PART 8 Ch i St tPART 8 – Changes in States, Colligative Properties and PhaseColligative Properties, and Phase

Diagram

Reference: Chapter 5, 12 in textbookp ,

1

States of MaterialsSolidSolid LiquidLiquid GasGas

2

Gas and Ideal Gas

Properties of a Gasope es o a GasLow density, No fixed shape

Can be significantly expanded and compressed

Can be mixed by any ratiosCan be mixed by any ratios

Ideal Gas – AssumptionIdeal Gas AssumptionGas molecules have no volume;

No intermolecular interaction

3

Ideal Gas: P, V, T

P s VP vs. VP * V = constant or P1 * V1 = P2 * V21 1 2 2

V vs. TV / T = constant or V1 / T1 = V2 / T2

P vs. TP / T t t P / T P / TP / T = constant or P1 / T1 = P2 / T2

Note: the units of P V TNote: the units of P, V, T4

Ideal Gas Law

Ideal Gas EquationqP * V = n * R * T

P 101 3 kPa = 1 atm = 760 mmHg = 760 torrP： 101.3 kPa = 1 atm = 760 mmHg = 760 torrV： 1 m3 = 1 × 103 dm3 ( L ) = 1 × 106 cm3 (mL)T： K = C° + 273.15n ： # of moles of all gasesR = 8.31 J • mol-1•K-1 = 0.082 atm • L • mol-1 • K-1

Avogadro’s LawUnder same P T 1 mole of gas occupies the same VUnder same P, T, 1 mole of gas occupies the same V.

At the Standard Temperature & Pressure (STP), 1

mole of any ideal gas is 22.4 liter.5

Practice

Q1 N H is a fuel prepared by the following reaction:Q1：N2H4 is a fuel, prepared by the following reaction:2NH3 (g) + NaOCl (aq) → N2H4 (aq) + NaCl (aq) + H2O (l) To synthesize 15.0 kg N2H4, how much volume of 10oC, 3.63 atm of NH3 (g) is needed?

Solution:n = m / MW = 1 50 x 104 g / 32 g/mol = 469 moln (N2H4) = m (N2O4) / MW (N2O4) = 1.50 x 104 g / 32 g/mol = 469 mol

V(NH3) = n (NH3) RT/ P

= 2n (N2H4) RT/ P

= 2 x 469 x 0.082 x (273 + 10) / 3.63

= 5.99 x 103 L6

Partial Pressure

Dalton’s LawUnder constant temperature & volume, the total

f i t l th ti lpressure of a gas mixture equals the partial pressure

of each gas component.

P = P + P + P + + PPtotal = PA + PB + PC + …… + Pi

Gas Partial Pressure (Pi)P of a gas component solely occupying whole volume

7

Molar FractionEach gas component is valid for the Ideal Gas Law:

PA = nART/ V, PB = nBRT/ V …… ， Pi= niRT/ V

P = n RT/ V + n RT/ V + + n RT/ VPtotal = nART/ V + nBRT/ V + …… + niRT/ V

= (nA + nB + … … + ni )RT/ V( A B i )

= ntotal RT/ V

Pi / Ptotal = (niRT/ V) / (ntotalRT/ V) = ni / ntotal

ni / ntotal = Xi (molar fraction)

8Pi = Ptotal × Xi

Partial Volume

Partial Vol me of a gas componentPartial Volume of a gas componentVolume of a gas component under T and Ptotaltotal

Vtotal = VA + VB + VC + …… + Vi

Molar Fraction (Xi)Molar Fraction (Xi)Xi = ni / ntotal = Pi / Ptotal = Vi / Vtotal

Pi ×Vtotal= Ptotal × Vi

9

Practice

Q2: At 25°C, 1 atm, we collected a mixture of H2 and , , 2saturated water vapor 152 mL. If the saturated water vapor is 23.76 mmHg, calculate: (a) Partial pressure of H2 ( ) (b) A t f b t f H ( ) ( ) V l fH2 (g); (b) Amount of substance of H2 (g); (c) Volume of this H2 (g) if it is dry.

Solution:(1) PH2 = Ptotal- PH2O = 760 - 23.76 = 736.2 mmHg ( ) H2 total H2O g(2) PH2Vtotal = nH2RT

nH2 = PH2Vtotal / RT= 736.2 x 0.152 / 62.4 x 298.15 = 6.2 x 10-3 mol

(3) PH2Vtotal= VH2 Ptotal

VH2 = PH2Vtotal / Ptotal = 736.2 x 152 / 760 = 147 ml10

PracticeQ3: At 250°C, PCl5 completaly evaporates and then partially dissociates into PCl and Cl If 2 98 g PCl ispartially dissociates into PCl3 and Cl2。If 2.98 g PCl5 is placed in a 1.00 dm3 container and completely evaporates into gas, the total pressure is 113 kPa. Calculate the g , ppartial pressure of each gas component.

