Chapter 5

c° 2011 Ismail Tosun

Chapter 5

Fugacity of a Pure Component

This chapter introduces a new concept called fugacity, which is the basis for establishing equi-librium criteria between different phases of a pure substance. After showing how to calculatethis quantity for gases, liquids, and solids, the variation of temperature with pressure (or viceversa) during phase transformation of pure substances will be discussed.

5.1 MOLAR GIBBS ENERGY OF A PURE IDEAL GAS

For a pure substance, molar Gibbs energy is given in differential form as

d eG = eV dP − eS dT (5.1-1)

The ideal gas equation of state relates the molar volume to pressure as

eV IG =RT

P(5.1-2)

Substitution of Eq. (5.1-2) into Eq. (5.1-1) leads to

d eGIG =RT

PdP − eS dT = RT d lnP − eS dT (5.1-3)

At constant temperature, Eq. (5.1-3) reduces to

d eGIG = RT d lnP at constant T (5.1-4)

Integration of Eq. (5.1-4) at constant temperature from the standard state pressure, P o, tothe state of interest at which the pressure is P gives

ln

µP

P o

¶=eGIG(T, P )− eGIG(T,P o)

RT(5.1-5)

5.2 DEFINITION OF FUGACITY AND FUGACITY COEFFICIENT

Equation (5.1-5) is valid only for pure, ideal gases. In order to generalize it, in 1901 GilbertNewton Lewis1 introduced a function fi, called the fugacity of pure component i, in terms ofthe departure function for molar Gibbs energy as

ln

∙fi(T, P )

P

¸=( eGi − eGIG

i )T,P

RT

=( eHi − eHIG

i )T,P

RT−(eSi − eSIG

i )T,P

R

(5.2-1)

1Professor of physical chemistry and the dean of the College of Chemistry at University of California, Berkeley.

93

subject to the following constraint

fi/P → 1 as P → 0 (ideal gas state) (5.2-2)

which applies to solid, liquid, and gas.The fugacity coefficient of pure component i, φi, is defined as the ratio of the fugacity to

the pressure, i.e.,

φi =fiP

(5.2-3)

It is a measure of the deviation of the system from ideal gas behavior, i.e., φi = 1 for an idealgas. Note that both fugacity and the fugacity coefficient are intensive properties of the systemand are a function of any two other intensive properties, such as temperature and pressure, fora single-phase pure material. Fugacity coefficient may be greater or less than unity as shownin Figure 5.1.

0 0 P

iφ 1

Figure 5.1 Variation of the fugacity coefficient with pressure.

Addition of Eqs. (5.1-5) and (5.2-1) leads to an alternative expression for the fugacity ofpure component i in the form

eGi(T, P ) = eGIGi (T, P

o)−RT lnP o +RT ln fi(T, P ) (5.2-4)

or eGi(T,P ) = λi(T ) +RT ln fi(T,P ) (5.2-5)

where λi(T ) is the molar Gibbs energy of pure component i at unit fugacity, and is defined by

λi(T ) = eGIGi (T, P

o)−RT lnP o (5.2-6)

Differentiation of Eq. (5.2-5) at constant temperature gives

d eGi = RTd ln fi at constant T (5.2-7)

When two phases of a pure substance are in equilibrium, we have

Tα = Tβ Pα = Pβ eGαi = eGβ

i (5.2-8)

Letting Tα = Tβ = T and Pα = Pβ = P , equality of molar Gibbs energies leads to

λi(T ) +RT ln fαi (T, P ) = λi(T ) +RT ln fβi (T, P ) ⇒ fαi (T, P ) = fβi (T, P ) (5.2-9)

indicating that fugacities of a pure component must be equal to each other under the conditionsof equilibrium.

94

The word fugacity comes from the same root as "fugitive". It can be thought of as a degreeof "discomfort". The higher the fugacity, the more uncomfortable a molecule is in a givenphase. To ease the measure of discomfort, a molecule escapes to another phase in which itsfugacity is lower. For this reason, fugacity is also called "escaping tendency", the measure ofwhich is molar Gibbs energy. All spontaneous physical and/or chemical processes take placein the direction of decreasing fugacity and molar Gibbs energy. For example, the driving forcefor mass transfer is the difference in fugacity.

At constant temperature, Eq. (5.1-1) reduces to

d eGi = eVi dP (5.2-10)

Substitution of Eq. (5.2-10) into Eq. (5.2-7) gives

RT d ln fi = eVi dP (5.2-11)

Subtraction of the term RTd lnP from both sides of Eq. (5.2-11) and integration from an idealgas state to any state lead to

ln

µfiP

¶= lnφi =

Z P

0

Ã eViRT− 1

P

!dP T = constant (5.2-12)

which is applicable to gases, liquids, and solids. Fugacity (or fugacity coefficient) can becalculated from either Eq. (5.2-1) or Eq. (5.2-12).

5.3 FUGACITY OF A PURE GASEOUS SPECIES

Fugacity of a pure gas at a given temperature and pressure can be calculated from Eq. (5.2-12),i.e.,

lnφVi =

Z P

0

Ã eViRT− 1

P


or

lnφVi =

Z P

0

ÃZVi − 1P


At low pressures, gases behave ideally, i.e., ZVi = 1, and Eq. (5.3-2) leads to

lnφVi = 0 ⇒ φVi = 1 or fVi (T, P ) = P (5.3-3)

indicating that for a pure gas at low pressures the fugacity coefficient is unity and the fugacityis equal to the total pressure.

5.3.1 Fugacity from Volume-Explicit Equations of State

Substitution of eVi = eVi(P, T ) into Eq. (5.3-1) and integration of the resulting equation eitheranalytically or numerically give the fugacity of a gas at a specified temperature and pressure.

Example 5.1 Calculate the fugacity coefficient of ammonia at 350K and 50 bar if the equa-tion of state is given as

P (eV − b) = RT

where b = 37.3 cm3/mol.

95

Solution

From the given equation of state eVNH3 = b+RT

P(1)

Substitution of Eq. (1) into Eq. (5.3-1) gives

lnφNH3 =

Z 50

0

µb

RT+1

P− 1

P

¶dP =

bP

RT

¯̄̄̄500

=(37.3)(50)

(83.14)(350)= 0.064

Therefore,φNH3 = 1.066

5.3.2 Fugacity from Property Tables

If property tables are available, then fugacity can be calculated directly without using anequation of state as shown in the following example.

Example 5.2 Calculate the fugacity of steam at 873K and 50 bar using the following data:

T (K) P ( bar) bH ( kJ/ kg) bS ( kJ/ kg.K)873 50 3666.5 7.2589873 0.1 3705.4 10.1608

Solution

Let the values of temperature and pressure at states 1 and 2 be (873K, 0.1 bar) and (873K, 50 bar),respectively. From Eq. (5.2-5) eG1 = λ (873K) +RT ln f1 (1)eG2 = λ (873K) +RT ln f2 (2)

Subtraction of Eq. (1) from Eq. (2) results in

eG2 − eG1 = RT ln

µf2f1

¶(3)

Note that Eq. (3) can also be obtained from Eq. (5.2-1). Since eG = eH − T eS, Eq. (3) becomes( eH2 − eH1)− T (eS2 − eS1) = RT ln

µf2f1

¶(4)

or

Mh( bH2 − bH1)− T (bS2 − bS1)i = RT ln

µf2f1

¶(5)

where M is the molecular weight of water. At 873K and 0.1 bar, it is plausible to assume thatsteam behaves as an ideal gas, i.e.,

f1 = P1 = 0.1 bar

96

Substitution of the numerical values into Eq. (5) gives

18h(3666.5− 3705.4)− (873)(7.2589− 10.1608)

i= (8.314)(873) ln

µf20.1

¶The solution yields

f2 = 48.6 bar

5.3.3 Fugacity from the Virial Equation of State

At low to moderate pressures, it is possible to truncate the virial equation of state after thesecond term as given by Eq. (3.1-9), i.e.,

ZVi = 1 +

BiP

RT(5.3-4)

where Bi, the second virial coefficient for pure i, is calculated from Eq. (3.1-10). Substitutionof Eq. (5.3-4) into Eq. (5.3-2) and integration give

lnφVi =BiP

RT(5.3-5)

5.3.4 Fugacity from Cubic Equations of State

The cubic equations of state are pressure-explicit, i.e., P = P (eV , T ), and the use of Eq. (5.3-1)is not practical for this case. Therefore, it is useful to have an equation relating fugacity to anintegral over volume rather than pressure.

For this purpose, the starting point is the equation of state for a real gas, i.e.,

P eVi = ZVi RT (5.3-6)

At constant temperature, differentiation of Eq. (5.3-6) gives

deVieVi +dP

P=

dZVi

ZVi

(5.3-7)

Substitution of Eqs. (5.3-6) and (5.3-7) into Eq. (5.3-2) yields

lnφVi = ZVi − 1− lnZV

i +1

RT

Z ∞

eVi⎛⎝P − RTeVi

⎞⎠ deV T = constant (5.3-8)

Substitution of the cubic equation of state into Eq. (5.3-8) and integration lead to the followinggeneral expression for calculating fugacity coefficients in the vapor phase:

lnφVi = ZVi − 1− ln(ZV

i −Bi)−ΘVi (5.3-9)

where the term ΘVi is given in Table 5.1 depending on the equation of state.

