Thermodynamic Properties of Fluids

1

Purpose of this Chapter

• To develop from the first and second laws the fundamental property relations which underlie the mathematical structure of thermodynamics.

– Derive equations which allow calculation of enthalpy and entropy values from PVT and heat capacity data.

– Discuss diagrams and tables by which property values are presented for convenient use.

– Develop generalized correlations which provide estimated of property values in he absence of complete experimental information.

2

Property Relations for Homogeneous Phases

Fundamental Properties

Although this equation 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 form one equilibrium state to another.

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

nVPdnSTdnUd

dWdQnUd revrev)(

3

…(6.1)

Define

H = Enthalpy

A = Helmholtz energy

G = Gibbs energy

PVUH

TSUA

4

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

SdTVdPdG

SdTPdVdA

VdPTdSdH

PdVTdSdU

5

Maxwell’s equaitons

TP

TV

PS

VS

P

S

T

V

V

S

T

P

S

V

P

T

S

P

V

T

6

Enthalpy and Entropy as Functions of T and P

Temperature derivatives:

T

C

T

S

CT

H

P

P

P

P

7

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

P

P

P

p

8

…(6.21)

…(6.20)

Property Relations for Homogeneous Phases Internal Energy as Function of P

• The pressure dependence of the internal energy is shown as

PdVTdsdU

TPTP

VP

T

VT

P

U

9

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 to as follows:

(6.24)

(6.23)

P

dPR

T

dTCdS

dTCdH

P

R

T

VRTPV

ig

P

ig

ig

P

ig

P

ig

ig

10

Property Relations for Homogeneous Phases Alternative 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

11

Property Relations for Homogeneous Phases Alternative 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

12

Example 6.1

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

T( C) P(bar) Cp(Jmol-1K-1 V(cm3mol-1) (K-1)

25

25

50

50

1

1,000

1

1,000

75.305

…

75.314

…

18.071

18.012

18.234

18.174

256x10-6

366x10-6

458x10-6

568x10-6

13

14

15

166

13

11

10513102

568458

204.182

174.18234.18

,50

310.752

314.75305.75

,1

K

molcmV

CTforand

KJmolC

barPFor

p

12

1

2

12212

ln

1

PPVT

TCS

PPVTTTCH

p

p

16

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

12

1

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 25 C.

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

17

• 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

V

V

V

V

dVT

dTCdS

dVPTdTCdU

V

V

18

Property Relations for Homogeneous Phases The Gibbs Energy as a Generating Function

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

• 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.

P

T

P

RTGT

RT

H

P

RTG

RT

V

dTRT

HdP

RT

V

RT

Gd

2

19

Residual Properties

• The definition for the generic residual property is:

• M is the molar value of any extensive thermodynamics property: V, U, H, S, G.

• M, Mig = the actual and ideal gas properties which are at the same temperature and pressure.

igR MMM

20

• Residual gibbs energy:

• G, Gig = the actual and ideal gas values of the Gibbs energy at the same temperature and pressure.

• Residual volume:

igR GGG

1ZP

RTV

P

RTVVVV

R

igR

21

Fundamental property relation for residual properties

• The fundamental property relation for residual preperties 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

22

23

)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

24

25

2

2112

21

21

2

12

00

0

00

2

1

2

1 43

536

526

TT

DCTTBTA

R

C

)T/Tln(

T

dTC

C

TT

D)TTT(

CBTA

R

C

TT

dTC

C

).(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.

They are important because they provide a convenient base

for the calculation of real-gas properties.

Example 6.3

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

1. Table 6.1 gives compressibility-factor data

2. The vapor pressure of isobutane at 630 K 15.46 bar

3. Set H0ig = 18,115 Jmol-1 and S0

ig = 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)

26

27

28

Solution 6.3

Eqs. (6.46) and (6.48) are used to calculate HR and SR.

Plot ( Z/ T)P/P and (Z-1)/P vs. P

From the compressibility-factor data at 360 K (Z-1)/P

The slope of a plot of Z vs. T ( Z/ T)P/P

Data for the required plots are shown in Table 6.2.

P

dPZ

P

dP

T

Z P

P

P

)1(00

29

30

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

31

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

32

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

33

• 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

34

Application of these equations, useful for gases up to

moderate pressure, requires data for both the second and

third virial coefficients.

