chapter 3

c° 2011 Ismail Tosun

Chapter 3

Calculation of Changes in Internal Energy,Enthalpy, and Entropy

In the previous chapter, general expressions for calculating changes in internal energy, enthalpy,and entropy are developed. These expressions contain partial derivatives involving temperature,pressure, and molar volume. The purpose of this chapter is to show how to evaluate thesederivatives in a systematic manner.

3.1 EQUATIONS OF STATE

Any mathematical relationship between temperature, pressure, and molar volume is called anequation of state, i.e.,

f(T, P, eV ) = 0 (3.1-1)

Equations of state can be expressed either in pressure-explicit form

P = P (T, eV ) (3.1-2)

or in volume-explicit form eV = eV (T,P ) (3.1-3)

Moreover, an equation of state can also be expressed in terms of the dimensionless compress-ibility factor, Z, as

Z =P eVRT

(3.1-4)

3.1.1 Ideal Gas Equation of State

The ideal gas model is dependent on the following assumptions:

• Molecules occupy no volume,• Collisions of the molecules are totally elastic, i.e., energy of the molecules before a collision isequal to the energy of the molecules after a collision. In other words, there are no interactionsbetween the molecules.

The equation of state for an ideal gas is given by

P eV = RT (3.1-5)

Since Z = 1 for an ideal gas, the compressibility factor shows the deviation from ideal behavior.A representative plot of the compressibility factor as a function of pressure with temperatureas a parameter is shown in Figure 3.1.

29

1.0

1.5

Z

P

T2

0.5

T3

T1

T1>T2>T3

Ideal gas

Figure 3.1 Variation of Z as a function of P and T .

3.1.2 The Virial Equation of State

The virial equation of state is useful for calculating thermodynamic properties in the gas phase.It can be derived from statistical mechanics and is given by an infinite series in molar volume,eV , as

Z =P eVRT

= 1 +B(T )eV +

C(T )eV 2 + ... (3.1-6)

The coefficient, B(T ), is called the second virial coefficient, C(T ) is called the third virialcoefficient, and so on. All virial coefficients are dependent on temperature. In practice, termsabove the third virial coefficient are rarely used in chemical thermodynamics.

An equivalent form of the virial expansion is an infinite series in pressure expressed as

Z = 1 +B0(T )P + C 0(T )P 2 + ... (3.1-7)

The coefficients B0 and C 0 can be expressed in terms of B and C as

B0 =B

RTand C 0 =

C −B2

(RT )2(3.1-8)

In practice, it is recommended to consider only the second virial coefficient for pressures up to15 bar. Under these circumstances, Eq. (3.1-7) takes the form

Z = 1 +BP

RT(3.1-9)

Van Ness and Abbott (1982) proposed the following equation to estimate the second virialcoefficient for nonpolar fluids:

B =RTcPc

hB(0) + ωB(1)

i(3.1-10)

30

whereB(0) = 0.083− 0.422

T 1.6r

(3.1-11)

B(1) = 0.139− 0.172T 4.2r

(3.1-12)

The terms Tr, reduced temperature, Pr, reduced pressure, and ω, acentric factor, are defined by

Tr =T

Tcand Pr =

P

Pc(3.1-13)

ω = − 1.0− log(P vapr |Tr=0.7) (3.1-14)

The definition of the acentric factor is based on the observation that logP vapr versus 1/Tr is

approximately a straight line (Pitzer et al., 1955)1. For noble gases (Ar, Kr, Xe) logP vapr = −1

at 1/Tr = 1/0.7 ' 1.43, indicating that ω = 0. With reference to Figure 3.2, Pitzer (1977)explained the conceptual basis of the acentric factor as follows:

"Intermolecular forces operate between the centers of regions of substantial elec-tron density. These centers are the molecular centers for Ar and (approximately)for CH4, but are best approximated by the separate CH3 and CH2 groups in C3H8- hence the name acentric factor for the forces arising from points other than mole-cular centers."

Figure 3.2 The conceptual basis of the acentric factor.

Example 3.1 Calculate the molar volume of ethylene at 350K and 10 bar if it obeys thevirial equation of state.

Solution

From Appendix ATc = 282.5K Pc = 50.6 bar ω = 0.089

1See Problem 5.21.

31

The reduced temperature and pressure values are

Tr =350

282.5= 1.239 Pr =

10

50.6= 0.198

The terms B(0) and B(1) are calculated from Eqs. (3.1-11) and (3.1-12) as

B(0) = 0.083− 0.422

(1.239)1.6= − 0.217

B(1) = 0.139− 0.172

(1.239)4.2= 0.069

The use of Eq. (3.1-10) in Eq. (3.1-9) leads to

Z = 1 +PrTr

hB(0) + ωB(1)

i(1)

Substitution of numerical values into Eq. (1) results in

Z = 1 +0.198

1.239

h− 0.217 + (0.089)(0.069)

i= 0.966

Therefore, the molar volume is

eV =ZRT

P=(0.966)(83.14)(350)

10= 2811 cm3/mol

3.1.3 Cubic Equations of State

The cubic equations of state are expressed in pressure-explicit form, i.e., P = P (eV , T ), someof which are given in Table 3.1. The cubic equations of state have been extensively used overthe last three decades2, Soave-Redlich-Kwong and Peng-Robinson3 being the most popularones. The reason for this is twofold: (i) they are applicable over a wide range of pressures andtemperatures, (ii) they are capable of describing substances in both liquid and vapor phases.

Table 3.1 Cubic equations of state.

Equation P a b

van der WaalsRTeV − b

− aeV 2 27

64

ÃR2T 2c

Pc

!1

8

µRTcPc

¶

Redlich-KwongRTeV − b

− aeV (eV + b)√T

0.42748

ÃR2T 2.5c

Pc

!0.08664

µRTcPc

¶

Soave-Redlich-Kwong

RTeV − b− aeV (eV + b)

0.42748

ÃR2T 2c

Pc

!α 0.08664

µRTcPc

¶

Peng-RobinsonRTeV − b

− aeV (eV + b) + b(eV − b)0.45724

ÃR2T 2c

Pc

!α 0.07780

µRTcPc

¶2For a comprehensive review on the state of the cubic equations of state, see Valderrama (2003).3Peng-Robinson equation of state is one of the most widely used correlations in chemical engineering. To

celebrate the twentieth anniversary of its publication, a symposium was organized in Edmonton, Canada, in 1996.Papers presented at the symposium were published in a special issue of Industrial and Engineering ChemistryResearch (May 1998).

32

These equations are called "cubic" because they are of the third degree in molar volume,i.e., eV 3 + c1 eV 2 + c2 eV + c3 = 0 (3.1-15)

In each equation given in Table 3.1, the first term is identical, i.e., RT/(eV − b), and accountsfor the repulsive forces between the molecules. The second term, on the other hand, is differentin each equation and accounts for the attractive forces between the molecules. The parameters"a" and "b" are representative of attractive forces and volume of a molecule, respectively4.Note that the parameter "a" in the van der Waals and Redlich-Kwong equations of state is aconstant. In the case of Soave-Redlich-Kwong and Peng-Robinson equations of state, however,it is dependent on temperature through the term α defined by

α =

⎧⎪⎨⎪⎩£1 +

¡0.480 + 1.574ω − 0.176ω2

¢ ¡1−√Tr¢¤2

Soave-Redlich-Kwong

£1 +

¡0.37464 + 1.54226ω − 0.26992ω2

¢ ¡1−√Tr¢¤2

Peng-Robinson

(3.1-16)

A typical isotherm of a cubic equation of state is shown in Figure 3.3-a. The horizontalline joining the points L and V is drawn so that the areas N and M are equal, known as theMaxwell equal area rule. The points L and V represent the saturated liquid and saturatedvapor, respectively, at the given temperature T . The intersection of the horizontal line onthe pressure axis gives the vapor (or saturation) pressure5 of the substance at T . When thisprocedure is repeated for other isotherms as well, the resulting pressure versus specific volumediagram is shown in Figure 3.3-b. Vapor and liquid phases exist in equilibrium within thedome-shaped region. The critical isotherm is tangent and has a point of inflection at thecritical point. The resulting mathematical expressions, i.e.,µ

∂P

∂ eV¶Tc

=

µ∂2P

∂ eV 2¶Tc

= 0 (3.1-17)

are used to evaluate the parameters "a" and "b" appearing in the cubic equations of state givenin Table 3.1.

P

V̂

VAPOR-LIQUID REGION

T1

T2

TcPvap

P

VV̂ LV̂ V̂

V L

N

M

T

(a) (b)

Critical Point

Figure 3.3 Pressure versus specific volume diagram for a pure substance.

At a given temperature and pressure, the solution of Eq. (3.1-15) gives three roots. WhenT > Tc, the only meaningful root is the one greater than b, the other two being complex

4The value of eV must be greater than b. Why?5Vapor pressure can be thought of a measure of the escaping tendency of a pure substance.

33

conjugate numbers. On the other hand, when T < Tc, depending on the value of pressure wecan conclude the following: (i) P = P vap, the liquid and vapor phases are in equilibrium, thelargest and the smallest roots correspond to the molar volumes of the saturated vapor andliquid phases, respectively; the intermediate root has no physical meaning, (ii) P < P vap, thefluid is superheated vapor and the largest root gives its molar volume, (iii) P > P vap, the fluidis compressed liquid and the smallest root gives its molar volume.

It is also possible to express Eq. (3.1-15) in terms of the compressibility factor Z as

Z3 + pZ2 + q Z + r = 0 (3.1-18)

where the coefficients p, q, and r are given in Table 3.2. Since the value of Z generally liesbetween 0 and 1, Eq. (3.1-18) is preferred over Eq. (3.1-15).

Table 3.2 Parameters in Eq. (3.1-18)6.

Eq. ofState

A B p q r

VDW27

64

ÃPr

T2r

!1

8

µPr

Tr

¶− 1−B A −AB

RK 0.42748

ÃPr

T2.5r

!0.08664

µPr

Tr

¶− 1 A−B −B2 −AB

SRK 0.42748

ÃPr

T2r

!α 0.08664

µPr

Tr

¶− 1 A−B −B2 −AB

PR 0.45724

ÃPr

T2r

!α 0.07780

µPr

Tr

¶− 1 +B A− 2B − 3B2 −AB +B2 +B3

In engineering calculations, it is always more convenient to work with dimensionless quan-tities. For this reason, in Table 3.2, the parameters a and b are expressed in terms of thedimensionless parameters A and B defined by

A =

⎧⎪⎪⎪⎨⎪⎪⎪⎩aP

(RT )2van der Waals, Soave-Redlich-Kwong, Peng-Robinson

aP

R2T 2.5Redlich-Kwong

(3.1-19)

B =bP

RT(3.1-20)

Example 3.2 Estimate the molar volume of methane at 340K and 30 bar using the Peng-Robinson equation of state.

6VDW - van der Waals, RK - Redlich-Kwong, SRK - Soave-Redlich-Kwong, PR - Peng-Robinson

34

Solution

An algorithm to obtain the molar volume is as follows:

1. Obtain Tc, Pc, and ω from Appendix A,2. Calculate reduced temperature and reduced pressure,3. Use Eq. (3.1-16) to determine the term α,4. Determine the dimensionless parameters A and B (Table 3.2),5. Evaluate the parameters p, q, and r appearing in Eq. (3.1-18) (Table 3.2),6. Solve Eq. (3.1-18), i.e., Z3 + pZ2 + q Z + r = 0,7. Use eV = ZRT/P and determine the molar volume.


The reduced temperature is

Tr =T

Tc=

340

190.6= 1.784

The reduced pressure is

Pr =P

Pc=

30

46.1= 0.651

The use Eq. (3.1-16) gives the term α as

α =n1 +

h0.37464 + (1.54226)(0.011)− (0.26992)(0.011)2

i ³1−√1.784

´o2= 0.754

The dimensionless parameters A and B are calculated as

A = 0.45724

ÃPr

T2r

!α =

(0.45724)(0.651)(0.754)

(1.784)2= 0.071

B = 0.07780

µPr

Tr

¶=(0.07780)(0.651)

1.784= 0.028

The parameters p, q, and r are calculated as

p = − 1 +B = − 1 + 0.028 = − 0.972q = A− 2B − 3B2 = 0.071− (2)(0.028)− (3)(0.028)2 = 0.011r = −AB +B2 +B3 = − (0.071)(0.028) + (0.028)2 + (0.028)3 = − 1.182× 10−3

Equation (3.1-18) takes the form

Z3 − 0.972Z2 + 0.011Z − 1.182× 10−3 = 0 ⇒ Z = 0.962

Therefore, the molar volume is given by

eV =ZRT

P=(0.962)(83.14)(340)

30= 906.4 cm3/mol

Example 3.3 A 1m3 rigid tank contains 25 kg of propylene at 298K. Determine the pressureand the state of propylene in the tank using the Peng-Robinson equation of state. The vaporpressure of propylene at 298K is 11.53 bar.

