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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 these derivatives in a systematic manner. 3.1 EQUATIONS OF STATE Any mathematical relationship between temperature, pressure, and molar volume is called an equation of state, i.e., f (T, P, Ve ) = 0 (3.1-1) Equations of state can be expressed either in pressure-explicit form

or in volume-explicit form

P = P (T, Ve )

Ve = Ve (T, P )

(3.1-2) (3.1-3)

Moreover, an equation of state can also be expressed in terms of the dimensionless compressibility factor, Z, as P Ve Z= (3.1-4) RT 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 is equal to the energy of the molecules after a collision. In other words, there are no interactions between the molecules. The equation of state for an ideal gas is given by P Ve = 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 temperature as a parameter is shown in Figure 3.1. 29

1.5

T1>T2>T3

Z

T1 1.0 T2

Ideal gas

T3 0.5

P

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, Ve , as B(T ) C(T ) P Ve =1+ + ... (3.1-6) Z= + RT Ve Ve 2

The coefficient, B(T ), is called the second virial coefficient, C(T ) is called the third virial coefficient, and so on. All virial coefficients are dependent on temperature. In practice, terms above 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 + B 0 (T ) P + C 0 (T ) P 2 + ...

(3.1-7)

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

B RT

and

C0 =

C − B2 (RT )2

(3.1-8)

In practice, it is recommended to consider only the second virial coefficient for pressures up to 15 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 virial coefficient for nonpolar fluids: i RTc h (0) B + ω B (1) (3.1-10) B= Pc 30

where

0.422 Tr1.6 0.172 = 0.139 − 4.2 Tr

B (0) = 0.083 −

(3.1-11)

B (1)

(3.1-12)

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

T Tc

and

Pr =

P Pc

ω = − 1.0 − log( Prvap |Tr =0.7 )

(3.1-13) (3.1-14) log Prvap

The definition of the acentric factor is based on the observation that versus 1/Tr is approximately a straight line (Pitzer et al., 1955)1 . For noble gases (Ar, Kr, Xe) log Prvap = −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 electron 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 C3 H8 - hence the name acentric factor for the forces arising from points other than molecular centers."

Figure 3.2 The conceptual basis of the acentric factor. Example 3.1 Calculate the molar volume of ethylene at 350 K and 10 bar if it obeys the virial equation of state. Solution From Appendix A Tc = 282.5 K 1

Pc = 50.6 bar

See Problem 5.21.

31

ω = 0.089

The reduced temperature and pressure values are 350 10 = 1.239 Pr = = 0.198 Tr = 282.5 50.6 The terms B (0) and B (1) are calculated from Eqs. (3.1-11) and (3.1-12) as 0.422 B (0) = 0.083 − = − 0.217 (1.239)1.6 0.172 = 0.069 B (1) = 0.139 − (1.239)4.2 The use of Eq. (3.1-10) in Eq. (3.1-9) leads to i Pr h (0) Z =1+ B + ω B (1) Tr Substitution of numerical values into Eq. (1) results in i 0.198 h − 0.217 + (0.089)(0.069) = 0.966 Z =1+ 1.239 Therefore, the molar volume is

(1)

(0.966)(83.14)(350) ZRT = = 2811 cm3 / mol Ve = P 10 3.1.3 Cubic Equations of State The cubic equations of state are expressed in pressure-explicit form, i.e., P = P (Ve , T ), some of which are given in Table 3.1. The cubic equations of state have been extensively used over the last three decades2 , Soave-Redlich-Kwong and Peng-Robinson3 being the most popular ones. The reason for this is twofold: (i) they are applicable over a wide range of pressures and temperatures, (ii) they are capable of describing substances in both liquid and vapor phases. Table 3.1 Cubic equations of state. Equation van der Waals

Redlich-Kwong

Soave-RedlichKwong

Peng-Robinson 2

P RT a − Ve − b Ve 2

27 64

RT a − √ Ve − b Ve (Ve + b) T RT a − Ve − b Ve (Ve + b)

RT a − Ve − b Ve (Ve + b) + b(Ve − b)

Ã

a

b

! 2

R2 Tc Pc

0.42748

Ã

R2 Tc2.5

0.42748

Ã

R2 Tc2

0.45724

Ã

Pc

Pc R2 Tc2 Pc

1 8 !

! !

α

α

µ

RTc Pc

¶

0.08664

µ

RTc Pc

¶

0.08664

µ

RTc Pc

¶

0.07780

µ

RTc Pc

¶

For a comprehensive review on the state of the cubic equations of state, see Valderrama (2003). Peng-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 Chemistry Research (May 1998). 3

32

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

In each equation given in Table 3.1, the first term is identical, i.e., RT /(Ve − b), and accounts for the repulsive forces between the molecules. The second term, on the other hand, is different in 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 a constant. 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 ⎧£ ¢¡ ¡ √ ¢¤2 ⎪ Soave-Redlich-Kwong ⎨ 1 + 0.480 + 1.574 ω − 0.176 ω2 1 − Tr (3.1-16) α= ¡ ¢¡ ⎪ √ ¢¤2 ⎩£ 2 1 + 0.37464 + 1.54226 ω − 0.26992 ω 1 − Tr Peng-Robinson

A typical isotherm of a cubic equation of state is shown in Figure 3.3-a. The horizontal line joining the points L and V is drawn so that the areas N and M are equal, known as the Maxwell equal area rule. The points L and V represent the saturated liquid and saturated vapor, respectively, at the given temperature T . The intersection of the horizontal line on the pressure axis gives the vapor (or saturation) pressure5 of the substance at T . When this procedure is repeated for other isotherms as well, the resulting pressure versus specific volume diagram is shown in Figure 3.3-b. Vapor and liquid phases exist in equilibrium within the dome-shaped region. The critical isotherm is tangent and has a point of inflection at the critical point. The resulting mathematical expressions, i.e., ¶ µ 2 ¶ µ ∂ P ∂P = =0 (3.1-17) ∂ Ve Tc ∂ Ve 2 Tc are used to evaluate the parameters "a" and "b" appearing in the cubic equations of state given in Table 3.1.

P Critical Point

P

vap

M

L

V

P

Tc N

T1

T VAPOR-LIQUID REGION

ˆL V

ˆV V

ˆ V

ˆ V

(a)

Figure 3.3

T2

(b)

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. When T > Tc , the only meaningful root is the one greater than b, the other two being complex 4 5

The value of Ve must be greater than b. Why? Vapor 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 we can conclude the following: (i) P = P vap , the liquid and vapor phases are in equilibrium, the largest and the smallest roots correspond to the molar volumes of the saturated vapor and liquid phases, respectively; the intermediate root has no physical meaning, (ii) P < P vap , the fluid is superheated vapor and the largest root gives its molar volume, (iii) P > P vap , the fluid is 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 + p Z2 + 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 lies between 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. of State VDW

RK

SRK

PR

A Ã ! 27 Pr 64 Tr2 0.42748

Ã

0.42748

Ã

0.45724

Ã

Pr 2.5 Tr

Pr 2 Tr

Pr 2 Tr

B 1 8 !

! !

α

α

µ

Pr Tr

¶

p

q

r

−1 − B

A

− AB

0.08664

µ

Pr Tr

¶

−1

A − B − B2

− AB

0.08664

µ

Pr Tr

¶

−1

A − B − B2

− AB

0.07780

µ

Pr Tr

¶

−1 + B

A − 2B − 3B 2

− AB + B 2 + B 3

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

A=

⎧ aP ⎪ ⎪ ⎪ ⎨ (RT )2

⎪ ⎪ ⎪ ⎩ aP R2 T 2.5

van der Waals, Soave-Redlich-Kwong, Peng-Robinson (3.1-19) Redlich-Kwong

B=

bP RT

(3.1-20)

Example 3.2 Estimate the molar volume of methane at 340 K and 30 bar using the PengRobinson equation of state. 6

VDW - 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. 2. 3. 4. 5. 6. 7.

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

From Appendix A

Tc = 190.6 K

Pc = 46.1 bar

ω = 0.011

The reduced temperature is Tr =

T 340 = 1.784 = Tc 190.6

Pr =

P 30 = 0.651 = Pc 46.1

The reduced pressure is

The use Eq. (3.1-16) gives the term α as n h i³ ´o2 √ α = 1 + 0.37464 + (1.54226)(0.011) − (0.26992)(0.011)2 1 − 1.784 = 0.754

The dimensionless parameters A and B are calculated as à ! Pr (0.45724)(0.651)(0.754) A = 0.45724 = 0.071 α= 2 (1.784)2 Tr µ ¶ Pr (0.07780)(0.651) = 0.028 B = 0.07780 = Tr 1.784 The parameters p, q, and r are calculated as p = − 1 + B = − 1 + 0.028 = − 0.972

q = A − 2 B − 3 B 2 = 0.071 − (2)(0.028) − (3)(0.028)2 = 0.011

r = − AB + B 2 + B 3 = − (0.071)(0.028) + (0.028)2 + (0.028)3 = − 1.182 × 10−3 Equation (3.1-18) takes the form Z 3 − 0.972 Z 2 + 0.011 Z − 1.182 × 10−3 = 0

⇒

Z = 0.962

Therefore, the molar volume is given by (0.962)(83.14)(340) ZRT = = 906.4 cm3 / mol Ve = P 30 Example 3.3 A 1 m3 rigid tank contains 25 kg of propylene at 298 K. Determine the pressure and the state of propylene in the tank using the Peng-Robinson equation of state. The vapor pressure of propylene at 298 K is 11.53 bar. 35

Solution From Appendix A Tc = 365.2 K

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 intensive properties. The state of propylene lies on the 298 K isotherm as shown in the figure below. To identify the exact location, one more independent intensive property must be known besides temperature. P

Pvap

T = 298 K

V% L

V% V

V%

Since the tank volume and the amount of propylene are given, this additional intensive property is the molar volume, calculated as Ve =

1 = 1.683 m3 / kmol 25 42.08

To determine the state of propylene, this value must be compared with the molar volumes of saturated liquid and vapor. To determine Ve L and Ve V at 298 K and 11.53 bar, it is first necessary to calculate Z L and Z V . The reduced temperature and pressure values are Tr =

T 298 = 0.816 = Tc 365.2

and

Pr =

P 11.53 = 0.251 = Pc 46

The use Eq. (3.1-16) gives the term α as n h i³ ´o2 √ α = 1 + 0.37464 + (1.54226)(0.144) − (0.26992)(0.144)2 1 − 0.816 = 1.118

The dimensionless parameters A and B are calculated as à ! Pr (0.45724)(0.251)(1.118) A = 0.45724 = 0.193 α= 2 (0.816)2 Tr µ ¶ Pr (0.07780)(0.251) = 0.024 = B = 0.07780 Tr 0.816

The cubic equation in terms of the compressibility factor, Eq. (3.1-18), takes the form Z 3 − 0.976 Z 2 + 0.143 Z − 4.042 × 10−3 = 0 The solution gives Z L = 0.037

and

Z V = 0.804

Therefore, molar volumes of the saturated liquid and vapor are (0.037)(8.314 × 10−2 )(298) Z L RT Ve L = = = 0.080 m3 / kmol P 11.53 36

(0.804)(8.314 × 10−2 )(298) Z V RT = = 1.728 m3 / kmol Ve V = P 11.53 Since Ve L < Ve < Ve V , propylene exists as a two-phase mixture and the pressure within the tank is 11.53 bar. The mole fraction of vapor in the tank, x, is given by x=

