ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt

Using the real gas state equations we are able to de-termine differences in fluid behavior from its ideal state for properties, such as u, h, s, cv and cp.. PARTIAL MOLAL PROPERTY In Cha

jj Example 36

temperature?

Solution

µJ T = (∆h/cp)/∆P = ∆T/∆P = (Te –203)/(40 – 60) = 0.31 K bar–1 Therefore, Te =–73.2ºC

There-If v < T(∂v/∂T)P or T βP < 1, µJT > 0, and vice versa

At the inversion point, µJT = 0 Inversion occurs when TinvβP = (T/v)(∂v/∂T)P = 1 For

a real gas Pv = ZRT, i.e., ∂v/∂T = ZR/P + (RT/P)∂Z/∂T If (T/v)(∂v/∂T)P = 1, this plies that ZRT/Pv + (RT2/Pv)∂Z/∂T = 1, or (∂Z/∂T)P = 0

im-For cooling to occur, the inlet pressure must be lower than the inversion pressure.Cooling of a gas can also be accomplished using isentropic expansion We define µs =(∂T/∂P)s that is related to the temperature decrease due to the work delivered during

an isentropic process Recall that

isen-For substances that expand upon heating βP > 0 and, hence, µs > 0 If βPT > 1, then

µJT < µs, i.e., the isentropic expansion results in greater cooling than isenthalpic pansion for the same pressure ratio

ex-The values of Tinv, βP, βT, and Psat can be directly obtained from the state equations,while µJT, h, u, cv, and other such properties depend both on the equations of state andThe experimentally determined value of µJT for N2 is 0.31 K bar–1 at–70°C and 5MPa If the gas is throttled from 6 MPa and –67ºC to 4 MPa, what is the final exit

the ideal gas properties Therefore, the accuracy of the state equations of state can bedirectly inferred by comparing the predicted values of Tinv, βP, βT, and Psat with ex-perimental data.

The entropy change during an adiabatic throttling process equals the entropy ated Therefore, its value must be positive, since throttling is an inherently irreversibleprocess We can determine ds by using the following relation:

(b)

(a) Exit

Figure 20: (a) Illustration of a throttling process (b) The

variation of temperature with respect to pressure during

throt-tling

(∂s/∂P)h = – (v/T) (160)During throttling, dPh < 0 and v/T > 0; hence Eq.(160) dictates that dsh>0 From secondlaw for fixed mass adiabatic system, dsh (= δσ) =-(v/T) dPh> 0.

Since µJT < 0, incompressible liquids will heat up upon throttling

In context of Eq.(F), if b « v, (v–b)2≈ v2 Consequently, using Eq.(F),

In the limit T→∞ (when the fluid approaches ideal gas behavior), Eq (F) implies that

µJT = –b/cp, i.e., µJTcp/vc´ = –b/vc´ = –1/8 This limiting value is based on the real gasequation of state However, for ideal gases, µJT = 0, since b→0 for the ideal gas pointmass molecules Recall that b equals the geometrical free volume available for mole-cules to move around The corresponding result using the RK equation is µJT cP/vc´ ≈–b/vc´≈ –0.08664

2 Temperature Change During Throttling

Recall from Chapter 2 that the internal energy of unit mass of any fluid can bechanged by frictional process and by performing boundary deformation work (Pdv) For in-compressible fluid, dv=0 and hence boundary deformation work is zero; thus "u" can changeonly with a frictionless process (e.g., flow of liquids through pipes) where the mechanical part

of energy “vdP” is converted into internal energy using frictional processes Thus a nation of both of the processes Pdv and vdP result in temperature change during the throttlingprocess

combi-a Incompressible Fluid

Assume that an incompressible fluid is throttled from a higher to a lower pressure der steady state conditions Let us follow unit mass as it enters and exits the throttling device.Since dv=0, there is no deformation work and hence u cannot change due to Pdv However theunit mass is pushed into the throttling device with a pump work of vPi, while the same mass ispushed out with a pump work of vPe Thus there must have been destruction of mechanicalenergy from vpi to vPe which is converted into thermal energy (see Chapter 2) resulting in anincrease of u and hence an increase of T during throttling The energy increases by an amount

un-du = –vdP ≈ –v(Pe – Pi) (Alternately dh =du+ pdv+ vdp = du +0+ vdp=0.) Also, recall that µJT

