Single Component Systems

Single Com ponent System s

1

First Order Transition

Gibbs Free Energy is the sam e for water and ice at 0°C. There is an enthalpy of fusion DH^f and an entropy change on

m elting DS^f. These balance G = H –TS. C^p= (dH/dT)^pThere is a change in the slope of the H vs. T plot at the m elting point. Ice holds less heat than water.

DG^f= 0 = DH^f – T^fDS^f T^f = DH^f/DS^f

3

Clausius-Clapeyron Equation

Consider two phases at equilibrium , a and b

dµ^a = dµ^b dG = Vdp –SdT

so

V^adp – S^adT = V^bdp – S^bdT so

dp/dT = DS/DV and

DG = 0 = DH – TDS so DS = DH/T and

dp/dT = DH/(TDV) Clapeyron Equation For transition to a gas phase, DV ~ V^gas and for low density gas (ideal) V = RT/p

d(lnp)/dT = DH/(RT²) Clausius-Clapeyron Equation -S U V

H A -p G T

Clausius Clapeyron Equation d(ln p)/dT = DH/(RT²) Clausius-Clapeyron Equation

d(ln p^Sat) = (-DH^vap/R) d(1/T)

ln[p^Sat/ p^RSat] = (-DH^vap/R) [1/T – 1/T^R] Shortcut Vapor Pressure Calculation:

5

Clausius Clapeyron Equation d(ln p^Sat) = (-DH^vap/R) d(1/T)

ln[p^Sat/ p^RSat] = (-DH^vap/R) [1/T – 1/T^R]

This is a kind of Arrhenius Plot

Clapeyron Equation predicts linear T vs p for transition

dp/dT = DH/(TDV) Clapeyron Equation

7

Clausius-Clapeyron Equation

Consider absorption of a gas on a surface

First order transition from a vapor to an absorbed layer

Find the equilibrium pressure and tem perature for a m onolayer of absorbed hydrogen on a m esoporous carbon storage m aterial

Use Clausius –Clapeyron Equation to determ ine the enthalpy of absorption

9

W hat About a Second Order Transition?

For Exam ple: Glass Transition T^g versus P?

There is only one “phase” present. A flowing phase and a “locked-in” phase for T^g. There is no discontinuity in H, S, V

dV = 0 = (dV/dT)^pdT + (dV/dp)^T dp = VadT – Vk^Tdp dp/dT^g = Da/Dk^T

T^g should be linear in pressure.

a = (1/V) (dV/dT)^p kT = (1/V) (dV/dP)^T

dp/dT^g = Da/Dk^T

Soft Matter, 2020, 16, 4625

x is the dielectric relaxation tim e Glass transition depends on the rate of observation, so you need to fix a rate of observation to determ ine the transition tem perature.

From L. H . Sperling, "Introduction to Physical Polym er Science, 2'nd Ed." 11

The glass transition occurs w hen the free volum e reaches a fixed percent of the total volum e according to the iso-free volum e theory. This figure show s this value to be 11.3% . The bottom dashes line is the occupied volum e of m olecules, w hich increases w ith tem perature due to vibration of atom s. The right solid line is the liquid line w hich decreases w ith tem perature due to reduced translational and rotational m otion (free volum e) as w ell as m olecular vibrations (occupied volum e). A t about 10% the translational and rotational m otion is locked out and the m aterial becom es a glass. The free volum e associated w ith these m otions is locked in at Tg.

Flory-Fox Equation

Fox Equation

Num ber of end-groups = 2/Mⁿ

This indicates that the param eter of interest is 1/T^g

T^gis the tem perature where a certain free volum e is found due to therm al expansion, V = V^occupied + V^free = V⁰ + Va^TdT

T^gis the tem perature where V^free/V = 0.113

End groups have m ore free volum e T^g occurs when the free volum e reaches less than V^free ≤ 0.113V

13

Second order transition Curie and Neel Tem peratures Ferro to Para M agnetic

Ferri to Para M agnetic

Second order transition Neel Tem perature (like Curie Tem p for antiferrom agnetic)

Inden M odel t = T/T^tr For t <1

15

Landau theory for 2’nd order transitions based on a Taylor series expansion of the Gibbs free energy in the “Order Param eter” G

-The free energy is analytic (there is a function)

-The free energy is sym m etric (only even powers of T)

The order param eter is originally the m agnetization, m For liquid crystals it is the director

For binary blends it can be the com position

Curie Tem perature is the critical point for ordering. Above T^c no order and m = 0 in the absence of a m agnetic field param agnetism

Below T^c m has a value.

At constant T and p

”a” is a bias associated with the direction of m agnetization, this is 0 above T^c

“b” is positive above T^c and changes sign at T^c

At the Curie transition (second order transition)

Order param eter is 1 at 0 K so and

17

Single Com ponent Phase Diagram s

For a single com ponent an equation of state relates the variables of the system , PVT PV = RT or Z = 1 Ideal Gas at low p or high T or low r

Virial Equation of State

So a phase diagram will involve two free variables, such as P vs T or T versus r.

