Chemical Engineering Thermodynamics-I

Lecture 4Prof.Dr.Mahmood Saleem

Institute of Chemical Engineering & Technology

University of the Punjab, Lahore

Contents

• Introduction• Property relations for Homogeneous Phase• Residual Properties• Residual Properties by Equation of State• Property Relations for Two Phase Systems• Thermodynamic Property Diagrams and Tables• Generalized property Correlations for Gases• Extension of Generalized property Correlations

for Gases to Mixtures• Exercises

Introduction

• Processing of materials (pure and mixtures) is key affair in chemical processes

• Estimation of many properties (and change) other than volumetric properties (H,U,s,G etc.) are required.

• How to know complete property set when volumetric properties are not available

• What tools can one use to handle such issues?

Property Relations for Homogeneous Phases

Fundamental Properties

Although equation (6.1) is derived from the special case of a reversible process, it not restricted in application to reversible process.

It applies to any process in a system of constant mass that results in a differential change from one equilibrium state to another.

The system may consist of a single phase or several phases; may be chemically inert or may undergo chemical reaction.

nVdPnSdTnUd

dWdQnUd revrev

..

)(

…(6.1)

Definitions

H = Enthalpy

A = Helmholtz energy

G = Gibbs energy

PVUH

TSUA

TSHG

…(2.11)

…(6.2)

…(6.3)

Based on one mole (or to a unit mass) of a homogeneous fluid of constant composition, they simplified to

Function Change

SdTVdPdGTPG

SdTPdVdATVA

VdPTdSdHPsH

PdVTdSdUVsU

....).........,(

....).........,(

....).........,(

.....).........,(

Maxwell’s equations

TP

TV

PS

VS

P

S

T

V

V

S

T

P

S

V

P

T

S

P

V

T

TP

PT

P

S

T

V

dTT

GdP

P

GdG

dTSdPVdG

TPG

.

..................

),(

Enthalpy and Entropy as Functions of T and P

Temperature derivatives:

T

C

T

S

CT

H

P

P

PP

Pressure derivatives:

PT

PT

T

VTV

P

H

T

V

P

S

The most useful property relations for the enthalpy and entropy of a homogeneous phase result when these properties are expressed as functions of T and P (how H and S vary with T and P).

dPT

V

T

dTCdS

dPT

VTVdTCdH

PP

Pp

…(6.21)

…(6.20)

Property Relations for Homogeneous Phases

Internal Energy as Function of P (U(P))

• The pressure dependence of the internal energy is shown as

PdVTdsdU

TPT P

VP

T

VT

P

U

TP

TV

PS

VS

P

S

T

V

V

S

T

P

S

V

P

T

S

P

V

T

Property Relations for Homogeneous Phases

The Ideal Gas State

• For ideal gas, expressions of dH and dS (eq.6.20-6.21) as functions of T and P can be simplified as follows using ideal gas law:

(6.24)

(6.23)

P

dPR

T

dTCdS

dTCdH

P

R

T

VRTPV

igP

ig

igP

ig

P

igig

dPT

V

T

dTCdS

dPT

VTVdTCdH

PP

Pp

Property Relations for Homogeneous PhasesAlternative Forms for Liquids

• Relations of liquids can be expressed in terms of and as follows:

VTPP

U

VTP

H

VP

S

T

T

T

1

TP

TV

PS

VS

P

S

T

V

V

S

T

P

S

V

P

T

S

P

V

T

dPT

V

T

dTCdS

dPT

VTVdTCdH

PP

Pp

PdVTdsdU

TPT P

VP

T

VT

P

U

dPdTV

dV

P

V

V

T

V

V

dPP

VdT

T

VdV

T

P

TP

1

1

Property Relations for Homogeneous PhasesAlternative Forms for Liquids

• Enthalpy and entropy as functions of T and P as follows:

and are weak functions of pressure for liquids, they are usually assumed constant at appropriate average values for integration.

)....(VdPT

dTCdS

)....(VdPTdTCdH

P

P

296

2861

Practice 1

Determine the enthalpy and entropy changes of liquid water for a change of stage from 1 bar 25C to 1,000 bar 50C.

