Limitations of Flory-Huggins theory: 1. Does not apply to dilute solutions (ϕ2 ≤ ~0.1). This is why dilute solution data is not included in the above figure. Theory assumes that polymer molecules overlap in solution and are not isolated. This is the basis for saying probability a cell adjacent to a monomer unit is occupied by solvent ≈ volume fraction of solvent. If dilute, this probability is near 100% and is not equal to the volume fraction of solvent. 2. Only considers configurational entropy. Later refinements of this theory show that χ must also have an entropic contribution. 3. Use of the same lattice for solvent and polymer is questionable.

~~~~~~~~~~~~~~~~~~

A CLOSER LOOK AT THE SIGNIFICANCE OF χ

Δµ1 = RT lnφ1 +φ2 1−

1 x

% &

( )+ χφ2

2 +

, -

/ 0

Δµ1 = RT ln(1−φ2 )+φ2 + χφ2

2% &

' (

we have shown:

ln(1− x) = − x −

2 −

3 −

4 − i i i ≈ − x −

assuming high molecular weight (x ! ∞) and using ϕ1 = 1 – ϕ2:

expanding the ln term using:

Δµ1 = RT −φ2 −

1 2 φ2

2 +φ2 + χφ2 2%

& '

(

) *

Δµ1 = RTφ2

2 χ− 1 2

& '

(

) *

When χ = ½ , Δµ1 = 0. Since Δµ1 = RTlna1 ! a1 = 1.0 ! p1 = p1o. i.e. the solution behaves as if there is no polymer (component 2) present.

MSE 708: THERMODYNAMICS OF MATERIALS Page 147

In other words, the tendency of a solvent molecule to escape into the vapor phase is not altered by the presence of the polymer molecules. ! the solution behaves as if there are no polymer-solvent interactions. ! solution behaves ideally when χ = ½. When χ < ½ ! a1 < 1.0, i.e. the solvent has less tendency to escape into the vapor phase, meaning that the polymer-solvent interactions must be strong. As χ continues to decrease below ½, (negative χ's are possible), the strength of the polymer-solvent interactions increase. ! good solvents for a particular polymer have χ < ½ (χ may be negative). If χ > ½ ! a1 > 1.0. This would require p1 > p1º, which is not too likely. In this case polymer-solvent interactions are very weak and a solution should not occur. ! poor solvents for the polymer have χ > ½. ! a solution for which χ = ½ is on the verge of precipitation. Temperature dependence: Increasing T causes χ to decrease, meaning that the solvent quality is improved as T increases. The temperature at which χ = ½ is referred to as the "theta temperature", T = Θ.

~~~~~~~~~~~~~~~~~~~

χ ∝

1 T

MSE 708: THERMODYNAMICS OF MATERIALS Page 148

THERMODYNAMIC THEORY OF RUBBER ELASTICITY

The unique elastic properties of rubber can be derived from thermodynamics. Consider a piece of rubber in uniaxial tension:

A f = (restoring) force L, Lo = stretched, initial length A, Ao = stretched, initial cross-sectional area λ = extension ratio = L/Lo λ is related to the engineering strain, ε:

ε =

L − Lo Lo

= L Lo

− 1= λ − 1

Assume: ΔV = 0 upon stretching (works well) V = Vo AL = AoLo

Now consider the types of work involved when the rubber is stretched by dL: + fdL = work done on system ! positive – PdV = work by system to change the volume of the surroundings ! negative Since dV = 0 ! PdV = 0 and we have for the differential work: dW = fdL Assuming the stretching is done reversibly, using the first and second laws:

dU = dQ + dW = TdS + fdL

f =

dU dL

− T dS dL

solve for f:

Finally, assume stretching is isothermal ! impose constant T:

f =

∂U ∂L "

# $

& ' T,V

− T ∂S ∂L "

# $

& ' T,V

Equation of state for rubber

MSE 708: THERMODYNAMICS OF MATERIALS Page 149

Two terms contribute to the restoring force f, both have units of force or energy/ length (from energy = force x length).

∂U ∂L "

# $

& ' T,V

= "energy elasticity"

Physical origin: - Elastic straining of bond lengths and bond angles - Changes in intermolecular interactions (secondary bonding) with stretching Both of the above are very small. Stretching rubber involves rotation about backbone bonds only, no straining of bond lengths and angles. No significant changes in intermolecular interactions during stretching.

∂U ∂L "

# $

& ' T,V

≈ 0so:

This result is analogous to an ideal gas:

∂U ∂V "

# $

& ' T ≈ 0

The term “ideal rubber” is adopted in analogy to an ideal gas. So for an ideal rubbery polymer:

∂U ∂L "

# $

& ' T = 0

Second term: −T

∂S ∂L

$ %

' ( T,V

= "entropy elasticity"

First term:

Physical origin: orientation of polymer chains that occurs during stretching.

stretch

unoriented high S

oriented low S

As L #, S $. So

∂S ∂L "

# $

& ' T < 0 and –T

∂S ∂L "

# $

& ' T

> 0

MSE 708: THERMODYNAMICS OF MATERIALS Page 150

The entropy elasticity is a positive contribution to the restoring force and is very large compared to the energy elasticity. Thus, the equation of state can be well- approximated using only the entropy term:

Equation of state ideal rubber

f = −T

∂S ∂L #

$ %

' ( T,V

Thus, the restoring force in rubber is primarily due to the tendency for oriented polymer molecules to regain lost entropy. Rubber is often referred to as an “entropy spring”. This behavior is very unique! Note that rubber must be lightly crosslinked to prevent molecules from permanently sliding past each other when stretched. Crosslinking also ensures 100% recovery when stress is released. This type of mechanical behavior is different from purely elastic materials and is referred to as rubbery elastic. Comparison of rubber with conventional elastic materials: For metals, ceramics, some polymers; origin of elasticity is the change in internal energy that occurs upon elastic deformation.

i.e. ΔU is large ΔS is small For rubber: ΔU is small (~ 0) ΔS is large

~~~~~~~~~~~~~~~~~

MSE 708: THERMODYNAMICS OF MATERIALS Page 151

MOLECULAR THEORY

To put the equation of state to practical use, must find an expression for S as a function of deformation. For generality, we will allow for any type of deformation, e.g. uniaxial, shear, biaxial, etc. Model: A crosslinked network of polymer molecules (network chains). Each network chain adopts a randomly coiled conformation in the unstretched state.

r = end-to-end distance n = number of backbone bonds in

network chain l = carbon-carbon bond length Assumptions:

1. ΔV = 0 during deformation 2. 3. affine deformation: macroscopic deformation = molecular deformation 4. network chains have a distribution of lengths (i.e. different values of n) As the rubber network is stretched, the r values for each network chain will change. S decreases as the chains are stretched, so S = S(r). Consider a single network chain. For an individual chain, n is fixed. But many values of r are still possible, depending on how the chain is coiled. So there will be a distribution of r values, even for a network chain of fixed n. We desire the equilibrium or most probable distribution of end-to-end distances.

ideal rubber ⇒ ∂U ∂L( )T,V = 0

• • network

chain

crosslinks

• • • •

• network chain

crosslinks

MSE 708: THERMODYNAMICS OF MATERIALS Page 152

Define p(n,r) = probability of finding the end of a chain having n bonds at a distance r from the other end, taken as the origin.

This is essentially the 3-D random walk problem. A drunk takes n steps in random directions, each step has the same length l. The probability that he will end up a distance r away from the starting point after n steps is given by the Gaussian probability distribution:

p(n,r) =

3 2πnl2 "

# $

& '

3/2

exp −3r 2

2nl2 "

# $$

& ''

p(n,r) has units of (volume)-1 and is a probability density (probability/unit volume). To obtain a probability, we must multiply by a volume element. In this case, the volume element is a spherical shell:

• r y

dr dV = 4πr 2dr p(n,r)(4πr2dr) = probability that a network chain of

n bonds has its end between r and r + dr if the other end is at r = 0. This probability is normalized, i.e.

p(n,r)dV

∞ ∫ = 4πr 2

∞ ∫ p(n,r)dr = 1

4πr 2p(n,r)

rmax r •

rmax

2 = 2 3

nl2 Plot of this distribution:

Maximum value occurs at the most probable value of r: This is not the same as the mean-square (average) value: The mean-square value comes from:

r 2 = nl2

r 2 = (r 2 )(4

∞ ∫ πr 2 ) p(n,r)dr

rmax

2 = 2 3

nl2

Properties of this distribution:

MSE 708: THERMODYNAMICS OF MATERIALS Page 153

Limitations of the Gaussian distribution (or random walk model): 1. Random walk steps can occur in any direction, but fixed bond angles and steric

effects in polymer chains limit the directions in which "steps" can occur. 2. In a random walk, the same location can be visited more than once. This

corresponds to the "phantom chain". This cannot happen for real polymer chains. 3. The distribution of end-to-end distances is no longer Gaussian at λ > ~ 5 (more

later). These limitations can be accounted for to some extent, but the equations become more complicated.

~~~~~~~~~~~~~~~~~~~~~~~~

ENTROPY OF A SINGLE NETWORK CHAIN, s Define this as s (lower case). We know that s = k ln Ω. Ω is the number of ways the system can be arranged. In this case, Ω is related to the number of conformations available to a network chain having a fixed n and r.

p(n,r) =

3 2πnl2 "

# $

& '

3/2

exp −3r 2

2nl2 "

# $$

& ''i.e. Ω is proportional to

b2 =

3 2nl2

Let and let the constant of proportionality = c' (includes (1/π)3/2)

now: Ω = c 'b 3 exp(−b2r 2 )

so: s = k ln c 'b3 exp(−b2r 2 )⎡⎣

⎤ ⎦ = k ln c 'b

3( ) − kb2r 2

or: s = c − kb2r 2 with c = k ln c 'b3( )

If the network chain is on a Cartesian coordinate system with one end at the origin, the other end a distance r from the origin:

r 2 = x2 + y2 + z2

So the entropy of a single, unstretched network chain becomes:

s = c −kb 2(x2 + y2 + z2 )

The "o" subscripts indicate the unstretched dimensions. The extension ratios may be different in all 3 directions, allowing for many types of deformation. Because we assumed an affine deformation, the dimensions of the individual network chains will change by the same ratios along x, y and z:

MSE 708: THERMODYNAMICS OF MATERIALS Page 154

s ' = c −kb2(λx

2x2 +λy 2y2 +λz

2z2 )

We now stretch the rubber network so that the macroscopic dimensions change by the following extension ratios along the x, y and z axes:

So the change in entropy of a single chain (of fixed n) upon stretching becomes:

Δs = s '− s = −kb2 x2(λx

2 −1)+ y2(λy 2 −1)+ z2(λz

2 −1)⎡⎣ ⎤ ⎦

~~~~~~~~~~~~~~~~~~~~~~

ENTROPY CHANGE FOR THE ENTIRE RUBBER NETWORK Divide all the network chains into groups of equal n: Group # 1 2 . . . . . . . . . . p # bonds per chain n1 n2 np # chains per group N1 N2 Np parameter b b1 b2 bp contribution to total ΔS ΔS1 ΔS2 ΔSp Let N = ΣNp = total number of network chains

λx =

Lx Lo,x

λy = Ly Lo,y

λz = Lz Lo,z

So the end of a network chain moves from (x, y, z) to (λxx, λyy, λzz) upon stretching. The entropy for the stretched chain (= s') will be:

• •

•

0,0,0

λxx, λyy, λzz

x,y,z stretched

unstretched

MSE 708: THERMODYNAMICS OF MATERIALS Page 155

Consider now the chains in group p. These all have np bonds and there are Np chains in the group. The entropy change upon stretching for group p will be:

ΔSp = Δsp

i=1

Np ∑ = −kbp

2 xi 2(λx

2 −1)+ yi 2(λy

2 −1)+ zi 2(λz

2 −1)⎡⎣ ⎤ ⎦

i=1

Np ∑

ΔSp = −kbp

2 (λx 2 −1) xi

i ∑ +(λy

2 −1) yi 2

i ∑ +(λz

2 −1) zi 2

i ∑

& '

(

) *

where xi, yi and zi are the unstretched coordinates of the end of chain i. Breaking the summation into separate terms:

For each chain: xi

2 + yi 2 + zi

2 = ri 2

i ∑ + yi

i ∑ + zi

i ∑ = ri

i ∑

The mean-square values of the coordinates are: Subscripts are not needed here since these quantities represent the collection of network chains.

Due to the symmetry of the random walk model, excursions in the x, y and z directions are equally likely.

x2 = y2 = z2

i ∑ = yi

i ∑ = zi

i ∑

i ∑ + yi

i ∑ + zi

i ∑ = ri

i ∑

This allows us to write:

using the 3 relationships above:

we also know that:

from the last two equations:

xi 2

i ∑ = yi

i ∑ = zi

i ∑ =

1 3

ri 2

i ∑

x2 =

1 Np

xi 2

i ∑

y2 =

1 Np

yi 2

i ∑

z2 =

1 Np

zi 2

i ∑

MSE 708: THERMODYNAMICS OF MATERIALS Page 156

Substituting this last result into the ΔSp equation and factoring it out of the "[ ]":

ΔSp = −

1 3

kbp 2 (λx

2 −1)+(λy 2 −1)+(λz

2 −1)⎡⎣ ⎤ ⎦ ri

i ∑

We can also define a mean-square r value as we did for the mean-square x, y and z values:

where the subscript "n" indicates that this value refers to network chains, as opposed to "free" (outside the network) polymer chains. For "free" polymer chains with np bonds of length l undergoing a random walk, the mean-square r value is:

n =

1 Np

ri 2

i ∑ or ri

i ∑ = Np rp

o = npl

where the subscript "o" indicates that the chain is outside of the network. We can use this equation to rewrite the definition of the "b" parameter:

bp 2 =

3 2npl

2 =

3 2

rp 2

(1)

(2)

Incorporating equations (1) and (2) into the ΔSp equation:

ΔSp = − 1 2

kNp rp

rp 2

λx 2 +λy

2 +λz 2 − 3$%

& '

This is ΔS for group p. To get ΔS for the entire network, we simply sum over all the groups.

ΔS = ΔSp p ∑ = −

1 2

k λx 2 +λy

2 +λz 2 − 3%&

' ( Np

rp 2

o p ∑

is different for each group p since np is different for each group. rp

MSE 708: THERMODYNAMICS OF MATERIALS Page 157

To simplify this sum further, one of two assumptions can be made: 1. All chains in the network have the same contour length (same number of bonds).

i.e. np = n = constant

n = r 2

n = constant

o = r 2

o = constant

This leads to:

ΔS = − 1 2

Nk r 2

r 2 o

λx 2 +λy

2 +λz 2 − 3$%

& '

2. The end-to-end distance of a chain in the network is the same is if it were out of the network, i.e.

n = rp

This is the most common assumption. It holds as long as the crosslink density is low (np is high) and the network is not subject to significant dilation due to solvent swelling or thermal expansion. This assumption also allows us to retain the distribution of network chain lengths (more realistic than 1 above). The result is:

ΔS = −

1 2

Nk λx 2 +λy

2 +λz 2 − 3$%

& '

Note that ΔS = 0 in the undeformed state (λx = λy = λz = 1).