35

Solution

From Appendix A

Tc = 365.2K Pc = 46.0 bar ω = 0.144 M = 42.08 g/mol

The state of a single-phase, single-component system can be specified by two independent in-tensive properties. The state of propylene lies on the 298K isotherm as shown in the figurebelow. To identify the exact location, one more independent intensive property must be knownbesides temperature.

V% VV% LV%

T = 298 K

Pvap

P

Since the tank volume and the amount of propylene are given, this additional intensive propertyis the molar volume, calculated as

eV =1

25

42.08

= 1.683m3/ kmol

To determine the state of propylene, this value must be compared with the molar volumes ofsaturated liquid and vapor. To determine eV L and eV V at 298K and 11.53 bar, it is firstnecessary to calculate ZL and ZV . The reduced temperature and pressure values are

Tr =T

Tc=

298

365.2= 0.816 and Pr =

P

Pc=11.53

46= 0.251

The use Eq. (3.1-16) gives the term α as

α =n1 +

h0.37464 + (1.54226)(0.144)− (0.26992)(0.144)2

i ³1−√0.816

´o2= 1.118

The dimensionless parameters A and B are calculated as

A = 0.45724

ÃPr

T2r

!α =

(0.45724)(0.251)(1.118)

(0.816)2= 0.193

B = 0.07780

µPr

Tr

¶=(0.07780)(0.251)

0.816= 0.024

The cubic equation in terms of the compressibility factor, Eq. (3.1-18), takes the form

Z3 − 0.976Z2 + 0.143Z − 4.042× 10−3 = 0

The solution givesZL = 0.037 and ZV = 0.804

Therefore, molar volumes of the saturated liquid and vapor are

eV L =ZLRT

P=(0.037)(8.314× 10−2)(298)

11.53= 0.080m3/ kmol

36

eV V =ZVRT

P=(0.804)(8.314× 10−2)(298)

11.53= 1.728m3/ kmol

Since eV L < eV < eV V , propylene exists as a two-phase mixture and the pressure within the tankis 11.53 bar. The mole fraction of vapor in the tank, x, is given by

x =eV − eV LeV V − eV L

=1.683− 0.0801.728− 0.080 = 0.97

3.2 CALCULATION OF THE CHANGE IN INTERNAL ENERGY

In the previous chapter, the general equation to calculate the change in internal energy wasgiven by Eq. (2.2-26), i.e.,

deU = eCV dT +

∙T

µ∂P

∂T

¶eV − P

¸deV (3.2-1)

This equation is valid for gases, liquids, and solids.

3.2.1 Change in Internal Energy for Liquids and Solids

For solids and liquids, it is more convenient to express Eq. (3.2-1) in terms of the coefficientof thermal expansion, β, and isothermal compressibility, κ. As shown in Example 2.1, the useof the triple product rule leads to µ

∂P

∂T

¶eV =

β

κ(3.2-2)

Substitution of Eq. (3.2-2) into Eq. (3.2-1) and using eCV ' eCP give7

deU = eCP dT +

µβT

κ− P

¶deV (3.2-3)

The variation of internal energy with pressure at constant temperature isÃ∂ eU∂P

!T

=

µβT

κ− P

¶Ã∂ eV∂P

!T

(3.2-4)

The use of Eq. (2.2-54) in Eq. (3.2-4) results inÃ∂ eU∂P

!T

= eV (κP − βT ) (3.2-5)

3.2.2 Change in Internal Energy for Gases

The term (∂P/∂T )eV in Eq. (3.2-1) can be evaluated once the equation of state is known. Itshould be noted, however, that the heat capacity data are available only at ideal gas conditions,i.e., as either P → 0 or eV →∞. Therefore, it is necessary to develop an equation relating theactual value of eCV to an ideal one, eC∗V . Since eU is a state function, i.e., exact differential, theuse of Eq. (1.2-2) in Eq. (3.2-1) leads toÃ

∂ eCV

∂ eV!T

=

½∂

∂T

∙T

µ∂P

∂T

¶eV − P

¸¾eV (3.2-6)

7As shown in Problem 2.3, eCV ' eCP if Tβ2/(ρ bCPκ)¿ 1.

37

Carrying out the differentiation leads toÃ∂ eCV

∂ eV!T

= T

µ∂2P

∂T 2

¶eV (3.2-7)

Under isothermal conditions, integration of Eq. (3.2-7) from ideal gas state to real gas stategives Z real gas

ideal gasd eCV = T

Z real gas

ideal gas

µ∂2P

∂T 2

¶eV deV (3.2-8)

or

eCV = eC∗V + T

Z eV∞

µ∂2P

∂T 2

¶eV deV T = constant (3.2-9)

Substitution of Eq. (3.2-9) into Eq. (3.2-1) gives the change in internal energy as

deU = " eC∗V + T

Z eV∞

µ∂2P

∂T 2

¶eV deV # dT +

∙T

µ∂P

∂T

¶eV − P

¸deV (3.2-10)

Appendix B gives the molar heat capacities of gases in the ideal gas state as a function oftemperature in the form eC∗P = a+ b T + c T 2 + dT 3 + e T 4 (3.2-11)

The value of eC∗V can be found from the relationship

eC∗V = eC∗P −R (3.2-12)

Let us consider a process in which the system goes from an initial state (eV1, T1) to a finalstate (eV2, T2). Since internal energy is a state function, calculation of ∆eU is independent ofthe path chosen in going from state 1 to state 2. The most convenient paths to be followed forthe integration of Eq. (3.2-10) in going from state 1 to state 2 are shown in Figure 3.4.

APathBPath

T

V~

2T 1T

1~V

2~V

Figure 3.4 Paths of integration to calculate the change in internal energy.

Thus, the change in internal energy is given by

∆eU = Z eV2eV1∙T

µ∂P

∂T

¶eV − P

¸T1

deV + Z T2

T1

" eC∗V + T

Z eV2∞

µ∂2P

∂T 2

¶eV deV # dT Path A

(3.2-13)

∆eU = Z T2

T1

" eC∗V + T

Z eV1∞

µ∂2P

∂T 2

¶eV deV # dT +

Z eV2eV1∙T

µ∂P

∂T

¶eV − P

¸T2

deV Path B

(3.2-14)

38

The use of Eqs. (3.2-13) and (3.2-14) requires the value of (∂2P/∂T 2)eV to be evaluated.Unless the equation of state is very simple, evaluation of this term may lead to a mathematicallydifficult expression to integrate. Such a difficulty can be avoided by choosing an alternativepath for the integration of Eq. (3.2-1) as shown in Figure 3.5, where eV =∞ indicates the idealgas state.

T 2T 1T

1~V

2~V

V~ ∞=V~

Figure 3.5 Alternative path of integration to calculate the change in internal energy.

The evaluation of Eq. (3.2-1) along the path shown in Figure 3.3 gives

∆eU = Z ∞

eV1∙T

µ∂P

∂T

¶eV − P

¸T1

deV + Z T2

T1

eC∗V dT +

Z eV2∞

∙T

µ∂P

∂T

¶eV − P

¸T2

deV (3.2-15)

Depending on the equation of state, equations for calculating the change in internal energyare given below.

• van der Waals equation of state

The use of the van der Waals equation of state in Eq. (3.2-15) leads to

∆eU = R

µA1T1Z1− A2T2

Z2

¶+

Z T2

T1

eC∗V dT (3.2-16)

where the dimensionless parameter A is given in Table 3.2.

• Redlich-Kwong equation of state

The use of the Redlich-Kwong equation of state in Eq. (3.2-15) leads to

∆eU = 3R

2

∙A1T1B1

ln

µ1 +

B1Z1

¶− A2T2

B2ln

µ1 +

B2Z2

¶¸+

Z T2

T1

eC∗V dT (3.2-17)

where the dimensionless parameters A and B are given in Table 3.2.

• Peng-Robinson equation of state

The use of the Peng-Robinson equation of state in Eq. (3.2-15) leads to

∆eU = "T2 (da/dT )T2 − a2√8 b

#ln

"Z2 +

¡1 +√2¢B2

Z2 +¡1−√2¢B2

#

−"T1 (da/dT )T1 − a1

√8 b

#ln

"Z1 +

¡1 +√2¢B1

Z1 +¡1−√2¢B1

#+

Z T2

T1

eC∗V dT (3.2-18)

39

whereda

dT= − aΓ

T(3.2-19)

in which

Γ =¡0.37464 + 1.54226ω − 0.26992ω2

¢rTrα

(3.2-20)

Substitution of Eq. (3.2-19) into Eq. (3.2-18) and rearrangement give

∆eU = RT1A1(1 + Γ1)√8B1

ln

"Z1 +

¡1 +√2¢B1

Z1 +¡1−√2¢B1

#− RT2A2(1 + Γ2)√

8B2ln

"Z2 +

¡1 +√2¢B2

Z2 +¡1−√2¢B2

#

+

Z T2

T1

eC∗V dT

(3.2-21)where the dimensionless parameters A and B are given in Table 3.2.

Example 3.4 Show that Eqs. (3.2-14) and (3.2-15) are equivalent to each other.

Solution

Equation (3.2-14) is given as

∆eU = Z T2

T1

eC∗V dT +

Z T2

T1

"T

Z eV1∞

µ∂2P

∂T 2

¶eV deV # dT| {z }

X

+

Z eV2eV1∙T

µ∂P

∂T

¶eV − P

¸T2

deV (1)

Let us consider the term X in Eq. (1), i.e.,

X =

Z T2

T1

"Z eV1∞

T

µ∂2P

∂T 2

¶eV deV # dT (2)

Note that

T

µ∂2P

∂T 2

¶eV =

∂

∂T

∙T

µ∂P

∂T

¶eV − P

¸eV (3)

Substitution of Eq. (3) into Eq. (2) gives

X =

Z T2

T1

(Z eV1∞

∂

∂T

∙T

µ∂P

∂T

¶eV − P

¸deV) dT (4)

Changing the order of integration leads to

X =

Z eV1∞

½Z T2

T1

∂

∂T

∙T

µ∂P

∂T

¶eV − P

¸dT

¾deV

=

Z eV1∞

∙T

µ∂P

∂T

¶eV − P

¸T2

deV − Z eV1∞

∙T

µ∂P

∂T

¶eV − P

¸T1

deV (5)

40


∆eU = Z T2

T1

eC∗V dT +

Z eV1∞

∙T

µ∂P

∂T

¶eV − P

¸T2

deV − Z eV1∞

∙T

µ∂P

∂T

¶eV − P

¸T1

deV+

Z eV2eV1∙T

µ∂P

∂T

¶eV − P

¸T2

deVor

∆eU = Z T2

T1

eC∗V dT +

Z eV2∞

∙T

µ∂P

∂T

¶eV − P

¸T2

deV − Z eV1∞

∙T

µ∂P

∂T

¶eV − P

¸T1

deVwhich is identical with Eq. (3.2-15).

Example 3.5 Two moles of acetylene is compressed isothermally but irreversibly in a piston-cylinder assembly from 0.03m3 and 350K to 0.002m3. Calculate the work required if the heatremoved from the system is 1.8 kJ. Acetylene is represented by the van der Waals equation ofstate.

Solution

From Appendix ATc = 308.3K Pc = 61.4 bar

System: Acetylene within the piston-cylinder assembly

Application of the first law givesW = ∆U −Q (1)

Since the process is isothermal, i.e., T1 = T2 = T , Eq. (3.2-16) reduces to

∆eU = RT

µA1Z1− A2

Z2

¶(2)

The total change in internal energy is

∆U = n∆eU = nRT

µA1Z1− A2

Z2

¶(3)

Noting that

A1 =27

64

Pr1T 2r

A2 =27

64

Pr2T 2r

Z1 =P1V1nRT

Z2 =P2V2nRT

(4)

Eq. (3) becomes

∆U = n2

Ã27

64

R2T 2c

Pc

!µ1

V1− 1

V2

¶(5)

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

∆U = (2)2∙27

64

(8.314× 10−5)2(308.3)261.4

¸µ1

0.03− 1

0.002

¶= − 8.427×10−3 bar.m3 = − 842.7 J

The work done on the system can now be calculated from Eq. (1) as

W = − 842.7 + 1800 = 957.3 J

41

Example 3.6 A well-insulated rigid tank is divided into two equal parts of 3× 10−4m3 eachby a rigid partition. One side contains 10 moles of methane at 500K, and the other side isevacuated. The partition is removed and methane fills the entire tank. Using the van der Waalsequation of state, determine the final temperature of the gas.