97

Table 5.1 The term ΘVi in Eq. (5.3-9)

2.

Equation of State ΘVi

van der WaalsAi

ZVi

(A)

Redlich-KwongAi

Biln

Ã1 +

Bi

ZVi

!(B)

Soave-Redlich-KwongAi

Biln

Ã1 +

Bi

ZVi

!(C)

Peng-RobinsonAi√8Bi

ln

"ZVi + (1 +

√2)Bi

ZVi + (1−

√2)Bi

#(D)

Equation (5.3-9) can also be obtained by substituting departure function expressionsdeveloped in Section 3.6.1 into Eq. (5.2-1).

Example 5.3 Calculate the fugacity coefficient of acetylene at 373K and 1, 10, and 50 barif it obeys

a) van der Waals equation of state,b) Redlich-Kwong equation of state,c) Soave-Redlich-Kwong equation of state,d) Peng-Robinson equation of state.

Solution

From Appendix ATc = 308.3K Pc = 61.4 bar ω = 0.190

The reduced temperature is

Tr =T

Tc=

373

308.3= 1.210

The values of the reduced pressures are

P ( bar) 1 10 50

Pr 0.016 0.163 0.814

a) For the van der Waals equation of state, the values of A, B, Z, and φ are given in thefollowing table:

P( bar)

A B Z φ

1 4.61× 10−3 1.653× 10−3 0.997 0.99710 0.047 0.017 0.969 0.97050 0.235 0.084 0.830 0.852

2The dimensionless parameters Ai and Bi can be calculated from the expressions given in Table 3.2.

98

b) For the Redlich-Kwong equation of state, the values of A, B, Z, and φ are given in thefollowing table:

P( bar)

A B Z φ

1 4.247× 10−3 1.146× 10−3 0.997 0.99710 0.043 0.012 0.968 0.96950 0.216 0.058 0.833 0.850

c) For the Soave-Redlich-Kwong equation of state, the values of α, A, B, Z, and φ are givenin the following table:

P( bar)

α A B Z φ

1 0.853 3.985× 10−3 1.146× 10−3 0.997 0.99710 0.853 0.041 0.012 0.971 0.97150 0.853 0.203 0.058 0.850 0.863

d) For the Peng-Robinson equation of state, the values of α, A, B, Z, and φ are given in thefollowing table:

P( bar)

α A B Z φ

1 0.873 4.362× 10−3 1.029× 10−3 0.997 0.99710 0.873 0.044 0.010 0.966 0.96650 0.873 0.222 0.052 0.829 0.843

5.3.5 Fugacity from the Principle of Corresponding States

In terms of reduced temperature and pressure, Eq. (5.2-1) takes the form

lnφVi =1

Tr

( eHi − eHIGi )T,P

RTc−(eSi − eSIG

i )T,P

R(5.3-10)

in which the enthalpy and entropy departure functions are calculated from Eqs. (3.6-24) and(3.6-25), respectively.

5.4 FUGACITY OF A PURE LIQUID

When liquid and vapor phases of a pure component i are in equilibrium with each other, thenumber of degrees of freedom is unity. In other words, the equilibrium state can be determinedby specifying one independent intensive variable, for example, temperature. At any given tem-perature, the corresponding pressure is the vapor pressure (or saturation pressure), P vap

i , asshown in Figure 5.2.

99

P

Vapor

iG~

Liquid

vapiP

T = constant

Figure 5.2 Variation of Gibbs energy with pressure at constant temperature.

The vapor and liquid phases are in equilibrium with each other at the point of intersectionof eGi versus P curves in Figure 5.2. Thus,

eGVi (T,P

vapi ) = eGL

i (T, Pvapi ) ⇒ fVi (T,P

vapi ) = fLi (T, P

vapi ) (5.4-1)

In other words, there is no variation in fugacity during a phase change, i.e., from liquid tovapor or vice versa.

Using Eq. (5.2-5), molar Gibbs energy of a pure liquid at any temperature and pressure isexpressed as eGL

i (T, P ) = λi(T ) +RT ln fLi (T, P ) (5.4-2)

On the other hand, molar Gibbs energy of a pure liquid at the same temperature but at thecorresponding vapor pressure is given by

eGLi (T, P

vapi ) = λi(T ) +RT ln fLi (T, P

vapi ) (5.4-3)

Subtraction of Eq. (5.4-3) from Eq. (5.4-2) gives

eGLi (T, P )− eGL

i (T, Pvapi ) = RT ln

"fLi (T, P )

fLi (T, Pvapi )

#(5.4-4)

The use of Eq. (5.4-1) in Eq. (5.4-4) leads to

eGLi (T, P )− eGL

i (T,Pvapi ) = RT ln

"fLi (T, P )

fVi (T, Pvapi )

#(5.4-5)

At constant temperature, integration of Eq. (5.2-10) from P vapi to P yields

eGLi (T,P )− eGL

i (T, Pvapi ) = eV L

i (P − P vapi ) (5.4-6)

in which the molar volume of liquid is considered constant. Substitution of Eq. (5.4-6) intoEq. (5.4-5) and rearrangement result in

fLi (T, P ) = fVi (T, Pvapi ) exp

" eV Li (P − P vap

i )

RT

#(5.4-7)

100

or

fLi (T, P ) = P vapi φVi (T,P

vapi ) exp

" eV Li (P − P vap

i )

RT

#(5.4-8)

The exponential term in Eqs. (5.4-7) and (5.4-8) is commonly referred to as the Poyntingcorrection factor. It accounts for the compression of the liquid to a pressure P greater thanits vapor pressure, P vap

i . Table 5.2 indicates the variation in Poynting correction factor as afunction of pressure. Examination of the values clearly reveals that the Poynting correctionfactor (PCF) is almost unity at low pressures but may become large at extremely high pressuresor low temperatures.

Table 5.2 Variation of the Poynting correction factor as a function of pressure andtemperature when eV L = 50 cm3/mol.

T (K) P −P vap ( bar) Poynting Correction Factor

298 1 1.002

298 100 1.224

298 1000 7.524

300 50 1.105

100 50 1.351

50 50 1.825

Example 5.4 Specify the conditions under which the fugacity of liquid is almost equal to thevapor pressure at a given temperature, i.e., fLi (T, P ) ' P vap

i .

Solution

When pressure is not very high, the Poynting correction factor is unity and Eq. (5.4-8) sim-plifies to

fLi (T, P ) = P vapi φVi (T, P

vapi ) (1)

When the vapor pressure is not very high, the vapor phase may be considered an ideal gas andφVi (T, P

vapi ) = 1. Thus, Eq. (1) becomes

fLi (T, P ) = P vapi (2)

Comment: Using steam tables, Rittmann et al. (1982) expressed the fugacity of saturatedwater and saturation pressure of water as

ln fsatH2O = 8.9672−1138.26

T− 10.6317× 10

5

T 2+8.8730× 107

T 3

lnP vapH2O = 13.8365−

6.6075× 103T

+10.1855× 105

T 2− 17.7039× 10

7

T 3

where fsatH2O and P vapH2O are in bar, and T is in K. These equations are valid when

300 ≤ T ≤ 600

The calculated values of P vapH2O and fsatH2O are given in the following table:

101

T (K) P vapH2O ( bar) fsatH2O ( bar)

300 0.032 0.035350 0.425 0.413400 2.503 2.392450 9.395 8.760500 26.560 23.496

Thus, at higher values of vapor pressure, the difference between vapor pressure and fugacitybecomes appreciable.

Evaluation of liquid fugacities requires molar volumes of saturated liquids to be known.Molar volumes of various liquids as a function of temperature are given in Table 5.3.

Table 5.3 Molar volumes of various liquids (Holmes and van Winkle, 1970).