~~~~~~~~~~~~~~~~~~~~

MSE 708: THERMODYNAMICS OF MATERIALS Page 158

ΔS FOR VARIOUS STRESS STATES 1. Uniaxial extension and compression Assume the tensile axis is z. Then: and

λz =

Lz Lo,z

= λ λx = λy

Lx Ly Lz = Lo,x Lo,y Lo,z ⇒ λx λy λz = 1

ΔV = 0

λx λy = 1 λz

⇒ λx = λy = 1

λz =

Putting this into ΔS: ΔS = −

1 2

Nk λ2 + 2 λ − 3

% &

( )

To incorporate this result into the equation of state, we need

ΔS = S − So ⇒ ∂ΔS ∂L

⎛

⎝ ⎜

⎞

⎠ ⎟ T,V

= ∂S ∂L

⎛

⎝ ⎜

⎞

⎠ ⎟ T,V

− ∂So ∂L

⎛

⎝ ⎜

⎞

⎠ ⎟ T,V

= ∂S ∂L

⎛

⎝ ⎜

⎞

⎠ ⎟ T,V

f = −T

∂S ∂L

⎛ ⎝⎜

⎞ ⎠⎟ T

= NkT Lo

λ − 1 λ2

⎛ ⎝⎜

⎞ ⎠⎟

subbing into the eqn. of state for an ideal rubber

We can rewrite the number of network chains, N as: N =

ρVo Mc

where Mc = number avg. MW of network chains, g/mol ρ = rubber density, g/cm3

Vo = volume, cm3

NA = 6.023 x 1023 mol-1

ΔS for uniaxial extension or compression

∂S ∂L ⎛

⎝ ⎜

⎞

⎠ ⎟ T,V

∂S ∂L

⎛

⎝ ⎜

⎞

⎠ ⎟ T,V

= 1

∂S ∂λ

⎛

⎝ ⎜

⎞

⎠ ⎟ T

= − Nk Lo

λ − 1 λ2

⎛

⎝ ⎜

⎞

⎠ ⎟

MSE 708: THERMODYNAMICS OF MATERIALS Page 159

OTHER QUANTITIES

σ =

f Ao

= ρRT Mc

λ − 1 λ2

⎛ ⎝⎜

⎞ ⎠⎟

σT =

f A =

fλ Ao

= ρRT Mc

λ2 − 1 λ

⎛ ⎝⎜

⎞ ⎠⎟

Engineering stress, σ True stress, σT Young's modulus, Y Initial modulus, Yi

Y =

dσ dε

= dσ dλ

= ρRT Mc

1+ 2 λ3

& '

(

) *

Yi = Y(λ = 1) =

3ρRT Mc

W = Vo σ dλ

λ ∫ =

ρRTVo 2Mc

λ2 + 2 λ − 3

⎛ ⎝⎜

⎞ ⎠⎟

Isothermal, reversible work, W, done upon stretching from λ = 1 to λ:

f =

ρAoRT Mc

λ − 1 λ2

⎛ ⎝⎜

⎞ ⎠⎟

Finally: f =

ρVoNAkT McLo

λ − 1 λ2

⎛ ⎝⎜

⎞ ⎠⎟

equation of state, ideal rubber uniaxial extension. Note non-linear behavior with λ

slope of σ vs. λ at λ = 1 Mc obtained from σ vs. λ measurements

MSE 708: THERMODYNAMICS OF MATERIALS Page 160

2. Shear in the x-y plane

No strain in z direction, so λz = 1.0 and λx = λ, then λy =

1 λ

(from λx λy λz = 1)

ΔS = −

1 2

Nk λ2 + 1 λ2

− 2 $

% &

( )

ΔS = −

1 2

Nk λ − 1 λ

% &

( )

N =

ρVo Mc

NAusing ΔS = −

ρRVo 2Mc

λ − 1 λ

⎛ ⎝⎜

⎞ ⎠⎟

It can be shown that the shear strain, γ, is related to λ through

γ ≈ λ −

1 λ

(OK for γ ≤ 0.6)

Isothermal, reversible work done upon shearing an ideal rubber

W = −Q = −TΔS = ρRTVo

2Mc λ −

1 λ

⎛ ⎝⎜

⎞ ⎠⎟

Shear stress, τ: or

W =

ρRTVo 2Mc

γ 2So W reduces to:

Thus, rubber obeys Hooke's Law in shear (t vs. λ is linear). Other equations above that involve the quantity (ρRT/Mc) can be re-cast in terms of G.

τ = Gγ with G =

ρRT Mc

= shear modulus

τ =

1 Vo

dW dγ

= ρRT Mc

Yi =

3ρRT Mc

⇒ Yi = 3Ge.g.

ΔU = 0 = Q + W !

(G varies with T)

MSE 708: THERMODYNAMICS OF MATERIALS Page 161

This last equation also follows from the relationship between Y and G that involves Poisson's ratio, ν. For rubber, ν = 0.5.