S l tiSolution:MWPCl5 = 208.5 nPCl5 = 2.98 / 208.5 = 1.43 x 10-2 mol

When PCl5 completely evaporates but before dissociation:

P = nRT/V = 1.43 x 10-2 x 8.314 x (250 + 273.15) / 1.00 x 10-3 = 62.14 kPa

C ( ) C ( ) C ( )When partially dissociate: PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

PPCl5 = 62.14 – x, PPCl3 = x, PCl2 = x

P P + P + P 62 14 x + x + x 113 kPa x 50 86 kPaPtotal = PPCl5 + PPCl3 + PCl2 = 62.14 – x + x + x = 113 kPa x = 50.86 kPa

So: PPCl5 = 62.14 – x = 11.28 kPa, PPCl3 = PCl2 = 50.86 kPa 11

PhasePhase

A material or material fraction with (macroscopically)

homogeneous physical & chemical propertiesg p y p p

e.g. Gas, Liquid, (water vs. oil), Solid, (different solids)

Phase ChangeA same material changes from one phase to another.

Ph E ilib iPhase EquilibriumFor a material, when its all fractions of different ,

phases remain unchanged under a certain condition.12

LiquidLiquid

Li id i ki d f "C d d Ph "Liquid is a kind of "Condensed Phase"

Short-range ordered; Long-range disorderedg ; g g

Liquid Evaporation

VaporVapor：：gas molecules above liquid

Open Open

LiquidSystemSystem

13Molecules with higher energy will escape from liquid phase to vapor phase first

Liquid Evaporation

Closed Closed SystemSystem

v Evaporation

Condensation

14t

(Saturated) Vapor Pressure( )Vapor Pressure / Saturated Vapor Pressure

P of a vapor phase at equilibrium with its liquid phase

Vapor pressure increases with temperatureVapor pressure increases with temperature

At a liquid’s boiling point (Tb), Vapor pressure = 1 atm

Artificial Piston

15

How Vapor Pressure Changes with Tg

Vapor Pressure ~ Temperaturep p

constH

P vap +Δ−

=lg

ΔHvap: Heat of vaporization (Enthalpy of vaporization)

constRT

P +303.2

lg

ΔHvap: Heat of vaporization (Enthalpy of vaporization)

O)

or o

f H2

⎞⎛Δ TTH

P (v

ap

⎟⎟⎠

⎞⎜⎜⎝

⎛×−

×Δ

=12

12

1

2

303.2lg

TTTT

RH

pp vap

161/Tlg

⎠⎝

PracticeQ: For phenol (C6H5OH), ΔHvap = 48.139 kJ/mol, and its vapor pressure is 100 mmHg at 392.5 K. What is the vapor pressure of phenol at 350 K?What is the vapor pressure of phenol at 350 K?

Solution:For phenol: ΔHvap = 48.139 kJ/mol T1 = 392.5 K P1 = 100 mmHg

T2 = 350 K P2 = ？mmHg

KK

392 53505.392350

K)l8 314J/(2 303J/mol1000139.48

100lg 2

2 ⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ −×

=p

1668.0100

,778.0100

lg

K392.5350K)l8.314J/(mo2.303100

22

2

=−=

⎠⎝⎠⎝ ×⋅×pp

17mmHg68.16100100

2 =p

PracticeQ: At 40oC, compress 1.00 L air containing saturated bezene (C H ) vapor from 1 atm to 5 atm If the vaporbezene (C6H6) vapor from 1 atm to 5 atm. If the vapor pressure of bezene is 181.7 mmHg at 40oC, how many

f b i d d t li id i thi ?grams of benzene is condensed to liquid in this process?