Solution

From Appendix ATc = 190.6K Pc = 46.1 bar

From Appendix B

eC∗P = 36.155− 0.511× 10−1 T + 2.215× 10−4 T 2 − 1.824× 10−7 T 3 + 4.899× 10−11 T 4System: Contents of the tank

Application of the first law gives

∆U = Q|{z}0

+ W|{z}0

⇒ ∆eU = 0 (1)

If methane were an ideal gas, the temperature would remain constant at 500K. In this case,however, methane is represented by the van der Waals equation of state and the change ininternal energy is given by Eq. (3.2-16), i.e.,

∆eU = R

µA1T1Z1− A2T2

Z2

¶+

Z T2

T1

eC∗V dT = 0 (2)

where subscripts 1 and 2 represent the initial and final states, respectively. Noting thatn1 = n2 = n and making use of the substitution

A1 =27

64

Pr1T 2r1

A2 =27

64

Pr2T 2r2

Z1 =P1V1nRT1

Z2 =P2V2nRT2

(3)

Eq. (2) takes the form

n

Ã27

64

R2T 2c

Pc

!µ1

V1− 1

V2

¶+

Z T2

T1

eC∗V dT = 0 (4)

The heat capacity at constant volume is

eC∗V = eC∗P −R

= 27.841− 0.511× 10−1 T + 2.215× 10−4 T 2 − 1.824× 10−7 T 3 + 4.899× 10−11 T 4


(10)

∙27

64

(8.314× 10−5)2(190.6)246.1

¸µ1

3× 10−4 −1

6× 10−4¶µ

1

10−5J

bar.m3

¶+

Z T2

500

¡27.841− 0.511× 10−1 T + 2.215× 10−4 T 2 − 1.824× 10−7 T 3 + 4.899× 10−11 T 4

¢dT = 0

The solution of the above equation by MATHCADR°gives T2 = 390K.

42

Example 3.7 Suppose that the methane in Example 3.6 is replaced by gas A, which is rep-resented by the following equation of state:

P =RTeV − b

− a

T eV 2where

a = 35J.K.m3/mol2 and b = 4.8× 10−5m3/mol

If the ideal gas heat capacity of gas A is eC∗V = 1.5R, determine the final temperature.Solution

Since both eQ and fW are zero, then∆eU = 0 (1)

The change in internal energy is given by Eq. (3.2-15), i.e.

∆eU = Z ∞

eV1∙T

µ∂P

∂T

¶eV − P

¸T1

deV + Z T2

T1

eC∗V dT +

Z eV2∞

∙T

µ∂P

∂T

¶eV − P

¸T2

deV (2)

The integrand in Eq. (2) is evaluated using the given equation of state asµ∂P

∂T

¶eV =

ReV − b+

a

T 2 eV 2 ⇒ T

µ∂P

∂T

¶eV − P =

2 a

T eV 2 (3)

Substitution of Eqs. (1) and (3) into Eq. (2) gives

0 =2 a

T1

Z ∞

eV11eV 2 deV +

Z T2

T1

1.5RdT +2 a

T2

Z eV2∞

1eV 2 deV (4)

Carrying out the integration leads to

2 a

T1 eV1 + 1.5R(T2 − T1)−2 a

T2 eV2 = 0 (5)

Substitution of the numerical values into Eq. (5) results in

(2)(35)

(500)(3× 10−4/10) + (1.5)(8.314)(T2 − 500)−(2)(35)

T2(6× 10−4/10)= 0 (6)

The solution of Eq. (6) gives T2 = 375.2K.

Alternate solution: The solution is also possible by using either Eq. (3.2-13) or Eq. (3.2-14).Note that µ

∂2P

∂T 2

¶eV = −

2 a

T 3 eV 2 (7)

Substitution of Eqs. (1), (3), and (7) into Eq. (3.2-13) leads to

0 =2 a

T1

Z eV2eV1

1eV 2 deV +Z T2

T1

Ã1.5R− 2 a

T 2

Z eV2∞

1eV 2 deV!dT (8)

43

or

0 =2 a

T1

Z eV2eV1

1eV 2 deV +Z T2

T1

Ã1.5R+

2 a

T 2 eV2!dT (9)

Once the integration is carried out, the result is equivalent to Eq. (5).On the other hand, substitution of Eqs. (1), (3), and (7) into Eq. (3.2-14) leads to

0 =

Z T2

T1

Ã1.5R− 2 a

T 2

Z eV1∞

1eV 2 deV!dT +

2 a

T2

Z eV2eV11eV 2 deV (10)

or

0 =

Z T2

T1

Ã1.5R+

2 a

T 2eV1!dT +

2 a

T2

Z eV2eV1

1eV 2 deV (11)

Integration of Eq. (11) is also equivalent to Eq. (5).

Example 3.8 A piston-cylinder assembly contains carbon monoxide at 283K and 1 bar. Itis compressed to a pressure of 80 bar by a reversible and isothermal process. Calculate the workand heat interactions with the surroundings if carbon monoxide obeys the following equation ofstate:

P eVRT

= 1 +BeV +

CeV 2with

B = − 23.1 cm3/mol and C = 1560 cm6/mol2

Solution

System: CO within the piston-cylinder assembly

The reversible work done in compressing the gas is given by

fW = −Z eV2eV1 P deV (1)

Once fW is determined, the heat interaction is calculated indirectly from the first law as

eQ = ∆eU − fW (2)

Substitution of the given equation of state into Eq. (1) gives

fW = −RT

Z eV2eV1µ1eV +

BeV 2 + CeV 3¶deV

= −RT

"ln

Ã eV2eV1!−B

Ã1eV2 −

1eV1!− C

2

Ã1eV 22 −

1eV 21!#

(3)

Using MATHCADR°, the initial and final molar volumes are calculated from the equation of

state as(1) eV1

(83.14)(283)= 1 − 23.1eV1 +

1560eV 21 ⇒ eV1 = 23, 506 cm3/mol44

(80) eV2(83.14)(283)

= 1 − 23.1eV2 +1560eV 22 ⇒ eV2 = 275 cm3/mol

Substitution of the numerical values into Eq. (3) results in

fW = − (8.314)(283)½ln

µ275

23, 506

¶+ 23.1

µ1

275− 1

23, 506

¶− 1560

2

∙1

(275)2− 1

(23, 506)2

¸¾= 10, 295 J/mol

Since the process is isothermal Eq. (3.2-13) reduces to

∆eU = Z eV2eV1∙T

µ∂P

∂T

¶eV − P

¸T1

deV (4)

From the given equation of stateµ∂P

∂T

¶eV =

ReVµ1 +

BeV +CeV 2¶

⇒ T

µ∂P

∂T

¶eV − P = 0 (5)

Therefore, ∆eU = 0 and the heat interaction of the system with its surroundings can be calculatedfrom Eq. (2) as eQ = − 10, 295 J/mol

Example 3.9 Solve Example 3.5 if acetylene obeys the Peng-Robinson equation of state.

Solution


The change in internal energy is calculated from Eq. (3.2-21). Since the process is isothermal,i.e., T1 = T2 = T = 350K and Γ1 = Γ2 = Γ, it simplifies to

∆eU = RT (1 + Γ)√8

(A1B1ln

"Z1 +

¡1 +√2¢B1

Z1 +¡1−√2¢B1

#− A2

B2ln

"Z2 +

¡1 +√2¢B2

Z2 +¡1−√2¢B2

#)(1)

The reduced temperature is

Tr =T

Tc=

350

308.3= 1.135

and the parameter α is calculated from Eq. (3.1-16) as

α =h1 +

¡0.37464 + (1.54226)(0.190)− (0.26992)(0.190)2

¢ ³1−√1.135

´i2= 0.916

Therefore,

a = 0.45724

µR2T 2cPc

¶α =

(0.45724)(8.314× 10−5)2(308.3)2(0.916)61.4

= 4.482× 10−6 bar.m6

mol2

b = 0.07780RTcPc

=(0.0778)(8.314× 10−5)(308.3)

61.4= 3.249× 10−5m3/mol

45

The use of Eq. (3.2-20) gives

Γ1 = Γ2 =£0.37464 + (1.54226)(0.190)− (0.26992)(0.190)2

¤r1.1350.916

= 0.732

State 1 (T1 = 350K, eV1 = 0.03/2 = 0.015m3/mol)The pressure is calculated from the Peng-Robinson equation of state as

P1 =(8.314× 10−5)(350)0.015− 3.249× 10−5

− 4.482× 10−6(0.015)(0.015 + 3.249× 10−5) + (3.249× 10−5)(0.015− 3.249× 10−5) = 1.924 bar

The reduced pressure is

Pr1 =1.924

61.4= 0.031

The dimensionless parameters are calculated as

A1 = 0.45724

ÃPr1

T2r

!α =

(0.45724)(0.031)(0.916)

(1.135)2= 0.010

B1 = 0.07780

µPr1Tr

¶=(0.07780) (0.031)

1.135= 0.002

The cubic equation for Z, Eq. (3.1-18), takes the form

Z31 − 0.998Z21 + 5.988× 10−3 Z1 − 1.599× 10−5 = 0 ⇒ Z1 = 0.992

State 2 (T2 = 350K, eV2 = 0.002/2 = 0.001m3/mol)The pressure is

P2 =(8.314× 10−5)(350)0.001− 3.249× 10−5

− 4.482× 10−6(0.001)(1× 10−3 + 3.249× 10−5) + (3.249× 10−5)(0.001− 3.249× 10−5) = 25.863 bar

Therefore, the reduced pressure is

Pr2 =25.863

61.4= 0.421


A2 = 0.45724

ÃPr2

T2r

!α =

(0.45724)(0.421)(0.916)

(1.135)2= 0.137

B2 = 0.07780

µPr2Tr

¶=(0.07780) (0.421)

1.135= 0.029


Z32 − 0.971Z22 + 0.076Z2 − 3.108× 10−3 = 0 ⇒ Z2 = 0.889

46


∆eU = (8.314)(350)(1 + 0.732)√8

(0.010

0.002ln

"0.992 + (1 +

√2)(0.002)

0.992 + (1−√2)(0.002)

#

− 0.1370.029

ln

"0.889 + (1 +

√2)(0.029)

0.889 + (1−√2)(0.029)

#)= − 702 J/mol

Therefore, the work interaction is

W = ∆U −Q = (2)(− 702) + 1800 = 396 J

which is quite different from the one calculated in Example 3.5.

Comment: The results are strongly dependent on the equation of state used in the calculations.

3.3 CALCULATION OF THE CHANGE IN ENTHALPY

In the previous chapter, the general equation to calculate the change in enthalpy was given byEq. (2.2-31), i.e.,

d eH = eCP dT +

"eV − T

Ã∂ eV∂T

!P

#dP (3.3-1)

This equation is valid for gases, liquids, and solids.

3.3.1 Change in Enthalpy for Liquids and Solids

For solids and liquids, it is more convenient to express Eq. (3.3-1) in terms of the coefficientof thermal expansion, β. The use of Eq. (2.2-52) in Eq. (3.3-1) leads to

d eH = eCP dT + eV (1− βT ) dP (3.3-2)

Example 3.10 One mole of methanol (β = 12.0× 10−4K−1, eV = 40.47 cm3/mol) at 293Kis compressed isothermally from 1 bar to 500 bar. Calculate the change in enthalpy.

Solution

Integration of Eq. (3.3-2) at constant temperature gives

∆ eH = eV (1− βT ) ∆P (1)

in which variations in eV and β with pressure are considered negligible. Substitution of thevalues into Eq. (1) results in

∆ eH = (40.47× 10−6)h1− (12.0× 10−4)(293)

i(500− 1)× 105 = 1309 J/mol

47

3.3.2 Change in Enthalpy for Gases

Since the heat capacity data are available only at ideal gas conditions, i.e., as P → 0 or eV →∞,it is first necessary to develop an equation relating actual eCP value to an ideal one, eC∗P . SinceeH is a state function, i.e., exact differential, the use of Eq. (1.2-2) in Eq. (3.3-1) leads toÃ

∂ eCP

∂P

!T

=

(∂

∂T

"eV − T

Ã∂ eV∂T

!P

#)P

(3.3-3)

Carrying out the differentiation leads toÃ∂ eCP

∂P

!T

= −T

Ã∂2 eV∂T 2

!P

(3.3-4)

Under isothermal conditions, integration of Eq. (3.3-4) from ideal gas state to real gas stategives Z real gas

ideal gasd eCP = −T

Z real gas

ideal gas

Ã∂2eV∂T 2

!P

dP (3.3-5)

or

eCP = eC∗P − T

Z P

0

Ã∂2 eV∂T 2

!P

dP T = constant (3.3-6)

Substitution of Eq. (3.3-6) into Eq. (3.3-1) gives the change in enthalpy as

d eH =

" eC∗P − T

Z P

0

Ã∂2 eV∂T 2

!P

dP

#dT +

"eV − T

Ã∂ eV∂T

!P

#dP (3.3-7)

Let us consider a process in which the system goes from an initial state (P1, T1) to a finalstate (P2, T2). Since enthalpy is a state function, calculation of ∆ eH is independent of the pathchosen in going from state 1 to state 2. The most convenient paths to be followed for theintegration of Eq. (3.3-7) in going from state 1 to state 2 are shown in Figure 3.6 as paths Aand B.

APath BPath

1P

2P

P

1T 2T T

Figure 3.6 Paths of integration to calculate the change in enthalpy.