Yi = 2G(1+ ν) ⇒ ν = 0.5 ⇒ Yi = 3G

Other relationships can be derived for other stress states by identifying λx, λy and λz subject to the constraint that λx λy λz = 1.0.

~~~~~~~~~~~~~~~~

TEST OF THE THEORY - UNIAXIAL EXTENSION/COMPRESSION

To apply the theory, must measure Mc. Can be obtained from σ - λ tests (Yi) or other means (swelling measurements). Note non-linear behavior (rubbery elastic).

Theory works very well in compression (λ < 1.0). Works in tension up to λ = 1.2, then over-estimates stress. Deviation at low λ is most likely due to physical entanglements which act like crosslinks but allow some relative movement of the chains. Less stress is required for a given extension when the chains start to slide through the entanglements. At high λ ~ 6.0, the experimental data markedly exceeds the theoretical prediction. This is due to a breakdown of the assumption of a Gaussian distribution of end-to- end distances used to derive the ΔS equation.

solid line - theory

uniaxial defomation

solid line - theory

uniaxial deformation

MSE 708: THERMODYNAMICS OF MATERIALS Page 162

f =

ρRTAo Mc

λ − 1 λ2

⎛ ⎝⎜

⎞ ⎠⎟

Reason: Gaussian chains ! p(n,r) ! 0 only when r ! ∞, Real chains ! p(n,r) ! 0 when r ! nl (the contour length)

Real chains reach a limiting extension at a much smaller r than Gaussian chains.

~~~~~~~~~~~~~~~~~~~~~~ Some interesting predictions from the theory:

1. At constant λ: #T #f !! - Restoring force increases with heating if deformation is held fixed.

2. At constant f: #T $λ (ignore 1/λ2 term) - Stretched rubber shrinks when heated. Hang constant weight from strip of rubber and heat. Weight rises.

- (1) and (2) are the result of the tendency for entropy to increase with increasing

temperature. - Molecules move from an oriented state toward a more coiled-up (random)

conformation. - End-to-end distance decreases. Since ends are crosslinked, this causes the rubber

network to contract. Unstretched rubber (λ = 1) behaves normally: Heat ! expands. Cool ! contracts. 3. #λ adiabatically (Q = 0) ! rubber gets warm !! If $λ adiabatically, rubber cools. Reason for this not evident from above equations. It can be shown that (homework):

Adiabatic ! Q = 0 ! ΔS = 0 ! S = constant Typical temperature change: at λ = 6 ! ΔT = 10 ºC

∂T ∂λ

$ %

' ( S = σ ρc

σ = engineering stress ρ = density c = specific heat (at constant L)

MSE 708: THERMODYNAMICS OF MATERIALS Page 163

REFINEMENTS TO THE THEORY 1. Account for network defects such as free chain ends and loops which do not contribute to the restoring force.

• •

• • •

• free ends

loop

2. Alternative distribution for p(n,r) that is valid at larger λ (see work by Treloar). This distribution shows an upturn around λ ~ 5 – 6 as seen in experimental data. 3. Mooney-Rivlin equation Derived from continuum mechanics- not a molecular theory. Result for work is:

W = C1 λx 2 +λy

2 +λz 2 − 3( )+ C2 1

λx 2 +

1 λy

2 +

1 λz

2 − 3

% %

( (

C1, C2 = empirical constants. Note similarity to molecular theory. The two coincide when C1 = ½GVo and C2 = 0. Fits data to higher λ than molecular theory. Probably due to 2 adjustable parameters.

~~~~~~~~~~~~~~~~~~~

"ineffective" crosslink

MSE 708: THERMODYNAMICS OF MATERIALS Page 164

CAPILLARITY EFFECTS IN THERMODYNAMICS (DeHoff Chapter 12, sections 1 – 5.3)

Capillarity - addresses effects of interfacial curvature on thermodynamic properties - early efforts used small capillaries to create high curvature to study these effects - for bulk liquids and solids, effects are usually small. When size of particle or phase gets into nanometer range, effects can be significant.

These effects are driven by the curvature of the interface and the surface energy. Local curvature For an arbitrary curve in 2 dimensions, the local curvature at point P is defined in DeHoff Fig. 12.1. Select two points A and B on the curve on either side of P. Construct the unique circle that contains A, P and B on its circumference. Now let A and B approach P, and reduce the size of the circle so A, B and P remain on the circumference.

2H =

1 r1 +

1 r2

where r1 and r2 are the local curvatures in two perpendicular directions on the curved surface. If the interface has the shape of a sphere, then

In the limit, as A and B reach P, a unique circle is obtained (12.1b) with radius r. r is defined as the radius of curvature at P. 1/r is defined as the local curvature at P.

local curvature at P =

1 r

Extending this to 3 dimensions, consider a small slice of a 2-D curved surface, representing an interface. The local mean curvature at point P is defined as H:

• P

r1 = r2 = r and H =

1 r

MSE 708: THERMODYNAMICS OF MATERIALS Page 165

The surface tension or surface free energy, γ Consider a soap film stretched over a wire frame with a moveable bar:

L soap film f

The soap film exerts a force on the moveable bar in a direction opposite to f. Define a quantity γ as the force per unit length:

γ =

f L

The work necessary to extend the bar a distance dx is then

work = fdx = γLdx

Since Ldx = dA, where A = area, the work could also be written as

work = γdA

This shows that γ could equally well be described as an energy per unit area.

! surface tension is used when γ is a force/length ! surface free energy is used when γ is an energy/area

These two quantities are identical for isotropic fluid phases. They are not necessarily the same when dealing with solid phases. Why is γ a free energy? To see this, incorporate the work for deformation of the soap film into the thermodynamic equations. An extra work term must be added (work on system is positive):

δW = – PdV + γdA

For an open, single-component system, film stretched reversibly:

dU = δQ + δW + µdn = TdS – PdV + γdA + µdn

from definition of G: G = H – TS = U + PV – TS

MSE 708: THERMODYNAMICS OF MATERIALS Page 166

differentiating: dG = dU + PdV + VdP – TdS – SdT inserting dU: dG = TdS – PdV + γdA + µdn + PdV + VdP – TdS – SdT

dG = VdP – SdT + γdA + µdn From the equations for dU and dG, we see that γ can be expressed as

γ =

∂U ∂A #

$ %

' ( S,V,n

= ∂G ∂A #

$ %

' ( T,P,n

For processes that occur at constant T and P (the usual case), γ is interpreted as a (Gibbs) free energy per unit area. Equilibrium between two phases separated by a curved interface

units: energy/area

β α

α Define the system as small particles, droplets, etc. of β phase surrounded by α phase. The curvature is defined to be positive when the interface is convex relative to the β phase, as shown.

Conditions for equilibrium (thermal and chemical): T α = Tβ and µk

α =µk β

These are the same conditions as for systems without curved interfaces. The requirement for mechanical equilibrium (i.e. pressure) is different. To determine the requirement for pressure equilibrium, consider again a soap film, this time in the shape of a spherical bubble of radius r (analogous to the picture above).

total surface energy = γA = γ(4πr2) Assume the radius of the bubble decreases by dr. The total surface energy then decreases by

d(γA) = d(4γπr2) = 8γπrdr Decreasing the size of the bubble compresses the air inside, creating a pressure difference across the soap film. To stay in equilibrium, the work associated with this pressure difference must exactly balance the decrease in surface energy.

MSE 708: THERMODYNAMICS OF MATERIALS Page 167

work against ΔP = decrease in surface energy

ΔPdV = ΔP(4πr2)(dr) = 8γπrdr

ΔP =

2γ r

As bubble size decreases ! pressure inside the bubble increases. ΔP is insignificant for large bubbles (ΔP = ~ 0.1 torr for r = 1 cm, γ = 30 ergs/cm2).

~~~~~~~~~~~~~~~~~~~~~~ Applying this concept to a spherical particle of β phase inside the α phase: The effect of the surface free energy is to increase the pressure inside the β particle relative to outside the particle, i.e. in the α phase. The condition for equilibrium is given by the equation above:

ΔP = Pβ −Pα =

2γ r

incorporating the curvature, H = 1/r : P β −Pα = 2γH DeHoff eqn. 12.27

The same equation holds for the more general definition of H for a non-spherical interface described by the two local radii of curvature:

2H =

1 r1 +

1 r2

" #

% &

Pβ −Pα = γ

1 r1 +

1 r2

& ''

(

) **

Surface free energy and surface tension are equal when both phases are isotropic fluids. When one or both phases are crystalline solids, they are not necessarily equal. Surface tension ! becomes a surface stress and is a tensor quantity Surface free energy ! varies by location on the surface (not a constant) and depends on which crystal planes are at the surface and their orientation. Eqn. 12.27 still describes the condition for local mechanical equilibrium, but evaluation of γ and H become more complicated.

dr • r

For a filled droplet with one interface. For a hollow bubble, replace γ with 2γ.

MSE 708: THERMODYNAMICS OF MATERIALS Page 168

Phase boundary shifts in unary systems due to capillarity In what follows, we assume the conditions at the interface are such that eqn. 12.27 applies. For a unary two-phase system (α + β) in which interfacial curvature plays a role, the conditions for equilibrium are:

Tα = Tβ

µα = µβ

Pβ = Pα + 2γH A change of state may involve changes in H, as well as in P and T. If the α and β phases are taken through an arbitrary change of state:

dµ α = dGα = –SαdTα + VαdPα molar quantities,

so µ = G dµ

β = dGβ = –SβdTβ + VβdPβ

If the process is reversible, and both phases stay in equilibrium, then the conditions above must hold, so

dTα = dTβ = dT

dµα = dµβ = dµ

dPβ = dPα + 2γdH The last relationship indicates that the pressure in the β phase may be changed either by changing Pα or by changing H. Setting the equations for dµα and dµβ equal, and substituting for dPβ :

–S αdT+ VαdPα = –SβdT+ VβdPβ

–SαdT+ VαdPα = –SβdT+ Vβ dPα + 2γdH( )

Sα – Sβ( )dT − Vα − Vβ( )dPα + 2γVβdH = 0

Define: ΔS = Sα – Sβ( ) and ΔV = Vα − Vβ( )

These definitions define a change from β ! α (α is state 2, β is state 1).