1.683 − 0.080 Ve − Ve L = 0.97 = 1.728 − 0.080 Ve V − Ve L

3.2 CALCULATION OF THE CHANGE IN INTERNAL ENERGY In the previous chapter, the general equation to calculate the change in internal energy was given by Eq. (2.2-26), i.e., ¶ ∙ µ ¸ ∂P e e dU = CV dT + T − P dVe (3.2-1) ∂T Ve

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 coefficient of thermal expansion, β, and isothermal compressibility, κ. As shown in Example 2.1, the use of the triple product rule leads to ¶ µ ∂P β (3.2-2) = ∂T Ve κ

eV ' C eP give7 Substitution of Eq. (3.2-2) into Eq. (3.2-1) and using C ¶ µ βT e e − P dVe dU = CP dT + κ

The variation of internal energy with pressure at constant temperature is ! à ¶Ã e ! µ e ∂V βT ∂U −P = ∂P κ ∂P T

(3.2-3)

(3.2-4)

T

The use of Eq. (2.2-54) in Eq. (3.2-4) results in à ! e ∂U = Ve (κP − βT ) ∂P

(3.2-5)

T

3.2.2 Change in Internal Energy for Gases

The term (∂P/∂T )Ve in Eq. (3.2-1) can be evaluated once the equation of state is known. It should be noted, however, that the heat capacity data are available only at ideal gas conditions, i.e., as either P → 0 or Ve → ∞. Therefore, it is necessary to develop an equation relating the e∗ . Since U e is a state function, i.e., exact differential, the eV to an ideal one, C actual value of C V use of Eq. (1.2-2) in Eq. (3.2-1) leads to à ! ∙ µ ¶ ½ ¸¾ eV ∂C ∂P ∂ T = −P (3.2-6) ∂T ∂T Ve ∂ Ve T Ve 7

eV ' C eP if T β 2 /(ρC bP κ) ¿ 1. As shown in Problem 2.3, C

37

Carrying out the differentiation leads to ! Ã µ 2 ¶ eV ∂ P ∂C =T ∂T 2 Ve e ∂V

(3.2-7)

T

Under isothermal conditions, integration of Eq. (3.2-7) from ideal gas state to real gas state gives Z real gas Z real gas µ 2 ¶ ∂ P eV = T dC dVe (3.2-8) ∂T 2 Ve ideal gas ideal gas or Z Ve µ 2 ¶ ∂ P ∗ eV = C e +T C dVe T = constant (3.2-9) V 2 ∞ ∂T Ve

Substitution of Eq. (3.2-9) into Eq. (3.2-1) gives the change in internal energy as # " ¶ ∙ µ ¸ Z Ve µ 2 ¶ P ∂ ∂P ∗ e= C eV + T dVe dT + T − P dVe (3.2-10) dU ∂T 2 Ve ∂T Ve ∞

Appendix B gives the molar heat capacities of gases in the ideal gas state as a function of temperature in the form eP∗ = a + b T + c T 2 + d T 3 + e T 4 (3.2-11) C e∗ can be found from the relationship The value of C V

eV∗ = C eP∗ − R C

(3.2-12)

Let us consider a process in which the system goes from an initial state (Ve1 , T1 ) to a final e is independent of state (Ve2 , T2 ). Since internal energy is a state function, calculation of ∆U the path chosen in going from state 1 to state 2. The most convenient paths to be followed for the integration of Eq. (3.2-10) in going from state 1 to state 2 are shown in Figure 3.4. ~ V

Path A

Path B

~ V2 ~ V1

T1

Figure 3.4

T2

T

Paths of integration to calculate the change in internal energy.

Thus, the change in internal energy is given by e= ∆U e= ∆U

Z Z

¶ ¸ ∙ µ Z ∂P e −P dV + T ∂T Ve Ve1 T1 Ve2

T2 T1

"

e∗ + T C V

Z

Ve1 µ ∞

∂2P ∂T 2

¶

Ve

dVe

T2 T1

#

"

e∗ + T C V

dT +

38

Z

Z

Ve2

∞

µ

∂2P ∂T 2

¶

Ve

dVe

#

dT

¶ ¸ Ve2 ∙ µ ∂P T −P dVe ∂T e e V1 V T2

Path A (3.2-13) Path B (3.2-14)

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

~ V =∞

~ V2 ~ V1

T1

T2

T

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

Figure 3.5

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

Z

∞∙

Ve1

T

µ

∂P ∂T

¶

Ve

−P

¸

T1

dVe +

Z

T2 T1

e∗ dT + C V

Z

Ve2

∞

¶ ∙ µ ¸ ∂P T −P dVe ∂T Ve T2

(3.2-15)

Depending on the equation of state, equations for calculating the change in internal energy are 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 e =R ∆U

µ

A1 T1 A2 T2 − Z1 Z2

¶

+

Z

T2 T1

where the dimensionless parameter A is given in Table 3.2.

e∗ dT C V

(3.2-16)

• Redlich-Kwong equation of state The use of the Redlich-Kwong equation of state in Eq. (3.2-15) leads to e= ∆U

µ µ ∙ ¶ ¶¸ Z T2 3R A1 T1 B1 B2 A2 T2 e∗ dT ln 1 + ln 1 + − + C V 2 B1 Z1 B2 Z2 T1

(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 " # " √ ¢ # ¡ T2 (da/dT )T2 − a2 Z + 1 + 2 B2 2 e= √ √ ¢ ¡ ∆U ln 8b Z2 + 1 − 2 B2 " # " √ ¢ # Z T ¡ T1 (da/dT )T1 − a1 2 Z1 + 1 + 2 B1 eV∗ dT √ √ ¢ ¡ − C ln + 8b Z1 + 1 − 2 B1 T1 39

(3.2-18)

where

da aΓ =− dT T

in which

(3.2-19)

¡ ¢ Γ = 0.37464 + 1.54226 ω − 0.26992 ω2

r

Tr α

(3.2-20)

Substitution of Eq. (3.2-19) into Eq. (3.2-18) and rearrangement give " " √ ¢ # √ ¢ # ¡ ¡ Z Z + 1 + 2 B + 1 + 2 B2 A (1 + Γ ) A (1 + Γ ) RT RT 1 1 2 1 1 1 2 2 2 e= √ ¢ √ ¢ √ √ ¡ ¡ ∆U ln ln − 8 B1 Z1 + 1 − 2 B1 8 B2 Z2 + 1 − 2 B2 +

Z

T2

T1

e∗ dT C V

(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 e= ∆U

Z

T2

T1

eV∗ dT + C

Z |

" Z T

T2

T1

Ve1

∞

µ

∂2P ∂T 2 {z

¶

X

Let us consider the term X in Eq. (1), i.e., Z "Z e T2

V1

X=

T1

Note that T

µ

∂2P ∂T 2

Ve

∞

¶

=

Ve

T

dVe

µ

#

∂2P ∂T 2

dT + } ¶

Ve

Z

dVe

Ve2

Ve1

#

¶ ∙ µ ¸ ∂P T −P dVe ∂T Ve T2

dT

∙ µ ¶ ¸ ∂P ∂ T −P ∂T ∂T Ve Ve

Substitution of Eq. (3) into Eq. (2) gives ) ∙ µ ¶ ¸ Z T2 (Z Ve1 ∂P ∂ T X= − P dVe dT ∂T Ve T1 ∞ ∂T

(1)

(2)

(3)

(4)

Changing the order of integration leads to

∙ µ ¶ ¸ ¾ ∂P ∂ T X= − P dT dVe ∂T Ve ∞ T1 ∂T ¶ ¶ ¸ ¸ Z Ve1 ∙ µ Z Ve1 ∙ µ ∂P ∂P e T T = −P dV − −P dVe ∂T Ve ∂T e ∞ ∞ V T2 T1 Z

Ve1

½Z

T2

40

(5)

Substitution of Eq. (5) into Eq. (1) gives e= ∆U

Z

T2

T1

eV∗ dT + C

or e= ∆U

Z

T2

T1

Z

Ve1

∞

¶ ¶ ∙ µ ¸ ¸ Z Ve1 ∙ µ ∂P ∂P e T T −P dV − −P dVe ∂T Ve ∂T e ∞ V T2 T1 ¶ ¸ Z Ve2 ∙ µ ∂P + −P dVe T ∂T Ve Ve1 T2

eV∗ dT + C

Z

Ve2

∞

¶ ¶ ¸ ¸ ∙ µ Z Ve1 ∙ µ ∂P ∂P −P dVe − −P dVe T T ∂T Ve ∂T Ve ∞ T2 T1

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

Example 3.5 Two moles of acetylene is compressed isothermally but irreversibly in a pistoncylinder assembly from 0.03 m3 and 350 K to 0.002 m3 . Calculate the work required if the heat removed from the system is 1.8 kJ. Acetylene is represented by the van der Waals equation of state. Solution From Appendix A Tc = 308.3 K

Pc = 61.4 bar

System: Acetylene within the piston-cylinder assembly Application of the first law gives W = ∆U − Q

(1)

Since the process is isothermal, i.e., T1 = T2 = T , Eq. (3.2-16) reduces to ¶ µ A1 A2 e − ∆U = RT Z1 Z2

(2)

The total change in internal energy is

e = nRT ∆U = n ∆U

Noting that A1 =

27 Pr1 64 Tr2

A2 =

Eq. (3) becomes 2

∆U = n

27 Pr2 64 Tr2 Ã

2 2 27 R Tc 64 Pc

µ

A1 A2 − Z1 Z2

Z1 = !µ

¶

P1 V1 nRT

1 1 − V1 V2

(3)

Z2 = ¶

P2 V2 nRT

(4)

(5)

Substitution of the numerical values into Eq. (5) gives ¸µ ¶ ∙ −5 2 2 1 1 2 27 (8.314 × 10 ) (308.3) − = − 8.427×10−3 bar. m3 = − 842.7 J ∆U = (2) 64 61.4 0.03 0.002 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 × 10−4 m3 each by a rigid partition. One side contains 10 moles of methane at 500 K, and the other side is evacuated. The partition is removed and methane fills the entire tank. Using the van der Waals equation of state, determine the final temperature of the gas. Solution From Appendix A Tc = 190.6 K

Pc = 46.1 bar

From Appendix B eP∗ = 36.155 − 0.511 × 10−1 T + 2.215 × 10−4 T 2 − 1.824 × 10−7 T 3 + 4.899 × 10−11 T 4 C

System: Contents of the tank

Application of the first law gives W ∆U = Q + |{z} |{z} 0

0

e =0 ∆U

⇒

(1)

If methane were an ideal gas, the temperature would remain constant at 500 K. In this case, however, methane is represented by the van der Waals equation of state and the change in internal energy is given by Eq. (3.2-16), i.e., ¶ Z T2 µ A1 T1 A2 T2 eV∗ dT = 0 e + C − (2) ∆U = R Z1 Z2 T1

where subscripts 1 and 2 represent the initial and final states, respectively. n1 = n2 = n and making use of the substitution A1 =

27 Pr1 64 Tr21

A2 =

27 Pr2 64 Tr22

Z1 =

P1 V1 nRT1

Z2 =

Noting that

P2 V2 nRT2

(3)

Eq. (2) takes the form n

Ã

2 2 27 R Tc 64 Pc

!µ

1 1 − V1 V2

¶

+

Z

T2

T1

The heat capacity at constant volume is

eV∗ dT = 0 C

(4)

eP∗ − R eV∗ = C C

= 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

Substitution of the numerical values into Eq. (4) gives ¸µ ¶µ ¶ 1 J 1 1 27 (8.314 × 10−5 )2 (190.6)2 − (10) 64 46.1 3 × 10−4 6 × 10−4 10−5 bar. m3 Z T2 ¡ ¢ 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 + ∙

500

The solution of the above equation by MATHCAD

42

R °

gives T2 = 390 K.