SubstanceT(K)

eV L

( cm3/mol)Substance

T(K)

eV L

( cm3/mol)

228.15 67.380 273.15 143.045Acetone 273.15 71.483 n-Heptane 323.15 152.303

323.15 76.826 373.15 163.619

273.15 51.092 273.15 127.301Acetonitrile 303.15 53.214 n-Hexane 323.15 136.388

355.15 57.4 373.15 148.211

273.15 86.783 273.15 39.556Benzene 323.15 92.263 Methanol 373.15 44.874

373.15 98.537 473.15 57.939

273.15 89.873 273.15 87.3n-Butanol 343.15 97.8 Methyl ethyl ketone 333.15 94.5

413.15 108.7 373.15 100.0

293.15 96.518 273.15 158.970Carbon tetrachloride 353.15 104.192 n-Octane 333.15 170.630

413.15 114.379 393.15 185.182

273.15 78.218 273.15 111.8Chloroform 303.15 81.185 n-Pentane 333.15 122.9

333.15 84.5 373.15 131.4

288.15 107.470 293.15 74.785Cyclohexane 306.30 109.841 n-Propanol 343.15 78.962

352.35 116.630 393.15 84.515

273.15 57.141 303.15 107.415Ethanol 323.15 60.356 Toluene 353.15 113.717

373.15 64.371 400.00 120.879

273.15 95.3 277.13 18.060Ethyl acetate 323.15 102.1 Water 323.15 18.278

373.15 110.5 373.15 18.844

102

If experimental data are unavailable, molar volumes of saturated liquids can be estimatedby the modified Rackett equation (Spencer and Danner, 1972)

eV L =RTcPc

Z1+(1−Tr)2/7RA (5.4-9)

where ZRA is the Rackett parameter, given by

ZRA = 0.29056− 0.08775ω (5.4-10)

Example 5.5 Estimate the molar volume of liquid carbon dioxide at 274K and 70 barusing the Soave-Redlich-Kwong equation of state, the Peng-Robinson equation of state, andthe Rackett equation.

Solution

From Appendix ATc = 304.2K Pc = 73.8 bar ω = 0.239

The values of the reduced temperature and pressure are

Tr =274

304.2= 0.901 and Pr =

70

73.8= 0.949

The molar volume is calculated from

eV L =ZLRT

P

For the Soave-Redlich-Kwong and Peng-Robinson equations of state the results are given in thefollowing table:

Equation of State A B ZL eV L ( cm3/mol)

Soave-Redlich-Kwong 0.544 0.091 0.160 52.1

Peng-Robinson 0.575 0.082 0.142 46.2

The Rackett parameter is

ZRA = 0.29056− (0.08775)(0.239) = 0.270

The use of Eq. (5.4-9) gives

eV L =(83.14)(304.2)

73.8(0.270)1+(1−0.901)

2/7= 47.1 cm3/mol

Comment: In general, the Soave-Redlich-Kwong equation of state overpredicts the liquid molarvolume by 10% to 35%.

Example 5.6 Calculate the fugacity of benzene at 320K and 100 bar. The vapor pressure ofbenzene at 320K is 0.308 bar.

103

Solution

From Appendix ATc = 562K Pc = 48.9 bar ω = 0.212

Since T < Tc and P > P vap, benzene exists as a liquid at the given temperature and pressure.Application of Eq. (5.4-8) gives

fLbenzene (320K, 100 bar) = 0.308φVbenzene (320K, 0.308 bar) exp

" eV Lbenzene(100− 0.308)(83.14)(320)

#(1)

Since the vapor pressure is low, vapor may be considered an ideal gas, i.e.,

φVbenzene (320K, 0.308 bar) ' 1

From Table 5.3 eV Lbenzene ' 92.263 cm3/mol

Substitution of the numerical values into Eq. (1) gives the liquid fugacity as

fLbenzene (320K, 100 bar) = 0.308 exp

∙(92.263)(100− 0.308)

(83.14)(320)

¸= 0.435 bar

If liquid fugacity is known at (T, P1), the value of the liquid fugacity at (T,P2) can becalculated from Eq. (5.4-8) as

fLi (T, P2) = fLi (T, P1) exp

" eV Li (P2 − P1)

RT

#(5.4-11)

5.4.1 Fugacity from Cubic Equations of State

Since cubic equations of state can describe pure substances in both the vapor and liquid phases,Eq. (5.3-9) is also applicable to liquids, i.e.,

lnφLi = ZLi − 1− ln(ZL

i −Bi)−ΘLi (5.4-12)

where the term ΘLi is given in Table 5.4 depending on the equation of state.

Example 5.7 Estimate the vapor pressure of ethanol at 313K by using the condition ofchemical equilibrium. Assume that ethanol is represented by the Peng-Robinson equation ofstate.

Solution

From Appendix ATc = 514K Pc = 63bar ω = 0.644

The condition of equilibrium is given by Eq. (5.4-1), i.e.,

fVethanol(T,Pvapethanol) = fLethanol(T,P

vapethanol) (1)

104

Table 5.4 The term ΘLi in Eq. (5.4-12)

3.

Equation of State ΘLi

van der WaalsAi

ZLi

(A)

Redlich-KwongAi

Biln

Ã1 +

Bi

ZLi

!(B)

Soave-Redlich-KwongAi

Biln

Ã1 +

Bi

ZLi

!(C)

Peng-RobinsonAi√8Bi

ln

"ZLi + (1 +

√2)Bi

ZLi + (1−

√2)Bi

#(D)

Dividing Eq. (1) by P vapethanol gives

φVethanol(T,Pvapethanol) = φLethanol(T, P

vapethanol) (2)

An algorithm to estimate the vapor pressure at a given temperature is given as follows:

1. Obtain Tc, Pc, and ω from Appendix A,2. Determine the reduced temperature,3. Use Eq. (3.1-16) to determine the term α,4. Assume P vap

ethanoland determine the reduced pressure,5. Determine the dimensionless parameters A and B (Table 3.2),6. Evaluate the parameters p, q, and r appearing in Eq. (3.1-18) (Table 3.2),7. Solve Eq. (3.1-18), i.e., Z3 + pZ2 + q Z + r = 0. The largest and the smallest rootscorrespond to ZV and ZL, respectively.

8. Use ZV in Eq. (5.3-9) and calculate φVethanol. Use ZL in Eq. (5.4-12) and calculateφLethanol. Repeat steps (4)-(8) until φ

Vethanol = φLethanol.

The reduced temperature is

Tr =T

Tc=313

514= 0.609

The term α is calculated from Eq. (3.1-16) as

α =h1 +

¡0.37464 + (1.54226)(0.644)− (0.26992)(0.644)2

¢ ³1−√0.609

´i2= 1.628

The results of the iterative procedure are given below :

P vapethanol ( bar) A× 103 B × 104 ZL × 104 ZV φL φV

0.1 3.186 2.028 2.379 0.997 1.877 0.9970.2 6.372 4.056 4.759 0.994 0.939 0.9940.1888 6.015 3.829 4.492 0.994 0.994 0.994

Therefore, the vapor pressure of ethanol at 313K is 0.1888 bar (or 141.6mmHg). In otherwords, ethanol exists in liquid form at 313K when P > 0.1888 bar.

3The dimensionless parameters Ai and Bi can be calculated from the expressions given in Table 3.2.

105

5.5 FUGACITY OF A PURE SOLID

When solid and vapor phases of a pure component i are in equilibrium with each other, thenumber of degrees of freedom is unity. In other words, the equilibrium state can be determinedby specifying one independent intensive variable, for example temperature. At any giventemperature, the corresponding pressure is the sublimation pressure (or vapor pressure), P sub

i ,as shown in Figure 5.3.

P

Vapor

iG~

Solid

T = constant

subiP

Figure 5.3 Variation of Gibbs energy with pressure at constant temperature.

The vapor and solid phases are in equilibrium with each other at the point of intersectionof eGi versus P curves in Figure 5.3. Thus,

eGVi (T,P

subi ) = eGS

i (T, Psubi ) ⇒ fVi (T,P

subi ) = fSi (T, P

subi ) (5.5-1)

Using Eq. (5.2-5), molar Gibbs energy of a pure solid at any temperature and pressure isexpressed as eGS

i (T, P ) = λi(T ) +RT ln fSi (T,P ) (5.5-2)

On the other hand, molar Gibbs energy of a pure solid at the same temperature but at thecorresponding sublimation pressure is given by

eGSi (T, P

subi ) = λi(T ) +RT ln fSi (T,P

subi ) (5.5-3)

Subtraction of Eq. (5.5-3) from Eq. (5.5-2) gives

eGSi (T, P )− eGS

i (T, Psubi ) = RT ln

"fSi (T, P )

fSi (T, Psubi )

#(5.5-4)


eGSi (T,P )− eGS

i (T, Psubi ) = RT ln

"fSi (T, P )

fVi (T, Psubi )

#(5.5-5)

At constant temperature, integration of Eq. (5.2-10) from P subi to P yields

eGSi (T, P )− eGS

i (T,Psubi ) = eV S

i (P − P subi ) (5.5-6)

106

Substitution of Eq. (5.5-6) into Eq. (5.5-5) and rearrangement result in

fSi (T, P ) = P subi φVi

¡T,P sub

i

¢exp

" eV Si (P − P sub

i )

RT

#(5.5-7)

Sublimation pressure of solids is normally low. Therefore, ideal gas behavior for the gas phaseover the pure solid can be assumed, i.e., φVi

¡T, P sub

i

¢' 1, and Eq. (5.5-7) simplifies to

fSi (T, P ) = P subi exp

" eV Si (P − P sub

i )

RT

#(5.5-8)

If solid fugacity is known at (T, P1), the value of solid fugacity at (T, P2) can be calculatedfrom Eq. (5.5-8) as

fSi (T,P2) = fSi (T, P1) exp

" eV Si (P2 − P1)

RT

#(5.5-9)

Example 5.8 Calculate the fugacity of ice at 250K and 100 bar using the following data:

lnP subice ( bar) = −

2378.6

T+ 51.06 lnT − 0.225T + 1.62× 10−4T 2 − 233.45

eVice( cm3/mol) = 19.655 + 0.0022364 (T − 273.15)Solution

At 250KP subice = 7.743× 10−4 bar and eVice = 19.603 cm3/mol

The use of Eq. (5.5-8) gives

fice = 7.743× 10−4 exp∙(19.603)(100)

(83.14)(250)

¸= 8.51× 10−4 bar

5.6 PHASE TRANSITIONS AND EQUILIBRIUM CRITERIA

The pressure-temperature diagram of a pure component includes three curves representingliquid-vapor, solid-vapor, and liquid-solid equilibrium as shown in Figure 5.4. Such plots arealso referred to as the phase diagrams.

