Chapter 4: Physical Transformation of Pure Substances

Major Topics: Review:

Phase, Phase Transition Dependence of G on T, P Phase Diagrams Gibb’s Phase Rule Chemical Potential Dependence on T, P Clapyeron Equation Clausius-Clapyeron Equation Phases Phase: state of matter that is uniform throughout in chemical composition and physical state

# of phases, P

Examples of P=1: • Pure gas • Gas mixture • Sodium chloride solution • Ice Example of P=3: Examples of P=2: • 퐶푎퐶푂 푠 → 퐶푎푂 푠 + 퐶푂 푔 • Ice & water mixture 3 2 • Oil & water Must distinguish between Components and Constituents Phase Transition

Phase transition: spontaneous conversion of one phase to another

Transition temperature, Tm: T at which the 2 phases are in equilibrium for a given P Chemical Potential

For a pure substance, 휇 = 퐺푚 For a multi-component system,   i i Chemical potential, 휇, is the potential for moving material.

In this chapter, from one phase to another (phase change) In Ch. 6, from one substance to another (reaction)

Criterion of phase equilibrium: At equilibrium the chemical potential is the same throughout the sample. Phase Boundaries Phase boundary: pressure and temperature line where two phases can coexist. Phase transition of a stable phase occurs when P,T cross the phase boundaries. a  b

Triple point: pressure and temperature where all three phases can coexist: vapor  liq  solid Vapor Pressure - pressure of a vapor in equilibrium with its condensed phase Vapor pressure increases with temperature. Boiling point: temperature at which the vapor pressure becomes equal to the external pressure. Gibbs Phase Rule Phase rule: F = C – P + 2 F = degrees of freedom, i,e,, the number of parameters that can be varied independently while preserving the number of phases in equilibrium. P = the number of phases C = the number of components, where components are chemically independent constituents of a system. Number of components: Ethanol water mixture: two constituents & two components + - Aqueous NaCl solution: three constituents (Na , Cl , H2O); only 2 components because Na+ and Cl- can’t vary independently For a pure substance, C=1 If P=1, then F = 1 - 1+ 2 = 2. Thus, 2 variables, typically T and P, may be changed independently without changing the number of phases. If P = 2, then F = 1, and only 1 variable can be changed independently. Example: 1-component phase diagram

F = C – P + 2 Which row has the correct number of phases present at each point?

a b c d

A 1 2 3 4

B 1 3 3 2

C 2 2 2 3

D 1 2 2 3

Phase Diagram of H2O

Phase Diagram of CO2

Phase Diagram of He Temperature dependence of chemical potential

휕휇 휕퐺 = 푚 = −푆 휕푇 휕푇 푚 푃 푃

• Temperature dependence of chemical potential is determined by the entropy.

• The phase with larger entropy shows steeper change of chemical potential vs. temperature: At the temperature where the relative value of chemical potential changes, a phase 푆푚 푔 > 푆푚 ℓ > 푆푚 푠 transition occurs. Change in Chemical Potential with Pressure dG   SdT VdP (basic equation)

Change in  with P at constant T:

휕휇 휕퐺푚 = = 푉푚 휕푃 푇 휕푃 푇

(In practice, the linear lines depicted are o curved, since Gm = Gm + ln P )

In most cases 푉푚 ℓ > 푉푚 푠 , so an increase in P will increase melting Temp. Water is different! Melting point Melting point increases at higher decreases at higher pressure. (typical pressure. (e.g. H2O) material) Vapor pressure of a liquid under pressure Effect of external pressure on vapor pressure of a liquid is given by: 푝 = 푝∗e푉m(l)∆푃Τ푅푇 Where p is the observed vapor pressure, ∆P is the applied pressure,

Vm is the molar volume, and p* is the V.P. in absence of applied press.

Density of water is 0.997 g/cm-3 at 298 K. Molar volume = 18.1 cm-3/mol

Vapor pressure is increased by 0.73% When pressure is increased by 10 bar. (Brief Illustration 4B.2) Slope of phase boundary (Clapeyron equation) On a phase boundary, two phases are in equilibrium: 푑퐺 = 푑μ = −푆푑푇 + 푉푑푃 a (P,T)  b (P,T) For infinitesimal changes in P and T: phase a

da  Sa dT V a dP  S b dT V b dP  db p (V b V a )dP  (S b  Sa )dT phase b dP  S Clapeyron equation:  trs dT trsV

T Thus, the expected slope of a phase boundary can be obtained from ΔS and ΔV of the phase transition.