Example 3.7 Suppose that the methane in Example 3.6 is replaced by gas A, which is represented by the following equation of state: P = where a = 35 J. K. m3 / mol2

RT a − T Ve 2 Ve − b

b = 4.8 × 10−5 m3 / mol

and

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

Solution

e and W f are zero, then Since both Q

e =0 ∆U

(1)

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

Z

∞∙

Ve1

T

µ

∂P ∂T

¶

Ve

−P

¸

T1

dVe +

Z

T2 T1

eV∗ dT + C

Z

¶ ¸ ∙ µ ∂P −P dVe T ∂T Ve T2

Ve2

∞

The integrand in Eq. (2) is evaluated using the given equation of state as µ ¶ ¶ µ ∂P R ∂P 2a a = ⇒ T −P = + ∂T Ve ∂T 2 2 e e Ve T V T Ve 2 V −b

(2)

(3)

Substitution of Eqs. (1) and (3) into Eq. (2) gives 2a 0= T1

Z

∞

Ve1

1 e dV + Ve 2

Z

T2

T1

2a 1.5 R dT + T2

Z

Ve2

∞

Carrying out the integration leads to

1 e dV Ve 2

2a 2a + 1.5 R(T2 − T1 ) − =0 T1 Ve1 T2 Ve2

(4)

(5)

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

(2)(35) (2)(35) + (1.5)(8.314)(T2 − 500) − =0 −4 (500)(3 × 10 /10) T2 (6 × 10−4 /10)

(6)

The solution of Eq. (6) gives T2 = 375.2 K. Alternate solution: The solution is also possible by using either Eq. (3.2-13) or Eq. (3.2-14). Note that µ 2 ¶ 2a ∂ P =− (7) ∂T 2 Ve T 3 Ve 2

Substitution of Eqs. (1), (3), and (7) into Eq. (3.2-13) leads to ! Z T2 Ã Z e Z e 2 a V2 1 e 2 a V2 1 e 0= dV + dV dT 1.5 R − 2 T1 Ve1 Ve 2 T ∞ Ve 2 T1 43

(8)

or 2a 0= T1

Z

Ve2

Ve1

Z

1 e dV + Ve 2

T2

T1

Ã

2a 1.5 R + T 2 Ve

2

!

dT

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 ! Z e Z e Z T2 Ã 2 a V2 1 e 2 a V1 1 e dV dT + dV 1.5 R − 2 0= T T2 Ve1 Ve 2 e2 T1 ∞ V or

0=

Z

T2

T1

Ã

!

2a 1.5 R + T 2 Ve

1

dT +

2a T2

Z

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

Ve2

Ve1

1 e dV e V2

(9)

(10)

(11)

Example 3.8 A piston-cylinder assembly contains carbon monoxide at 283 K and 1 bar. It is compressed to a pressure of 80 bar by a reversible and isothermal process. Calculate the work and heat interactions with the surroundings if carbon monoxide obeys the following equation of state: B P Ve C =1+ + RT Ve Ve 2 with

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 f=− W

Z

Ve2

Ve1

P dVe

f is determined, the heat interaction is calculated indirectly from the first law as Once W e = ∆U e−W f Q

(1)

(2)

Substitution of the given equation of state into Eq. (1) gives ¶ 1 C B dVe + + 2 3 e e e e V " 1 Ã V! V Ã V ! !# Ã 1 Ve2 1 1 1 C = − RT ln − − −B − 2 2 Ve 2 Ve1 Ve2 Ve1 Ve1 2

f = − RT W

R °

Z

Ve2

µ

(3)

Using MATHCAD , the initial and final molar volumes are calculated from the equation of state as 23.1 (1) Ve1 1560 =1 − + ⇒ Ve1 = 23, 506 cm3 / mol (83.14)(283) 2 Ve1 Ve1 44

23.1 (80) Ve2 1560 =1 − + 2 (83.14)(283) Ve2 Ve2

⇒

Ve2 = 275 cm3 / mol

Substitution of the numerical values into Eq. (3) results in ¶ µ ¶ ∙ ½ µ ¸¾ 1 1 275 1 1560 1 f + 23.1 − − W = − (8.314)(283) ln − 23, 506 275 23, 506 2 (275)2 (23, 506)2 = 10, 295 J/ mol Since the process is isothermal Eq. (3.2-13) reduces to e= ∆U

Z

Ve2

Ve1

¸ ∙ µ ¶ ∂P −P dVe T ∂T Ve T1

From the given equation of state ¶ ¶ µ µ R ∂P B C = 1+ + ∂T Ve Ve Ve Ve 2

⇒

T

µ

∂P ∂T

(4)

¶

Ve

−P =0

(5)

e = 0 and the heat interaction of the system with its surroundings can be calculated Therefore, ∆U from Eq. (2) as e = − 10, 295 J/ mol Q Example 3.9 Solve Example 3.5 if acetylene obeys the Peng-Robinson equation of state. Solution From Appendix A Tc = 308.3 K

Pc = 61.4 bar

ω = 0.190

The change in internal energy is calculated from Eq. (3.2-21). Since the process is isothermal, i.e., T1 = T2 = T = 350 K and Γ1 = Γ2 = Γ, it simplifies to " " ( √ ¢ # √ ¢ #) ¡ ¡ Z Z + 1 + 2 B + 1 + 2 B2 A RT (1 + Γ) A 1 1 2 1 2 e= √ ¢ √ ¢ √ ¡ ¡ ∆U ln ln − (1) B1 B2 8 Z1 + 1 − 2 B1 Z2 + 1 − 2 B2

The reduced temperature is

Tr =

T 350 = 1.135 = Tc 308.3

and the parameter α is calculated from Eq. (3.1-16) as h ´i2 √ ¡ ¢³ α = 1 + 0.37464 + (1.54226)(0.190) − (0.26992)(0.190)2 1 − 1.135 = 0.916

Therefore,

a = 0.45724

µ

R2 Tc2 Pc

¶

b = 0.07780

α=

bar. m6 (0.45724)(8.314 × 10−5 )2 (308.3)2 (0.916) = 4.482 × 10−6 61.4 mol2

RTc (0.0778)(8.314 × 10−5 )(308.3) = 3.249 × 10−5 m3 / mol = Pc 61.4 45

The use of Eq. (3.2-20) gives ¤ £ Γ1 = Γ2 = 0.37464 + (1.54226)(0.190) − (0.26992)(0.190)2

r

1.135 = 0.732 0.916

State 1 (T1 = 350 K, Ve1 = 0.03/2 = 0.015 m3 / 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 = 1.924 bar (0.015)(0.015 + 3.249 × 10−5 ) + (3.249 × 10−5 )(0.015 − 3.249 × 10−5 )

The reduced pressure is 1.924 = 0.031 61.4 The dimensionless parameters are calculated as à ! Pr1 (0.45724)(0.031)(0.916) = 0.010 A1 = 0.45724 α= 2 (1.135)2 Tr Pr1 =

B1 = 0.07780

µ

Pr1 Tr

¶

=

(0.07780) (0.031) = 0.002 1.135

The cubic equation for Z, Eq. (3.1-18), takes the form Z13 − 0.998 Z12 + 5.988 × 10−3 Z1 − 1.599 × 10−5 = 0

⇒

Z1 = 0.992

State 2 (T2 = 350 K, Ve2 = 0.002/2 = 0.001 m3 / mol) The pressure is P2 = −

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

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

Therefore, the reduced pressure is Pr2 =

25.863 = 0.421 61.4

The dimensionless parameters are calculated as à ! Pr2 (0.45724)(0.421)(0.916) = 0.137 α= A2 = 0.45724 2 (1.135)2 Tr B2 = 0.07780

µ

Pr2 Tr

¶

=

(0.07780) (0.421) = 0.029 1.135

The cubic equation for Z, Eq. (3.1-18), takes the form Z23 − 0.971 Z22 + 0.076 Z2 − 3.108 × 10−3 = 0 46

⇒

Z2 = 0.889

Substitution of the numerical values into Eq. (1) gives + 0.732) e = (8.314)(350)(1 √ ∆U 8

(

" # √ 0.992 + (1 + 2)(0.002) 0.010 √ ln 0.002 0.992 + (1 − 2)(0.002) " #) √ 0.889 + (1 + 2)(0.029) 0.137 √ ln − = − 702 J/ mol 0.029 0.889 + (1 − 2)(0.029)

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 by Eq. (2.2-31), i.e., " Ã ! # e ∂ V e =C eP dT + Ve − T dH (3.3-1) dP ∂T P

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 coefficient of thermal expansion, β. The use of Eq. (2.2-52) in Eq. (3.3-1) leads to e =C eP dT + Ve (1 − βT ) dP dH

(3.3-2)

Example 3.10 One mole of methanol (β = 12.0 × 10−4 K−1 , Ve = 40.47 cm3 / mol) at 293 K is compressed isothermally from 1 bar to 500 bar. Calculate the change in enthalpy. Solution Integration of Eq. (3.3-2) at constant temperature gives e = Ve (1 − βT ) ∆P ∆H

(1)

in which variations in Ve and β with pressure are considered negligible. Substitution of the values into Eq. (1) results in h i e = (40.47 × 10−6 ) 1 − (12.0 × 10−4 )(293) (500 − 1) × 105 = 1309 J/ mol ∆H

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 Ve → ∞, e∗ . Since eP value to an ideal one, C it is first necessary to develop an equation relating actual C P e is a state function, i.e., exact differential, the use of Eq. (1.2-2) in Eq. (3.3-1) leads to H Ã ( ! " Ã ! #) eP ∂C ∂ e ∂ Ve = (3.3-3) V −T ∂P ∂T ∂T T

P

P

Carrying out the differentiation leads to ! Ã ! Ã eP ∂ 2 Ve ∂C = −T ∂P ∂T 2 T

(3.3-4)

P

Under isothermal conditions, integration of Eq. (3.3-4) from ideal gas state to real gas state gives Z real gas Z real gas à 2 e ! ∂ V eP = − T dC dP (3.3-5) ∂T 2 ideal gas ideal gas P

or

e∗ − T eP = C C P

Z

P 0

Ã

∂ 2 Ve ∂T 2

!

dP

T = constant

(3.3-6)

P

Substitution of Eq. (3.3-6) into Eq. (3.3-1) gives the change in enthalpy as # " Ã " ! # Z P Ã 2e ! e ∂ V ∂ V ∗ e= C eP − T dP dT + Ve − T dH dP ∂T 2 ∂T 0 P

(3.3-7)

P

Let us consider a process in which the system goes from an initial state (P1 , T1 ) to a final e is independent of the path state (P2 , T2 ). Since enthalpy is a state function, calculation of ∆H chosen in going from state 1 to state 2. The most convenient paths to be followed for the integration of Eq. (3.3-7) in going from state 1 to state 2 are shown in Figure 3.6 as paths A and B. Path A

P

Path B P2

P1

T1

T2

T

Paths of integration to calculate the change in enthalpy.

Figure 3.6

Thus, the change in enthalpy is given by e= ∆H

Z

P2 P1

"

Ve − T

Ã

∂ Ve ∂T

! #

P T1

dP +

Z

T2 T1

"

48

e∗ − T C P

Z

P2 0

Ã

∂ 2 Ve ∂T 2

!