When P < Pc, a substance in the gaseous state is called either a gas (T > Tc) or a vapor(T < Tc). Under isothermal conditions, while a vapor can be liquefied by exerting pressure, agas cannot be liquefied no matter what pressure is applied to it.

The triple point is the only point on the phase diagram where the solid, liquid, and vaporphases coexist in equilibrium4. In other words, it is the intersection of the liquid-vapor (vaporpressure curve), solid-liquid (fusion or melting curve), and solid-vapor (sublimation pressurecurve) coexistence curves. Note that the number of degrees of freedom, F , is zero at the triplepoint. Triple point temperatures and pressures of some substances are given in Table 5.5. If the

4 In general, the triple point is the point of intersection of three different phases. If a substance exists indifferent forms of solid, e.g., graphite and diamond for carbon, it can have more than one triple point.

107

Solid line normal subs tancesDashed line water

−⎛ ⎞⎜ ⎟−⎝ ⎠

LIQUID

Triple point

Sublimation curve

P

T

Vaporization curve

Fusion curve

Critical point

SOLID

VAPOR

GAS

FLUID

Figure 5.4 Temperature-pressure diagram.

Table 5.5 Normal boiling point, normal melting point, and triple point temperature andpressure for various substances5.

Temperature ( ◦C) at P = 1atm Triple Point

Substance Boiling Point Melting (Freezing) Point T ( ◦C) P ( atm)

Ammonia − 33.33 − 77.74 − 77.74 0.06Bromine 58.75 − 7.25 7.25 0.058Carbon dioxide − 78.45 − 56.57 − 56.57 5.1172Naphthalene 217.99 80.15 80.27 0.0099Sulfur dioxide − 10.05 − 75.55 − 75.47 0.0165Water 100 0 0.01 0.006

triple point pressure is less than 1 atm, as in the case of water, it is possible to have all threephases of a substance under atmospheric pressure depending on temperature. If the triplepoint pressure is higher than 1 atm, as in the case of carbon dioxide, then a substance cannotexist in the liquid form under atmospheric pressure, and the transition from the solid to vaporform, i.e., sublimation, takes place with an increase in temperature.

At temperatures and pressures higher than the critical values, substances exist in the fluid(or supercritical) region and are called supercritical fluids. They possess both the gaseousproperties (viscosity, diffusivity, surface tension) of being able to diffuse into substances easily,and the liquid property (density) of being able to dissolve substances.

The slope of the vaporization curve gives the rate of change of vapor pressure of liquid withtemperature. The slope of the sublimation curve gives the rate of change of sublimation (orvapor) pressure of solid with temperature. Finally, the slope of the fusion (or melting) curvegives the rate of change of the melting (or freezing) pressure of solid with temperature. Whilethe fusion curve has a positive slope for most substances, the slope becomes negative for water.Since the fusion curve generally has a very steep slope, the triple point temperature for mostsubstances is close to their melting (or freezing) temperature at atmospheric pressure, knownas normal melting (or freezing) temperature.

5Compiled from Korea Thermophysical Properties Data Bank (www.cheric.org/research/kdb).

108

Phase diagrams can also be expressed in the form of pressure versus specific volume asshown in Figure 5.5. Contrary to the case of the pressure versus temperature diagram, transi-tion from one phase to another is shown in the form of area.

Pc

Pt

VAPOR

P

LIQ

UID

SOLI

D

SOLI

D +

LIQ

UID

V̂

Triple point

Tc

SOLID + VAPOR

GAS

LIQUID + VAPOR

Isotherms

Figure 5.5 Pressure-specific volume diagram.

5.6.1 Clapeyron Equation

The starting point for obtaining quantitative information on the slopes of these equilibriumcurves, i.e., dP/dT , is the fact that molar Gibbs energies of the phases are equal to each other

eGα = eGβ (5.6-1)

where subscripts α and β represent different phases. The differential form of Eq. (5.6-1) is

d eGα = d eGβ (5.6-2)


eVα dP − eSα dT = eVβ dP − eSβ dT (5.6-3)

Rearrangement of Eq. (5.6-3) yields µdP

dT

¶eq

=eSα − eSβeVα − eVβ (5.6-4)

where the subscript eq indicates that the path of integration is on the vaporization curve, thefusion curve, or the sublimation curve.

The entropy difference in Eq. (5.6-4) must be expressed in terms of measurable quantities.For this purpose, the use of the identity eG = eH − T eS in Eq. (5.6-1) leads to

eHα − T eSα = eHβ − T eSβ ⇒ eSα − eSβ = eHα − eHβ

T(5.6-5)

109

Substitution of Eq. (5.6-5) into Eq. (5.6-4) givesµdP

dT

¶eq

=eHα − eHβ

T (eVα − eVβ) (5.6-6)

which is known as the Clapeyron6 equation.

5.6.2 Vapor-Liquid Equilibrium

The amount of heat to convert saturated liquid to saturated vapor is called the enthalpy changeon vaporization or simply the heat of vaporization, ∆ eHvap. If the α-phase is considered to bevapor and the β-phase to be liquid, theneHα − eHβ = eHV − eHL = ∆ eHvap (5.6-7)

and eVα − eVβ = eV V − eV L ' eV V

=RT

P vapif the vapor is an ideal gas (5.6-8)

Substitution of Eqs. (5.6-7) and (5.6-8) into Eq. (5.6-6) results in

dP vap

P vap=∆ eHvap

R

dT

T 2(5.6-9)

ord lnP vap

dT=∆ eHvap

RT 2(5.6-10)

which is referred to as the Clausius7-Clapeyron equation. Integration of Eq. (5.6-10) dependson whether ∆ eHvap is constant or changes with temperature.

• ∆ eHvap is independent of temperature

When ∆ eHvap is constant, Eq. (5.6-10) can be rearranged as

d lnP vap = − ∆eHvap

Rd

µ1

T

¶(5.6-11)

Integration of Eq. (5.6-11) gives

lnP vap = − ∆eHvap

RT+ C (5.6-12)

where C is an integration constant. Equation (5.6-12) indicates that a plot of lnP vap versus1/T is a straight line with a slope equal to (−∆ eHvap/R). Between any two states 1 and 2 onthe vapor pressure curve, Eq. (5.6-12) yields

ln

ÃP vap2

Pvap1

!= − ∆

eHvap

R

µ1

T2− 1

T1

¶(5.6-13)

which provides accurate estimates of vapor pressures only over small temperature ranges.6Benoit Paul Emile Clapeyron, French engineer and physicist.7Rudolf Julius Emanuel Clausius, German physicist and mathematician.

110

• ∆ eHvap varies with temperature

The variation of ∆ eHvap with temperature is expressed in the form

d∆ eHvap

dT=

d eHV

dT− d eHL

dT= eCV

P − eCLP = ∆ eCV L

P (5.6-14)


∆ eHvap =

Z T

Tref

∆ eCV LP dT +∆ eHvap

ref (5.6-15)

where ∆ eHvapref is the value of ∆

eHvap at the reference temperature, Tref . Once ∆ eHvap is ex-pressed as a function of temperature from Eq. (5.6-15), substitution of the resulting expressioninto Eq. (5.6-10) and integration yield the relationship between vapor pressure and tempera-ture.

If ∆ eCV LP is independent of temperature, then Eq. (5.6-15) reduces to

∆ eHvap = ∆ eCV LP (T − Tref ) +∆ eHvap

ref (5.6-16)

Substitution of Eq. (5.6-16) into Eq. (5.6-10) and integration yield

lnP vap =

Ã∆ eCV L

P

R

!lnT −

Ã∆ eHvap

ref −∆ eCV LP Tref

R

!1

T+ C (5.6-17)

where C is an integration constant.In the literature, among the various equations relating vapor pressure to temperature, the

most widely used is the Antoine equation, expressed in the form

lnP vap = A− B

T +C(5.6-18)

The constants A, B, and C are given in Appendix C. Note that when C = 0, Eq. (5.6-18)reduces to Eq. (5.6-12).