Solid - Liquid phase boundary

For a solid-liquid phase transition, i.e., melting:  H S fus fus T Thus, dP SH fus fus dTfus V T fus V

Because ∆fusV is typically very small and usually positive, the slope of the P-T phase boundary is expected to be large.

When ∆fusV is negative (e.g., water), the slope of the P-T phase boundary is large, but negative. Liquid/Vapor phase boundary  S  H trsH dP vap vap trsS  and   T dT vapV TvapV For the case of a gas-liquid transition: ___ V( g ) V ( ) so  Vvap ( g )  V ( g ) Thus, dP  H  H  vap  vap (for an ideal gas) dT TV (g) T(RT / P) This rearranges to:

Clausius-Clapeyron equation

dPln  H dPln vap H  vap  dT RT 2 d(1/ T ) R Problems with Clausius-Clapyeron

• Many approaches to problems to find vapH, Tb, Pvapor dln P  H   v 1 1. Assume Constant Enthalpy, i.e., ∆H ≠ f(T) d T  R a. Given P vs. T data. Plot lnP vs. 1/T; fit points to a straight line, and get H from slope. b. Given 2 (P,T) points. Use definite integration:

pT 22vap H 1 P vap H 11   H dln P dT ln 2    vap 11 pT2     PP21exp   11RT PRTT1   2 1   RTT21 What information would you need for a more accurate calculation? C-C continued c. Given a Formula for ln p (T)

Example: The vapor pressure of benzene in the range 10 - 30 ºC is: 1780 logPtorr 7.960 TK

Calculate Hvap: p2  H T2 1 d ln P  vap dT  R  T 2 p1 T1  H  1780 ln P  A  v 1  2.3037.960  R T  TK

 H v  2.3031780  H  1780 2.303 8.31451  34.08 kJ R v mole C-C given H(T) 2. For accurate work account, must for T dependence of H, i.e., ∆H = f(T). a. With a formula for HT), can integrate the C-C Eqn:

Example: Find the boiling temperature for N2 at P = 100 Torr. (The B.P. at P=1 atm is 77.33K) given: 1  vHJ  mole  8070  32.07TK

8070 32.07TK Substituting in CC eqn: dln P  dT RT2 970.6 3.857 Simplifying: dln P  dT  dT T2 T   P2 1 1 T2 Integrating: ln  970 .6  3.857 ln P1 T1 T2  T1 100  1 1  T Putting in Nos: ln  970 .6   3.857 ln 2 760 77.33 T2  77.33

 970 .6 Simplify: 2.028  12.551   3.857 ln T2 16.771  T2 

970.6 Rearrange: 0 31.350   3.857lnT2 T2

Can use Excel solver to get: T2=63.2 K C-C Finding H(T) b. If know vapH at one T, can find a general formula for vapH(T), assuming constant heat capacity: vH  Hgas  Hliq

 H H  H  Since: v   gas   liq    C   v P T P  T P  T P then vCP  CP,gas  CP,liq

Therefore, the vapH(T), relative to vapH(Tr) is: T vH(T)  vH(Tr )   vCPdT Tr Then use this formula in the C-C equation. -1 -1 Example: Estimate the Pv of CCl4 at 400 K w/ Cp(gas) = 83.30 J-K mole , -1 -1 -1 Cp(l) = 131.74 J-K mole , vapH = 30.0 kJ-mole , and Tb = 349.9 K.

To find a formula for the T-dependence of vapH, use Cp’s, Tr, and rH: T 3 vH(T)  30.010  83.3131.74dT 349.9

3 vH  30.010  48.44T 349.9 30.0103  48.44T  349.9 dln P  dT RT2 6.65103dT 5.83 dln P   dT T2 T

P2 3 1 1  T2 ln  6.6510     5.83ln P1  T1 T2  T1

Substituting P1 = 1 atm, T1 = 349.9 K, and T2 = 400 K yields lnP2=1.600, which gives P2 = 4.95 atm, the vapor pressure of CCl4 at 400K Recap

휕휇 휕휇 = −푆 = 푉 휕푇 푚 휕푃 푚 푃 푇

Clapeyron equation: phase boundary dP  S  H     dT V TV Clausius-Clapeyron equation: vapor phase boundary

d ln P  H dln P  H vap   v  2 1 dT RT d T  R Constant enthalpy: (P, T) data; (P, T) formula

H(T): given formula; find formula from Cp Solid-Vapor Boundary

Vapor pressure of solids is called sublimation pressure

e.g., I2 gas above solid I2 ; CO2 solid sublimes directly to gas at atmospheric pressure

Transition from solid to vapor is called sublimation.