P

dP

#

dT

Path A (3.3-8)

e= ∆H

Z

T2 T1

"

e∗ − T C P

Z

P1 0

Ã

∂ 2 Ve ∂T 2

!

dP

P

#

dT +

Z

P2 P1

"

Ve − T

Ã

∂ Ve ∂T

! #

dP

Path B

P T2

(3.3-9)

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

where b = 3.73 × 10−5 m3 / mol.

P (Ve − b) = RT

Solution From Appendix B eP∗ = 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 C

System: Compressor

For a steady-state flow system, the first law gives e = ∆H e −W fs Q

(1)

e it is first necessary to evaluate the partial derivatives appearing in Eqs. To calculate ∆H, (3.3-8) and (3.3-9) Ã ! ! Ã ! Ã e ∂ V ∂ 2 Ve R ∂ Ve = =0 Ve − T =b (2) ∂T P ∂T 2 ∂T P

P

P

Substitution of Eq. (2) either into Eq. (3.3-8) or into Eq. (3.3-9) gives Z T2 e eP∗ dT ∆H = b (P2 − P1 ) + C

(3)

T1

or

£ ¤ e = (3.73 × 10−5 ) (50 − 1) × 105 ∆H Z 400 ¡ ¢ + 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 350

= 2078.2 J/ mol

The heat interaction is calculated from Eq. (1) as e = 2078.2 − 1500 = 578.2 J/ mol Q Example 3.12 An evacuated gas cylinder of 50 L volume is connected to a main line that supplies CO2 at the constant conditions of 400 K and 2 MPa. The valve between the line and the cylinder is opened and the cylinder quickly fills with CO 2 until the pressure is 2 MPa, and then the valve is closed. Calculate the temperature immediately after the valve is closed. The equation of state for CO2 is given by

where b = 4.41 × 10−5 m3 / mol.

P (Ve − b) = RT 49

Solution From Appendix B eP∗ = 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 C

System: Contents of the cylinder

The unsteady-state material and energy balances become dnin − dnout = dnsys | {z }

(1)

0

e in dnin − H e dn + δQ + δW = d(nU e )sys H | out{z out} |{z} |{z}

(2)

e )sys e in dnsys = d(nU H

(3)

e in (n2 − n1 ) = n2 U e2 − n1 U e1 H

(4)

e2 e in = U H

(5)

e2 − H e in P2 Ve2 = H

(6)

e 2 (2 MPa, T2 ) − H e in (2 MPa, 400 K) RT2 + bP2 = H

(7)

0

0

0

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

e in is constant, integration of Eq. (3) gives Since H

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

e 2 − P2 Ve2 , reduces Eq. (5) to e2 = H Expressing internal energy in terms of enthalpy, i.e., U The use of the equation of state results in

The right-hand side of Eq. (7) can be calculated from either Eq. (3.3-8) or Eq. (3.3-9). Both equations simplify to Z T2 e e eP∗ dT (8) H2 (2 MPa, T2 ) − Hin (2 MPa, 400 K) = C 400

Substitution of Eq. (8) into Eq. (7) gives

RT2 + bP2 =

Z

T2

400

or

eP∗ dT C

(9)

8.314 T2 + (4.41 × 10−5 )(2 × 106 ) = Z T2 (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 400

(10)

50

Solution of Eq. (10) by MATHCAD

R °

gives T2 = 498 K.

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

Since equations of state are not explicit in molar volume, evaluation of the ³ the cubic ´ ³ ´ 2 2 e e terms ∂ V /∂T and ∂ V /∂T in Eqs. (3.3-8) and (3.3-9) causes difficulty. The term P P ³ ´ ∂ Ve /∂T can be evaluated by the use of the triple product rule as P

Ã

∂ Ve ∂T

!

=−

P

(∂P/∂T )Ve

(3.3-10)

(∂P/∂ Ve )T

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

P = then Ã

∂ Ve ∂T

!

RT a − √ Ve − b Ve (Ve + b) T

=

(3.3-11)

R a + e e e V − b 2 V (V + b) T 3/2

(3.3-12)

a (2Ve + b) − √ (Ve − b)2 Ve 2 (Ve + b)2 T RT

P

Substitution of Eq. (3.3-12) together with the term (∂ 2 Ve /∂T 2 )P into Eqs. (3.3-8) and (3.3-9) leads to a formidable equation. To circumvent mathematical complexity, an alternative path is chosen for the integration of Eq. (3.3-1) as shown in Figure 3.7, where P = 0 indicates the ideal gas state. P P2

P1 0

Figure 3.7

T1

T2

T

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 e= ∆H

Z |

0 P1

"

Ã

∂ Ve Ve − T ∂T {z X

! #

P T1

dP + }

Z

T2 T1

e∗ dT + C P

Z |

P2 0

"

Ã

∂ Ve Ve − T ∂T {z Y

! #

P T2

dP }

(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) ³ is not´straightforward. While integrations in Eq. (3.3-13) are with respect to P , leads to an expression containing Ve as can be seen from Eq. (3.3-12). the term ∂ Ve /∂T P Therefore, it is useful to have an integration over volume rather than pressure. For this purpose consider the integral à ! # Z P" e ∂ V dP (3.3-14) Ve − T ∂T 0 P T

Since P = P (T, Ve ), then

dP =

µ

∂P ∂ Ve

¶

T

dVe +

µ

∂P ∂T

¶

dT

(3.3-15)

Ve

Note that the integrand in Eq. (3.3-14) is evaluated at constant temperature. Under the condition of constant temperature, Eq. (3.3-15) reduces to µ ¶ ∂P dVe (3.3-16) dP = e ∂V T

Substitution of Eq. (3.3-16) into Eq. (3.3-14) gives à à ! # ! µ ¶ ¶ # Z P" Z Ve " µ e e ∂P ∂ V ∂ V ∂P dP = −T dVe Ve − T Ve ∂T ∂T e T e T ∞ 0 ∂ V ∂ V P T P T

(3.3-17)

From the triple product rule

Ã

∂ Ve ∂T

! µ P

∂P ∂ Ve

¶

T

=−

µ

∂P ∂T

¶

(3.3-18)

Ve

The use of Eq. (3.3-18) in Eq. (3.3-17) leads to à ! # µ ¶ ¶ ¸ Z P" Z Ve ∙ µ e ∂P ∂P ∂ V dP = +T dVe Ve Ve − T ∂T ∂T Ve T ∞ 0 ∂ Ve T

(3.3-19)

P T

Thus, Eq. (3.3-13) takes the form e= ∆H

Z

∞∙

Ve1

+

Ve

Z

µ

Ve2

∞

∂P ∂ Ve

¶

T

+T

µ

∂P ∂T

¶ ¸

Ve T1

dVe +

Z

T2 T1

¶ ¸ µ ∙ µ ¶ ∂P ∂P +T dVe Ve ∂T Ve T ∂ Ve T 2

e∗ dT C P

(3.3-20)

Depending on the equation of state, equations for calculating the change in enthalpy are given below. • van der Waals equation of state The use of the van der Waals equation of state in Eq. (3.3-20) leads to µ ¶ µ ¶ Z T2 A A 2 1 e = RT2 Z2 − 1 − e∗ dT − RT1 Z1 − 1 − + ∆H C P Z2 Z1 T1 52

(3.3-21)

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. (3.3-20) leads to µ µ ∙ ¶¸ ∙ ¶¸ Z T2 3 A2 B2 3 A1 B1 e e∗ dT ∆H = RT2 Z2 − 1 − ln 1 + ln 1 + C − RT1 Z1 − 1 − + P 2 B2 Z2 2 B1 Z1 T1 (3.3-22) 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. (3.3-20) leads to (

" # " √ ¢ #) ¡ T2 (da/dT )T2 − a2 Z + 1 + 2 B2 1 2 e = RT2 Z2 − 1 + √ ¢ √ ¡ ∆H ln RT2 8b Z2 + 1 − 2 B2 ( " # " √ ¢ #) Z T ¡ T1 (da/dT )T1 − a1 2 Z1 + 1 + 2 B1 1 eP∗ dT √ √ ¢ ¡ − RT1 Z1 − 1 + C + ln RT1 8b Z1 + 1 − 2 B1 T1

(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 e = RT2 ∆H

" √ ¢ #) ¡ Z2 + 1 + 2 B2 A2 (1 + Γ2 ) √ ¢ ¡ ln Z2 − 1 − √ 8 B2 Z2 + 1 − 2 B2

(

− RT1

"

(

√ ¢ #) Z T ¡ 2 Z1 + 1 + 2 B1 A1 (1 + Γ1 ) e∗ dT √ ¢ ¡ ln C Z1 − 1 − √ + P 8 B1 Z1 + 1 − 2 B1 T1

(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 323 K to 15 bar and 343 K by an adiabatic compressor in a petrochemical plant. Calculate the work input per mole 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 A Tc = 369.9 K

Pc = 42.5 bar

ω = 0.153

From Appendix B eP∗ = 29.595 + 0.838 × 10−1 T + 3.256 × 10−4 T 2 − 3.958 × 10−7 T 3 + 13.129 × 10−11 T 4 C

System: Compressor

53

This is a steady-state flow process and from the first law fs e= Q e +W ∆H |{z}

(1)

0

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

Tr

Pr

1 2

0.873 0.927

0.118 0.353

a) The enthalpy change is given by Eq. (3.3-22). Therefore, it is first necessary to determine A, B, and Z values at the compressor inlet and outlet. • State 1 [T1 = 323 K (P vap = 17.4 bar), P1 = 5 bar] The dimensionless parameters are calculated as à ! Pr1 (0.42748)(0.118) = 0.071 = A1 = 0.42748 2.5 (0.873)2.5 Tr1 ¶ µ Pr1 (0.08664)(0.118) = 0.012 = B1 = 0.08664 Tr1 0.873 The cubic equation for Z, Eq. (3.1-18), takes the form Z13 − Z12 + 0.059 Z1 − 8.52 × 10−4 = 0

⇒

Z1 = 0.938

• State 2 [T2 = 343 K (P vap = 25.8 bar), P2 = 15 bar] The dimensionless parameters are calculated as à ! Pr2 (0.42748)(0.353) = 0.182 A2 = 0.42748 = 2.5 (0.927)2.5 Tr2 ¶ µ Pr2 (0.08664)(0.353) = 0.033 B2 = 0.08664 = Tr2 0.927 The cubic equation for Z, Eq. (3.1-18), takes the form Z23 − Z22 + 0.148 Z2 − 6 × 10−3 = 0

⇒

Z2 = 0.830

The third term on the right-hand side of Eq. (3.3-22) is calculated as Z 343 eP∗ dT = 1612 J/ mol C 323

Thus, substitution of numerical values into Eq. (3.3-22) gives µ ¶ µ ¶¸ ∙ 0.033 3 0.182 e ln 1 + ∆H = (8.314)(343) 0.830 − 1 − 2 0.033 0.830 ∙ µ ¶ µ ¶¸ 3 0.071 0.012 − (8.314)(323) 0.938 − 1 − ln 1 + + 1612 = 676.9 J/ mol 2 0.012 0.938 54

b) The enthalpy change is given by Eq. (3.3-24). Therefore, it is first necessary to determine A, B, Γ, and Z values at the compressor inlet and outlet. • State 1 [T1 = 323 K (P vap = 17.4 bar), P1 = 5 bar] The parameter α is calculated from Eq. (3.1-16) as h ´i2 √ ¡ ¢³ = 1.081 α1 = 1 + 0.37464 + (1.54226)(0.153) − (0.26992)(0.153)2 1 − 0.873