Example 5.9 Calculate the heat of vaporization of n-heptanol [CH3(CH2)6OH] at the normalboiling temperature.

Solution

From Appendix C

lnP vap = 8.6866− 2626.42

T − 146.6 (1)

At the normal boiling temperature, the vapor pressure of liquid equals 1 atm (∼ 1.013 bar).Therefore, the normal boiling temperature can be calculated from the Antoine equation as

ln 1.013 = 8.6866− 2626.42

T − 146.6 ⇒ T = 449.4K

The heat of vaporization can be determined from Eq. (5.6-10). Differentiation of Eq. (1) withrespect to temperature gives

d lnP vap

dT=

2626.42

(T − 146.6)2 (2)

111

Thus, comparison of Eq. (2) with Eq. (5.6-10) results in

∆ eHvap

RT 2=

2626.42

(T − 146.6)2

or

∆ eHvap =2626.42RT 2

(T − 146.6)2 =(2626.42)(8.314)(449.4)2

(449.4− 146.6)2 = 48, 098 J/mol

5.6.3 Solid-Liquid Equilibrium

The amount of heat required to melt a solid is called the heat of fusion, ∆ eHfus. If the α-phaseis considered to be liquid and the β-phase to be solid, then

eHα − eHβ = eHL − eHS = ∆ eHfus > 0 (5.6-19)

The volume change on melting, ∆eV fus, is given by

∆eV fus = eV L − eV S (5.6-20)

Therefore, the slope of the solid-liquid equilibrium curve, i.e., Eq. (5.6-6), becomes

dP

dTm=

∆ eHfus

Tm∆eV fus(5.6-21)

where Tm is the melting temperature. Since∆ eHfus > 0, the slope of the solid-liquid equilibriumcurve is either positive or negative depending on the sign of ∆eV fus. For almost all substances∆eV fus > 0 and thus the slope of the solid-liquid equilibrium curve has a positive slope. Wateris an exception as the density of ice, 917 kg/m3, is less than that of liquid water, 1000 kg/m3.In this case

∆eV fus =

µ1

1000− 1

917

¶18 = − 1.63× 10−3m3/ kmol

and the slope of the solid-liquid equilibrium curve is negative.Integration of Eq. (5.6-21) depends on whether ∆ eHfus is constant or changes with tem-

perature.

• ∆ eHfus is independent of temperature

When ∆ eHfus is constant, integration of Eq. (5.6-21) gives

P =

Ã∆ eHfus

∆eV fus

!lnTm + C (5.6-22)

where C is an integration constant. Between any two states 1 and 2 on the solid-liquid equi-librium curve, Eq. (5.6-22) yields

P2 − P1 =

Ã∆ eHfus

∆eV fus

!ln

µTm2

Tm1

¶=

Ã∆ bHfus

∆bV fus

!ln

µTm2

Tm1

¶(5.6-23)

Example 5.10 Explain quantitatively why a 55 kg ice skater glides on solid ice but not ondry ice (solid CO2) under atmospheric conditions? For water ∆ eHfus = 6012 kJ/ kmol.

112

Solution

The normal melting (or freezing) temperature of water is 0 ◦C. If the length and the thicknessof the skating blade are 150mm and 3mm, respectively, then the pressure exerted by the weightof the skater is

P =mg

A=

(55)(9.8)

(150× 10−3)(3× 10−3) = 1.198× 106 Pa = 1.198× 103 kPa

As a result of the increase in pressure, the melting temperature can be calculated from Eq.(5.6-23) as

Tm2 = Tm1 exp

"Ã∆eV fus

∆ eHfus

!(P2 − P1)

#

= 273 exp

∙µ− 1.63× 10−3

6012

¶¡1.198× 103 − 101.3

¢¸= 272.92K

The decrease in the melting temperature leads to liquefaction of water, which in turn acts as alubricating layer during skating. In the case of dry ice, on the other hand, dP/dTm is positiveand the melting temperature increases with increasing pressure. As a result, carbon dioxideremains as a solid.

Comment: Melting under pressure and refreezing again with a release of pressure is knownas "regelation". Another typical example of this phenomenon is the movement of a weightedwire through an ice block without cutting it into two pieces. The applied pressure lowers themelting temperature of ice and a film of water forms. As the wire moves downward, the releaseof pressure on the meltwater causes refreezing of the melted portion.

• ∆ eHfus varies with temperature

The variation of ∆ eHfus with temperature is expressed in the form

d∆ eHfus

dT=

d eHL

dT− d eHS

dT= eCL

P − eCSP = ∆ eCLS

P (5.6-24)


∆ eHfus =

Z Tm

Tref

∆ eCLSP dT +∆ eHfus

ref (5.6-25)

where ∆ eHfusref is the value of ∆ eHfus at the reference temperature, Tref . Once ∆ eHfus is

expressed as a function of melting temperature from Eq. (5.6-25), substitution of the resultingexpression into Eq. (5.6-21) and integration yield the relationship between pressure and meltingtemperature on the solid-liquid equilibrium curve.

If ∆ eCLSP is independent of temperature, then Eq. (5.6-25) reduces to

∆ eHfus = ∆ eCLSP (Tm − Tref ) +∆ eHfus

ref (5.6-26)

Substitution of Eq. (5.6-26) into Eq. (5.6-21) and integration yield

P = ∆ eCLSP Tm +

Ã∆ eHfus

ref −∆ eCLSP Tref

∆eV fus

!lnTm + C (5.6-27)

where C is an integration constant.

113

Example 5.11 The normal melting temperature of lead (atomic weight = 207) is 600.65Kand the enthalpy of fusion is 4812 J/mol. Estimate the melting temperatures of lead at 50, 100,150, and 200 bar using the following data:

ρL = 10, 215 kg/m3 ρS = 11, 350 kg/m3

eCLP = 32.4− 3.1× 10−3T eCS

P = 23.6 + 9.75× 10−3Twhere eCP is in J/mol.K and T is in K.

Solution

The difference between the heat capacities of liquid and solid is

∆ eCLSP = 8.8− 12.85× 10−3T (1)

Taking Tref = 600.65K and ∆ eHfusref = 4812 J/mol, substitution of Eq. (1) into Eq. (5.6-25)

gives∆ eHfus = 8.8T − 6.425× 10−3 T 2 + 1844.3 (2)

The volume change on melting, ∆eV fus, is given by

∆eV fus =

µ1

10, 215− 1

11, 350

¶207 = 2.0264× 10−3m3/ kmol = 2.0264× 10−6m3/mol (3)

Thus, substitution of Eqs. (2) and (3) into Eq. (5.6-21) and integration result in

(P − 1.013)× 105 = 1

2.0264× 10−6Z Tm

600.65

µ8.8T − 6.425× 10−3 T 2 + 1844.3

T

¶dT

Simplification gives

P = 43.427Tm + 9101.4 lnTm − 1.5853× 10−2 T 2m − 78, 594 (4)

Melting temperatures calculated at different pressures are given in the table below.

P ( bar) 50 100 150 200

Tm (K) 601.9 603.1 604.4 605.7

5.6.4 Solid-Vapor Equilibrium

The amount of heat required to convert solid to vapor is called the heat of sublimation, ∆ eHsub.In general, heat of sublimation is indirectly determined from heat of vaporization and heat offusion as

∆ eHsub = eHV − eHS = ( eHV − eHL) + ( eHL − eHS) = ∆ eHvap +∆ eHfus (5.6-28)

The volume change on sublimation is given by

∆eV sub = eV V − eV S ' eV V

=RT

P subif the vapor is an ideal gas (5.6-29)

114

Thus, the slope of the solid-vapor equilibrium line, i.e., Eq. (5.6-6), becomes

d lnP sub

dT=∆ eHsub

RT 2(5.6-30)

which is similar to Eq. (5.6-19). Since ∆ eHsub > ∆ eHvap, the slope of the sublimation curve issteeper than that of the vaporization curve.

If ∆ eHsub is independent of temperature, integration of Eq. (5.6-30) gives

lnP sub = − ∆eHsub

RT+ C (5.6-31)

The constant of integration in Eq. (5.6-31), C, can be evaluated using the triple point data as

C =∆ eHsub

RTt+ lnPt (5.6-32)

where Pt and Tt represent triple point pressure and temperature, respectively. Substitution ofEq. (5.6-32) into Eq. (5.6-31) gives

ln

µP sub

Pt

¶= − ∆

eHsub

R

µ1

T− 1

Tt

¶(5.6-33)

5.7 ANALYSIS OF PHASE DIAGRAMS USING STABILITY CRITERIA

It was stated in Section 4.3 that a substance, at a given temperature and pressure, is most stablein a phase in which its Gibbs energy is minimum. For example, under normal conditions, i.e.,298K and 1 bar (0.1MPa), the difference between the Gibbs energies of graphite and diamondis given as eGdiamond(298K, 10

5 Pa)− eGgraphite(298K, 105 Pa) = 2870 J/mol

Since eGgraphite < eGdiamond, graphite is the more stable phase under these conditions. If it isdesired to produce diamond from graphite at room temperature, what should be the minimumpressure imposed for such a transformation?