T

( C)

P

(bar)

Cp

(Jmol-1K-1

V

(cm3mol-1)

(K-1)

25 1 75.305 18.071 256x10-6

25 1000 … 18.012 366x10-6

50 1 75.314 18.234 458x10-6

50 1,000 … 18.174 568x10-6

Constant P

Constant T

166

13

11

10513102

568458

204.182

174.18234.18

,50

310.752

314.75305.75

,1

K

molcmV

CTforand

KJmolC

barPFor

p

121

2

12212

ln

1

PPVT

TCS

PPVTTTCH

p

p

Integrated forms of equations for change in Enthalpy and Entropy

Average value at T1 and T2 at 1 bar

Average value at P1 and P2 at 50 oC

11

6

1

6

13.593.006.6

10

1000,1204.1810513

15.298

15.323ln310.75

400,3517,1883,1

10

1000,1204.1815.32310513115.29815.323310.75

KJmol

S

Jmol

H

121

2

12212

ln

1

PPVT

TCS

PPVTTTCH

p

p

Note that the effect of P of almost 1,000 bar on H and S of liquid water is less than that of T of only 25C.

On substitution of values

Property Relations for Homogeneous Phases

Internal Energy and Entropy as Function of T and V

• Useful property relations for T and V as independent variables are

V

VT

V

V

T

P

PT

PT

V

U

T

C

T

S

• The Partial derivatives dU and ds of homogeneous fluids of constant composition to temperature and volume are

• Alternative forms of the above equations are

dVT

P

T

dTCdS

dVPT

PTdTCdU

VV

VV

dVT

dTCdS

dVPTdTCdU

V

V

dPdTV

dV

P

V

V

T

V

V

dPP

VdT

T

VdV

T

P

TP

1

1

Property Relations for Homogeneous PhasesThe Gibbs Energy

• An alternative form of a fundamental property relation as defined in dimensionless terms:

P

T

T

RTGT

RT

H

P

RTG

RT

V

dTRT

HdP

RT

V

RT

Gd

2

•The Gibbs energy when given as a function of T and P therefore serves as a generating function for the

other thermodynamic properties, and implicitly represents complete information.

Residual Properties

• The definition for the generic residual property is:

igR MMM M is the actual molar value of any extensive thermodynamics property: V,

U, H, S, G.

Mig = the ideal gas properties

which are at the same

temperature and pressure.

M = the Residual Molar gas

properties which are at the same temperature and

pressure.

• Residual Gibbs energy:

• Residual volume:

igR GGG

1

ZP

RTV

P

RTVVVV

R

igR PV=ZRT

Fundamental property relation for residual properties

• The fundamental property relation for residual properties applies to fluids of constant composition.

).(

T

RT/GT

RT

H

).(P

RT/G

RT

V

).(dTRT

HdP

RT

V

RT

Gd

P

RR

T

RR

RRR

446

436

4262

)48.6()()1(,

)46.6();44.645.6.(

0)45.6()1(

)(),43.6.(

00

0

000

0

TconstP

dPZ

P

dP

T

ZT

R

SSo

RT

G

RT

H

RT

SFrom

P

dP

T

ZT

RT

HEq

RT

Ggasidealfor

P

dPZdP

RT

V

RT

G

RT

G

TconstdPRT

V

RT

GdEqFrom

P

P

PR

RRR

P

PR

P

RP

RP

P

RR

RR

Enthalpy and Entropy from Residual Properties

).(SP

PlnRdTCSS

).(HdTCHH;onSubstituti

P

PlnRdTCSSdTCHH

);.(and)..(EqofnIntegratio

SSSHHH;SandHtoApplied

RT

T

ig

P

ig

RT

T

ig

P

ig

T

T

ig

P

igigT

T

ig

P

igig

RigRig

516

506

246236

0

0

0

0

00

0

0

00

2

2112

21

21

2

12

00

0

00

2

1

2

1 43

536

526

TT

DCTTBTA

R

C

)T/Tln(TdT

CC

TT

D)TTT(

CBTA

R

C

TT

dTCC

).(SP

PlnR

T

TlnCSS

).(H)TT(CHH

lmamlm

SP

T

T

ig

P

SP

amam

HP

T

T

ig

P

HP

R

SP

igig

R

HP

igig

The true worth of the Eq. for ideal gases is now evident. The true worth of the Eq. for ideal gases is now evident. They are important because they provide a convenient base They are important because they provide a convenient base for the calculation of real-gas properties. for the calculation of real-gas properties.