The use of Eq. (3.2-20) gives h

2

Γ1 = 0.37464 + (1.54226)(0.153) − (0.26992)(0.153)

i r 0.873 1.081

= 0.543

The dimensionless parameters are calculated as à ! Pr1 (0.45724)(0.118)(1.081) A1 = 0.45724 = 0.077 α1 = 2 (0.873)2 Tr1 à ! Pr1 (0.07780)(0.118) = 0.011 B1 = 0.07780 = 0.873 Tr 1

The cubic equation for Z, Eq. (3.1-18), takes the form Z13 − 0.989 Z12 + 0.055 Z1 − 7.247 × 10−4 = 0

⇒

Z1 = 0.931

• State 2 [T2 = 343 K (P vap = 25.8 bar), P2 = 15 bar] The parameter α is calculated from Eq. (3.1-16) as h ´i2 √ ¢³ ¡ α2 = 1 + 0.37464 + (1.54226)(0.153) − (0.26992)(0.153)2 1 − 0.927 = 1.046

The use of Eq. (3.2-20) gives h

2

Γ2 = 0.37464 + (1.54226)(0.153) − (0.26992)(0.153)

i r 0.927 1.046

= 0.569

The dimensionless parameters are calculated as à ! Pr2 (0.45724)(0.353)(1.046) A2 = 0.45724 = 0.196 α2 = 2 (0.927)2 Tr2 à ! Pr2 (0.07780) (0.353) = 0.030 B2 = 0.07780 = 0.927 Tr 2

The cubic equation for Z, Eq. (3.1-18), takes the form Z23 − 0.97 Z22 + 0.133 Z2 − 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 e = − 1570 + 524 + 1612 = 566 J/ mol ∆H 55

3.4 CALCULATION OF THE CHANGE IN ENTROPY In the previous chapter, the general equations to calculate the change in entropy are given by Eqs. (2.2-38) and (2.2-41), i.e., eV C

dSe =

dT +

T

eP C

dSe =

dT −

T

µ Ã

∂P ∂T ∂ Ve ∂T

¶

Ve

!

dVe

(3.4-1)

dP

(3.4-2)

P

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 dSe = dSe =

eV C T

eP C T

dT +

β e dV κ

(3.4-3)

dT − β Ve dP

(3.4-4)

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

P ( bar)

298 298 323 323

1 1000 1 1000

eP ( J/ mol. K) C 75.305

Ve ( cm3 / mol) 18.071 18.012 18.234 18.174

75.314

β ( K−1 ) 256 × 10−6 366 × 10−6 458 × 10−6 568 × 10−6

Solution Since entropy is a state function, the path of integration is arbitrary. Let us consider the integration path as shown in the figure below: P (bar) 1000

1

298

323

T (K)

Integration of Eq. (3.4-4) gives ∆Se =

Z

T2

T1

Z P2 eP C dT − β Ve dP T P1 56

(1)

eP is a weak function of temperature, it is possible to take the arithmetic average of C eP , Since C eP i, in going from 298 K to 323 K at a constant pressure of 1 bar. Similarly, it is possible hC to take the arithmetic averages of β and Ve in going from 1 bar to 1000 bar at a constant temperature of 323 K. Thus, Eq. (1) reduces to µ ¶ T2 e e − hβi hVe i (P2 − P1 ) (2) ∆S = hCP i ln T1 The average values are

75.305 + 75.314 = 75.310 J/ mol. K 2 ¶ µ 458 + 568 hβi = × 10−6 = 513 × 10−6 K−1 2 18.234 + 18.174 hVe i = = 18.204 cm3 / mol 2

eP i = hC

Substitution of the numerical values into Eq. (2) leads to ¶ µ i h 323 e − (513 × 10−6 )(18.204 × 10−6 ) (1000 − 1) × 105 = 5.13 J/ mol. K ∆S = (75.310) ln 298 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) once eP values to C e∗ and C e∗ , it is necessary to eV and C the equation of state is known. To relate C V P use Eqs. (3.2-9) and (3.3-6), respectively. The result is ⎤ ⎡ ¶ µ e∗ Z Ve µ 2 ¶ C V ∂ P ∂P e e ⎦ ⎣ + dV dT + dVe (3.4-5) dS = T ∂T 2 Ve ∂T Ve ∞ ⎡

dSe = ⎣

e∗ C P T

−

Z

P 0

Ã

∂ 2 Ve ∂T 2

!

P

⎤

dP ⎦ dT −

Ã

∂ Ve ∂T

!

dP

(3.4-6)

P

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

Z Z

Ve2

∙µ

T2

⎡

Ve1

T1

⎣

∂P ∂T

e∗ C V T

¶ ¸

+

Ve T1

Z

Ve1

∞

dVe +

µ

∂2P ∂T 2

Z

T2 T1

¶

Ve

⎡ ⎣

e∗ C V

+

T

Z

⎤

dVe ⎦ dT +

Ve2

∞

Z

µ

Ve2

Ve1

∂2P ∂T 2 ∙µ

⎤

¶

∂P ∂T

dVe ⎦ dT

Ve

¶ ¸

Ve T2

dVe

Path A

(3.4-7)

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 ∆Se = −

Z

P2 P1

"Ã

∂ Ve ∂T

! #

P T1

dP +

Z

T2 T1

⎡ ⎣

e∗ C P T

−

57

Z

P2 0

Ã

∂ 2 Ve ∂T 2

!

P

⎤

dP ⎦ dT

Path A

(3.4-9)

∆Se =

Z

T2 T1

⎡ ⎣

e∗ C P T

−

Z

P1 0

Ã

∂ 2 Ve

∂T 2

⎤

!

dP ⎦ dT −

P

Z

P2 P1

"Ã

∂ Ve ∂T

! #

dP

Path B

(3.4-10)

P T2

Example 3.15 A 0.7 m3 insulated, rigid tank contains ammonia at 350 K and 50 bar. A valve on the tank is opened, and the pressure inside quickly drops to 10 bar, at which point the valve is closed. Calculate the mass of ammonia vented to the atmosphere if the equation of state is given by P (Ve − b) = RT where b = 3.73 × 10−2 m3 / kmol. Solution From Appendix B eP∗ = 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 C

The difference between the initial and final mass of ammonia gives the amount of ammonia vented to the atmosphere. For this purpose, it is first necessary to determine the final temperature 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 adiabatic expansion then ⇒ ∆Se = 0 Se1 = Se2

The entropy change can be calculated from either Eq. (3.4-9) or Eq. (3.4-10). From the given equation of state, i.e., RT +b Ve = P the following partial derivatives are obtained: Ã ! ! Ã ∂ Ve R ∂ 2 Ve = =0 and ∂T P ∂T 2 P

P

Thus, Eqs. (3.4-9) and (3.4-10) both reduce to 0=

Z

T2

T1

e∗ C P T

dT −

Z

P2

P1

R dP P

Substitution of the numerical values results in ¶ Z T2 µ 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 T 350 µ ¶ 10 − (8.314) ln =0 50 R °

The solution by MATHCAD gives T2 = 240.6 K. The initial and final molar volumes are calculated from the equation of state as (8.314 × 10−2 )(350) + 3.73 × 10−2 = 0.6193 m3 / kmol Ve1 = 50 (8.314 × 10−2 )(240.6) + 3.73 × 10−2 = 2.0376 m3 / kmol Ve2 = 10 58

The initial and final number of moles of ammonia in the tank are n1 =

V 0.7 = 1.13 kmol = 0.6193 Ve 1

V 0.7 = 0.34 kmol = n2 = 2.0376 e V 2

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 complicated expressions to integrate. To eliminate such a problem, Eq. (3.4-1) can be integrated along the path shown in Figure 3.3. The result is ∆Se =

Z

∞ ∙µ

Ve1

∂P ∂T

¶ ¸

Ve T1

dVe +

Z

e∗ C V

T2

dT +

T

T1

Z

Ve2

∞

∙µ

∂P ∂T

¶ ¸

Ve T2

dVe

(3.4-11)

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

Z

0 P1

"Ã

∂ Ve ∂T

! #

dP +

P T1

Z

e∗ C P

T2

T

T1

dT −

Z

P2 0

"Ã

∂ Ve ∂T

! #

dP

(3.4-12)

P T2

Depending on the equation of state, equations for calculating the change in entropy are given below. • van der Waals equation of state The use of the van der Waals equation of state in Eq. (3.4-11) leads to8 ∆Se = R ln

µ

Z2 − B2 Z1 − B1

¶

+

Z

T2 T1

e∗ C P T

dT − R ln

µ

P2 P1

¶

(3.4-13)

where the dimensionless parameter B is given in Table 3.2. • Redlich-Kwong equation of state The use of the Redlich-Kwong equation of state in Eq. (3.4-11) leads to ∆Se = R ln + 8

µ Z

Z2 − B2 Z1 − B1 T2

T1

e∗ C P T

¶

∙ ¶ ¶¸ µ µ R A2 A1 B2 B1 − − ln 1 + ln 1 + 2 B2 Z2 B1 Z1

dT − R ln

µ

P2 P1

¶

See Problem 3.21.

59

(3.4-14)

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. (3.4-11) leads to " √ ¢ # ¡ µ ¶ (da/dT ) Z + 1 + 2 B2 T Z − B 2 2 2 2 √ ¢ √ ¡ ln ∆Se = R ln + Z1 − B1 8b Z2 + 1 − 2 B2 " √ ¢ # Z T e∗ ¡ µ ¶ (da/dT )T1 2 C Z1 + 1 + 2 B1 P2 P √ √ ¡ ¢ ln dT − R ln − (3.4-15) + T P 8b Z1 + 1 − 2 B1 1 T1

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

µ

Z2 − B2 Z1 − B1

¶ "

" √ ¢ # ¡ RA2 Γ2 Z2 + 1 + 2 B2 √ ¢ ¡ − √ ln 8 B2 Z2 + 1 − 2 B2

√ ¢ # Z T e∗ µ ¶ 2 C RA1 Γ1 Z1 + 1 + 2 B1 P2 P √ ¢ ¡ + √ dT − R ln ln + P1 8 B1 Z1 + 1 − 2 B1 T1 T ¡

(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) ⎛ ⎞ µ ¶ Z T2 C Z T2 C e∗ e∗ Ve2 P P V 2 dT − R ln dT + R ln ⎝ ⎠ (3.4-17) = P1 T1 T T1 T e V1 Example 3.16 Three moles of isobutane at 40 bar and 450 K is contained in a piston-cylinder assembly. Calculate the work interaction with the surroundings using the Redlich-Kwong equation of state if a) Isobutane expands reversibly and isothermally to 10 bar, b) Isobutane expands reversibly and adiabatically to 10 bar. Solution From Appendix A Tc = 407.7 K

Pc = 36.5 bar

ω = 0.183

From Appendix B eP∗ = 24.108 + 2.091 × 10−1 T + 2.166 × 10−4 T 2 − 3.372 × 10−7 T 3 + 11.619 × 10−11 T 4 C

System: Isobutane within the piston-cylinder assembly

a) The work done by isobutane during reversible expansion is calculated as ⎤ ⎡ Z Ve2 Z Ve2 a ⎦ dVe ⎣ RT − P dVe = − n W = −n √ e Ve1 Ve1 V −b Ve (Ve + b) T 60

(1)

Since the process is isothermal, i.e., temperature remains constant at 450 K throughout the process, integration of Eq. (1) is straightforward, with the result " Ã Ã Ã ! !# ! Ve1 − b b b a a W = n RT ln − √ ln 1 + (2) + √ ln 1 + b T b T Ve − b Ve Ve 2

1

2

In terms of the dimensionless parameters, A and B, defined in Table 3.2, Eq. (2) takes the form µ µ ¶ µ ¶ ¶ ¶¸ ∙ µ P2 Z1 − B1 B1 B2 A1 A2 ln 1 + ln 1 + + ln + − (3) W = nRT ln Z2 − B2 P1 B1 Z1 B2 Z2 Note that the work interaction can also be calculated indirectly from the first law as e − Q) e W = n (∆U

or

(4)

e − T ∆S) e W = n (∆U

(5)

Substitution of Eqs. (3.2-17) and (3.4-14) into Eq. (5) also leads to Eq. (3). • State 1 (T1 = 450 K, P1 = 40 bar)

The reduced temperature and pressure are Tr1 =

T1 450 = 1.104 = Tc 407.7

Pr1 =

P1 40 = 1.096 = Pc 36.5

The dimensionless parameters are calculated as A1 = 0.42748

Ã

B1 = 0.08664

!