The starting point for answering such a question is the fact that spontaneous transitionfrom graphite to diamond takes place when the two phases are in equilibrium with each other,i.e., eGdiamond(298K, P ) = eGgraphite(298K, P ) (5.7-1)

At constant temperature, Eq. (5.1-1) gives the variation of Gibbs energy with pressure as

d eG = eV dP (5.7-2)

Therefore,

eGdiamond(298K, P ) = eGdiamond(298K, 105 Pa) + eVdiamond(P − 105) (5.7-3)

eGgraphite(298K, P ) = eGgraphite(298K, 105 Pa) + eVgraphite(P − 105) (5.7-4)

Substitution of Eqs. (5.7-3) and (5.7-4) into Eq. (5.7-1) and simplification lead to

P = 105 +eGdiamond(298K, 10

5 Pa)− eGgraphite(298K, 105 Pa)eVgraphite − eVdiamond

(5.7-5)

115

The densities of graphite and diamond are approximately 2300 and 3530 kg/m3, respectively.Thus, substitution of the numerical values into Eq. (5.7-5) gives

P = 105 +2870× 103µ1

2300− 1

3530

¶(12)

= 1.58× 109 Pa

Thus, at room temperature, graphite can be converted into diamond when P ≥ 1580MPa.

5.7.1 Gibbs-Helmholtz Equation

In some cases, it is much more convenient to determine the variation of eG/T with temperatureunder constant pressure. Mathematically speaking, we are interested in

∂

∂T

Ã eGT

!P

=1

T

Ã∂ eG∂T

!P

−eGT 2

(5.7-6)

From Eq. (5.1-1) Ã∂ eG∂T

!P

= − eS (5.7-7)

Substitution of Eq. (5.7-7) into Eq. (5.7-6) gives

∂

∂T

Ã eGT

!P

= −eHT 2

(5.7-8)

which is known as the Gibbs-Helmholtz equation.

Example 5.12 Estimate the normal melting temperature of lead using the following data:

P( bar)

T(K)

PhaseeH

( J/mol)

eS( J/mol.K)

1.013 400 Solid 2770 72.81.013 1200 Liquid 31, 160 113.1

The heat capacities at constant pressure are given by

eCSP = 23.6 + 9.75× 10−3T and eCL

P = 32.4− 3.1× 10−3T

where eCP is in J/mol.K and T is in K.

Solution

A representative phase diagram of lead is shown below. At the normal melting temperature,Tm, solid and liquid phases are in equilibrium with each other (State B), i.e.,

eGS¯̄̄Tm

Tm=

eGL¯̄̄Tm

Tm(1)

116

Tm

C B A

1200 K 400 K

1.013 bar

P

T

LS

V

At constant pressure, Eq. (5.7-8) is expressed in the form

d

Ã eGT

!= −

eHT 2

dT (2)

Integration of Eq. (2) requires the enthalpy to be expressed as a function of temperature, i.e.,

d eHdT

= eCP (3)

For the solid phase

d eHS

dT= 23.6 + 9.75× 10−3T ⇒ eHS = 23.6T + 4.875× 10−3 T 2 + C1

From the value of enthalpy at 400K, the integration constant C1 is evaluated as − 7450. ThuseHS = 23.6T + 4.875× 10−3 T 2 − 7450 (4)

Substitution of Eq. (4) into Eq. (2) and integration from 400K to Tm give

eGS¯̄̄Tm

Tm= 96.1− 7450

Tm− 23.6 lnTm − 4.875× 10−3 Tm (5)

For the liquid phase

d eHL

dT= 32.4− 3.1× 10−3T ⇒ eHL = 32.4T − 1.55× 10−3 T 2 + C2

Since the value of enthalpy at 1200K is known, the integration constant C2 is evaluated as− 5488. Thus, eHL = 32.4T − 1.55× 10−3 T 2 − 5488 (6)

Substitution of Eq. (6) into Eq. (2) and integration from Tm to 1200K give

eGL¯̄̄Tm

Tm= 145.3− 5488

Tm− 32.4 lnTm + 1.55× 10−3 Tm (7)

Substitution of Eqs. (5) and (7) into Eq. (1) and simplification lead to

49.2 +1962

Tm− 8.8 lnTm + 6.425× 10−3Tm = 0 (8)

117

The solution of Eq. (8) by MATHCADR°gives Tm = 602.3K.

Comment: The actual value of the normal melting temperature of lead is 600.65K as givenin Example 5.11.

5.8 VARIATION OF FUGACITY WITH PRESSURE AND TEMPERATURE

Variation of fugacity with pressure under isothermal conditions can be easily obtained fromEq. (5.2-7) as µ

∂ ln fi

∂P

¶T

=1

RT

Ã∂ eGi

∂P

!T| {z }eVi=

eViRT

(5.8-1)

On the other hand, differentiation of Eq. (5.2-1) with respect to temperature by keepingpressure constant gives µ

∂ ln fi

∂T

¶P

=1

R

∂

∂T

Ã eGi − eGIGi

T

!P

(5.8-2)

The use of the Gibbs-Helmholtz equation, Eq. (5.7-8), in Eq. (5.8-2) yields

µ∂ ln fi

∂T

¶P

= −eHi − eHIG

i

RT 2(5.8-3)

Note that the term ( eHi − eHIGi ) is simply the departure function for molar enthalpy.

Variation in ln fi is expressed in the form of an exact differential as

d ln fi =

µ∂ ln fi

∂P

¶T

dP +

µ∂ ln fi

∂T

¶P

dT (5.8-4)

Substitution of Eqs. (5.8-1) and (5.8-3) into Eq. (5.8-4) gives

d ln fi =

Ã eViRT

!dP −

Ã eHi − eHIGi

RT 2

!dT (5.8-5)

Example 5.13 Use Eq. (5.8-5) to express the fugacity of pure solid in terms of the fugacityof pure liquid.

Solution

For a liquid, Eq. (5.8-5) takes the form

d ln fLi =

Ã eV Li

RT

!dP −

Ã eHLi − eHIG

i

RT 2

!dT (1)

For a solid, on the other hand, Eq. (5.8-5) becomes

d ln fSi =

Ã eV Si

RT

!dP −

Ã eHSi − eHIG

i

RT 2

!dT (2)

118

Subtraction of Eq. (1) from Eq. (2) gives

d ln

ÃfSi

fLi

!= −

∆eV fusi

RTdP +

∆ eHfusi

RT 2dT (3)

Integration of Eq. (3) from the normal melting temperature, Tmi, to T , and from 1 atm ( ˜ 1 bar)to P , under the assumption that the heat of fusion is a constant leads to

ln

ÃfSi

fLi

!= −

∆eV fusi (P − 1)RT

+∆ eHfus

i

RTmi

µ1− Tmi

T

¶(4)

Since the difference between the liquid and solid molar volumes is usually negligible, it is plau-sible to neglect the first term on the right-hand side of Eq. (4) without loss of accuracy. Hence,Eq. (4) simplifies to

ln

ÃfSi

fLi

!=∆ eHfus

i

RTmi

µ1− Tmi

T

¶(5)

Comment: Equation (5) will be derived in Section 12.1 using a different approach.

REFERENCES

Bignell, N. and V.E. Bean, 1988, Metrologia, 25, 141-145.

Brunetti, B., V. Piacente and G. Portalone, 2000, J. Chem. Eng. Data, 45, 242-246.

Douglas, T.B., 1948, J. Am. Chem. Soc., 70 (6), 2001-2002.

Holmes, M.J. and M. van Winkle, 1970, Ind. Eng. Chem., 62 (1), 21-31.

Lee, B. and W.C. Edmister, 1971, Ind. Eng. Chem. Fundam., 10 (2), 229-236.

Rittmann, B., H. Knapp and J.M. Prausnitz, 1982, Ind. Eng. Chem. Process Des. Dev., 21,695-698.

Spencer, C.F. and R.P. Danner, 1972, J. Chem. Eng. Data, 17, 236-241.

Stein, S.E. and R.L. Brown, 1994, J. Chem. Inf. Comput. Sci., 34, 581-587.

Trouton, F.T., 1883, Nature, 27, 292.