Thus, the change in enthalpy is given by

∆ eH =

Z P2

P1

"eV − T

Ã∂ eV∂T

!P

#T1

dP +

Z T2

T1

" eC∗P − T

Z P2

0

Ã∂2 eV∂T 2

!P

dP

#dT Path A

(3.3-8)

48

∆ eH =

Z T2

T1

" eC∗P − T

Z P1

0

Ã∂2 eV∂T 2

!P

dP

#dT +

Z P2

P1

"eV − T

Ã∂ eV∂T

!P

#T2

dP Path B

(3.3-9)

Example 3.11 Ammonia is to be compressed from 350K and 1 bar to 400K and 50 bar. Ifthe work supplied to the compressor is 1500 J/mol, calculate the heat that must be supplied permole of ammonia passing through the compressor. The equation of state is given by

P (eV − b) = RT

where b = 3.73× 10−5m3/mol.

Solution

From Appendix BeC∗P = 34.236− 0.221× 10−1 T + 1.213× 10−4 T 2 − 1.088× 10−7 T 3 + 3.203× 10−11 T 4System: Compressor

For a steady-state flow system, the first law giveseQ = ∆ eH −fWs (1)

To calculate ∆ eH, it is first necessary to evaluate the partial derivatives appearing in Eqs.(3.3-8) and (3.3-9)Ã

∂ eV∂T

!P

=R

P

Ã∂2eV∂T 2

!P

= 0 eV − T

Ã∂ eV∂T

!P

= b (2)

Substitution of Eq. (2) either into Eq. (3.3-8) or into Eq. (3.3-9) gives

∆ eH = b (P2 − P1) +

Z T2

T1

eC∗P dT (3)

or

∆ eH = (3.73× 10−5)£(50− 1)× 105

¤+

Z 400

350

¡34.236− 0.221× 10−1 T + 1.213× 10−4 T 2 − 1.088× 10−7 T 3 + 3.203× 10−11 T 4

¢dT

= 2078.2 J/mol

The heat interaction is calculated from Eq. (1) aseQ = 2078.2− 1500 = 578.2 J/molExample 3.12 An evacuated gas cylinder of 50 L volume is connected to a main line thatsupplies CO2 at the constant conditions of 400K and 2MPa. The valve between the line andthe cylinder is opened and the cylinder quickly fills with CO2 until the pressure is 2MPa, andthen the valve is closed. Calculate the temperature immediately after the valve is closed. Theequation of state for CO2 is given by

P (eV − b) = RT

where b = 4.41× 10−5m3/mol.

49

Solution

From Appendix B

eC∗P = 29.268− 0.224× 10−1 T + 2.653× 10−4 T 2 − 4.153× 10−7 T 3 + 20.057× 10−11 T 4System: Contents of the cylinder

The unsteady-state material and energy balances become

dnin − dnout| {z }0

= dnsys (1)

eHin dnin − eHout dnout| {z }0

+ δQ|{z}0

+ δW|{z}0

= d(neU)sys (2)

Note that Q is considered zero since filling takes place quickly and heat transfer is a slowprocess. Combination of Eqs. (1) and (2) leads to

eHin dnsys = d(neU)sys (3)

Since eHin is constant, integration of Eq. (3) giveseHin(n2 − n1) = n2 eU2 − n1 eU1 (4)

The cylinder is initially evacuated, i.e., n1 = 0, and Eq. (4) simplifies toeHin = eU2 (5)

Expressing internal energy in terms of enthalpy, i.e., eU2 = eH2 − P2eV2, reduces Eq. (5) toP2 eV2 = eH2 − eHin (6)

The use of the equation of state results in

RT2 + bP2 = eH2(2MPa, T2)− eHin(2MPa, 400K) (7)

The right-hand side of Eq. (7) can be calculated from either Eq. (3.3-8) or Eq. (3.3-9). Bothequations simplify to

eH2(2MPa, T2)− eHin(2MPa, 400K) =

Z T2

400

eC∗P dT (8)


RT2 + bP2 =

Z T2

400

eC∗P dT (9)

or

8.314T2 + (4.41× 10−5)(2× 106) =Z T2

400(29.268− 0.224× 10−1 T + 2.653× 10−4 T 2 − 4.153× 10−7 T 3 + 20.057× 10−11 T 4) dT

(10)

50

Solution of Eq. (10) by MATHCADR°gives T2 = 498K.

Comment: It is implicitly assumed that there is no heat transfer between the cylinder surfaceand CO2. Under what conditions is this assumption valid?

Since the cubic equations of state are not explicit in molar volume, evaluation of theterms

³∂ eV /∂T´

Pand

³∂2 eV /∂T 2´

Pin Eqs. (3.3-8) and (3.3-9) causes difficulty. The term³

∂ eV /∂T´Pcan be evaluated by the use of the triple product rule as

Ã∂ eV∂T

!P

= −(∂P/∂T )eV(∂P/∂ eV )T (3.3-10)

For example, if a gas is represented by the Redlich-Kwong equation of state, i.e.,

P =RTeV − b


(3.3-11)

then Ã∂ eV∂T

!P

=

ReV − b+

a

2 eV (eV + b)T 3/2

RT

(eV − b)2− a (2eV + b)eV 2 (eV + b)2

√T

(3.3-12)

Substitution of Eq. (3.3-12) together with the term (∂2eV /∂T 2)P into Eqs. (3.3-8) and(3.3-9) leads to a formidable equation. To circumvent mathematical complexity, an alter-native path is chosen for the integration of Eq. (3.3-1) as shown in Figure 3.7, where P = 0indicates the ideal gas state.

0 1T 2T T

1P

2P

P

Figure 3.7 Alternative path of integration to calculate the change in enthalpy.

The evaluation of Eq. (3.3-1) along the path shown in Figure 3.5 gives

∆ eH =

Z 0

P1

"eV − T

Ã∂ eV∂T

!P

#T1

dP| {z }X

+

Z T2

T1

eC∗P dT +

Z P2

0

"eV − T

Ã∂ eV∂T

!P

#T2

dP| {z }Y

(3.3-13)

It is apparent that the use of Eq. (3.3-13) is much simpler than that of Eqs. (3.3-8) and(3.3-9). However, in the case of cubic equations of state, evaluation of the terms X and Y in

51

Eq. (3.3-13) is not straightforward. While integrations in Eq. (3.3-13) are with respect to P ,

the term³∂ eV /∂T´

Pleads to an expression containing eV as can be seen from Eq. (3.3-12).

Therefore, it is useful to have an integration over volume rather than pressure. For this purposeconsider the integral Z P

0

"eV − T

Ã∂ eV∂T

!P

#T

dP (3.3-14)

Since P = P (T, eV ), thendP =

µ∂P

∂ eV¶T

deV +µ∂P∂T

¶eV dT (3.3-15)

Note that the integrand in Eq. (3.3-14) is evaluated at constant temperature. Under thecondition of constant temperature, Eq. (3.3-15) reduces to

dP =

µ∂P

∂ eV¶T

deV (3.3-16)

Substitution of Eq. (3.3-16) into Eq. (3.3-14) givesZ P

0

"eV − T

Ã∂ eV∂T

!P

#T

dP =

Z eV∞

"eV µ ∂P

∂ eV¶T

− T

Ã∂ eV∂T

!P

µ∂P

∂ eV¶T

#T

deV (3.3-17)

From the triple product rule Ã∂ eV∂T

!P

µ∂P

∂ eV¶T

= −µ∂P

∂T

¶eV (3.3-18)

The use of Eq. (3.3-18) in Eq. (3.3-17) leads toZ P

0

"eV − T

Ã∂ eV∂T

!P

#T

dP =

Z eV∞

∙eV µ ∂P

∂ eV¶T

+ T

µ∂P

∂T

¶eV¸T

deV (3.3-19)

Thus, Eq. (3.3-13) takes the form

∆ eH =

Z ∞

eV1∙eV µ ∂P

∂ eV¶T

+ T

µ∂P

∂T

¶eV¸T1

deV + Z T2

T1

eC∗P dT

+

Z eV2∞

∙eV µ ∂P

∂ eV¶T

+ T

µ∂P

∂T

¶eV¸T2

deV(3.3-20)

Depending on the equation of state, equations for calculating the change in enthalpy are givenbelow.


The use of the van der Waals equation of state in Eq. (3.3-20) leads to

∆ eH = RT2

µZ2 − 1−

A2Z2

¶−RT1

µZ1 − 1−

A1Z1

¶+

Z T2

T1

eC∗P dT (3.3-21)

52

where the dimensionless parameter A is given in Table 3.2.



∆ eH = RT2

∙Z2 − 1−

3

2

A2B2ln

µ1 +

B2Z2

¶¸−RT1

∙Z1 − 1−

3

2

A1B1ln

µ1 +

B1Z1

¶¸+

Z T2

T1

eC∗P dT

(3.3-22)where the dimensionless parameters A and B are given in Table 3.2.



∆ eH = RT2

(Z2 − 1 +

1

RT2

"T2 (da/dT )T2 − a2

√8 b

#ln

"Z2 +

¡1 +√2¢B2

Z2 +¡1−√2¢B2

#)

−RT1

(Z1 − 1 +

1

RT1

"T1 (da/dT )T1 − a1

√8 b

#ln

"Z1 +

¡1 +√2¢B1

Z1 +¡1−√2¢B1

#)+

Z T2

T1

eC∗P dT (3.3-23)

where the derivative da/dT is given by Eq. (3.2-19). Substitution of Eq. (3.2-19) into Eq.(3.3-23) and expressing the resulting equation in dimensionless form yield

∆ eH = RT2

(Z2 − 1−

A2(1 + Γ2)√8B2

ln

"Z2 +

¡1 +√2¢B2

Z2 +¡1−√2¢B2

#)

−RT1

(Z1 − 1−

A1(1 + Γ1)√8B1

ln

"Z1 +

¡1 +√2¢B1

Z1 +¡1−√2¢B1

#)+

Z T2

T1

eC∗P dT

(3.3-24)

The dimensionless parameters A and B are given in Table 3.2, and Γ is defined by Eq. (3.2-20).

Example 3.13 Propane gas is to be compressed irreversibly from 5 bar and 323K to 15 barand 343K by an adiabatic compressor in a petrochemical plant. Calculate the work input permole of propane passing through the compressor if

a) Propane obeys the Redlich-Kwong equation of state,b) Propane obeys the Peng-Robinson equation of state.

Solution


From Appendix B

eC∗P = 29.595 + 0.838× 10−1 T + 3.256× 10−4 T 2 − 3.958× 10−7 T 3 + 13.129× 10−11 T 4System: Compressor

53

This is a steady-state flow process and from the first law

∆ eH = eQ|{z}0

+ fWs (1)

Therefore, the work input to the compressor is equal to the change in enthalpy. Determinationof enthalpy at the compressor inlet (state 1) and outlet (state 2) requires reduced temperatureand pressure values to be known:

State Tr Pr

1 0.873 0.1182 0.927 0.353

a) The enthalpy change is given by Eq. (3.3-22). Therefore, it is first necessary to determineA, B, and Z values at the compressor inlet and outlet.

• State 1 [T1 = 323K (P vap = 17.4 bar), P1 = 5bar]


A1 = 0.42748

ÃPr1

T2.5r1

!=(0.42748)(0.118)

(0.873)2.5= 0.071

B1 = 0.08664

µPr1Tr1

¶=(0.08664)(0.118)

0.873= 0.012


Z31 − Z21 + 0.059Z1 − 8.52× 10−4 = 0 ⇒ Z1 = 0.938



A2 = 0.42748

ÃPr2

T2.5r2

!=(0.42748)(0.353)

(0.927)2.5= 0.182

B2 = 0.08664

µPr2Tr2

¶=(0.08664)(0.353)

0.927= 0.033


Z32 − Z22 + 0.148Z2 − 6× 10−3 = 0 ⇒ Z2 = 0.830

The third term on the right-hand side of Eq. (3.3-22) is calculated asZ 343

323

eC∗P dT = 1612 J/mol

Thus, substitution of numerical values into Eq. (3.3-22) gives

∆ eH = (8.314)(343)

∙0.830− 1− 3

2

µ0.182

0.033

¶ln

µ1 +

0.033

0.830

¶¸− (8.314)(323)

∙0.938− 1− 3

2

µ0.071

0.012

¶ln

µ1 +

0.012

0.938

¶¸+ 1612 = 676.9 J/mol

54

b) The enthalpy change is given by Eq. (3.3-24). Therefore, it is first necessary to determineA, B, Γ, and Z values at the compressor inlet and outlet.