=

(0.42748)(1.096) = 0.366 (1.104)2.5

¶

=

(0.08664)(1.096) = 0.086 1.104

Pr1 2.5 Tr1

µ

Pr1 Tr1

The cubic equation for Z, Eq. (3.1-18), takes the form Z13 − Z12 + 0.273 Z1 − 0.031 = 0

⇒

Z1 = 0.656

• State 2 (T2 = 450 K, P2 = 10 bar) The reduced temperature and pressure are Tr2 =

T2 450 = 1.104 = Tc 407.7

Pr2 =

P2 10 = 0.274 = Pc 36.5

The dimensionless parameters are calculated as à ! Pr2 (0.42748)(0.274) = 0.091 A2 = 0.42748 = 2.5 (1.104)2.5 Tr2 ¶ µ Pr2 (0.08664)(0.274) = 0.022 B2 = 0.08664 = Tr2 1.104 61

The cubic equation for Z, Eq. (3.1-18), takes the form Z23 − Z22 + 0.069 Z2 − 2 × 10−3 = 0

⇒

Z2 = 0.928

Substitution of the numerical values into Eq. (3) gives ∙ µ ¶ µ ¶ µ ¶ 0.656 − 0.086 10 0.366 0.086 W = (3)(8.314)(450) ln + ln + ln 1 + 0.928 − 0.022 40 0.086 0.656 µ ¶¸ 0.091 0.022 − ln 1 + = − 15, 964 J 0.022 0.928 b) Since temperature does not remain constant in this case, integration of Eq. (1) can only be possible numerically. Hence, it is much more convenient to calculate work indirectly from the first law, i.e., (6) W = ∆U − Q |{z} 0

To calculate the change in internal energy, it is first necessary to calculate the temperature at the final state. Since the process is reversible and adiabatic (or isentropic), the entropy remains constant, i.e., ∆Se = 0 (7) or, from Eq. (3.4-14), R ln

µ

Z2 − B2 Z1 − B1

¶

∙ µ µ ¶ ¶¸ Z T2 e∗ CP B2 B1 R A2 A1 dT ln 1 + ln 1 + − − + 2 B2 Z2 B1 Z1 T T1

− R ln

µ

P2 P1

¶

= 0 (8)

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

A2

B2

Z2

425 400 390 396

0.106 0.123 0.131 0.126

0.023 0.024 0.025 0.024

0.913 0.894 0.885 0.891

∆Se ( J/ mol. K) 9.282 1.312 − 1.926 0.020

Thus, T2 ' 396 K and the change in internal energy is calculated from Eq. (3.2-17) as ( ∙ µ ¶ µ ¶¸ 3(8.314) (0.366)(450) 0.086 (0.126)(396) 0.024 ln 1 + − ln 1 + ∆U = (3) 2 0.086 0.656 0.024 0.891 ¾ Z 396 ∗ e + (9) CV dT 450

Noting that

Z

396

450

eV∗ dT = C

Z

396

450

eP∗ − R) dT = − 6544 J/ mol (C

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 fluids is dependent only on the reduced temperature and pressure (or volume), i.e., Z = Z(Tr , Pr )

Z = Z(Tr , Ver )

or

(3.5-1)

In other words, fluids must have the same compressibility factor at the same reduced temperature and pressure (or volume). This generalization comes from the fact that the van der Waals equation of state can be expressed in the form given by Eq. (3.5-1) as will be shown in the following example. Example 3.17 Show that the van der Waals equation of state can be expressed in the form given by Eq. (3.5-1). Solution The van der Waals equation of state is P = The use of Eq. (3.1-17) gives ¶ µ ∂P =0 ∂ Ve Tc µ

∂2P ∂ Ve 2

¶

a RT − Ve − b Ve 2 RTc 2a = 3 (Vec − b)2 Vec

⇒

=0

⇒

Tc

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

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

1e Vc 3

RTc 3a = 4 (Vec − b)3 Vec

9 RTc Vec 8

(1)

(2)

(3)

(4)

(5)

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

Ve P Ve a = − RT Ve − b Ve RT

(6)

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

Ver

1 Ver − 3

−

9 1 8 Ve Tr r

(7)

Comment: At the critical point, Ver = Tr = 1. Thus, the critical compressibility factor, Zc , is calculated from Eq. (7) as 3 Zc = = 0.375 8

63

The critical compressibility factor, Zc , is the value of Z at Pr = Tr = 1. If Eq. (3.5-1) were universally valid, all fluids would have to have the same Zc value. In reality, however, Zc values for real fluids range from 0.23 to 0.31. To make the principle of corresponding states applicable to 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 . These values were later improved by Lee and Kesler (1975) using the modified Benedict-Webb-Rubin equation 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 3.1 Tr = 1.239

Pr = 0.198

ω = 0.089

The use of Eq. (D-2) in Appendix D gives Ver(0) = 6.0522

and

Z (0) = 0.9654

and

Ver(R) = 6.0776

The values of Z (0) and Z (R) are calculated from Eqs. (D-6) and (D-7), respectively, as Z (R) = 0.9695

From Eq. (D-8) 0.9695 − 0.9654 = 0.0103 0.3978 The compressibility factor is calculated from Eq. (3.5-3) as Z (1) =

Z = 0.9654 + (0.089)(0.0103) = 0.9663 which is almost the same as the Z value obtained in Example 3.1. Therefore, Ve = 2811 cm3 / mol. 3.6 DEPARTURE FUNCTIONS Internal energy, enthalpy, and entropy are all state functions. When a system goes from an initial state of (T1 ,P1 ) to a final state of (T2 ,P2 ), calculations of ∆U , ∆H, and ∆S are independent of the path chosen. Such changes can be easily calculated if one chooses the path shown in Figure 3.8. For an ideal gas, the changes in enthalpy and entropy are given by Z T2 IG eP∗ dT e (3.6-1) C ∆H = T1

64

(T1, P1 )

(T2 ,P2 )

(T1, P1 )

(T2 ,P2 )

Ideal Gas

Real Gas

The path followed in the solution of thermodynamic problems.

Figure 3.8

∆SeIG =

Z

T2

e∗ C P

dT − R ln

T

T1

µ

P2 P1

¶

(3.6-2)

Using the path shown in Figure 3.8, the changes in enthalpy and entropy are given by e = −(H e −H e IG )T ,P + ∆H 1 1 ∆Se = − (Se − SeIG )T1 ,P1 +

Z

T2 T1

e∗ C P T

Z

T2 T1

e −H e IG )T ,P e∗ dT + (H C 2 2 P

dT − R ln

µ

P2 P1

¶

+ (Se − SeIG )T2 ,P2

(3.6-3)

(3.6-4)

The term departure function is defined as the change in property in going from the real state to the ideal gas state or vice versa at constant temperature and pressure, i.e., (ψ − ψ IG )T,P . In a similar fashion, the change in internal energy is given by e = −(U e −U e IG )T ,P + ∆U 1 1

Z

T2 T1

e −U e IG )T ,P e∗ dT + (U C 2 2 V

(3.6-5)

e IG − U e )T,P , is calculated indirectly from the The departure function for internal energy, (U departure function for enthalpy as follows. From the definition of enthalpy e =U e + P Ve = U e + ZRT H

(3.6-6)

e IG + RT e IG = U H

(3.6-7)

e −U e IG )T,P = (H e −H e IG )T,P + (1 − Z)RT (U

(3.6-8)

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

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

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 function for enthalpy as ³ ´ e −H e IG H

T,P

=

Z

P 0

"

Ve − T

Ã

∂ Ve ∂T

! #

dP =

P T

65

Z

Ve

∞

¶ ¸ µ ∙ µ ¶ ∂P ∂P e +T dVe V ∂T e e V T ∂V T (3.6-9)

On the other hand, comparison of Eq. (3.6-4) with Eq. (3.4-12) gives the departure function for entropy as à ! # Z P" ³ ´ e ∂ V R − = dP (3.6-10) Se − SeIG P ∂T T,P 0 P T

The use of Eqs. (3.3-16) and (3.3-18) in Eq. (3.6-10) also expresses the departure function for entropy in the form ³ ´ Se − SeIG

T,P

=

Z

Ve

∞

∙µ

∂P ∂T

¶

Ve

−

R Ve

¸

T

dVe + R ln Z

(3.6-11)

Departure functions for various equations of state are given below. • van der Waals equation of state

The use of the van der Waals equation of state in Eqs. (3.6-9) and (3.6-11) leads to ³ ´ e −H e IG H

T,P

=Z −1−

A Z

(3.6-12)

T,P

= ln (Z − B)

(3.6-13)

RT

³ ´ IG e e S−S R

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 ³ ´ e −U e IG U A T,P =− (3.6-14) RT Z • Redlich-Kwong equation of state The use of the Redlich-Kwong equation of state in Eqs. (3.6-9) and (3.6-11) leads to ³ ´ e −H e IG H

µ ¶ 3 A B =Z −1− ln 1 + 2B Z

(3.6-15)

µ ¶ 1 A B = ln (Z − B) − ln 1 + 2B Z

(3.6-16)

T,P

RT

³ ´ Se − SeIG

T,P

R

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 ³ ´ e −U e IG µ ¶ U 3 A B T,P =− ln 1 + (3.6-17) RT 2B Z • Peng-Robinson equation of state The use of the Peng-Robinson equation of state in Eqs. (3.6-9) and (3.6-11) leads to 66

³ ´ e −H e IG H

T,P

RT

´ ³ Se − SeIG

1 =Z −1+ RT

T,P

R

∙

√ ¢ # ¡ ¸ " Z + 1+ 2 B T (da/dT ) − a √ √ ¢ ¡ ln 8b Z + 1− 2 B

1 = ln(Z − B) + R

µ

da/dT √ 8b

¶

"

√ ¢ # ¡ Z + 1+ 2 B √ ¢ ¡ ln Z + 1− 2 B

(3.6-18)

(3.6-19)

where the dimensionless parameters A and B are given in Table 3.2, and the derivative term da/dT is given by Eq. (3.2-19). Substitution of Eq. (3.6-18) into Eq. (3.6-8) gives ´ ³ √ ¢ # ¡ e −U e IG ∙ ¸ " U Z + 1 + 2 B T (da/dT ) − a T,P √ ¢ √ ¡ = ln (3.6-20) RT 8 bRT Z+ 1− 2 B Equations (3.6-18)-(3.6-20) can be further simplified with the help of Eq. (3.2-19). The resulting equations are given as ³ ´ " √ ¢ # ¡ e −H e IG H Z + 1+ 2 B A (1 + Γ) T,P √ ¢ ¡ ln =Z −1− √ (3.6-21) RT 8B Z + 1− 2 B ³ ´ Se − SeIG

T,P

R

³ ´ e −U e IG U

" √ ¢ # ¡ Z+ 1+ 2 B AΓ √ ¢ ¡ ln = ln(Z − B) − √ 8B Z+ 1− 2 B

T,P

RT

where Γ is defined by Eq. (3.2-20).