Other unusual variables m ight also be involved such as m agnetic field, electric field.

Then a 2D phase diagram would require specification of the fixed free variables.

19

Single Com ponent Phase Diagram s

For a single com ponent an equation of state relates the variables of the system , PVT

Isochoric phase diagram

Gibbs Phase Rule

21

Field Induced Transitions

Consider constant volum e (isochoric) and subject to a m agnetic field -S U V

H A -p G T

dU = -pdV + TdS + Bdm = TdS + Bdm dA = -SdT – pdV +Bdm = -SdT + Bdm

dA = -SdT + Bdm

So A is naturally broken into functions of T and m

(dA/dT)^m = -S (dA/dm )^T = B

dA = (dA/dT)^m dT + (dA/dm )^T dm dA = (dA/dT)^m dT + (dA/dm )^T dm

Take the second derivative

d²A/(dTdm ) =(d/dT(dA/dm )^T)^m = (d/dm (dA/dT)^m)^T = d²A/(dm dT) Using the above expressions and the m iddle two term s

Legendre Transform ation

M agnetic Field Strength B M agnetic M om ent, m

(strength of a m agnet) M agnetic m om ent drops with T Torque = m x B

Assum e constant volum e, V

23

Consider constant volum e (isochoric) and subject to a m agnetic field -S U V

H A -p G T

dU = -pdV + TdS + Bdm = TdS + Bdm dA = -SdT – pdV +Bdm = -SdT + Bdm

Legendre Transform ation

M agnetic Field Strength B M agnetic M om ent, m

(strength of a m agnet) M agnetic m om ent drops with T Torque = m x B

Assum e constant volum e, V

We want to know how the m agnetic m om ent, m , changes with tem perature at constant volum e and field strength, B, (dm /dT)^B,V. Intuitively, we know that this decreases.

Define a Helm holtz free energy m inus the m agnetic field energy, A’,

A’ = A – Bm , and set its derivative to 0. This is the com plete HFE for a m agnetic field, (see the Alberty paper section 4, probably need to read the whole paper or just believe it) dA’ = 0 = dA – Bdm – m dB = -SdT + Bdm –Bdm –m dB = -SdT –m dB = 0

We can perform a Legendre Transform on this equation yielding:

(dm /dT)^B,V = (dS/dB)^T,V

So the change in m agnetic m om ent with tem perature (which decreases) is equal to the reduction in entropy with m agnetic field (as the m aterial orders).

W ith this extension the four M axwell relations expand to 27 with the norm al param eters and a very large num ber if you include the different fields in slide 16

M agnetic field strength decreases with tem perature

The Curie tem perature is where m agnets lose their perm anent m agnetic field

(dm /dT)^B,V = (dS/dB)^T,V

The rate of change of m agnetic m om ent in tem perature at constant field reflects the isotherm al change in

entropy with m agnetic field. At the Curie Tem perature entropy doesn’t change with field at constant

tem perature.

Ising M odel

25

Gibbs Phase Rule with n additional com ponents

Degrees of freedom , F plus num ber of phases Ph, equals the num ber of com ponents, C ,plus 2 plus the num ber of

additional com ponents considered, n.

Equations of State for Gasses

Ideal Gas: pV = RT p = rRt Z = 1

27

-S U V H A

-p G T dG = -SdT + VdP

At constant T (dG = Vdp)^T

For an ideal gas V = RT/p

DG = RTln(pf/pⁱ) Ideal Gas at constant T, no Enthalpic Interactions For single com ponent m olar G = µ

µ0 is at p = 1 bar µ = µ0 + RT ln p

Chem ical Potential of an Ideal Gas

µ = µ0 + RT ln p i.g.

At equilibrium between two phases the chem ical potentials are equal and the fugacities of the two phases are also equal.

Real Gas

29

P= RT/V

Cubic Equation of State

Cubic Equation of State Solve cubic equations (3 roots)

Ideal G as Equation of State

Van der Waals Equation of State

Virial Equation of State

Peng-R obinson Equation of State (PR EO S) Z = 1

Law of corresponding states P = RTr/(1-br) – a r²

Single Com ponent Phase Diagram s

Isochoric phase diagram

31

Compound Tc(K) Pc(MPa)

METHANE 190.6 4.604 rc= 0.0104 mol/cm³

Gas Tc(K) Pc(MPa)

ISOPENTANE 460.4 3.381 rc= 0.00287 mol/cm³

At 0.8 * 460.4K = 368K And 0.64 M Pa 2 phases Higher pressure liquid Lower vapor

F(Z)=

33

CALculation of PHAse Diagram s, CALPHAD For m etal alloys to construct phase diagram s

Calculate the Gibbs Free Energy

Use a Taylor Series in Tem perature

Determ ine the phase equilibria using the chem ical potentials Calculate the derivatives of the free energy expression

Get H^{m SER} from H^{m 0} for the com ponents

35