Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

This course is approximately at this level

CHEMISTRY E182019 CH7

Phase equilibriaG=0Clausius Clapeyron

T-S diagrams CH7

T-s diagrams

Gas steam

L+GLiquid

Saturated vapour curve

p=1000 bar technically

realizable maximum

Saturated liquid curve

T-S diagrams evaporation CH7

T-s diagrams

Saturated steam

s’’=6 kJ/kgK

Saturated liquids’=2 kJ/kgK

Enthalpy of evaporation hLG=T(s’’-s’)=5004=2MJ/kg

Solid-Liquid-Gas CH7

Phase diagrams

S-solid

L-liquid

G-gas

T

p G-gas

L-liquid

S-solid

S+G

L+G

v

p

The reason why the regions L+G, S+G appear in the p-v diagram is that the specific volume v (unlike T,p) varies during phase transformations.

Solid-Liquid-Gas CH7

Phase diagram ice-water-steam

TRIPLE POINT

CRITICAL POINT

Hexagonal ice

Cubic iceliquid-like

hydrogen-bonded clusters dispersed within a gas-like

phase

Solid-Liquid-Gas CH7

S-solid

L-liquid

G-gas

T

p

Evaporation hLG>0, sLG>0, GLG=0, dp=dT=0

Sublimation hSG>0, sSG>0, GSG=0,

dp=dT=0

Melting hSL>0, sSL>0, GSL=0, dp=dT=0

During phase transitions the pressure and temperature are constant. Also Gibbs energy remains constant as follows from its definition g=h-Ts=0. Only specific volume increases or decreases.

Clausius Clapeyron Solid-Liquid-GasCH7

S-solid

L-liquid

G-gas

T

p

Slopes dp/dT are given by Clausius Clapeyron

equation

Phase transition lines in the p-T diagram are described by the Clausius Clapeyron equation

dp h

dT T v

Enthalpy of phase changes, e.g. hLG

Specific volume changes, e.g. vG-vL

Clausius Clapeyron Solid-Liquid-Gas CH7

S-solid

L-liquid

G-gas

T

p

dp/dt>0 because hLG>0 vLG>0 (volume of steam is

greater than volume of liquid)

The slope dp/dT is negative because specific volume of

ice is greater than volume of liquid

Melting point temperature of ice decreases with pressure – therefore ice under skates melts and forms a liquid film

T+dT

T

s=s’’-s’

Clausius Clapeyron derivationCH7

Closed loop (evaporation, expansion, condensation,

compression)

dT s dp v

dp s h

dT v T v

Heat added-difference between evaporation enthalpy at T+dT and condensation enthalpy at temperature T

Mechanical work-received in one cycle

Clausius Clapeyron equation can be derived from energy balance of a closed cycle in Ts diagram:

0

500

1000

1500

2000

2500

3000

0 100 200 300 400

T [C]

h

[kJ

/kg

]

Giving expression for ΔhLG

State equationAntoine’s equation

Clausius Clapeyron and hLG CH7

TC

BAp

ln

( '' ')LGhdp

dT T v v

''RT

pvM

2 2

2

ln

( )LG

RT d p RT Bh

M dT M C T

Clausius Clapeyron equation is exact, because follows from thermodynamic principles. Individual terms (dp/dT,v’’) can be approximated by semiempirical equations (different state equations, Antoine’s equation)

Result can be improved when using Van der

Waals

or Redlich Kwong state equation

Multicomponent equilibrium CH7

pA=pA"xAAnswer: Yes, Raoult’s law applicable to ideal liquids

HEAVY

LIGHTA

B

xA=

yA=

Liquid phase

Gaseous phase

Question: Is there a relationship between composition of binary mixture in the liquid phase xA and gaseous phase yA?

Binary mixture

Raoult’s law CH7

HEAVYB

p AB

C TB BB

B

,, exp( )

Fact: It does not matter, how much liquid is in the vessel, pressure of vapours is the same, and given by Antoine’s equation

Therefore also the molar volume nB/V is independent of amount of liquid.

Raoult’s law CH7

HEAVY

LIGHTA

B

xA

Volume V

Volume xAV

"

" (1 ) "A A A

AA B A A A B

n x py

n n x p x p

after expansion of nA molecules to volume V

A Ap V n RT

"A A Ap p xgiving Raoult’s law

…and also answer to the previous question

''A A Ap x V n RT

Raoult’s law CH7 HEAVY

LIGHTA

B

p AB

C TA AA

A

,, exp( )

p AB

C TB BB

B

,, exp( )

1-xA=xBxA

Volume V

Volume xAV

''( ) ''( )(1 )A B A A B Ap p p p T x p T x

Distillation CH7

Initial composition of Liquid Mixture

Coolingcondensing

Liquid mixture enriched by heavy component

Liquid mixture enriched by light component

Liquid – noncondensable gas CH7

H2O+CO2

xCO2

H2O+CO2

yCO2

2 2 2

'' ( )CO CO COp p T x

What to do if T > Tcrit = 31 oC ?

Given temperature T and molar fraction of dissolved CO2

Henry’s law

2 2( )CO COp H T x

Henry’s constant can be found in tables

Tutorial CH7 HEAVY

LIGHTA

B

p AB

C TA AA

A

,, exp( )

p AB

C TB BB

B

,, exp( )

''( ) ''( )(1 )A B A A B Ap p p p T x p T x

Given total pressure p and molar fraction of liquid phase xA calculate

a) Equilibrium temperature T

b) Molar composition of vapoursNonlinear equation for

T (Excel solution)

"

" (1 ) "A A

AA A A B

x py

x p x p

0 xA 1

yA

p=const

Repeated distillation

Tutorial SYRINGE alcohol CH7

V0 V

1

VL

VG

Initial state: Syringe filled by liquid mixture H2O (B) + CH3OH (A) (methylalcohol). Initial volume V0, molar fraction of methylalcohol xA, number of moles nA, nB are given (approximated from density).

Final state: Volume is increased to V1. Temperature is constant (room temperature).

Calculate pressure p, molar fraction of methylalcohol in liquid and vapour phase.

Tutorial Syringe alcoholCH7

Unknown 9 variables

xA=? yA=? p=? VL=? VG=? nAL=? nAG=? nBL=? nBG=?

Equations

1G LV VV '' '' ''( ) AA B Bp pp x p

ALA

AL BL

nx

n n

AGA

AG BG

ny

n n

AL AGA nn n

BL BGB nn n

''A G GA Ax V np RT

'' (1 )A GB BGx V Tnp R

0AL B

B

LL

A

n nV V

n n

LINEAR NONLINEAR