MSE 708: THERMODYNAMICS OF MATERIALS Page 169

The change of state is defined this way since β was chosen as the reference phase when defining the curvature H. This yields

ΔSdT −ΔVdP α + 2γVβdH = 0

The first two terms are identical to the Clapeyron equation derived earlier for equilibrium between two phases. The third term adds the effect of a curved interface on the equilibrium. We now apply this equation to two important cases. 1. Effect of interfacial curvature on liquid-vapor equilibrium (vapor pressure) Consider the vapor pressure for the two cases shown below involving a single component and two phases, L and V. In the second case, the liquid is subdivided into small droplets which are surrounded by vapor. Assume the temperature is the same for both cases.

To be consistent with earlier conventions, β ! L phase; α ! V phase. Applying the modified Clapeyron equation with dT = 0 (isothermal):

ΔVdP V = 2γVLdH

ΔV = Vα − Vβ = VV − VL

We assume that VV >> VL, and treat the vapor as an ideal gas.

ΔV = VV =

RT PV

MSE 708: THERMODYNAMICS OF MATERIALS Page 170

substituting for ΔV:

RT PV

dPV = 2γVLdH

dPV

PVPo

P ∫ =

2γVL

RT dH

H ∫

Assuming γ and VL are don't vary with H or P, we can separate the variables and integrate:

P Po

= 2γVL

RT H or P = Po exp

2γVLH RT

# $$

& ''

Po = vapor pressure in a system with no curvature (flat interface) P = vapor pressure in a system with small spherical droplets Since the exponent is positive, ex > 1 ! so P > Po and the vapor pressure increases as the curvature increases (radius of droplets decreases). H has units of (length)-1, so (2γVL/RT) must have units of (length)+1. It is sometimes called the capillarity length scale for liquid-vapor equilibrium (λV):

λv =

2γVL

RT P = Po e

λvH = Po e λv /rso !

Typical values of λv are 1 – 2 nm (DeHoff Table 12.2). Not very sensitive to the type of material (water, metals, oxides, salts all in this range). If λvH << 1, the exponential function can be expanded as a series, and the higher- order terms can be ignored, so:

ex ≈ 1+ x and P = Po 1+ λvH( ) = Po 1+

λv r

⎛

⎝⎜ ⎞

⎠⎟

H = 1/r for spherical droplets

Experimental measurements confirm an exponential (or 1/r) dependence of vapor pressure on droplet size. DeHoff Fig. 12.9 shows Zn droplets at T = 900 K. The effect becomes significant below r = 100 nm.

(H = 0 ! no curvature, flat interface)

MSE 708: THERMODYNAMICS OF MATERIALS Page 171

The foregoing also implies that increasing the curvature of the interface will produce a shift in the liquid-vapor phase boundary on a unary phase diagram. At a given temperature, the boundary will shift to higher vapor pressures as the size of the droplets decreases (curvature increases). This can also be thought of as a decrease in the vaporization temperature at a fixed pressure.

droplet size

DeHoff Fig. 12.10

Effect of interfacial curvature on solid-liquid equilibrium (melting point) This is similar to the last example, but now we have solid and liquid phases of a single component. Assume that the pressure is the same in both cases.

Applying the modified Clapeyron equation with dP = 0:

β ! S phase α ! L phase

ΔSdT = −2γV SdH

Here dT represents the differential shift in the melting temperature produced by a differential change dH in the curvature of the solid phase. Assuming that ΔS, γ and VS do not change much with H or melting temperature, we can separate the variables and integrate:

Tm o

Tm ∫ = −

2γVS

ΔS dH

H ∫

Tm − Tm

o = − 2γVS

ΔS H

MSE 708: THERMODYNAMICS OF MATERIALS Page 172

rearranging: Tm = Tm

o − 2γVS

ΔS H

Gibbs-Thomson equation

Tmo = melting temperature (K) for a system with no curvature (flat interface) Tm = melting temperature (K) for a system with curvature = H Since ΔS = SL – SS > 0, the Gibbs-Thomson equation predicts that Tm decreases as H increases. Small crystals melt at lower temperatures than large crystals (at the same pressure). The ΔS term can be replaced in the Gibbs-Thomson equation as follows:

factor out Tmo :

Tm = Tm

o 1− 2γVS

Tm oΔS

H $

% & &

( ) )

At the equilibrium melting point: substituting:

ΔG = 0 = ΔHf − Tm oΔS ⇒ ΔHf = Tm

oΔS

λm =

2γVS

ΔHf a capillarity length scale for solid- liquid equilibrium can be defined: leading to:

Tm = Tm

o 1− 2γVS

ΔHf H

% & &

( ) )

Tm = Tm

o 1−λmH#$ %&

Typical values of λm are ca. 0.5 nm (DeHoff Table 12.2) Example (DeHoff ex. 12.3): melting point of silicon dendrites. Tmo (Si) = 1685 K Dendrites have tip radius = 0.5 µm ! ΔHf = 50.7 kJ/mole VS = 12 cm3/mole λm = 5 x 10-8 cm ! Tm = 1683 K γ = 1050 ergs/cm2

Not much change. But if the tip radius = 0.05 µm (50 nm) ! Tm = 1668 K .

H =

1 0.5 ×10−4 cm

= 2×104 cm−1

MSE 708: THERMODYNAMICS OF MATERIALS Page 173

This would also be the melting point of Si nanocrystals having a radius of 0.05 µm. The crystal size effect also shifts the solid-liquid phase boundary to lower temperatures on unary phase diagrams.