PROBLEMS

Problems related to Section 5.3

5.1 Derive Eq. (5.3-9) by substituting the van der Waals equation of state into Eq. (5.3-8).

5.2 Use Eq. (5.3-2) and calculate the fugacity coefficient of dimethyl ether at 420K and50 bar using the following data:

P( bar)

1 5 10 15 20 25 30 35 40 45 50

Z 0.994 0.969 0.937 0.905 0.871 0.836 0.799 0.761 0.721 0.678 0.631

(Answer: 0.72)

119

5.3 Calculate the fugacity of propylene at 400K and 50 bar if it obeys

a) van der Waals equation of state,b) Redlich-Kwong equation of state,c) Peng-Robinson equation of state.(Answer: a) 37.207 bar b) 36.597 bar c) 35.741 bar)

5.4 Calculate the fugacity of propane at 390K and 30 bar if it obeys

a) Soave-Redlich-Kwong equation of state,b) Peng-Robinson equation of state.(Answer: a) 24.141 bar b) 23.558 bar)

5.5 Using the following data for the fugacity coefficient of isobutane as a function of pressureat 500K, estimate its molar volume at 500K and 75 bar.

P ( bar) φC4H10 P ( bar) φC4H10 P ( bar) φC4H10

0 1.000 40 0.803 80 0.65310 0.947 50 0.761 90 0.62420 0.896 60 0.722 100 0.59930 0.848 70 0.686

(Answer: 349.3 cm3/mol)

5.6 In Example 5.2, Eq. (3) indicates that the change in Gibbs energy under isothermalconditions is given by eG2 − eG1 = RT ln

µf2f1

¶(1)

Combine Eq. (1) with Eq. (2.1-5) to show that

(fWs)rev = RT ln

µf2f1

¶(2)

which can be used to estimate the work interaction of a steady-flow system under reversibleand isothermal conditions.

Carbon dioxide is compressed in a reversible isothermal steady-state flow process from 3 barand 350K to 70 bar. Determine the work of compression per mole of carbon dioxide using thePeng-Robinson equation of state.

(Answer: 8509 J/mol)


5.7 Estimate the fugacity of 1-octene at 480K and 12 bar if it obeys the virial equation ofstate. The vapor pressure of 1-octene at 480K is 6.791 bar.

(Answer: 5.919 bar)

5.8 Estimate the pressure at which the fugacity of liquid ethanol at 293K is 5% higher thanthe fugacity of liquid at its saturation pressure.

(Answer: 23.5 bar)

120

5.9 Use the condition of vapor-liquid equilibrium and estimate the temperature of saturatedliquid water at 4.5 bar by using the Peng-Robinson equation of state. Compare the result withthe one obtained from the following equation:

lnP vapH2O = 13.8365−

6.6075× 103T

+10.1855× 105

T 2− 17.7039× 10

7

T 3300 ≤ T ≤ 600

in which P vapH2O is in bar and T is in K.

(Answer: 421.56K)

5.10 Estimate the vapor pressure of ethylene at 240K using the van der Waals equation ofstate.

(Answer: 25.5 bar)

5.11 Estimate the vapor pressure of liquid methane at 150K using the Redlich-Kwong equa-tion of state.

(Answer: 10.1 bar)

Problem related to Section 5.5

5.12 Pure A exists in solid-vapor equilibrium in a rigid container at 150K and 10 bar. Cal-culate the fugacity of pure solid if the vapor phase is described by the virial equation of state,i.e.,

Z = 1 +BP

RT

where B = 45× 10−5m3/mol.(Answer: 14.35 bar)


5.13 Consider transformation of saturated liquid to saturated vapor at constant temperatureand pressure by the addition of heat.

a) Show the process on a P -eV diagram.b) Show the process on a P -T diagram.c) Which plot is more informative for the given process? Explain.

5.14 A portion of the phase diagram of sulfur is shown below. There are two solid phases ofsulfur, rhombic and monoclinic. The standard Gibbs energy of formation of rhombic sulfur is0 kJ/mol and that of monoclinic sulfur is 0.33 kJ/mol at 25 ◦C.

a) On the phase diagram, label the four regions and indicate the triple points.b) Is it possible to completely melt rhombic sulfur at 200 ◦C? If no, explain the reason clearly.If yes, what pressure should be imposed on rhombic sulfur?c) It is required to convert rhombic sulfur to monoclinic sulfur and your friend suggests thefollowing options:

i) Heat rhombic sulfur at 1 atm,ii) Heat rhombic sulfur at 1× 10−6 atm,iii) Heat rhombic sulfur at 9× 103 atmiv) Decrease the pressure on rhombic sulfur at 65 ◦C,v) Decrease the pressure on rhombic sulfur at 140 ◦C,vi) Decrease the pressure on rhombic sulfur at 100 ◦C.

121

Which one(s) of these processes is possible?

1

104

103

102

101

10-1

10-2

10-3

10-4

10-5

10-6

20018016014012010080 60 40

Temperature (oC)

Pre

ssur

e (a

tm)

5.15 Trouton’s rule (Trouton, 1883) states that the entropy of vaporization of liquids at theirnormal boiling temperature, Tnb, is constant, i.e.,

∆eSvap =∆ eHvap

Tnb= 88J/mol.K (1)

It should be kept in mind that Trouton’s rule is an empirical relation and is not valid forhydrogen-bonded liquids.

a) Show that the combination of Trouton’s rule with the Clausius-Clapeyron equation leads to

P vap2 = P vap

1

µT2T1

¶10.6(2)

b) The normal boiling temperature of acetone is 56.5 ◦C. Calculate the vapor pressure ofacetone at various temperatures using Eq. (2) and compare the values with the ones given inthe literature.

5.16 An improved version of Trouton’s rule, known as the Trouton-Hildebrand-Everett rule,is given by

∆eSvap =∆ eHvap

Tnb= R (4 + lnTnb) (1)

The vapor pressure of chloroform at 323.15K is 0.703 bar. Use Eq. (1) to estimate the normalboiling temperature of chloroform. The experimental value is 334.3K.

(Answer: 335.2K)

5.17 As stated in Problem 3.13, an isotherm representing the cubic equation of state satisfiesthe following equation:

P vap(eV V − eV L) =

Z eV V

eV L

P deV (1)

a) Differentiate Eq. (1) with respect to temperature to obtain

dP vap

dT(eV V − eV L) =

Z eV V

eV L

µ∂P

∂T

¶eV deV (2)

Hint: Use Leibnitz’s rule to differentiate the right-hand side of Eq. (1).

122

b) Combine Eq. (2) with the Clapeyron equation and show that

∆ eHvap = T

Z eV V

eV L

µ∂P

∂T

¶eV deV (3)

c) For the van der Waals equation of state, show that the integration of the right-hand side ofEq. (3) leads to

∆ eHvap

RT= ln

Ã eZV −BeZL −B

!(4)

d) For the Redlich-Kwong equation of state, show that the integration of the right-hand sideof Eq. (3) leads to

∆ eHvap

RT= ln

Ã eZV −BeZL −B

!+

A

2Bln

" eZV ( eZL +B)eZL( eZV +B)

#(5)

e) Estimate ∆ eHvap for saturated benzene at 298.15K if it obeys the Redlich-Kwong equationof state.

(Answer: e) 32.65 kJ/mol)

5.18 Consider vapor-liquid equilibrium of a pure component.

a) Use P eV = ZRT and show that Eq. (5.6-6) takes the form

dP vap

P vap=∆ eHvap

R∆Zvap

dT

T 2(1)

where ∆Zvap = ZV − ZL.

b) Assume ∆ eHvap/∆Zvap to be a constant and show that the integration of Eq. (1) from thecritical pressure, Pc, to any pressure, P vap, leads to

log

µP vap

Pc

¶=

∆ eHvap

2.303R∆ZvapTc

µ1− 1

Tr

¶(2)

c) Use Eq. (3.1-14) and show that

∆ eHvap

2.303R∆ZvapTc=7

3(ω + 1) (3)

d) Combine Eqs. (2) and (3) to get

log

µP vap

Pc

¶=7

3(1 + ω)

µ1− 1

Tr

¶(4)

which can be used to estimate vapor pressures of liquids when Tr ≥ 0.6.e) Estimate the vapor pressure of carbon dioxide at 273.16K using Eq. (4) and compare itwith the literature value of 34.86 bar (Bignell and Bean, 1988).

5.19 Calculate the normal boiling temperature of n-propanol by the following two methods.

a) Use vapor pressure data in the form of the Antoine equation.

b) One way of estimating chemical and physical properties of pure compounds is the so-called group contribution method, which is based on the molecular structure of a compound.

123

The property in question is estimated by the additive contributions of the functional groupsconstituting the molecule. To estimate normal boiling temperatures, Tnb, of pure compounds,Stein and Brown (1994) proposed the following formula:

Tnb(K) = 198.2 +Xi

ni gi (1)

where ni is the number of groups of type i in the molecule and gi is the contribution of eachgroup. The normal boiling temperature predicted by Eq. (1) must be corrected by the followingequations:

Tnb(corrected) =½1.5577Tnb − 94.84− 7.705× 10−4T 2nb Tnb ≤ 700K0.4791Tnb + 282.7 Tnb > 700K

(2)

Take the gi values of 21.98, 24.22, and 88.46 for the CH3, CH2, and primary-OH groups.