" √ ¢ # ¡ Z+ 1+ 2 B A (1 + Γ) √ ¢ ¡ ln =− √ 8B Z+ 1− 2 B

(3.6-22)

(3.6-23)

Example 3.19 Isopentane is to be compressed from 3 bar and 500 K to 20 bar in a reversible isothermal compressor. Determine the work required per mole of isopentane as well as the heat interaction of the compressor with the surroundings. Assume that isopentane obeys the Peng-Robinson equation of state. Solution From Appendix A Tc = 461.0 K

Pc = 33.8 bar

ω = 0.227

System: Compressor The first law for a steady-state flow system is given by e =Q e+W fs ∆H

(1)

For a reversible and isothermal process, the heat interaction can be calculated from the second law as e = T ∆Se Q (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 2

1.085 1.085

0.089 0.592

0.942 0.942

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

State

P ( bar)

A

B

Z

Γ

1 2

3 20

0.033 0.217

6.382 × 10−3 0.042

0.973 0.812

0.763 0.763

e −H e IG H ( J/ mol) − 359.2 − 2647

Se − SeIG ( J/ mol. K) − 0.496 − 3.787

The change in enthalpy is calculated from Eq. (3.6-3) as e = − (H e −H e IG )T ,P + (H e −H e IG )T ,P = 359.2 − 2647 = − 2287.8 J/ mol ∆H 1 1 2 2

The change in entropy is calculated from Eq. (3.6-4) as µ ¶ P2 IG e e e + (Se − SeIG )T2 ,P2 ∆S = − (S − S )T1 ,P1 − R ln P1 µ ¶ 20 − 3.787 = − 19.064 J/ mol. K = 0.496 − (8.314) ln 3 The use of Eq. (2) gives the heat interaction as e = (500) (− 19.064) = − 9532 J/ mol Q

The work done can be calculated from Eq. (1) as

e −Q e = − 2287.8 + 9532 = 7244.2 J/ mol fs = ∆H W Example 3.20 A piston-cylinder assembly initially contains 0.1 m3 of ethylene at 393 K and 35 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 From Appendix A Tc = 282.5 K

Pc = 50.6 bar

ω = 0.089

System: Ethylene in the piston-cylinder assembly a) Since the internal energy of an ideal gas depends only on temperature, ∆U = 0 and the first law simplifies to Q = −W (1) 68

The work done on the gas can be determined from µ ¶ µ ¶ Z V2 Z V2 V2 nRT P1 dV = − nRT ln W =− = − nRT ln P dV = − V V1 P2 V1 V1

(2)

The number of moles of ethylene is n=

(35)(0.1) P1 V1 = 0.107 kmol = RT1 (8.314 × 10−2 )(393)

Substitution of the values into Eq. (2) gives W = − (0.107)(8.314)(393) ln

µ

35 205

¶

= 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 first law, 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 2

1.391 1.391

0.692 4.051

0.825 0.825

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

State

P ( bar)

A

B

Z

Γ

1 2

35 205

0.135 0.790

0.039 0.227

0.908 0.740

0.662 0.662

e −U e IG U ( kJ/ kmol) − 775 − 4611

Se − SeIG ( kJ/ kmol. K)

The number of moles of ethylene is n=

(35)(0.1) P1 V1 = 0.118 kmol = R T1 Z1 (8.314 × 10−2 )(393)(0.908)

Since the process is isothermal, i.e., T1 = T2 = T , Eq. (3.6-5) gives e −U e IG )T,P e = − (U e −U e IG )T,P + (U ∆U 1 2 = 775 − 4611 = − 3836 kJ/ kmol 69

− 1.953 − 10.222

Thus, the total change in internal energy is e = (0.118)(− 3836) = − 452.6 kJ ∆U = n ∆U

The change in entropy is calculated from Eq. (3.6-4) as µ ¶ P2 IG e e e ∆S = − (S − S )T1 ,P1 − R ln + (Se − SeIG )T2 ,P2 P1 ¶ µ 205 − 10.222 = − 22.965 kJ/ kmol. K = 1.953 − (8.314) ln 35 The amount of heat exchange is h i Q = T ∆S = (393) (0.118)(− 22.965) = − 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 as ³ ´ ´(0) ´(1) ³ ³ e −H e IG e −H e IG e −H e IG H H H T,P T,P T,P = +ω (3.6-24) RTc RTc RTc ³ ³ ´ ´(0) ´(1) ³ Se − SeIG Se − SeIG Se − SeIG T,P T,P T,P = +ω (3.6-25) R R R Appendix D gives the equations developed by Lee and Kesler (1975) to calculate the terms on the 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 From Appendix A Tc = 369.9 K

e fs = ∆H W

Pc = 42.5 bar

(1) ω = 0.153

From the equations given in Appendix D, the enthalpy departures are tabulated below. Therefore, the use of Eq. (3.6-24) gives ³ ´ e −H e IG H T1 ,P1 = − 0.159 + (0.153)(− 0.194) = − 0.189 RTc ´ ³ e −H e IG H T2 ,P2 = − 471 + (0.153)(− 527) = − 0.552 RTc 70

State 1 (Tr = 0.873, Pr = 0.118)

State 2 (Tr = 0.927, Pr = 0.353)

6.9658

2.1836

0.9385

0.8311

6.8739

2.1142

0.9261

0.8047

− 0.0311

− 0.0664

− 0.159

− 0.471

/RTc

− 0.236

− 0.681

/RTc

− 0.194

− 0.527

Parameters (0) Ver (0)

Z (R) Ve r (R)

Z

(1)

Z ´(0) ³ e −H e IG /RTc H

³ e −H e IG H ³ e −H e IG H

T,P ´(R) T,P ´(1)

T,P

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

343

323

Thus,

eP∗ dT = 1612 J/ mol C

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

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, one might think of how to choose the "best" equation of state among various alternatives. In other words, similar to the queen in Snow White who asked, "Mirror, mirror on the wall, who’s the fairest of them all?" (Martin, 1979). The truncated form of the virial equation of state, Eq. (3.1-9), can be used only for the vapor phase for pressures up to approximately 15 bar. For computational purposes, an equation of state representing both vapor and liquid phases would be ideal. In that respect, cubic equations of state have the advantage. However, no cubic equation of state can provide the P V T behavior in all regions of interest. For example, they usually fail near the critical region. Chapter 4 of the book "The Properties of Gases and Liquids" (Poling et al., 2004) provides the guidelines for selecting an equation of state for P V 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 Symposium 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, Doctoral Dissertation, 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 375 K. The experimental value is − 426.5 cm3 / mol (Gupta and Eubank, 1997). Also estimate the density of n-butane at 375 K and 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 coefficient for nonpolar fluids besides Eq. (3.1-10). For example, Meng et al. (2004) modified the wellknown Tsonopoulos (1979) correlation as i RTc h (0) B= f + ω f (1) (1) Pc 72

where

0.30252 0.15668 0.00724 0.00022 − − − (2) 2 3 8 Tr Tr Tr Tr 0.15581 0.38183 0.44044 0.00541 + − − (3) f (1) = 0.17404 − 2 3 8 Tr Tr Tr Tr Calculate the second virial coefficients for methane, ethane, and carbon dioxide at 300 K using Eq. (1). f (0) = 0.13356 −

(Answer: − 42.098 cm3 / mol, − 181.483 cm3 / mol, − 121.869 cm3 / mol) 3.3 Saturated isobutane vapor at 320 K is contained in a piston-cylinder assembly. It is expanded to 3 bar by a reversible isothermal process. At 320 K, the vapor pressure of isobutane is 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 and show that the Joule-Thomson coefficient is given by ¶ µ dB −B T dT Ã ! μ= (1) 2B d e∗ − T C P P dT 2 Substitute Eq. (3.1-10) into Eq. (1) to obtain ⎡ ⎞⎤ ⎛ Tc ⎣ 1.0972 0.8944 ⎠⎦ + ω ⎝− 0.139 + − 0.083 + 1.6 4.2 Pc Tr Tr μ= Ã ! e∗ C 1.75552 3.75648 P + Pr +ω 2.6 5.2 R Tr Tr 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 9

and

N=

9 pq − 27 r − 2 p3 54

If p, q, and r are real and if ∆ = M 3 + N 2 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

(2)

Case (i) Solutions for ∆ > 0 In this case the roots are given by x1 = S + T −

1 p 3

(3)

x2 = −

1 1 √ 1 (S + T ) − p + i 3 (S − T ) 2 3 2

(4)

x3 = −

1 1 √ 1 (S + T ) − p − i 3 (S − T ) 2 3 2

(5)

where S=

q √ 3 N+ ∆

and

Case (ii) Solutions for ∆ < 0

T =

q √ 3 N− ∆

(6)

The roots are given by ¸ ∙ √ 1 θ o + (i − 1)120 − p xi = ± 2 − M cos 3 3 where θ = arccos

s

N2 (− M )3

i = 1, 2, 3

(θ is in degrees)

(7)

(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 methanol vapor at 11 bar and 413 K analytically. (Answer: 0.01106 g/ cm3 ) 3.6 Estimate the molar volumes of saturated liquid and vapor for n-octane at 293 K using: a) Redlich-Kwong equation of state, b) Soave-Redlich-Kwong equation of state, c) Peng-Robinson equation of state. At 293 K, 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, c) 168.6 cm3 / mol, 14, 540 cm3 / mol)

14, 560 cm3 / mol

3.7 The maximum pressure, P , that a spherical tank can withstand is dependent on the tank radius, R, the wall thickness, tw , and the maximum tensile strength of the material, σ. The equation proposed by Strelzoff and Pan (1968) states that ⎧ 2t σ w ⎪ if P < 0.665 σ ⎪ ⎪ R + 0.2tw ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ µ ¶2 P = 2 σ tw + 1 − 2 σ ⎪ ⎪ R ⎪ ⎪ if P > 0.665 σ ¶2 µ ⎪ ⎪ ⎪ tw ⎪ ⎩ +1 +2 R where P is the internal gauge pressure.