Practice 2

Calculate H and S of saturated isobutane vapor at 630 K from the following information:

1. Table 6.1 gives compressibility-factor data2. The vapor pressure of isobutane at 630 K

15.46 bar3. Set H0

ig = 18,115 Jmol-1 and S0ig = 295.976

Jmol-1K-1 for the ideal-gas reference state at 300 K 1 bar

4. Cpig/R = 1.7765+33.037x10-3T (T/K)

Solution 6.3Solution 6.3

Eqs. (6.46) and (6.48) are used to calculate HEqs. (6.46) and (6.48) are used to calculate HRR and S and SRR..

Plot (Plot (Z/Z/T)T)PP/P and (Z-1)/P vs. P/P and (Z-1)/P vs. P From the compressibility-factor data at 360 K From the compressibility-factor data at 360 K (Z-1)/P(Z-1)/PThe slope of a plot of Z vs. T The slope of a plot of Z vs. T ((Z/Z/T)T)PP/P/P Data for the required plots are shown in Table 6.2.Data for the required plots are shown in Table 6.2.

P

dPZ

P

dP

T

Z P

P

P)1(

00

11

1

11

4

0

14

0

7345314868970

38412360314894930

3148

689702596094930486

94930103726360466

259601103726

KJmol...S

Jmol.,..H

KJmol.RFor

...R

S),.(.EqBy

.).)((RT

H),.(.EqBy

.P

dP)Z(K.

P

dP

T

Z

R

R

R

R

P

P

P

11

1

11

11

01

01

10

3

3

676286734541153148300

36016105576295

559821284123003604110511518

516506

16105314864912

41105314867912

09329300360

300360

3302

360300

2

100373377651

100373377651

KJmol...ln.)ln(..S

Jmol.,.,)(.,H

).(and)..(EqointSubstitute

KJmol.).(.C

KJmol.).(.C

K.)/ln()T/Tln(

TTT

KTT

T

T..BTAR

C

T..BTAR

C

S

ig

P

H

ig

P

lm

am

lmlm

S

ig

P

amam

H

ig

P

Residual Properties by Equations of State

Residual Properties from the Virial Equation of State

• The two-term virial eq. gives Z-1 = BP/RT.

)56.6(),47.6.(int

)55.6(/

),44.6.(

)54.6(,

dT

dB

R

P

R

SEqoonSubstituti

dT

dB

T

B

R

P

T

RTGT

RT

HEqBy

RT

BP

RT

GSo

R

P

RR

R

P

dPZ

RT

G PR

)1(0

In application is a more convenient variable than V,

PV = ZRT is written in the alternative form.

).(d)Z(

d

T

ZTZln

R

S),..(EqFrom

).(Zd

T

ZT

RT

H);.(and)..(EqoftionDifferenti

T

)RT/G(

T

P

P

Z

RT

H),.(and)..(EqFrom

).(ZlnZd)Z(

RT

G);..(EqointSubstitue

Z

dZd

P

dP)dZZd(RTdP),.(RTZP

R

R

RR

R

6061476

5961586576

1426406

58611496

576

00

0

2

0

• The three-term virial equation.

)63.6(2

1ln

)62.6(2

1

)61.6(ln2

32

).60.6()58.6.(int1

2

2

2

2

dT

dC

T

C

dT

dB

T

BTZ

R

S

dT

dC

T

C

dT

dB

T

BT

RT

H

ZCBRT

G

throughEqosubstituedisCBZ

R

R

R

Application of these equations, useful for gases up to Application of these equations, useful for gases up to moderate pressure, requires data for both the second and moderate pressure, requires data for both the second and third virial coefficients.third virial coefficients.

Residual Properties by Cubic Equations of State

IdT

dq

T

ZqIbsimplifyTo

bb

bd

dT

dq

T

Z

bb

bdq

b

bd

b

b

bb

b

dT

dq

T

Z

bb

bq

b

bZ

bb

bq

bZ

EqVeRTyEq

d,)1ln(

d1)-(Z;

)1)(1(

)(d

)1)(1(

)()(

1

d1)-(Z

);60.6(),58.6(Eqs.ofintegralsThe

)1)(1(

)64.6()1)(1(1

1

)1)(1(1

1

).51.3.(bygivenqas,/1substitutandbdevides)42.3.(

00

00

0 00

The generic equation of state presents two cases.The generic equation of state presents two cases.