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

α1 =h1 +

¡0.37464 + (1.54226)(0.153)− (0.26992)(0.153)2

¢ ³1−√0.873

´i2= 1.081


Γ1 =h0.37464 + (1.54226)(0.153)− (0.26992)(0.153)2

ir0.8731.081

= 0.543


A1 = 0.45724

ÃPr1

T2r1

!α1 =

(0.45724)(0.118)(1.081)

(0.873)2= 0.077

B1 = 0.07780

ÃPr1

Tr1

!=(0.07780)(0.118)

0.873= 0.011


Z31 − 0.989Z21 + 0.055Z1 − 7.247× 10−4 = 0 ⇒ Z1 = 0.931


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

α2 =h1 +

¡0.37464 + (1.54226)(0.153)− (0.26992)(0.153)2

¢ ³1−√0.927

´i2= 1.046


Γ2 =h0.37464 + (1.54226)(0.153)− (0.26992)(0.153)2

ir0.9271.046

= 0.569


A2 = 0.45724

ÃPr2

T2r2

!α2 =

(0.45724)(0.353)(1.046)

(0.927)2= 0.196

B2 = 0.07780

ÃPr2

Tr2

!=(0.07780) (0.353)

0.927= 0.030


Z32 − 0.97Z22 + 0.133Z2 − 4.953× 10−3 = 0 ⇒ Z2 = 0.814

Finally, substitution of the numerical values into Eq. (3.3-24) gives the change in enthalpy as

∆ eH = − 1570 + 524 + 1612 = 566 J/mol

55

3.4 CALCULATION OF THE CHANGE IN ENTROPY

In the previous chapter, the general equations to calculate the change in entropy are given byEqs. (2.2-38) and (2.2-41), i.e.,

deS = eCV

TdT +

µ∂P

∂T

¶eV deV (3.4-1)

deS = eCP

TdT −

Ã∂ eV∂T

!P

dP (3.4-2)

These equations are valid for gases, liquids, and solids.

3.4.1 Change in Entropy for Liquids and Solids

In terms of the coefficient of thermal expansion, β, and isothermal compressibility, κ, Eqs.(3.4-1) and (3.4-2) take the form

deS = eCV

TdT +

β

κdeV (3.4-3)

deS = eCP

TdT − β eV dP (3.4-4)

Example 3.14 Determine the change in entropy when liquid water at 1 bar and 298K iscompressed to 1000 bar and 323K. The following data for water are available:

T (K) P ( bar) eCP ( J/mol.K) eV ( cm3/mol) β (K−1)

298 1 75.305 18.071 256× 10−6298 1000 18.012 366× 10−6323 1 75.314 18.234 458× 10−6323 1000 18.174 568× 10−6

Solution

Since entropy is a state function, the path of integration is arbitrary. Let us consider theintegration path as shown in the figure below:

P (bar)

1000

1

T (K)323298

Integration of Eq. (3.4-4) gives

∆eS = Z T2

T1

eCP

TdT −

Z P2

P1

β eV dP (1)

56

Since eCP is a weak function of temperature, it is possible to take the arithmetic average of eCP ,h eCP i, in going from 298K to 323K at a constant pressure of 1 bar. Similarly, it is possibleto take the arithmetic averages of β and eV in going from 1 bar to 1000 bar at a constanttemperature of 323K. Thus, Eq. (1) reduces to

∆eS = h eCP i lnµT2T1

¶− hβi heV i (P2 − P1) (2)

The average values are

h eCP i =75.305 + 75.314

2= 75.310 J/mol.K

hβi =µ458 + 568

2

¶× 10−6 = 513× 10−6K−1

heV i = 18.234 + 18.174

2= 18.204 cm3/mol

Substitution of the numerical values into Eq. (2) leads to

∆eS = (75.310) lnµ323298

¶− (513× 10−6)(18.204× 10−6)

h(1000− 1)× 105

i= 5.13 J/mol.K

3.4.2 Change in Entropy for Gases

For gases, the change in entropy can be calculated from either Eq. (3.4-1) or Eq. (3.4-2) oncethe equation of state is known. To relate eCV and eCP values to eC∗V and eC∗P , it is necessary touse Eqs. (3.2-9) and (3.3-6), respectively. The result is

deS =⎡⎣ eC∗VT+

Z eV∞

µ∂2P

∂T 2

¶eV deV

⎤⎦ dT +

µ∂P

∂T

¶eV deV (3.4-5)

deS =⎡⎣ eC∗PT−Z P

0

Ã∂2 eV∂T 2

!P

dP

⎤⎦ dT −Ã∂ eV∂T

!P

dP (3.4-6)

Integration of Eq. (3.4-5) over the paths shown in Figure 3.2 leads to

∆eS = Z eV2eV1∙µ

∂P

∂T

¶eV¸T1

deV + Z T2

T1

⎡⎣ eC∗VT+

Z eV2∞

µ∂2P

∂T 2

¶eV deV

⎤⎦ dT Path A (3.4-7)

∆eS = Z T2

T1

⎡⎣ eC∗VT+

Z eV1∞

µ∂2P

∂T 2

¶eV deV

⎤⎦ dT +

Z eV2eV1∙µ

∂P

∂T

¶eV¸T2

deV Path B (3.4-8)

On the other hand, integration of Eq. (3.4-6) over the paths shown in Figure 3.4 results in

∆eS = −Z P2

P1

"Ã∂ eV∂T

!P

#T1

dP +

Z T2

T1

⎡⎣ eC∗PT−Z P2

0

Ã∂2eV∂T 2

!P

dP

⎤⎦ dT Path A (3.4-9)

57

∆eS = Z T2

T1

⎡⎣ eC∗PT−Z P1

0

Ã∂2eV∂T 2

!P

dP

⎤⎦ dT −Z P2

P1

"Ã∂ eV∂T

!P

#T2

dP Path B (3.4-10)

Example 3.15 A 0.7m3 insulated, rigid tank contains ammonia at 350K and 50 bar. Avalve on the tank is opened, and the pressure inside quickly drops to 10 bar, at which point thevalve is closed. Calculate the mass of ammonia vented to the atmosphere if the equation ofstate is given by

P (eV − b) = RT

where b = 3.73× 10−2m3/ kmol.

Solution

From Appendix BeC∗P = 34.236− 0.221× 10−1 T + 1.213× 10−4 T 2 − 1.088× 10−7 T 3 + 3.203× 10−11 T 4The difference between the initial and final mass of ammonia gives the amount of ammoniavented to the atmosphere. For this purpose, it is first necessary to determine the final temper-ature in the tank.

System: Ammonia remaining in the tank when the pressure reaches 10 bar

This is a closed system and if we assume that the system undergoes a reversible and adiabaticexpansion then eS1 = eS2 ⇒ ∆eS = 0The entropy change can be calculated from either Eq. (3.4-9) or Eq. (3.4-10). From the givenequation of state, i.e., eV =

RT

P+ b

the following partial derivatives are obtained:Ã∂ eV∂T

!P

=R

Pand

Ã∂2 eV∂T 2

!P

= 0

Thus, Eqs. (3.4-9) and (3.4-10) both reduce to

0 =

Z T2

T1

eC∗PT

dT −Z P2

P1

R

PdP

Substitution of the numerical values results inZ T2

350

µ34.236− 0.221× 10−1 T + 1.213× 10−4 T 2 − 1.088× 10−7 T 3 + 3.203× 10−11 T 4

T

¶dT

− (8.314) lnµ10

50

¶= 0

The solution by MATHCADR°gives T2 = 240.6K. The initial and final molar volumes are

calculated from the equation of state as

eV1 = (8.314× 10−2)(350)50

+ 3.73× 10−2 = 0.6193m3/ kmol

eV2 = (8.314× 10−2)(240.6)10

+ 3.73× 10−2 = 2.0376m3/ kmol

58

The initial and final number of moles of ammonia in the tank are

n1 =VeV1 =

0.7

0.6193= 1.13 kmol

n2 =VeV2 =

0.7

2.0376= 0.34 kmol

Therefore, the mass of ammonia vented to the atmosphere, m, is

m = (n1 − n2)M

= (1.13− 0.34)(17) = 13.43 kg

The second-order derivatives in Eqs. (3.4-7)-(3.4-10) may lead to mathematically compli-cated expressions to integrate. To eliminate such a problem, Eq. (3.4-1) can be integratedalong the path shown in Figure 3.3. The result is

∆eS = Z ∞

eV1∙µ

∂P

∂T

¶eV¸T1

deV + Z T2

T1

eC∗VT

dT +

Z eV2∞

∙µ∂P

∂T

¶eV¸T2

deV (3.4-11)

On the other hand, integration of Eq. (3.4-2) along the path shown in Figure 3.5 gives

∆eS = −Z 0

P1

"Ã∂ eV∂T

!P

#T1

dP +

Z T2

T1

eC∗PT

dT −Z P2

0

"Ã∂ eV∂T

!P

#T2

dP (3.4-12)

Depending on the equation of state, equations for calculating the change in entropy are givenbelow.


The use of the van der Waals equation of state in Eq. (3.4-11) leads to8

∆eS = R ln

µZ2 −B2Z1 −B1

¶+

Z T2

T1

eC∗PT

dT −R ln

µP2P1

¶(3.4-13)

where the dimensionless parameter B is given in Table 3.2.



∆eS = R ln

µZ2 −B2Z1 −B1

¶− R

2

∙A2B2ln

µ1 +

B2Z2

¶− A1

B1ln

µ1 +

B1Z1

¶¸

+

Z T2

T1

eC∗PT

dT −R ln

µP2P1

¶ (3.4-14)

8See Problem 3.21.

59

where the dimensionless parameters A and B are given in Table 3.2.



∆eS = R ln

µZ2 −B2Z1 −B1

¶+(da/dT )T2√

8 bln

"Z2 +

¡1 +√2¢B2

Z2 +¡1−√2¢B2

#

−(da/dT )T1√

8 bln

"Z1 +

¡1 +√2¢B1

Z1 +¡1−√2¢B1

#+

Z T2

T1

eC∗PT

dT −R ln

µP2P1

¶(3.4-15)

where the derivative term da/dT is given by Eq. (3.2-19). Substitution of Eq. (3.2-19) intoEq. (3.4-15) and expressing the resulting equation in dimensionless form yield

∆eS = R ln

µZ2 −B2

Z1 −B1

¶−

RA2Γ2√8B2

ln

"Z2 +

¡1 +√2¢B2

Z2 +¡1−√2¢B2

#

+RA1Γ1√8B1

ln

"Z1 +

¡1 +√2¢B1

Z1 +¡1−√2¢B1

#+

Z T2

T1

eC∗PT

dT −R ln

µP2P1

¶ (3.4-16)

The dimensionless parameters A and B are given in Table 3.2, and Γ is defined by Eq. (3.2-20).Depending on the problem statement, one can use the following identity in Eqs. (3.4-13),

(3.4-14), and (3.4-16)Z T2

T1

eC∗PT

dT −R ln

µP2P1

¶=

Z T2

T1

eC∗VT

dT +R ln

⎛⎝ eV2eV1⎞⎠ (3.4-17)

Example 3.16 Three moles of isobutane at 40 bar and 450K is contained in a piston-cylinderassembly. Calculate the work interaction with the surroundings using the Redlich-Kwong equa-tion of state if

a) Isobutane expands reversibly and isothermally to 10 bar,b) Isobutane expands reversibly and adiabatically to 10 bar.

Solution


From Appendix BeC∗P = 24.108 + 2.091× 10−1 T + 2.166× 10−4 T 2 − 3.372× 10−7 T 3 + 11.619× 10−11 T 4System: Isobutane within the piston-cylinder assembly

a) The work done by isobutane during reversible expansion is calculated as

W = −n

Z eV2eV1 P deV = −n

Z eV2eV1⎡⎣ RTeV − b


⎤⎦ deV (1)

60

Since the process is isothermal, i.e., temperature remains constant at 450K throughout theprocess, integration of Eq. (1) is straightforward, with the result

W = n

"RT ln

Ã eV1 − beV2 − b

!+

a

b√Tln

Ã1 +

beV1!− a

b√Tln

Ã1 +

beV2!#

(2)

In terms of the dimensionless parameters, A and B, defined in Table 3.2, Eq. (2) takes theform

W = nRT

∙ln

µZ1 −B1Z2 −B2

¶+ ln

µP2P1

¶+

A1B1ln

µ1 +

B1Z1

¶− A2

B2ln

µ1 +

B2Z2

¶¸(3)

Note that the work interaction can also be calculated indirectly from the first law as

W = n (∆eU − eQ) (4)

orW = n (∆eU − T ∆eS) (5)

Substitution of Eqs. (3.2-17) and (3.4-14) into Eq. (5) also leads to Eq. (3).