Residual Properties by Cubic Equations of State

IdT

dq

T

Z,qI)bln(;simplifyTo

)b)(b(

)b(d

dT

dq

T

Z

)b)(b(

)b(dq

b

)b(d

b

b

);.(),.(

)b)(b(

b

dT

dq

T

Z

).()b)(b(

bq

b

bZ

)b)(b(

bq

bZ

)...(Eq,/VeRTy)..(Eq

dd1)-(Z

d

d1)-(Z

Eqs.ofintegralsThe

bygivenqassubstitutandbdevides

0

0

0

00

0 0

1

11

111

606586

11

646111

1

111

1

5131423

35

36

The generic equation of state presents two cases.

Zb

bI:

)b.(Z

ZlnI

bZ

whenceRT

PZ

ρ

)a.(b

blnI:

1

6561

6561

11

IICase

RT

bP

Z.offavorineliminatedisWhen

ICase

37

)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

r

R

r

r

R

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

38

Ex. 6.4

39

40

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

41

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);..(EqofnIntegratio

V

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

42

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

43

Corresponding-States Correlations for Vapor Pressure

44 :

:

)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

sat

r

r

rr

rr

sat

r

rr

r

rr

rr

r

rr

rrrrr

sat

r

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. 6.6 Determine the vapor pressure for liquid n-hexane at 0, 30,

60 and 90 C: (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

tPsat

03350.025.30

01325.1,6736.0

6.507

9.341 sat

rr nnPT

45

t/ C Psat/kPa

(Antoine)

Psat/kPa

(Lee/Kesler)

t/ C Psat/kPa

(Antoine)

Psat/kPa

(Lee/Kesler)

0

60

6.052

76.46

5.835

76.12

30

90

24.98

189.0

24.49

190.0

46

47

Two-Phase Liquid/Vapor System

)82.6(

:.,,,,

)82.6()1(

:

1)1(

:

)(

bMxMM

formealternativAnetcSHUVMwhere

aMxMxM

equationgenericThe

xxVxVxV

fractionmassxVxVxV

molesnnnVnVnnV

ll

lv

vllv

ll

vlll

48

THERMODYNAMIC DIAGRAMS

49

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

r

r

R

r

r

P

P

r

r

c

R

rcrcrcrc

P

dPZ

P

dP

T

ZT

R

S

P

dP

T

ZT

RT

H

andEqsosubstitue

dTTdTTTTdPPdPPPP

r

r

r

r

r

50

51

)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

Pr

r

P

r

r

Pr

r

PR

c

R

c

R

c

R

r

r

P

Pr

r

r

r

P

Pr

r

c

R

PrPrPr

r

r

r

r

r

r

r

r

rrr

Table E.5 - E.12

52

Analytical correlation of the residual properties at low pressure

)88.6(

)87.6(

,);56.6()55.6.(

,

sec

10

11

00

1010

rr

r

R

r

r

r

rr

c

R

r

r

R

r

rr

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

formsncorrelatiotcoeff icienvirialonddgeneralizeThe

53

54

)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

2.4

1

6.1

0

rr

rr

r

r

TdT

dB

TdT

dB

TB

TB

HR and SR with ideal-gas heat capacities

R

T

ig

P

igR

T

ig

P

ig HdTCHHHdTCHH 1

0

012

0

02

12

)92.6(ln,

)91.6(

12

1

2

12

2

1

2

1

RR

T

T

ig

P

RR

T

T

ig

P

SSP

PRdTCSSimilarly

HHdTCH

)94.6(lnln

)93.6()(

12

1

2

1

2

1212

RR

S

ig

P

RR

H

ig

P

SSP

PR

T

TCS

HHTTCH

55

For a change from state 1 to 2:

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

Alternative form

56

A three-step calculational path

• Step 1 1ig: 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

ig

P

igigig

T

T

ig

P

igigig

57

RigRig SSSHHH 222222

Ex. 6.9

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

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

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

58

59

60

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

• The vapor pressure curve contains both

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

)75.6(ln T

BAPsat

11.699,2126.10,

0.42043.40lnint;

9.2660133.1lnint;

BAWhence

BApocriticalthe

BApoboilingnormalthe

61

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

lv

n

r

c

n

lv

n

n

62

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

rr

r

R

r

r

r

rr

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

2.4

1

6.1

0

rr

rr

r

r

TdT

dB

TdT

dB

TB

TB

63

11

1

1

1

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

64

• 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.

1

13

11

1

11

2

1

2

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

65

Extension to Gas Mixtures

)101.6()100.6(

)99.6()98.6()97.6(

PrPr

pcpc

i

ciipc

i

ciipc

i

ii

P

PP

T

TT

PyPTyTy

66

These replace Tr and Pr for reading entries from the table of

App. E, and lead to values of Z by Eq.(3.57), and HR/RTpc

by Eq.(6.85), and SR/R by Eq.(6.86).

Ex. 6.10

Estimate 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

67

68

69 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

70