Zb

bI

bZ

ZI

bZ

whenceRT

PZ

ρ

ab

bI

1:IICase

)65.6(ln1

RT

bP

Z.offavorineliminatedisWhen

)65.6(1

1ln

1:ICase

)68.6(ln

)(ln)ln(

)67.6(1ln

)(ln1

)66.6()ln(1

)66.6()1ln(1

qITd

TdZ

R

S

qITd

TdZ

RT

H

bqIZZZRT

G

aqIZbZRT

G

r

rR

r

rR

R

R

Find values for the HR and SR for n-butane gas at 500 K

50 bar as given by the Redlich/Kwong Eequation.

Solution

Tr = 500/425.1 = 1.176, Pr = 50/37.96 = 1.317

From Table 3.1:

8689.3176.108664.0

42748.0);54.3.(

09703.0176.1

317.108664.0);53.3.(

2/3

r

r

r

r

T

TqEq

T

PEq

Ex. 6.4Ex. 6.4

11

1

546.6)78735.0(314.8

505,40838.1500314.8,

78735.013247.0)8689.3(5.0)09703.06850.0ln(:)68.6.(

0838.1)13247.0)(8689.3)(15.0(16850.0:)67.6.(

:.2

1ln/)(ln,ln

2

1)(ln

13247.0ln

68500

)09703.0(

09703.009703.08689.309703.01

1:)52.3.(

KJmolS

JmolHThus

R

SEq

RT

HEq

ThenTdTdTTWith

Z

ZI

Then:..ZyieldsEq.thisofSolution

ZZ

Z

ZZ

ZqZEq

R

R

R

R

rrrr

These results are compared with those of other calculation in Table 6.3.

TWO-PHASE SYSTEMS

).(ZR

H

)T/(d

Plndor

).(ZRT

H

dT

PlndZ

P

RTVBut

).(VT

H

dT

dP;vaportoliquidfromtransitionPhase

equationClapeyronThe:).(VT

H

dT

dPT/HS,Thus

)transitionphaseofheatlatentThe(STH);..(EqofnIntegratioV

S

VV

SS

dT

dP,tarrangemenRe

dTSdPVdTSdPVdGdG,GG

l

lsat

l

lsat

l

sat

l

l

lsat

sat

sat

satsat

7461

736

726

716

86

2

The Clapeyron eq. for pure-species vaporization

Temperature Dependence of the Vapor Pressure of Liquids

r

.

r

sat

rr

sat

sat

Twhere

).(DCBA

)T(Pln;ToffunctionA

B.App,.BTableingivenaretstanconsAntoine

).(CT

BAPln:.eqAntoineThe

T

BAPln

1

7761

2

766

6351

Corresponding-States Correlations for Vapor Pressure

:

:

)81.6()(ln

)(lnln

)80.6(43577.0ln4721.136875.15

2518.15)(ln

)79.6(169347.0ln28862.109648.6

92714.5)(ln

)78.6()(ln)(ln)(ln

:/

1

0

61

60

10

satr

r

rr

rrsatr

rrr

rr

rrr

rr

rrrrrsatr

n

n

n

nn

P

T

where

TP

TPP

TTT

TP

TTT

TPwhere

TPTPTP

ncorrelatioKeslerLee

The reduced normal boiling point

The reduced vapor pressure corresponding to 1 atm

Ex. 6.6Determine the vapor pressure for liquid n-hexane at 0, 30,60 and 90C: (a) With constants from App. B.2.

(b) From the Lee/Kesler correlation for Prsat

Solution(a)

(b) Eq.(6.78);From Table B.1, From Eq.(6.81) =0.298

The average difference from the Antoine values is about 1.5%.