• State 1 (T1 = 450K, P1 = 40bar)The reduced temperature and pressure are

Tr1 =T1Tc=

450

407.7= 1.104 Pr1 =

P1Pc=

40

36.5= 1.096


A1 = 0.42748

ÃPr1

T2.5r1

!=(0.42748)(1.096)

(1.104)2.5= 0.366

B1 = 0.08664

µPr1Tr1

¶=(0.08664)(1.096)

1.104= 0.086


Z31 − Z21 + 0.273Z1 − 0.031 = 0 ⇒ Z1 = 0.656

• State 2 (T2 = 450K, P2 = 10bar)

The reduced temperature and pressure are

Tr2 =T2Tc=

450

407.7= 1.104 Pr2 =

P2Pc=

10

36.5= 0.274


A2 = 0.42748

ÃPr2

T2.5r2

!=(0.42748)(0.274)

(1.104)2.5= 0.091

B2 = 0.08664

µPr2Tr2

¶=(0.08664)(0.274)

1.104= 0.022

61


Z32 − Z22 + 0.069Z2 − 2× 10−3 = 0 ⇒ Z2 = 0.928


W = (3)(8.314)(450)

∙ln

µ0.656− 0.0860.928− 0.022

¶+ ln

µ10

40

¶+0.366

0.086ln

µ1 +

0.086

0.656

¶− 0.0910.022

ln

µ1 +

0.022

0.928

¶¸= − 15, 964 J

b) Since temperature does not remain constant in this case, integration of Eq. (1) can only bepossible numerically. Hence, it is much more convenient to calculate work indirectly from thefirst law, i.e.,

W = ∆U − Q|{z}0

(6)

To calculate the change in internal energy, it is first necessary to calculate the temperature atthe final state. Since the process is reversible and adiabatic (or isentropic), the entropy remainsconstant, i.e.,

∆eS = 0 (7)

or, from Eq. (3.4-14),

R ln

µZ2 −B2Z1 −B1

¶− R

2

∙A2B2ln

µ1 +

B2Z2

¶− A1

B1ln

µ1 +

B1Z1

¶¸+

Z T2

T1

eC∗PT

dT

− R ln

µP2P1

¶= 0 (8)

The final temperature T2 should be determined so as to satisfy Eq. (8). The results of thetrial-and-error solution are given below:

T2 (K) A2 B2 Z2 ∆eS ( J/mol.K)425 0.106 0.023 0.913 9.282400 0.123 0.024 0.894 1.312390 0.131 0.025 0.885 − 1.926396 0.126 0.024 0.891 0.020

Thus, T2 ' 396K and the change in internal energy is calculated from Eq. (3.2-17) as

∆U = (3)

(3(8.314)

2

∙(0.366)(450)

0.086ln

µ1 +

0.086

0.656

¶− (0.126)(396)

0.024ln

µ1 +

0.024

0.891

¶¸+

Z 396

450

eC∗V dT

¾(9)

Noting that Z 396

450

eC∗V dT =

Z 396

450( eC∗P −R) dT = − 6544 J/mol

the work interaction is calculated from Eq. (9) as

W = ∆U = − 12, 873 J

62

3.5 THE LAW OF CORRESPONDING STATES

According to the law (or principle) of corresponding states, the compressibility factor of fluidsis dependent only on the reduced temperature and pressure (or volume), i.e.,

Z = Z(Tr, Pr) or Z = Z(Tr, eVr) (3.5-1)

In other words, fluids must have the same compressibility factor at the same reduced tempera-ture and pressure (or volume). This generalization comes from the fact that the van der Waalsequation of state can be expressed in the form given by Eq. (3.5-1) as will be shown in thefollowing example.

Example 3.17 Show that the van der Waals equation of state can be expressed in the formgiven by Eq. (3.5-1).

Solution

The van der Waals equation of state is

P =RTeV − b

− aeV 2 (1)

The use of Eq. (3.1-17) givesµ∂P

∂ eV¶Tc

= 0 ⇒ RTc

(eVc − b)2=2 aeV 3c (2)

µ∂2P

∂ eV 2¶Tc

= 0 ⇒ RTc

(eVc − b)3=3 aeV 4c (3)

Division of Eq. (2) by Eq. (3) yields

b =1

3eVc (4)

Substitution of Eq. (4) into Eq. (2) results in

a =9

8RTceVc (5)

The compressibility factor for the van der Waals equation of state is

Z =P eVRT

=eVeV − b− aeV RT (6)

Substitution of Eqs. (4) and (5) into Eq. (6) leads to

Z =eVreVr − 13

− 98

1eVr Tr (7)

Comment: At the critical point, eVr = Tr = 1. Thus, the critical compressibility factor, Zc, iscalculated from Eq. (7) as

Zc =3

8= 0.375

63

The critical compressibility factor, Zc, is the value of Z at Pr = Tr = 1. If Eq. (3.5-1) wereuniversally valid, all fluids would have to have the same Zc value. In reality, however, Zc valuesfor real fluids range from 0.23 to 0.31. To make the principle of corresponding states applicableto all fluids, Pitzer et al. (1955) introduced a third parameter and modified Eq. (3.5-1) as

Z = Z(Pr, Tr, ω) (3.5-2)

where ω is the acentric factor defined by Eq. (3.1-14). It takes into account the geometry(deviation from spherical shape) and polarity of a molecule. For simple fluids, such as argon,krypton, and xenon, its value is zero. The functional form of Z is suggested in the form

Z(Pr, Tr, ω) = Z(0)(Pr, Tr) + ω Z(1)(Pr, Tr) (3.5-3)

Pitzer et al. (1955) tabulated the values of Z(0) and Z(1) as functions of Pr and Tr. Thesevalues were later improved by Lee and Kesler (1975) using the modified Benedict-Webb-Rubinequation of state as described in Appendix D.

Example 3.18 Solve Example 3.1 if ethylene obeys the principle of corresponding states.

Solution

From Example 3.1Tr = 1.239 Pr = 0.198 ω = 0.089

The use of Eq. (D-2) in Appendix D gives

eV (0)r = 6.0522 and eV (R)r = 6.0776

The values of Z(0) and Z(R) are calculated from Eqs. (D-6) and (D-7), respectively, as

Z(0) = 0.9654 and Z(R) = 0.9695

From Eq. (D-8)

Z(1) =0.9695− 0.9654

0.3978= 0.0103

The compressibility factor is calculated from Eq. (3.5-3) as

Z = 0.9654 + (0.089)(0.0103) = 0.9663

which is almost the same as the Z value obtained in Example 3.1. Therefore, eV = 2811 cm3/mol.

3.6 DEPARTURE FUNCTIONS

Internal energy, enthalpy, and entropy are all state functions. When a system goes froman initial state of (T1,P1) to a final state of (T2,P2), calculations of ∆U , ∆H, and ∆S areindependent of the path chosen. Such changes can be easily calculated if one chooses the pathshown in Figure 3.8.

For an ideal gas, the changes in enthalpy and entropy are given by

∆ eHIG =

Z T2

T1

eC∗P dT (3.6-1)

64

)P,T( 22

)P,T( 22

)P,T( 11

)P,T( 11

Real Gas

GasIdeal

Figure 3.8 The path followed in the solution of thermodynamic problems.

∆eSIG =

Z T2

T1

eC∗PT

dT −R ln

µP2P1

¶(3.6-2)

Using the path shown in Figure 3.8, the changes in enthalpy and entropy are given by

∆ eH = −( eH − eHIG)T1,P1 +

Z T2

T1

eC∗P dT + ( eH − eHIG)T2,P2 (3.6-3)

∆eS = − (eS − eSIG)T1,P1 +

Z T2

T1

eC∗PT

dT −R ln

µP2P1

¶+ (eS − eSIG)T2,P2 (3.6-4)

The term departure function is defined as the change in property in going from the real stateto the ideal gas state or vice versa at constant temperature and pressure, i.e., (ψ − ψIG)T,P .

In a similar fashion, the change in internal energy is given by

∆eU = −(eU − eU IG)T1,P1 +

Z T2

T1

eC∗V dT + (eU − eU IG)T2,P2 (3.6-5)

The departure function for internal energy, (eU IG − eU)T,P , is calculated indirectly from thedeparture function for enthalpy as follows. From the definition of enthalpy

eH = eU + P eV = eU + ZRT (3.6-6)

For an ideal gas, Eq. (3.6-6) reduces to

eHIG = eU IG +RT (3.6-7)

Subtraction of Eq. (3.6-7) from Eq. (3.6-6) gives

(eU − eU IG)T,P = ( eH − eHIG)T,P + (1− Z)RT (3.6-8)

3.6.1 Departure Functions from the Cubic Equations of State

Comparison of Eq. (3.6-3) with either Eq. (3.3-13) or Eq. (3.3-20) gives the departure functionfor enthalpy as

³ eH − eHIG´T,P

=

Z P

0

"eV − T

Ã∂ eV∂T

!P

#T

dP =

Z eV∞

∙eV µ ∂P

∂ eV¶T

+ T

µ∂P

∂T

¶eV¸T

deV(3.6-9)

65

On the other hand, comparison of Eq. (3.6-4) with Eq. (3.4-12) gives the departure functionfor entropy as ³eS − eSIG

´T,P

=

Z P

0

"R

P−Ã∂ eV∂T

!P

#T

dP (3.6-10)

The use of Eqs. (3.3-16) and (3.3-18) in Eq. (3.6-10) also expresses the departure function forentropy in the form

³eS − eSIG´T,P

=

Z eV∞

∙µ∂P

∂T

¶eV −

ReV¸T

deV +R lnZ (3.6-11)

Departure functions for various equations of state are given below.

• van der Waals equation of stateThe use of the van der Waals equation of state in Eqs. (3.6-9) and (3.6-11) leads to³ eH − eHIG

´T,P

RT= Z − 1− A

Z(3.6-12)

³eS − eSIG´T,P

R= ln (Z −B) (3.6-13)

where the dimensionless parameters A and B are given in Table 3.2. The use of Eq. (3.6-12)in Eq. (3.6-8) gives ³eU − eU IG

´T,P

RT= − A

Z(3.6-14)

• Redlich-Kwong equation of stateThe use of the Redlich-Kwong equation of state in Eqs. (3.6-9) and (3.6-11) leads to³ eH − eHIG

´T,P

RT= Z − 1− 3

2

A

Bln

µ1 +

B

Z

¶(3.6-15)

³eS − eSIG´T,P

R= ln (Z −B)− 1

2

A

Bln

µ1 +

B

Z

¶(3.6-16)

where the dimensionless parameters A and B are given in Table 3.2. The use of Eq. (3.6-14)in Eq. (3.6-8) gives ³eU − eU IG

´T,P

RT= − 3

2

A

Bln

µ1 +

B

Z

¶(3.6-17)

• Peng-Robinson equation of stateThe use of the Peng-Robinson equation of state in Eqs. (3.6-9) and (3.6-11) leads to

66

³ eH − eHIG´T,P

RT= Z − 1 + 1

RT

∙T (da/dT )− a√

8 b

¸ln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#(3.6-18)

³eS − eSIG´T,P

R= ln(Z −B) +

1

R

µda/dT√8 b

¶ln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#(3.6-19)

where the dimensionless parameters A and B are given in Table 3.2, and the derivative termda/dT is given by Eq. (3.2-19). Substitution of Eq. (3.6-18) into Eq. (3.6-8) gives³eU − eU IG

´T,P

RT=

∙T (da/dT )− a√

8 bRT

¸ln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#(3.6-20)

Equations (3.6-18)-(3.6-20) can be further simplified with the help of Eq. (3.2-19). The result-ing equations are given as³ eH − eHIG

´T,P

RT= Z − 1− A (1 + Γ)√

8Bln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#(3.6-21)

³eS − eSIG´T,P

R= ln(Z −B)− AΓ√

8Bln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#(3.6-22)

³eU − eU IG´T,P

RT= − A (1 + Γ)√

8Bln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#(3.6-23)

where Γ is defined by Eq. (3.2-20).

Example 3.19 Isopentane is to be compressed from 3 bar and 500K to 20 bar in a reversibleisothermal compressor. Determine the work required per mole of isopentane as well as theheat interaction of the compressor with the surroundings. Assume that isopentane obeys thePeng-Robinson equation of state.

Solution


System: Compressor

The first law for a steady-state flow system is given by

∆ eH = eQ+fWs (1)

For a reversible and isothermal process, the heat interaction can be calculated from the secondlaw as eQ = T ∆eS (2)

The values of reduced temperature, reduced pressure, and α at the compressor inlet (state 1)and outlet (state 2) are calculated as

67

State Tr Pr α

1 1.085 0.089 0.9422 1.085 0.592 0.942

The values of A, B, and Z, and the departure functions at the compressor inlet and outlet aregiven in the following table:

StateP

( bar)A B Z Γ

eH − eHIG

( J/mol)

eS − eSIG

( J/mol.K)

1 3 0.033 6.382× 10−3 0.973 0.763 − 359.2 − 0.4962 20 0.217 0.042 0.812 0.763 − 2647 − 3.787

The change in enthalpy is calculated from Eq. (3.6-3) as

∆ eH = − ( eH − eHIG)T1,P1 + (eH − eHIG)T2,P2 = 359.2− 2647 = − 2287.8 J/mol

The change in entropy is calculated from Eq. (3.6-4) as

∆eS = − (eS − eSIG)T1,P1 −R ln

µP2P1

¶+ (eS − eSIG)T2,P2

= 0.496− (8.314) lnµ20

3

¶− 3.787 = − 19.064 J/mol.K

The use of Eq. (2) gives the heat interaction as

eQ = (500) (− 19.064) = − 9532 J/molThe work done can be calculated from Eq. (1) as

fWs = ∆ eH − eQ = − 2287.8 + 9532 = 7244.2 J/molExample 3.20 A piston-cylinder assembly initially contains 0.1m3 of ethylene at 393K and35 bar. The pressure of the gas is increased to 205 bar by a reversible isothermal compression.Calculate the heat and work interactions with the surroundings by assuming

a) Ethylene is an ideal gas,b) Ethylene obeys the Peng-Robinson equation of state.