Solution:During the whole process, the amount of air remains same, which is: 1.00 × (750-181.7) = Vtotal (5 × 760-181.7) So: Vtotal = Vfinal= 0.157 LAt this time, moles of bezene is:

( ) 4.62273407.181

×+

×finalV

Initially, moles of bezene is:

So condensed bezene is:

( ) 4.62273407.18100.1

×+×

( )V 7181001 ×−So, condensed bezene is:

18

( )( ) g

VW final 612.078

4.62273407.18100.1

=××+

×=

Melting & Freezing Curvesg g

Melting Curve and Melting Point (Tm)g g ( m)

Freezing Curve & Freezing Pt Aee g Cu e & ee g te.g. Cooling curve of water at 1 atm

BC DSupercooling phenomenon

19

BC D

Sublimation & Deposition

Sublimation: direct change from solid phase to g pgas phase (e.g. I2 (s) I2 (g), CO2 (s) dry ice )

lid)

or o

f so

BHp sub +Δlg

P (v

ap BRT

p sub +−=303.2

lg

lg

1/TΔHsub: Heat of sublimation (Enthalpy of sublimation)

20Vapor pressure vs. Temperature

Phase Diagrams gPhase Diagram for water – Understand the

i f diff t Z Li d i tmeaning of different Zones, Lines, and points.

Melting point(s), Boiling point(s), at different T & P.

Triple point (for H2O: 0.006 atm, 0.01 oC)Triple point (for H2O: 0.006 atm, 0.01 C)

Critical point (for H2O: 218 atm, 374 oC) 21

New Slide of Practice

Te tbook page 472 12 54Textbook, page 472, 12.54Use the water diagram: (a) What is the g ( )

phase of water at 2 mmHg and -3C? (b) What term describe water at 400C andWhat term describe water at 400C and 225 atm? (c) What will happen if water at 20C and 3 mmHg is heated to 20C at-20C and 3 mmHg is heated to 20C at

same pressure? (d) Sketch part c in figure. ( ) D ib h ill h if(e) Describe what will happen if water at 200C and 1 atm is increased to 50 atm at constant temperature?

22

Solution Revisited (Chap. 13.5—13.7)( )

Different ways to represent concentration:y pSolubility, S (Mass percentage)

Molarity, C (# of solute moles per solution volume in liter)

Molality, m (# of solute moles per solvent mass in kg)Molality, m (# of solute moles per solvent mass in kg)

Mole fraction, X (Percentage of moles):

,+

=l tl t

solutesolute nn

nX

,+

=

+

solventsolvent

solventsolute

nnn

X

nn

231=+

+

solventsolute

solventsolute

XXnn

Colligative Propertiesg

Colligative Properties of Nonvolatile Solutesg pDo not depend on the intrinsic properties of solute itself

Depend ONLY on the number of solute particles

4 common Colligative PropertiesVapor pressure lowering

Boiling point elevationBoiling point elevation

Freezing point depression

Osmotic pressure24

1. Vapor Pressure Loweringg

Raoult’s Law:The solvent (volatile) vapor pressure is lowered when

it di l l til l tit dissolves a nonvolatile solute.

XPP *=

For a two component (i e solute + solvent) system:

solventXPP 0=

For a two-component (i.e. solute + solvent) system:

l tl t XPXPP )1(** 00 −==

solute

solutesolvent

XPPPXPXPP

*)1(

00

00

=−

25soluteXPPso *: 0=Δ

A Molecular View of Raoult’s LawSolvent (volatile) + Solid (nonvolatile)

XPP * solventXPP *0=

A mixture of two volatile materials

26

Vapor Pressure of 2 Volatile Molecules

Ptotal = P1,0 * X1 + P2,0 * X2

P1,0 and P2,0 : vapor pressure of pure substance

X X : mole fraction of each substance in the mixtureX1, X2 : mole fraction of each substance in the mixture,

X1 + X2 = 1 27

PracticeQ: A solution is composed by hexane (C6H14) and heptane (C7H16) The saturated vapor pressures forheptane (C7H16). The saturated vapor pressures for hexane and heptane are 0.198 atm and 0.06 atm, respectively. If the mole fraction of hexane in the solution is 0.4.(a) What is the vapor pressure of the solution?

(b) What are the mole fractions of hexane and heptane in the vapor

phase above the solution?phase above the solution?

(c) If we collect all vapor phase and condense into another empty

t i d l t th t t h ilib i Wh tcontainer, and let them re-evaporate to reach equilibrium. What

are the mole fractions in the new vapor?

(d) What can you conclude from the above results?