317.224

04.26968193.13ln

t

P sat

03350.025.30

01325.1,6736.0

6.507

9.341 sat

rr nnPT

t/C Psat/kPa(Antoine)

Psat/kPa(Lee/

Kesler)

t/C Psat/kPa(Antoine)

Psat/kPa(Lee/

Kesler)

060

6.05276.46

5.83576.12

3090

24.98189.0

24.49190.0

Two-Phase Liquid/Vapor System

)82.6(

:.,,,,

)82.6()1(

:

1)1(

:

)(

bMxMM

formealternativAnetcSHUVMwhere

aMxMxM

equationgenericThe

xxVxVxV

fractionmassxVxVxV

molesnnnVnVnnV

ll

lv

vllv

ll

vlll

THERMODYNAMIC DIAGRAMS

GENERALIZED PROPERTY CORRELATION FOR GASES

)84.6()1(

)83.6(

:)48.6()46.6.(int

,

00

0

2

r

r

P

r

r

P

P

rr

R

r

r

P

P

rr

c

R

rcrcrcrc

P

dPZ

P

dP

T

ZT

R

S

P

dP

T

ZT

RT

H

andEqsosubstitue

dTTdTTTTdPPdPPPP

r

r

r

r

r

)86.6(

1:)84.6.(

)85.6(

:)83.6.(

10

11

0

01

0

10

0

12

0

02

1010

R

S

R

S

R

S

P

dPZ

T

ZT

P

dPZ

T

ZT

R

SEq

RT

H

RT

H

RT

H

P

dP

T

ZT

P

dP

T

ZT

RT

HEq

T

Z

T

Z

T

ZZZZ

RRR

r

r

Prr

P

r

r

Prr

PR

c

R

c

R

c

R

r

r

P

Prr

r

r

P

Prr

c

R

PrPrPr

r

r

r

r

r

r

r

r

rrr

Table E.5 - E.12

Analytical correlation of the residual properties at low pressure

)88.6(

)87.6(

,);56.6()55.6.(

,

sec

10

11

00

1010

rrr

R

rr

rrr

c

R

rr

R

rrr

c

R

rrrc

c

dT

Bd

dT

BdP

R

S

dT

BdTB

dT

BdTBP

RT

H

dT

BdP

R

S

dT

BdTBP

RT

HandEqs

dT

dB

dT

dB

dT

BdBB

RT

BPB

formsncorrelatiotcoefficienvirialonddgeneralizeThe

)90.6(722.0

)89.6(675.0

)66.3(172.0

139.0

)65.3(422.0

083.0

2.5

1

6.2

0

2.41

6.10

rr

rr

r

r

TdT

dB

TdT

dB

TB

TB

HR and SR with ideal-gas heat capacities

RT

igP

igRT

igP

ig HdTCHHHdTCHH 1

0

012

0

02

12

)92.6(ln,

)91.6(

121

2

12

2

1

2

1

RRT

T

igP

RRT

T

igP

SSP

PRdTCSSimilarly

HHdTCH

For a change from state 1 to 2:

The enthalpy change for the process, H = H2 – H1

Alternative form

)94.6(lnln

)93.6()(

121

2

1

2

1212

RR

S

igP

RR

H

igP

SSP

PR

T

TCS

HHTTCH

A three-step calculational path• Step 11ig: A hypothetical process that transforms a real

gas into an ideal gas at T1 and P1.

• Step 1ig 2ig: Changes in the ideal-gas state from (T1,P1) to (T2,P2).

• Step 2ig 2: Another hypothetical process that transform the ideal gas back into a real gas at T2 and P2.

RigRig SSSHHH 111111

)96.6(ln

)95.6(

1

212

12

2

1

2

1

P

PR

T

dTCSSS

dTCHHH

T

T

igP

igigig

T

T

igP

igigig

RigRig SSSHHH 222222

Ex. 6.9Estimate V, U, H and S for 1-butane vapor at 200C, 70 bar

if H and S are set equal to zero for saturated liquid at 0C.

Assume: Tc=420.0 K, Pc=40.43 bar, Tn=266.9 K, =0.191

Cpig/R=1.967+31.630x10-3T-9.837x10-6T2 (T/K)

Solution

13

10

8.28770

)15.473)(14.83(512.0

512.0)142.0(191.0485.0

;4.3.)57.3.(

731.143.40

70127.1

0.420

15.273200

molcmP

ZRTV

ZZZ

EandETableandEq

PT rr

• Step (a): Vaporization of saturated liquid 1-butane at 0C

• The vapor pressure curve contains both

• The latent heat of vaporization, where Trn=266.9/420=0.636:

)75.6(ln T

BAP sat

11.699,2126.10,

0.42043.40lnint;

9.2660133.1lnint;

BAWhence

BApocriticalthe

BApoboilingnormalthe

1137,229.266314.8979.9

979.9636.0930.0

)013.143.40(ln092.1

930.0

)013.1(ln092.1

JmolH

T

P

RT

H

lvn

r

c

n

lvn

n

11

1

380

380

847915273

81021

810213680

350013722

1341

1

65004201527315273

KJmol..