Solution


System: Ethylene in the piston-cylinder assembly

a) Since the internal energy of an ideal gas depends only on temperature, ∆U = 0 and the firstlaw simplifies to

Q = −W (1)

68

The work done on the gas can be determined from

W = −Z V2

V1

P dV = −Z V2

V1

nRT

VdV = −nRT ln

µV2V1

¶= −nRT ln

µP1P2

¶(2)

The number of moles of ethylene is

n =P1V1RT1

=(35)(0.1)

(8.314× 10−2)(393) = 0.107 kmol

Substitution of the values into Eq. (2) gives

W = − (0.107)(8.314)(393) lnµ35

205

¶= 618 kJ

From Eq. (1), Q = − 618 kJ.

b) For a reversible isothermal process, the second law states that

Q = T ∆S (3)

Therefore, once Q and ∆U are calculated, the work done on the gas is calculated from the firstlaw, i.e.,

W = ∆U −Q (4)

Determination of internal energy and entropy values at the initial (state 1) and final (state 2)states requires reduced temperature, reduced pressure, and α values to be known:

State Tr Pr α

1 1.391 0.692 0.8252 1.391 4.051 0.825

The values of A, B, Z, and Γ, and the departure functions at the initial and final states aregiven in the table below:

StateP

( bar)A B Z Γ

eU − eU IG

( kJ/ kmol)

eS − eSIG

( kJ/ kmol.K)

1 35 0.135 0.039 0.908 0.662 − 775 − 1.9532 205 0.790 0.227 0.740 0.662 − 4611 − 10.222

The number of moles of ethylene is

n =P1V1RT1Z1

=(35)(0.1)

(8.314× 10−2)(393)(0.908) = 0.118 kmol

Since the process is isothermal, i.e., T1 = T2 = T , Eq. (3.6-5) gives

∆eU = − (eU − eU IG)T,P1 + (eU − eU IG)T,P2

= 775− 4611 = − 3836 kJ/ kmol

69

Thus, the total change in internal energy is

∆U = n∆eU = (0.118)(− 3836) = − 452.6 kJThe change in entropy is calculated from Eq. (3.6-4) as

∆eS = − (eS − eSIG)T1,P1 −R ln

µP2P1

¶+ (eS − eSIG)T2,P2

= 1.953− (8.314) lnµ205

35

¶− 10.222 = − 22.965 kJ/ kmol.K

The amount of heat exchange is

Q = T ∆S = (393)h(0.118)(− 22.965)

i= − 1065 kJ

The work done on the gas can be calculated from Eq. (4) as

W = − 452.6 + 1065 = 612.4 kJ

3.6.2 Departure Functions from the Principle of Corresponding States

Departure functions for enthalpy and entropy are given as³ eH − eHIG´T,P

RTc=

³ eH − eHIG´(0)T,P

RTc+ ω

³ eH − eHIG´(1)T,P

RTc(3.6-24)

³eS − eSIG´T,P

R=

³eS − eSIG´(0)T,P

R+ ω

³eS − eSIG´(1)T,P

R(3.6-25)

Appendix D gives the equations developed by Lee and Kesler (1975) to calculate the terms onthe right-hand sides of Eqs. (3.6-24) and (3.6-25).

Example 3.21 Solve Example 3.13 if propane obeys the principle of corresponding states.

Solution

Note that fWs = ∆ eH (1)


From the equations given in Appendix D, the enthalpy departures are tabulated below. Therefore,the use of Eq. (3.6-24) gives³ eH − eHIG

´T1,P1

RTc= − 0.159 + (0.153)(− 0.194) = − 0.189³ eH − eHIG

´T2,P2

RTc= − 471 + (0.153)(− 527) = − 0.552

70

ParametersState 1

(Tr = 0.873, Pr = 0.118)State 2

(Tr = 0.927, Pr = 0.353)eV (0)r 6.9658 2.1836

Z(0)

0.9385 0.8311eV (R)r 6.8739 2.1142

Z(R)

0.9261 0.8047

Z(1) − 0.0311 − 0.0664³ eH − eHIG

´(0)T,P

/RTc − 0.159 − 0.471³ eH − eHIG´(R)T,P

/RTc − 0.236 − 0.681³ eH − eHIG´(1)T,P

/RTc − 0.194 − 0.527

The change in enthalpy can be calculated from Eq. (3.6-3). From Example 3.13Z 343

323

eC∗P dT = 1612 J/mol

Thus,

∆ eH = (0.189) (8.314)(369.9) + 1612− (0.552)(8.314)(369.9) = 495.7 J/mol

From Eq. (1), the work done on the system is 495.7 J/mol.

3.7 WHICH EQUATION OF STATE TO USE?

In the low density region, i.e., when either pressure is very low or temperature is very high,gas behavior can be described by the ideal gas equation of state. Outside of this region, onemight think of how to choose the "best" equation of state among various alternatives. In otherwords, similar to the queen in Snow White who asked, "Mirror, mirror on the wall, who’s thefairest of them all?" (Martin, 1979).

The truncated form of the virial equation of state, Eq. (3.1-9), can be used only for thevapor phase for pressures up to approximately 15 bar.

For computational purposes, an equation of state representing both vapor and liquid phaseswould be ideal. In that respect, cubic equations of state have the advantage. However, no cubicequation of state can provide the PV T behavior in all regions of interest. For example, theyusually fail near the critical region.

Chapter 4 of the book "The Properties of Gases and Liquids" (Poling et al., 2004) providesthe guidelines for selecting an equation of state for PV T properties of pure substances.

71

REFERENCES

Gupta, D. and P.T. Eubank, 1997, J. Chem. Eng. Data, 42, 961-970.

Lee, B.I. and M.G. Kesler, 1975, AIChE Journal, 21, 510-525.

Martin, J.J., 1979, Ind. Eng. Chem. Fund., 18, 81-97.

Meng, L., Y.Y. Duan and L. Li, 2004, Fluid Phase Equilibria, 226, 109-120.

Peng, D.Y. and D.B. Robinson, 1976, Ind. Eng. Chem. Fund., 15, 59-64.

Pitzer, K.S., D.Z. Lippmann, R.F. Curl, C.M. Huggins and D.E. Peterson, 1955, J. Am. Chem.Soc., 77, 3433-3440.

Pitzer, K.S., 1977, Phase Equilibria and Fluid Properties in the Chemical Industry, ACS Sym-posium Series, Vol. 60, pp. 1-10.

Poling, B.E., J.M. Prausnitz and J.P. O’Connell, 2004, The Properties of Gases and Liquids,5th Ed., McGraw-Hill, New York.

Riedlich, O. and J.N.S. Kwong, 1949, Chem. Rev., 44, 233-244.

Soave, G., 1972, Chem. Eng. Sci., 4, 1197-1203.

Strelzhoff, S. and L.C. Pan, 1968, Chemical Engineering, 75 (Nov. 4), 191.

Tsonopoulos, C., 1979, Adv. Chem. Ser., 143-162.

Valderrama, J.O., 2003, Ind. Eng. Chem. Res., 42, 1603-1618.

van der Waals, J.D., 1873, Over de Continuiteit van de Gas-en Vloeistoftoestand, DoctoralDissertation, Leyden.

Van Ness H.C. and M.M. Abbott, 1982, Classical Thermodynamics of Nonelectrolyte Solutions,McGraw-Hill, New York.

PROBLEMS

Problems related to Section 3.1

3.1 Calculate the second virial coefficient for n-butane at 375K. The experimental value is− 426.5 cm3/mol (Gupta and Eubank, 1997). Also estimate the density of n-butane at 375Kand 5 bar.

(Answer: − 430.8 cm3/mol, 0.01 g/ cm3)

3.2 Various correlations are available in the literature to calculate the second virial coefficientfor nonpolar fluids besides Eq. (3.1-10). For example, Meng et al. (2004) modified the well-known Tsonopoulos (1979) correlation as

B =RTcPc

hf (0) + ω f (1)

i(1)

72

wheref (0) = 0.13356− 0.30252

Tr− 0.15668

T2r

− 0.00724T3r

− 0.00022T8r

(2)

f (1) = 0.17404− 0.15581Tr

+0.38183

T2r

− 0.44044T3r

− 0.00541T8r

(3)

Calculate the second virial coefficients for methane, ethane, and carbon dioxide at 300K usingEq. (1).

(Answer: − 42.098 cm3/mol, − 181.483 cm3/mol, − 121.869 cm3/mol)

3.3 Saturated isobutane vapor at 320K is contained in a piston-cylinder assembly. It isexpanded to 3 bar by a reversible isothermal process. At 320K, the vapor pressure of isobutaneis 6.243 bar. Using the virial equation of state, Eq. (3.1-9),

a) Estimate the molar volume of isobutane at the initial and final states,b) Determine the work done during the expansion process.(Answer: a) 3704 cm3/mol, 8310.7 cm3/mol b) − 1949.7 J/mol)

3.4 If a gas obeys the virial equation of state, Eq. (3.1-9), use Eq. (1) of Problem 2.1 andshow that the Joule-Thomson coefficient is given by

μ =

T

µdB

dT

¶−B

eC∗P − T

Ãd2B

dT 2

!P

(1)

Substitute Eq. (3.1-10) into Eq. (1) to obtain

μ =

TcPc

⎡⎣− 0.083 + 1.0972T1.6r

+ ω

⎛⎝− 0.139 + 0.8944T4.2r

⎞⎠⎤⎦eC∗PR+ Pr

Ã1.75552

T2.6r

+ ω3.75648

T5.2r

!

3.5 Analytical calculation of the roots of a quadratic equation

x3 + p x2 + q x+ r = 0 (1)

is as follows:

Let the terms M and N be defined as

M =3 q − p2

9and N =

9 pq − 27 r − 2 p354

(2)

If p, q, and r are real and if ∆ =M3 +N2 is the discriminant, then

• ∆ > 0; one root is real and two are complex conjugate,• ∆ = 0; all roots are real and at least two are equal,• ∆ < 0; all roots are real and unequal.

73

Case (i) Solutions for ∆ > 0

In this case the roots are given by

x1 = S + T − 13p (3)

x2 = −1

2(S + T )− 1

3p+

1

2i√3 (S − T ) (4)

x3 = −1

2(S + T )− 1

3p− 1

2i√3 (S − T ) (5)

where

S =3

qN +

√∆ and T =

3

qN −

√∆ (6)

Case (ii) Solutions for ∆ < 0

The roots are given by

xi = ± 2√−M cos

∙θ

3+ (i− 1)120o

¸− 13p i = 1, 2, 3 (7)

where

θ = arccos

sN2

(−M)3(θ is in degrees) (8)

In Eq. (7) the upper sign applies if N is positive, while the lower sign applies if N is negative.

If methanol obeys the van der Waals equation of state, predict the density of saturated methanolvapor at 11 bar and 413K analytically.

(Answer: 0.01106 g/ cm3)

3.6 Estimate the molar volumes of saturated liquid and vapor for n-octane at 293K using:

a) Redlich-Kwong equation of state,b) Soave-Redlich-Kwong equation of state,c) Peng-Robinson equation of state.

At 293K, the vapor pressure of n-octane is 1.387 bar.

(Answer: a) 196.9 cm3/mol, 15, 250 cm3/mol b) 189.6 cm3/mol, 14, 560 cm3/molc) 168.6 cm3/mol, 14, 540 cm3/mol)

3.7 The maximum pressure, P , that a spherical tank can withstand is dependent on the tankradius, R, the wall thickness, tw, and the maximum tensile strength of the material, σ. Theequation proposed by Strelzoff and Pan (1968) states that

P =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

2 twσ

R+ 0.2twif P < 0.665σ

2σ

µtwR+ 1

¶2− 2σµ

twR+ 1

¶2+ 2

if P > 0.665σ

where P is the internal gauge pressure.

74

Estimate the maximum amount of ethylene that can be stored in a spherical tank at 300K.The tank is made of stainless steel 316, has a radius of 1.5m, and has a wall thickness of13mm. Ethylene is represented by the Peng-Robinson equation of state. For stainless steel316, σ = 586MPa.

(Answer: )

3.8 A rigid tank contains propane at 350K and an unknown pressure. When the tank iscooled to 303K, the vapor starts condensing. Estimate the initial pressure in the tank usingthe Peng-Robinson equation of state. At 303K, the vapor pressure of propane is 10.84 bar.

(Answer: 13.234 bar)

3.9 A 0.8m3 rigid tank contains 2 kg of acetylene at 200K. Determine the state and pressureof acetylene in the tank using the Peng-Robinson equation of state. The vapor pressure ofacetylene at 200K is 1.905 bar.

(Answer: 1.544 bar)

3.10 A 0.05m3 piston-cylinder device contains 7moles of isobutane at 350K. It is compressedisothermally until the volume becomes 0.01m3. The vapor pressure of isobutane is given as afunction of temperature as

lnP vap = − 9.416723 lnT − 4577.132T

+ 73.51351 + 1.514279× 10−5 T 2

where P vap is in kPa and T is in K. Estimate the initial and final pressures using the Peng-Robinson equation of state.