74

Estimate the maximum amount of ethylene that can be stored in a spherical tank at 300 K. The tank is made of stainless steel 316, has a radius of 1.5 m, and has a wall thickness of 13 mm. Ethylene is represented by the Peng-Robinson equation of state. For stainless steel 316, σ = 586 MPa. (Answer: ) 3.8 A rigid tank contains propane at 350 K and an unknown pressure. When the tank is cooled to 303 K, the vapor starts condensing. Estimate the initial pressure in the tank using the Peng-Robinson equation of state. At 303 K, the vapor pressure of propane is 10.84 bar. (Answer: 13.234 bar) 3.9 A 0.8 m3 rigid tank contains 2 kg of acetylene at 200 K. Determine the state and pressure of acetylene in the tank using the Peng-Robinson equation of state. The vapor pressure of acetylene at 200 K is 1.905 bar. (Answer: 1.544 bar) 3.10 A 0.05 m3 piston-cylinder device contains 7 moles of isobutane at 350 K. It is compressed isothermally until the volume becomes 0.01 m3 . The vapor pressure of isobutane is given as a function of temperature as 4577.132 + 73.51351 + 1.514279 × 10−5 T 2 T where P vap is in kPa and T is in K. Estimate the initial and final pressures using the PengRobinson equation of state. (Answer: 3.812 bar, 12.469 bar) ln P vap = − 9.416723 ln T −

3.11 Suppose that a rigid cylinder contains propylene at 380 K and 150 bar. The contents of the cylinder are cooled to 285 K and part of the liquid that condensed at the bottom of the cylinder is removed. If the cylinder temperature is again increased to 380 K, will the pressure be 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 is heated at constant pressure until all the acetone vaporizes. Calculate the work done if the temperature at the final state is 400 K. Assume acetone is represented by the Peng-Robinson equation of state. The vapor pressure of acetone is given as a function of temperature as ln P vap = 10.0311 − where P vap is in bar and T is in K. (Answer: − 58, 598 J)

2940.46 T − 35.93

3.13 From Figure 3.3-a, conclude that the isotherm satisfies the following equation in the two-phase region: Z Ve V vap e V L e P dVe (1) P (V − V ) = Ve L

a) If a substance is represented by the Redlich-Kwong equation of state, show that Eq. (1) takes the form à ! # " eV − b e V (Ve L + b) V a V P vap (Ve V − Ve L ) = RT ln − √ ln (2) b T Ve L − b Ve L (Ve V + b) 75

b) Express Eq. (2) in terms of dimensionless quantities to obtain ∙ V L ¶ ¸ µ V Z (Z + B) A Z −B V L + ln =0 Z − Z − ln B ZL − B Z L (Z V + B)

(3)

Note that Eq. (3) can be used to estimate vapor pressures of pure substances at any given temperature. c) Using a trial-and-error procedure, estimate the vapor pressure of propane at 350 K if it obeys the Redlich-Kwong equation of state. (Answer: c) 30.9 bar) Problems related to Section 3.2 3.14 A rigid container is completely filled with 2 kg of mercury at 293 K and 1 bar and sealed. Estimate the amount of heat transferred so as to increase the pressure to 100 bar. The following data are provided for mercury ρ = 13.6 g/ cm3

bP = 0.139 J/ g. K C

β = 1.82 × 10−4 K−1

κ = 3.95 × 10−6 bar−1

bV 6= C bP . Hint: In this case C (Answer: 507.9 J)

3.15 A piston-cylinder assembly contains 5 moles of carbon monoxide at 300 K. The gas is compressed from 3 L to 1 L by a reversible isothermal process. Calculate the heat removed from 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 nonflow processes: (i) From an initial state of 300 K and 4 bar (state 1), it is compressed to 415 K and 15 bar (state 2); (ii) It is then cooled to 325 K at a constant pressure of 15 bar (state 3). For the overall process, the work done on the gas is 320 J. Estimate the heat interaction with the surroundings 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 330 K and 20 bar. The piston-cylinder assembly is well insulated and a mechanical lock prevents the piston from rising. The piston has a mass of 500 kg and a cross-sectional area of 0.02 m2 . The ambient pressure is 1 bar. The mechanical lock is now released. If acetylene obeys the Peng-Robinson equation of state, estimate the gas temperature when the system reaches equilibrium. Take g = 10 m/ s2 . (Answer: 267.7 K) 3.18 Consider a closed system undergoing a reversible adiabatic expansion from state 1 to state 2. a) Starting with

µ

∂T ∂ Ve

¶ Ã e ! Ã e! ∂V ∂S = −1 ∂T e e ∂ Se S T

76

V

(1)

show that

µ

∂T ∂ Ve

¶

e S

=−

T e C V

µ

∂P ∂T

¶

(2)

Ve

b) If an ideal gas with constant heat capacity undergoes an isentropic process, show that Eq. (2) reduces to ⎛ ⎞γ−1 T2 ⎝ Ve1 ⎠ = (3) T1 Ve2 e∗ . e∗ /C where γ = C P 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 equation of state, Eq. (3.1-9), with the parameter B dependent on temperature in the form B =α+βT e∗ is assumed constant, show that the outlet temperwhere α and β are known constants. If C P ature is given by α (P1 − P2 ) T2 = T1 + e∗ C P

3.20 If a gas is represented by the Peng-Robinson equation of state, use Eq. (3.3-10) and show that ⎧ ⎫ ⎪ ⎪ 1 A Γ ⎪ ⎪ ⎪ ⎪ ! Ã + ⎪ ⎪ ⎨ ⎬ Z − B Z (Z + B) + B(Z − B) ∂ Ve R = ⎪ ∂T P ⎪ 1 2 A (Z + B) ⎪ ⎪ ⎪ P ⎪ ⎪ ¤2 ⎪ ⎩ (Z − B)2 − £ ⎭ Z (Z + B) + B (Z − B) where Γ is defined by Eq. (3.2-20).

3.21 Calculate the change in molar enthalpy when isobutane gas is compressed from 350 K and 10 bar to 450 K and 90 bar in a steady-flow compressor. Use the Peng-Robinson equation of state. (Answer: 2431 J/ mol) 3.22 It is proposed to preheat a gas from 303 K to 398 K in a heat exchanger. The pressure of the gas at the inlet is 35 bar and the pressure drop in the heat exchanger may be considered negligible. Estimate the amount of heat required in J/ mol using the following data aP Z =1+ √ T

a = − 0.02 K1/2 / bar

with

eP∗ = 11.7 + 0.03 T C

e∗ is in J/ mol. K, and T is in K. where C P (Answer: 2103 J/ mol)

3.23 A vertical cylinder fitted with a frictionless piston contains 0.15 m3 of nitrogen at 350 K. The piston has a mass of 250 kg and a cross-sectional area of 0.1 m2 . The piston and cylinder 77

do not conduct heat, but heat can be added to the gas by a heating coil. The ambient pressure is 1 bar and the acceleration of gravity is 10 m/ s2 . Assume that nitrogen is described by the e∗ = 2.5R. If 2 kJ of energy is supplied to nitrogen Redlich-Kwong equation of state with C V through 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) 365 K, 1.32 bar b) 360.6 K, 1.25 bar) Problems related to Section 3.4 3.24 Using the van der Waals equation of state show that µ ¶ ∂P R = ∂T Ve Ve − b

e∗ − R into Eq. (3.4-11) to obtain e∗ = C Substitute Eq. (1) and C V P ⎤ ⎡ Z T2 C e∗ e T1 (V2 − b) P e ⎦ ⎣ dT ∆S = R ln + T T1 T2 (Ve1 − b)

(1)

(2)

Show that the substitution of

ZRT Ve = P

and

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 500 K 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 298 K. Is this process reversible, irreversible, or impossible? Assume that n-pentane obeys the Peng-Robinson equation of state. (Answer: Impossible) 3.26 Steam enters an adiabatic turbine at 70 bar and 873 K and leaves at a pressure of 4 bar. Estimate the maximum amount of work that can be delivered by the turbine if steam obeys the Peng-Robinson equation of state. (Answer: − 15, 190 J/ mol) 3.27 A chemical plant is in need of ethylene at 325 K and 15 bar. Since the available ethylene is at 480 K and 50 bar, two schemes have been suggested to decrease the temperature and pressure of ethylene to the desired values. In Scheme I, ethylene is first throttled from 50 bar to 15 bar and then sent to the heat exchanger to decrease its temperature to 325 K. In Scheme II, ethylene is first expanded in a reversible and adiabatic turbine to 15 bar and then sent to the heat exchanger. If ethylene obeys the Peng-Robinson equation of state, determine the heat transfer and work per unit mass of ethylene in each of the suggested schemes. b = − 131.4 J/ g) c=0 Q b = − 279.6 J/ g; Scheme II: W c = − 148.2 J/ g Q (Answer: Scheme I: W 78

Problem related to Section 3.5 3.28 Estimate the volume of a cylinder holding 150 moles of methane at 300 K and 40 bar using the principle of corresponding states. (Answer: 0.0876 m3 ) Problems related to Section 3.6 3.29 If a gas is represented by the virial equation of state, show that the enthalpy and entropy departures are given by ´ ³ Ã " !# e −H e IG H 0.139 0.083 1.0972 0.8944 T,P = Pr − +ω − (1) 2.6 5.2 RT Tr Tr Tr Tr ´ ³ Ã ! Se − SeIG 0.6752 0.7224 T,P = − Pr +ω (2) 2.6 5.2 R Tr Tr Ammonia at 350 K and 15 bar enters a throttling valve and leaves it at 1 bar. Estimate the exit temperature using the virial equation of state. (Answer: 331.13 K) 3.30 Calculate the departure functions for internal energy, enthalpy, and entropy for propane at 400 K 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: µ ¶ P b Z =1+ a− T RT where a = 15 × 10−6 m3 / mol and b = 0.06 m3 . K/ mol. a) Show that the departure functions for enthalpy and entropy take the form ³ ³ ´ ´ e −H e IG H Se − SeIG T,P T,P = A −2B and = −B RT R where the dimensionless parameters A and B are defined by A=

aP RT

and

B=

bP RT 2

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

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3.32 A well-insulated cylindrical tank of volume 0.1 m3 initially contains oxygen at 50 bar and 300 K. A relief valve is opened and oxygen is allowed to escape at a constant flow rate of 8 mol/ min. If oxygen is represented by the Peng-Robinson equation of state, estimate the temperature and pressure of oxygen in the tank after 10 minutes. Hint: Assume that the oxygen remaining in the tank undergoes a reversible and adiabatic expansion. (Why?) (Answer: 245.3 K, 25 bar) 3.33 A steady-flow turbine compresses ethane from 300 K and 1 bar to 350 K and 50 bar. Use Eqs. (3.6-3) and (3.6-4) to determine the changes in enthalpy and entropy, respectively, if ethane 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 constant volume is given by Z Ve µ 2 ¶ ´ ³ ∂ P ∗ e e =T dVe (1) CV − CV 2 ∂T T,P e ∞ V

Use Eq. (1) in Eq. (2.2-49) and show that the departure function for heat capacity at constant pressure is given by ´ ³ eP∗ eP − C C

T,P

=T

Z

Ve

∞

µ

∂2P ∂T 2

¶

Ve

dVe − T

(∂P/∂T )2Ve

(∂P/∂ Ve )T

−R

3.35 The departure function for molar heat capacity at constant volume is defined by ∙³ ³ ´ ´ ¸ ∂ ∗ IG e −U e eV eV − C U = C ∂T T,P T,P Ve

(2)

(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 to ´ ³ " √ ¢ # ¡ e∗ eV − C C V Z + 1+ 2 B A Γ(1 + Γ) T,P √ √ ¢ ¡ ln = (2) R 2 8B Z + 1− 2 B where Γ is defined by Eq. (3.2-20).

b) Combine Eq. (2) with Eqs. (2.2-51) and (3.2-12) to obtain ´ ³ " √ ¢ # ¡ e∗ eP − C C P Z+ 1+ 2 B A Γ(1 + Γ) T,P √ √ ¢ ¡ = ln R 2 8B Z+ 1− 2 B ∙ ¸2 1 AΓ + Z − B Z (Z + B) + B(Z − B) + − 1 (3) 1 2 A (Z + B) − (Z − B)2 £ Z (Z + B) + B (Z − B)¤2

c) Calculate the departure functions for molar heat capacities at constant volume and pressure for ethylene gas at 300 K 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 12 mol/ s of process steam at 2 bar with not less than 97% quality and not more than 15 K superheat. It is proposed to throttle the available steam at 1173 K and 20 bar through an adiabatic valve and then cool it to the required conditions for the process steam. Estimate the minimum duty of heat exchanger using the principle of corresponding states. The vapor pressure of steam is given as a function of temperature as 3816.44 ln P vap = 11.6834 − T − 46.13 where P vap is in bar and T is in K. (Answer: 394, 296 W)

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