28

PracticeAnswer: (a) VP of hexane: P0, hexane* Xhexane = 0.198 * 0.4 = 0.0792 atm,

VP of heptane: P0, heptane* Xheptane = 0.06 * 0.6 = 0.036 atmPtotal = P0, hexane* Xhexane + P0, heptane* Xheptane = 0.1152 atm

(b) Mole fraction in the vapor phase – using Y to represent:Y = 0 0792 / 0 1152 = 0 688 Y = 0 036 / 0 1152 = 0 312Yhexane = 0.0792 / 0.1152 = 0.688, Yheptane = 0.036 / 0.1152 = 0.312

(c) Liquid phase in the 2nd container = Vapor phase in the 1st containerVP of hexane: P0, hexane* Xhexane = 0.198 * 0.688 = 0.136 atm VP of heptane: P0, heptane* Xheptane = 0.06 * 0.312 = 0.0187 atm P P * X + P * X 0 155 atmPtotal = P0, hexane* Xhexane + P0, heptane* Xheptane = 0.155 atm

Yhexane = 0.136 / 0.155 = 0.879, Yheptane = 0.0187 / 0.155 = 0.121

(d) The one with higher vapor pressure (hexane) is enriched in the vapor phase – a phenomenon known as Distillation. 29

Fractional Distillation

Mixture of A & B(if: VP: P0,A > P0,B; or More A

than B

30Boiling pt: Tb, A < Tb,B)

than B

2. Boiling Point ElevationgSolventSolvent

mP/

atm

mKTTT bbbb ×=−′=Δ

SolutionSolution

T/K

SolutionSolution

T/KRecall: @ Boiling point (Tb), the vapor pressure = External pressure (1 atm);

So when the solvent dissolves some solute (with molality m) the vapor

31

So when the solvent dissolves some solute (with molality m), the vapor pressure drops, thus leading to an elevation of its boiling point.

PracticeQ: Dissolve 17.6 g of an unknown solute inside 250 g benzene this benzene solution’s boiling point increasesbenzene, this benzene solution s boiling point increases 1.00 oC. If Kb of benzene is 2.53 oC · kg / mol, what is the molecular weight of this unknown solute?g

Answer:w

Mw

KMw

KmKT soluteb

solute

solute

bbb1000

1000×

×=×

×=×=ΔwMw solventsolutesolvent ×

TwwK

Msobsolvent

solutebsolute

1000:

Δ×××

=

32molg /178

00.1250100060.1753.2

=×

××=

3. Freezing Point Depressiong

mKTTT ffff ×=−′=Δ

33

4. Osmotic Pressure

34

Osmotic Pressure

Osmotic Pressure: the pressure needed to stop p pthe net movement of solvent molecules through a semi-permeable membranep

n l t ****35

TRVnTRCsolution

solutesolute **** ==Π

Practice

Q: Dissolve 2 00 g of a protein in 0 100 L water andQ: Dissolve 2.00 g of a protein in 0.100 L water, and measure the osmotic pressure of this solution at 25 oC is 0.021 atm. What is the molar mass of this protein?p

Answer:Answer:

C = Π / RT

= 0.021 / 0.082 x 298 = 8.6 x 10-4 mol/L

That is: C = 8 6 x 10-4 mol/L= (2 00 g / M ) / 0 100 LThat is: C = 8.6 x 10 4 mol/L= (2.00 g / Mprotein ) / 0.100 L

So: Mprotein = 2.00 / (8.6 x 10-4 x 0.1) = 2.3 x 104 g/molp

36

Practice Electron microscopy image of

Fact: A method to estimate the

py gcells in high osmotic pressure solution

osmotic pressure of a cell, is to place the cells into a series of NaCl solutions with differentsolutions with different concentrations, and to find out the concentration in which the cells doconcentration in which the cells do not expand or shrink.

Q: If at 37oC cardiac muscle cells in a unknownQ: If at 37 C, cardiac muscle cells in a unknown concentration of NaCl solution do not expand or shrink. The equilibrium vapor pressure of water over this NaClq p psolution is 6.214 kPa. The pure water’s vapor pressure is 6.276 kPa. What is the osmotic pressure of this kind of

di l ll ?37

cardiac muscle cells?

Practice

Ans erAnswer:ΔP = P0,H2O * XNaCl = P0,H2O * nNaCl / (nNaCl + nH2O), ,

ΔP = P0,H2O – Psolution = 6.276 – 6.214 = 0.062 kPa

(nNaCl + nH2O) / nNaCl = P0,H2O / ΔP = 6.276 / 0.062 = 101

nH2O / nNaCl = 100H2O NaCl

Assume 1000 g H2O: nH2O = 1000 g / 18 g/mol = 55.5 mol,

nNaCl = 0.555 mol CNaCl = 0.555 mol/L

Osmotic pressure: Π = C R T = 0.555 x 0.082 x (273.15+37) = 14.11 atmp ( )

38