(Answer: a) 370.35K b) 363.14K)

5.20 As shown in Problem 4.1, the hydrostatic pressure distribution in fluids can be calculatedfrom the equation

dP

dz= − ρg (1)

where the distance z is measured in the direction opposite to gravity.The highest point on the earth’s surface is the top of Mount Everest, located in the Hi-

malayas on the border of Nepal and China. It is approximately 8900m above sea level. CombineEq. (1) with the Clausius-Clapeyron equation, Eq. (5.6-10), to estimate the boiling tempera-ture of water at the top of Mount Everest. The following data are provided:

• The average rate of decrease in air temperature with altitude is 6.5K/ km,• The air temperature at sea level is 293K,• The heat of vaporization of water is 40, 800 J/mol and it may be considered independent oftemperature,• The molecular weight of air is 29 g/mol.(Answer: 342.9K)

5.21 Based on Eq. (5.6-12), it is possible to express the vapor pressure of substances as afunction of temperature in the form

logP vapr = a− b

Tr(1)

which is valid from the triple point to the critical point. For monatomic gases (Ar, Kr, Xe),Pitzer et al. (1955) plotted logP vap

r versus 1/Tr and noted that logPvapr takes the value of − 1

at Tr = 0.7 (or 1/Tr = 1.43).

a) Using the following data for n-pentane, plot logP vapr versus 1/Tr.

T (K) 298 323 348 373 398 423 448

P vap ( bar) 0.6833 1.5923 3.2340 5.9447 9.9109 15.5454 23.0773

b) Using the method of least squares show that the data are represented by

logP vapr = 2.883− 2.894

Tr(2)

124

c) Use Eq. (3.1-14) and calculate the acentric factor for n-pentane.

5.22 Under atmospheric pressure, the heat of fusion and the melting temperature of p-xyleneare 17, 127 J/mol and 286.4K, respectively. Determine the entropy change on fusion.

(Answer: 59.8 J/mol.K)

5.23 The vapor pressure of n-heptanol is 84.398 kPa at 443.15K and the normal boilingtemperature is 448K. Estimate the heat of vaporization of n-heptanol.

(Answer: 62, 208 J/mol)

5.24 The vapor pressure of methyl sulfoxide is correlated by Douglas (1948) as a function oftemperature in the form

logP vap = 26.5− 3539.32T

− 6 logT 293.15 ≤ T ≤ 323.15

where P vap is in mmHg and T is in K. Estimate the heat of vaporization of methyl sulfoxideat 298.15K.

(Answer: 52.9 kJ/mol)

5.25 Uracil (C4H4N2O2) is an organic base of the pyrimidine family. The sublimation pressureof uracil as a function of temperature is given below (Brunetti et al., 2000)

T (K) P ( Pa) T (K) P ( Pa)

397.0 4.365× 10−2 428.0 5.370× 10−1404.0 8.128× 10−2 431.0 7.244× 10−1409.0 1.260× 10−1 435.5 1.047416.0 2.188× 10−1 440.0 1.479419.5 2.951× 10−1 443.0 1.950

424.0 4.266× 10−1

Calculate the heat of sublimation of uracil.


5.26 The following data are available for iodine

ρliquid = 4000 kg/m3 ρsolid = 4930 kg/m

3

lnP vapliquid = 47.83−

7381

T− 5.18 lnT lnP vap

solid = 34.16−8240

T− 2.51 lnT

where P vap is in atm and T is in K.

a) Calculate the values of temperature and pressure at the triple point and plot a representativephase diagram for iodine.

b)When iodine vapor, initially at 0.06 atm and 430K, is cooled at constant pressure, determinethe temperature at which the condensed phase first appears? What is this condensed phase?

c)When iodine vapor, initially at 0.21 atm and 430K, is cooled at constant pressure, determinethe temperature at which the condensed phase first appears? What is this condensed phase?

d) Calculate the heat of fusion of iodine at the triple point.e) Calculate the normal boiling temperature of iodine.(Answer: a) 385.5K, 0.115 atm b) 372.6K c) 403K d) 15, 700 J/mol e) 459K)

125

5.27 Using the following data provided for n-hexanol (C6H14O)

• Normal boiling temperature = 430K• Normal melting temperature = 226K• ∆ eHvap = 44.5 kJ/mol• ∆ eHfus = 15.48 kJ/mol

estimate its vapor pressure at

a) 200K,b) 415K.(Answer: a) 2.08× 10−7 atm b) 0.638 atm)

5.28 The vapor pressure of phenol as a function of temperature is given by

lnP vap = − 11.04 lnT − 10, 000T

+ 93.275 + 4.32× 10−6T 2

where P vap is in kPa and T is in K.

a) Calculate the heat of vaporization at the triple point temperature of 314.06K.b) If the heat of fusion is 11, 514 J/mol, show that the sublimation pressure of phenol isexpressed as

lnP sub = 31.34− 8181T

where P sub is in Pa and T is in K.

(Answer: a) 56, 501 J/mol)

5.29 The sublimation pressure of solid benzene is 299Pa at 243K. The vapor pressure ofliquid benzene is 6022Pa at 283K. If the triple point of benzene is at 278.5K and 4786Pa,estimate its heat of fusion.


5.30 One gram of copper is placed in a rigid tank having a volume of 1m3. Once the air isevacuated, the tank is sealed and the temperature is increased to 1600K. The melting pointof copper is 1356K and the vapor pressure of liquid copper is given by

lnP vap = − 39, 129T

+ 14.203

where P vap is in atm and T is in K.

a) How many phases are present within the tank?b) Estimate the heat of vaporization of copper.(Answer: a) 2 (liquid and vapor) b) 325.3 kJ/mol)

5.31 Using the Peng-Robinson equation of state, estimate the maximum amount of propanethat can be stored as vapor in a 0.1m3 tank at 250K.

(Answer: 0.5 kg)

5.32 What should be the minimum pressure inside the container of a disposable butanelighter at a temperature of 298K? Assume that n-butane is represented by the Peng-Robinsonequation of state.

(Answer: 2.43 bar)

126

5.33 At what temperature does the pure methane liquefy under atmospheric conditions?Assume that methane is represented by the Peng-Robinson equation of state.

(Answer: 111.55K)

5.34 Propylene is to be transported through a pipeline as a liquid. If the maximum temper-ature expected in the summer is around 330K, estimate the minimum pressure that should beimposed in the pipeline. Assume that propylene is represented by the Peng-Robinson equationof state.

(Answer: 23.6 bar)

5.35 Determine the state of methanol, i.e., liquid or vapor, under the following conditions:

a) T = 550K and P = 2000 barb) T = 550K and P = 1barc) T = 350K and P = 2bard) T = 350K and P = 1bar

5.36 A piston-cylinder assembly initially contains 4 moles of substance X in vapor phase at0.8 bar and 373K. The vapor is compressed isothermally and reversibly until the first dropletof liquid forms. Calculate the heat and work interactions with the surroundings if the vaporobeys the virial equation of state with B = − 1887 cm3/mol.

The normal boiling temperature and the heat of vaporization of substance X are 354.8K and29, 750 J/mol, respectively.

(Answer: Q = − 9032 JW = 9032 J)

5.37 When common gas-liquid-solid transitions take place, while temperature remains con-stant during vaporization, condensation, and sublimation, volume and entropy show abruptchange. Such transformations are called "first-order transitions". Noting thatµ

∂G

∂P

¶T

= V andµ∂G

∂T

¶P

= −S

the derivatives of Gibbs energy are discontinuous for a first-order transition. SinceH = G+TS,enthalpy is also discontinuous at the transition temperature.

A "second-order transition", on the other hand, is defined as one in which the first deriva-tive of Gibbs energy is continuous but the second derivative is discontinuous at the transitiontemperature. In other words, quantities such as volume, entropy, and enthalpy remain un-changed during phase change. Transition of a paramagnetic material into a ferromagnetic oneor vice versa is an example of a second-order transition.

How does the heat capacity at constant pressure vary during the first-order and second-ordertransitions?

Problem related to Section 5.8

5.38 Lee and Edmister (1971) proposed the following empirical equation for predicting thefugacity of pure liquid hydrocarbons:

ln

µf

P

¶= A1+

A2Tr+A3 lnTr+A4 T

2r +A5 T

6r +

µA6Tr+A7 lnTr +A8T

2r

¶Pr+A9T

3r P

2r −lnPr

+ ω

∙(1− Tr)

µA10 +

A11Tr

¶+A12

PrTr+A13T

3r P

2r

¸

127

whereA1 = 6.32873 A6 = − 0.018706 A11 = − 11.201A2 = − 8.45167 A7 = − 0.286517 A12 = − 0.05044A3 = − 6.90287 A8 = 0.18940 A13 = 0.002255A4 = 1.87895 A9 = − 0.002584A5 = − 0.33448 A10 = 8.7015

Estimate the specific enthalpy of liquid n-pentane at 433K and 18.5 bar relative to that of theideal gas at the same temperature.

(Answer: − 264.5 J/ g)

128

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