DeHoff Fig. 12.12

The crystal size effect is also responsible for "Ostwald ripening" or coarsening in binary alloys when annealed at elevated temperatures. Larger particles of β phase grow at the expense of smaller particles (DeHoff fig. 12.14). The total volume of β phase remains constant.

increasing time ! Explanation: the concentration of solute in the α phase is higher around small particles than around large particles due to the phase boundary shift caused by the higher curvature of the small particles (DeHoff fig. 12.13). This creates a concentration gradient of solute between small and large particles, and induces diffusion of solute from the small to the large particles.

MSE 708: THERMODYNAMICS OF MATERIALS Page 174

Effect of crystal size on melting point Tm for polymer crystals The size effect for melting of polymer crystals is especially important since polymer crystals always have one dimension that is very small, ~ 10 nm. From dilute solution, polymers crystallize via a chain-folding mechanism that produces lamellar single crystals with a thickness of ~ 10 nm and lateral dimensions of 10 – 100 µm.

The chain ends are excluded from the crystal. Crystal grows by addition of molecules to the crystal edges ({110} planes for PE). The nucleus is usually modeled as a rectangular parallelpiped. a,b,l = dimensions of nucleus γe, γs = end and side surface free energies (per unit area) The length dimension l is also known as the fold period. This dimension (crystal thickness) doesn't change as the crystal grows.

e.g. PE: Thickness ~ 100 Å ~ 80 – 100 C atoms ~ 40 – 50 unit cells

◠ ◠ ◠ ◠ ◠ ◠ ◠ ◠

γs

γe

Crystals observed in bulk polymers are similar. The thickness is about 100 Å while the lateral dimensions are much larger (tens of microns). The overall free energy change ΔGn for formation of a nucleus includes:

-- positive (destabilizing) terms resulting from the formation of new surfaces -- a negative (stabilizing) term resulting from the increased interactions between molecules that occur when they are in a crystalline arrangement (ΔGv).

ΔGn has units of energy. ΔGv has units of energy per unit volume.

ΔGn = 2alγs + 2blγs + 2abγe − ablΔGv

surface terms volume term

MSE 708: THERMODYNAMICS OF MATERIALS Page 175

ΔGc = 2alγs + 2blγs + 2abγe − ablΔGv

We can reinterpret this as the free energy change ΔGc for formation of a mature, rectangular crystal having dimensions of a, b and l, recognizing that l will be much smaller than a or b.

ΔGc abl

= 2γs b

+ 2γs a

+ 2γe

l −ΔGv

e.g. PE crystals: a, b ~ 10 – 100 µm, l ~ 10 – 50 nm. This is a difference of ~ 103. Divide this equation by the volume of the crystal, V = abl:

Since a,b >> l, the first two terms on the rhs are very small compared to the third.

ΔGc abl

= 2γe

l −ΔGv

ΔGv = ΔHv − TΔSv (all per unit volume)

ΔGv can be expressed in terms of more measurable quantities. At any temperature T at which the crystal is formed (below the equilibrium melting point):

ΔHf = enthalpy of fusion per unit volume ΔSf = entropy of fusion per unit volume

= equilibrium melting temperature for large crystals

ΔG = 0 = ΔHf − Tm oΔSf

ΔSf =

ΔHf Tm

If we now assume that ΔHf ~ ΔHv and ΔSf ~ ΔSv (a good approximation if T is not too far below equilibrium melting point), we can substitute into the ΔGv equation:

ΔGv = ΔHf − T

ΔHf Tm

o = ΔHf 1−

T Tm

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

ΔGv = ΔHf

Tm o − T Tm

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

where Tm o − T = (degree of) supercooling

Tm o

whereas at the equilibrium melting point for large crystals (ΔG ≠ ΔGv here):

and

MSE 708: THERMODYNAMICS OF MATERIALS Page 176

This gives the dependence of the free energy for crystal formation on T and the crystal thickness l. For a given value of l, the crystal will melt at some temperature T. At this temperature the crystal will be in equilibrium with the melt, so:

ΔGc = 0 and T = Tm = melting point for a crystal having thickness = l.

0 =

2γe l − ΔHf Tm

o Tm

o − Tm( )

Tm = Tm

o 1− 2γe lΔHf

% &&

( )) all T's in K

ΔGc abl

= 2γe

l − ΔHf Tm

o Tm

o − T( )

Substituting this for ΔGv in the earlier equation for ΔGc abl

This is essentially the Gibbs-Thomson equation with r (or 1/H) = l and γ = γe. This shows that crystals that are small in only one dimension also melt at a lower temperature than crystals that are large in all three dimensions. For PE: l = 50 – 500 Å ! Tm = 373 – 411 K = 100 – 138 ºC. Large range ! For PE crystals > 500 Å thick, Tm becomes independent of the crystal thickness. Polymers typically form very thin crystals, so they usually exhibit a size-dependent melting point. Annealing close to the melting point can increase the crystal thickness, thus raising Tm toward . Because of the way they are processed, bulk polymeric materials often contain a range of crystal thicknesses, therefore they melt over a relatively broad range of temperatures.

Tm o