Transition is governed by Hsub

Hsub = Hvap + Hfus because enthalpy is a state function.

Only difference in calculations: replace

Hvap by Hsub. Because Hsub > Hvap The slope of the P-T boundary for solid- vapor equilibria is greater than that for liquid-vapor equilibria.

Equilibria of Condensed Phases

Solid - Liquid or Solid - Solid equilibria.

From the Clapeyron Equation: dP H V dT or dP dT T V H T

To find the dependence of M.P. (Tf) on pressure, integrate:

*  fus H T PP ln *  fusVT

 f V  Vliq  Vsolid V  M  g mole g mL First-Order Transitions

• Many different types of phase transitions, e.g. solid-liquid; conducting-superconducting; solid- solid

• Can use chemical potential to classify phase transitions into different types using the so-called Ehrenfest Classification

• Phase changes are often accompanied by changes of H and V. These changes cause abrupt changes in the slope of vs. T and μ vs. P plots.

b        a   V  V   V P  P  m,b m,a   T T

b        a   S  S   S   H T T  T  m,b m,a    P P

In a first-order transition, ∂μ/∂T is discontinuous. Because H changes by a finite amount

for an infinitesimal change in T, Cp (= dH/dT) at the transition temperature is infinite. • Because V and ΔS are non-zero for both melting and vaporization, the slopes of  vs. P or  vs. T plots must be different on either side of the transition, i.e. d/dP and d/dT are discontinuous at the transition.

• A transition for which d/dT is discontinuous is classified as a first-order transition.

• All the phase transitions considered so far have been first-order, e.g. melting/freezing, vaporization/condensation, sublimation

• Distinctive features of 1st order phase transitions: • at the transition temperature, H, S, and V change discontinuously. • G is continuous, but its slope is not.

• Cp at transition temperature is infinite. First-Order vs. Second-Order Transitions

A transition for which d2/d2T is discontinuous is classified as a second-order transition. Examples: Characteristics of Second-Order Transitions + • NH4Cl -30 ºC onset of free rotation of NH4 • Finite discontinuity in Cp • MnO: alignment of the unpaired e- spins • S , V, and H are continuous, but not their 1st derivatives • Cu3Au: an order-disorder transition (Au & Cu • 1st deriv of G is continuous; 2nd deriv. is discontinuous ordered at low T, but mixed randomly at high T) More Transitions Characteristics of Lambda Transitions • Glass Phase Transitions Not a first-order transition, but has infinite Cp at glass: an amorphous solid that does not flow the "lambda point.“ readily, but the molecules are randomly H is finite and continuous and has a vertical distributed inflection point. No true latent heat. examples: glass, polymers, rubber Shape of heat capacity curve resembles l. glass transition: When glasses are warmed, they soften and become fluid

glass transition temperature: Tg (usually a range of temperatures)

Below Tg: rigid and brittle Above Tg: soft and rubbery e.g. Rubber ball frozen in LN2 Not first-order e.g., transition b/t two liquid phases of He: Shows hysteresis - depends on thermal history He-I and He-II (superfluid). of sample Molecular Interpretation of Phase Transitions

First order transitions involve relocation of atoms, molecules and changes energies of interactions. Second order involve changes in symmetry and order. Major Concept Review Property Equation Comment Equation number

Chemical potential  = Gm For a pure substance Phase rule F = C – P + 2 4A.1

Variation of  with (∂μ/∂T)p = –Sm 4B.1 temperature

Variation of  with (∂μ/∂p)T = Vm 4B.2 pressure ∗ 푉m(l)∆푃Τ푅푇 Vapour pressure in 푝 = 푝 e P = Papplied – p* 4B.3 the presence of applied pressure

Clapeyron equation dp/dT = ∆trsS/∆trsV 4B.5a 2 Clausius–Clapeyron d ln p/dT = ∆vapH/RT Assumes Vm(g) >>Vm(l) 4B.10 equation and vapour is a perfect gas