,

T

HS

Jmol,.

.),(H

)....(T

T

H

HFrom

./.TK.atheatlatentThe

lv

lv

.

lv

.

r

r

lv

n

lv

r

n

• Step (b): Transformation of saturated vapor into an ideal gas at (T1, P1).

• Tr = 0.650 and Pr = 1.2771/40.43 = 0.0316

)88.6(

)87.6(

10

11

00

rrr

R

rr

rrr

c

R

dT

Bd

dT

BdP

R

S

dT

BdTB

dT

BdTBP

RT

H

)90.6(722.0

)89.6(675.0

)66.3(172.0

139.0

)65.3(422.0

083.0

2.5

1

6.2

0

2.41

6.10

rr

rr

r

r

TdT

dB

TdT

dB

TB

TB

111

11

88.0)314.8)(1063.0(

344)420)(314.8)(0985.0(,

KJmolS

JmolHSo

R

R

• Step (c): Changes in the ideal gas state

• Tam = 373.15 K, Tlm = 364.04 K,

A = 1.967, B = 31.630x10-3, C = -9.837x10-6

Hig = 20,564 J mol-1

Sig = 22.18 J mol-1 K-1

1

2

1

2

1

2

12

1212

2

1

2

1

966

956

P

PlnR

T

TlnC

P

PlnR

T

dTCSSS:)..(Eq

)TT(CdTCHHH:)..(Eq

S

ig

P

T

T

ig

P

igigig

H

ig

P

T

T

ig

P

igigig

• Step (d): Transformation from the ideal gas to real gas state at T2 and P2.

Tr = 1.127 Pr = 1.731

• At the higher P; Eqs.(6.85) and (6.86) with interpolated values from Table E.7, E.8, E.11 and E.12.

113

11

1

112

12

2

2

218,3210

)8.287)(70(233,34

18.1418.22)88.0(84.79

233,34485,8564,20)344(810,21

18.14)314.8)(705.1(

485,8)0.420)(314.8)(430.2(

705.1)726.0)(191.0(566.1

430.2)713.0)(191.0(294.2

JmolbarJcm

PVHU

KJmolSS

JmolHH

KJmolS

JmolH

R

S

RT

H

R

R

R

C

R

Extension to Gas Mixtures

)101.6()100.6(

)99.6()98.6()97.6(

PrPr

pcpc

iciipc

iciipc

iii

P

PP

T

TT

PyPTyTy

These replace TThese replace Trr and P and Prr for reading entries from the table of for reading entries from the table of

App. E, and lead to values of Z by Eq.(3.57), and HApp. E, and lead to values of Z by Eq.(3.57), and HRR/RT/RTpcpc

by Eq.(6.85), and Sby Eq.(6.85), and SRR/R by Eq.(6.86). /R by Eq.(6.86).

Ex. 6.10Estimate V, HR, and SR for an equimolar mixture of

carbon dioxide(1) and propane(2) at 450 K and 140 bar by

the Lee/Kesler correlations.

Solution

From Table B.1,

41.215.58

140335.1

0.337

450,

15.58)48.42)(5.0()83.73)(5.0(

0.337)8.369)(5.0()2.304)(5.0(

188.0)152.0)(5.0()224.0)(5.0(

2211

2211

2211

prpr

ccpc

ccpc

PTWhence

barPyPyP

KTyTyT

yy

11

1

10

13

10

56.8)029.1)(314.8(

029.1)330.0)(188.0(967.0

:)86.6.(;12.11.

937,4762.1)337)(314.8(

762.1)169.0)(188.0(730.1

169.0730.1

:)85.6.(;8.7.

7.196140

)450)(14.83)(736.0(

736.0)205.0)(188.0(697.0

;4.3.

KJmolS

R

S

EqwithEandETableFrom

JmolH

RT

H

RT

H

RT

H

EqwithEandETableFrom

molcmP

ZRTV

ZZZ

EandETableFrom

R

R

R

pc

R

pc

R

pc

R