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
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