(Answer: 3.812 bar, 12.469 bar)

3.11 Suppose that a rigid cylinder contains propylene at 380K and 150 bar. The contents ofthe cylinder are cooled to 285K and part of the liquid that condensed at the bottom of thecylinder is removed. If the cylinder temperature is again increased to 380K, will the pressurebe greater than, equal to, or less than 150 bar? Explain clearly.

(Answer: Less than 150 bar)

3.12 A piston-cylinder assembly contains 3 kg of acetone with 60% quality. The system isheated at constant pressure until all the acetone vaporizes. Calculate the work done if thetemperature at the final state is 400K. Assume acetone is represented by the Peng-Robinsonequation of state. The vapor pressure of acetone is given as a function of temperature as

lnP vap = 10.0311− 2940.46

T − 35.93where P vap is in bar and T is in K.

(Answer: − 58, 598 J)

3.13 From Figure 3.3-a, conclude that the isotherm satisfies the following equation in thetwo-phase region:

P vap(eV V − eV L) =

Z eV V

eV L

P deV (1)

a) If a substance is represented by the Redlich-Kwong equation of state, show that Eq. (1)takes the form

P vap(eV V − eV L) = RT ln

Ã eV V − beV L − b

!− a

b√Tln

" eV V (eV L + b)eV L (eV V + b)

#(2)

75

b) Express Eq. (2) in terms of dimensionless quantities to obtain

ZV − ZL − lnµZV −B

ZL −B

¶+

A

Bln

∙ZV (ZL +B)

ZL(ZV +B)

¸= 0 (3)

Note that Eq. (3) can be used to estimate vapor pressures of pure substances at any giventemperature.

c) Using a trial-and-error procedure, estimate the vapor pressure of propane at 350K if it obeysthe Redlich-Kwong equation of state.

(Answer: c) 30.9 bar)


3.14 A rigid container is completely filled with 2 kg of mercury at 293K and 1 bar and sealed.Estimate the amount of heat transferred so as to increase the pressure to 100 bar. The followingdata are provided for mercury

ρ = 13.6 g/ cm3 bCP = 0.139 J/ g.K β = 1.82× 10−4K−1 κ = 3.95× 10−6 bar−1

Hint: In this case bCV 6= bCP .

(Answer: 507.9 J)

3.15 A piston-cylinder assembly contains 5 moles of carbon monoxide at 300K. The gas iscompressed from 3 L to 1 L by a reversible isothermal process. Calculate the heat removedfrom the system if carbon monoxide is represented by the van der Waals equation of state.

(Answer: − 15, 620 J)

3.16 One mole of carbon dioxide undergoes the following changes in a series of nonflowprocesses: (i) From an initial state of 300K and 4 bar (state 1), it is compressed to 415K and15 bar (state 2); (ii) It is then cooled to 325K at a constant pressure of 15 bar (state 3). Forthe overall process, the work done on the gas is 320 J. Estimate the heat interaction with thesurroundings assuming carbon dioxide obeys the Redlich-Kwong equation of state.

(Answer: 184 J)

3.17 A vertical cylinder fitted with a frictionless piston contains 1 mole of acetylene at 330Kand 20 bar. The piston-cylinder assembly is well insulated and a mechanical lock prevents thepiston from rising. The piston has a mass of 500 kg and a cross-sectional area of 0.02m2. Theambient pressure is 1 bar.

The mechanical lock is now released. If acetylene obeys the Peng-Robinson equation ofstate, estimate the gas temperature when the system reaches equilibrium. Take g = 10m/ s2.

(Answer: 267.7K)

3.18 Consider a closed system undergoing a reversible adiabatic expansion from state 1 tostate 2.

a) Starting with µ∂T

∂ eV¶eSÃ∂ eV∂ eS!T

Ã∂ eS∂T

!eV = − 1 (1)

76

show that µ∂T

∂ eV¶eS = −

TeCV

µ∂P

∂T

¶eV (2)

b) If an ideal gas with constant heat capacity undergoes an isentropic process, show that Eq.(2) reduces to

T2T1=

⎛⎝ eV1eV2⎞⎠γ−1

(3)

where γ = eC∗P/ eC∗V .Problems related to Section 3.3

3.19 A gas at T1 and P1 is throttled to a pressure P2. It is represented by the virial equationof state, Eq. (3.1-9), with the parameter B dependent on temperature in the form

B = α+ β T

where α and β are known constants. If eC∗P is assumed constant, show that the outlet temper-ature is given by

T2 = T1 +α (P1 − P2)eC∗P

3.20 If a gas is represented by the Peng-Robinson equation of state, use Eq. (3.3-10) andshow that Ã

∂ eV∂T

!P

=R

P

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩1

Z −B+

AΓ

Z (Z +B) +B(Z −B)

1

(Z −B)2− 2A (Z +B)£

Z (Z +B) +B (Z −B)¤2

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭where Γ is defined by Eq. (3.2-20).

3.21 Calculate the change in molar enthalpy when isobutane gas is compressed from 350Kand 10 bar to 450K and 90 bar in a steady-flow compressor. Use the Peng-Robinson equationof state.

(Answer: 2431 J/mol)

3.22 It is proposed to preheat a gas from 303K to 398K in a heat exchanger. The pressureof the gas at the inlet is 35 bar and the pressure drop in the heat exchanger may be considerednegligible. Estimate the amount of heat required in J/mol using the following data

Z = 1 +aP√T

with a = − 0.02K1/2/bar

eC∗P = 11.7 + 0.03Twhere eC∗P is in J/mol.K, and T is in K.

(Answer: 2103 J/mol)

3.23 A vertical cylinder fitted with a frictionless piston contains 0.15m3 of nitrogen at 350K.The piston has a mass of 250 kg and a cross-sectional area of 0.1m2. The piston and cylinder

77

do not conduct heat, but heat can be added to the gas by a heating coil. The ambient pressureis 1 bar and the acceleration of gravity is 10m/ s2. Assume that nitrogen is described by theRedlich-Kwong equation of state with eC∗V = 2.5R. If 2 kJ of energy is supplied to nitrogenthrough a heating coil, estimate the final temperature and pressure for the following two cases:

a) Stops placed at the initial equilibrium position of the piston prevent it from rising,

b) Piston moves freely within the cylinder.(Answer: a) 365K, 1.32 bar b) 360.6K, 1.25 bar)


3.24 Using the van der Waals equation of state show thatµ∂P

∂T

¶eV =

ReV − b(1)

Substitute Eq. (1) and eC∗V = eC∗P −R into Eq. (3.4-11) to obtain

∆eS = R ln

⎡⎣T1(eV2 − b)

T2(eV1 − b)

⎤⎦+ Z T2

T1

eC∗PT

dT (2)

Show that the substitution of

eV =ZRT

Pand b =

BRT

P(3)

into Eq. (2) leads to Eq. (3.4-13).

3.25 A piston-cylinder assembly initially contains 5 moles of n-pentane at 500K and 3 bar.The gas is compressed isothermally to 30 bar with a work input of 10 kJ. During this process,heat transfer takes place with the surroundings at a temperature of 298K. Is this process re-versible, irreversible, or impossible? Assume that n-pentane obeys the Peng-Robinson equationof state.

(Answer: Impossible)

3.26 Steam enters an adiabatic turbine at 70 bar and 873K and leaves at a pressure of 4 bar.Estimate the maximum amount of work that can be delivered by the turbine if steam obeysthe Peng-Robinson equation of state.

(Answer: − 15, 190 J/mol)

3.27 A chemical plant is in need of ethylene at 325K and 15 bar. Since the available ethyleneis at 480K and 50 bar, two schemes have been suggested to decrease the temperature andpressure of ethylene to the desired values.

In Scheme I, ethylene is first throttled from 50 bar to 15 bar and then sent to the heatexchanger to decrease its temperature to 325K. In Scheme II, ethylene is first expanded ina reversible and adiabatic turbine to 15 bar and then sent to the heat exchanger. If ethyleneobeys the Peng-Robinson equation of state, determine the heat transfer and work per unit massof ethylene in each of the suggested schemes.

(Answer: Scheme I:cW = 0 bQ = − 279.6 J/ g; Scheme II:cW = − 148.2 J/ g bQ = − 131.4 J/ g)78

Problem related to Section 3.5

3.28 Estimate the volume of a cylinder holding 150 moles of methane at 300K and 40 barusing the principle of corresponding states.

(Answer: 0.0876m3)


3.29 If a gas is represented by the virial equation of state, show that the enthalpy and entropydepartures are given by³ eH − eHIG

´T,P

RT= Pr

"0.083

Tr− 1.0972

T2.6r

+ ω

Ã0.139

Tr− 0.8944

T5.2r

!#(1)

³eS − eSIG´T,P

R= −Pr

Ã0.6752

T2.6r

+ ω0.7224

T5.2r

!(2)

Ammonia at 350K and 15 bar enters a throttling valve and leaves it at 1 bar. Estimate the exittemperature using the virial equation of state.

(Answer: 331.13K)

3.30 Calculate the departure functions for internal energy, enthalpy, and entropy for propaneat 400K and 80 bar using

a) van der Waals equation of state,b) Redlich-Kwong equation of state,c) Peng-Robinson equation of state.(Answer: a)− 5164 J/mol, − 7035 J/mol, − 12.597 J/mol.K b)− 6956 J/mol, − 8941 J/mol,− 17.245 J/mol.K c) − 7645 J/mol, − 9673 J/mol, − 18.861 J/mol.K)

3.31 A gas is represented by the following equation of state:

Z = 1 +

µa− b

T

¶P

RT

where a = 15× 10−6m3/mol and b = 0.06m3.K/mol.

a) Show that the departure functions for enthalpy and entropy take the form³ eH − eHIG´T,P

RT= A− 2B and

³eS − eSIG´T,P

R= −B

where the dimensionless parameters A and B are defined by

A =aP

RTand B =

bP

RT 2

b) The gas at 573K and 10 bar expands reversibly and adiabatically in a steady-flow turbineto 1 bar. Calculate the work output of the turbine. Take eC∗P = 3.5R.(Answer: b) − 7916.7 J/mol)

79

3.32 A well-insulated cylindrical tank of volume 0.1m3 initially contains oxygen at 50 barand 300K. A relief valve is opened and oxygen is allowed to escape at a constant flow rateof 8mol/min. If oxygen is represented by the Peng-Robinson equation of state, estimate thetemperature and pressure of oxygen in the tank after 10 minutes.

Hint: Assume that the oxygen remaining in the tank undergoes a reversible and adiabaticexpansion. (Why?)

(Answer: 245.3K, 25 bar)

3.33 A steady-flow turbine compresses ethane from 300K and 1 bar to 350K and 50 bar. UseEqs. (3.6-3) and (3.6-4) to determine the changes in enthalpy and entropy, respectively, ifethane obeys the Peng-Robinson equation of state.

(Answer: 101.7 J/mol, − 29.415 J/mol.K)

3.34 Use Eq. (3.2-9) and show that the departure function for heat capacity at constantvolume is given by ³ eCV − eC∗V ´

T,P= T

Z eV∞

µ∂2P

∂T 2

¶eV deV (1)

Use Eq. (1) in Eq. (2.2-49) and show that the departure function for heat capacity at constantpressure is given by

³ eCP − eC∗P´T,P

= T

Z eV∞

µ∂2P

∂T 2

¶eV deV − T

(∂P/∂T )2eV(∂P/∂ eV )T −R (2)

3.35 The departure function for molar heat capacity at constant volume is defined by³ eCV − eC∗V ´T,P

=∂

∂T

∙³eU − eU IG´T,P

¸eV (1)

a) If a gas is represented by the Peng-Robinson equation of state, show that the use of Eq.(3.6-20) in Eq. (1) leads to³ eCV − eC∗V ´

T,P

R=

AΓ(1 + Γ)

2√8B

ln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#(2)

where Γ is defined by Eq. (3.2-20).

b) Combine Eq. (2) with Eqs. (2.2-51) and (3.2-12) to obtain³ eCP − eC∗P´T,P

R=

AΓ(1 + Γ)

2√8B

ln

"Z +

¡1 +√2¢B

Z +¡1−√2¢B

#

+

∙1

Z −B+

AΓ

Z (Z +B) +B(Z −B)

¸21

(Z −B)2− 2A (Z +B)£

Z (Z +B) +B (Z −B)¤2− 1 (3)

c) Calculate the departure functions for molar heat capacities at constant volume and pressurefor ethylene gas at 300K and 150 bar if it obeys the Peng-Robinson equation of state.

(Answer: c) 6.470 J/mol.K, 49.506 J/mol.K)

80

3.36 In a chemical refinery, a certain process requires 12mol/ s of process steam at 2 barwith not less than 97% quality and not more than 15K superheat. It is proposed to throttlethe available steam at 1173K and 20 bar through an adiabatic valve and then cool it to therequired conditions for the process steam. Estimate the minimum duty of heat exchanger usingthe principle of corresponding states. The vapor pressure of steam is given as a function oftemperature as

lnP vap = 11.6834− 3816.44

T − 46.13where P vap is in bar and T is in K.

(Answer: 394, 296W)

81

chapter 3

Documents

van der waals

pc2 r2 tc

compressibility

ideal gas

attractive

acentric factor

molar volumes

innite series