≈ – v/cP < 0 Further throttling is inherently irreversible process and hence entropy always creases in adiabatic throttling process (e.g., increased T causes increased s)

in-b Ideal Gas

Now consider a compressible ideal gas Visualize the throttling process as a two stepprocedure First the specific volume is maintained as though fluid is incompressible and theenergy rises by the amount du = –vdP This leads to gas heating Now let the volume increaseduring the second stage (i.e expansion to low pressure) The volume increase cannot changethe intermolecular potential energy, since the gas is ideal, but Pdv work is performed Thisleads to decrease in the internal energy The total energy change is du = – vdP – Pdv = – d(Pv),i.e., or du + d(Pv) = dh = 0 For ideal gases, dh = cpdT, and, hence, dT = 0 In this case, theenergy decrease by the Pdv deformation work equals the energy increase due to pumping (=-vdP)

c Real Gas

In a real gas the additional work required to overcome the intermolecular attractionforces (or the increase in the molecular intermolecular potential energy, “ipe”) must be ac-counted for Then the temperature can decrease if increase in “ipe” is significant Consider therelation

µJT = – (v –T (∂v/∂T)P)/cp = – v/cp + {T (∂v/∂T)P}/cp

The first term on the RHS of this expression represents the heating effect due to flow work,while the second term accounts for the entire Pdv work and the energy required to overcomethe intermolecular forces Generally, both the terms are important for fluids.

3 Enthalpy Correction Charts

The Joule Thomson coefficient µJT = – (∂h/∂P)T/cP However,

– (∂h/∂P)T = –(∂(h–ho)/∂P)T – (∂ho/∂P)T = –(∂(h–ho)/∂P)T –0 = {∂hC/∂P}T

Therefore,

µJT = {∂hC/∂P}T/cP, and µJT,R = µJTcPPc/RTc = (∂hR,C/∂PR)T

where hR,c are given in enthalpy correction charts (Appendix, Figure B.3) The behavior of hR,C

is illustrated in Figure 22 Its value increases with a decrease in the pressure at a specifiedvalue of TR along the curve ABI Consequently, (∂hR,c/∂PR)T

R < 0, and µJT < 0 On the otherhand along curve ICD (∂hR,c/∂PR)T

R > 0 Point I represents the inversion point The ture change during throttling that accompanies a pressure change can be determined using theenthalpy correction charts Recall that ∆h ≈ 0, i.e., h2 – h1 = 0, i.e., (ho2 – hC,2) – (ho1 – hC,1) = 0.Assuming a constant specific heat and considering ideal gases, this leads to the relation (T2 –

Now R = (8.314/28.02)=0.297 kJ kg–1 K–1,cP =0.955×0.297+1.039=1.322 kJ kg–1 K–1,i.e.,

c Empirical Relations

For several gases, such as CO2, N2, CO, CH4, NH3, C3H8, Ar, and C2H4, the inversioncurve is approximately described by the expression

5 Throttling of Saturated or Subcooled Liquids

The cooling or heating of a vapor during throttling is partly due to the destruction ofthe mechanical part of the energy as well as boundary deformation work that occurs When asaturated liquid is throttled from (P1,T1) with enthalpy hf1, the final pressure P2 < P1 and tem-perature T2 < T1 As the pressure decreases, Tsat also decreases with decreased enthalpy ofsaturated liquid to hf2 Then the difference hf1- hf2 is used to evaporate a portion of the liquidsince h = hf1 Recall that ln P = A – B/T, (eq (153)) i.e.,

ln (P1/P2) = B/T1 (T1/T2 –1) or T2– T1 = – T1/(1 +B/(T1 ln (P1/P2)))

Therefore, defining the average Joule Thomson coefficient,

µJT = (T2– T1)/(P2 – P1) = –T1/(P2 – P1) (1 +B/(T1 ln (P1/P2))), or

≈ (T1/(P2 – P1) –BP1/T1) when (P1 – P2) « P1 (A)The quality x2 at the exit can be determined as follows Assuming state 1 is a saturated liquid,

P R

T R

RK Miller VW

Berth

Figure 22: Inversion curves for various state equations

6 Throttling in Closed Systems

Consider a large insulated tank that is divided into two sections A and B Section Aconsists of high pressure gases at the conditions (PA,1, TA,1) and section B consists of low pres-sure gases at the state (PB,1, TB,1) If the partition is ruptured, the tank will assume a new equi-librium state The state change occurs irreversibly and the entropy reaches a maximum value.The new equilibrium state can be obtained either by differentiating S = SA + SB with respect to

TA subject to the constraints that U,V and m are fixed, or by applying the energy balanceequations

Similarly, vB,1= 0.167 m3 kmole–1 (for which, PR,B,1 = 1.5 TR,B,1 = 1.2, and Z = 0.68).Using the relations, VA/vA,1 + VB/vB,1 = 8 and VB = 3VA, we obtain the expression

VA (1/vA,1 + 3/vB,1) = 8, i.e.,

VA = 0.277 m3, VB = 0.831 m3 so that V = 1.108 m3

Thereafter,

v2 = 0.139 m3 kmole–1, NA = 3.024 kmole, and NB = 4.976 kmole

Recall that the internal energy u – uo = (3/2 a/bT1/2) ln(v/(v+b)) In section A,

u – uo = –1684.7 kJ kmole–1, i.e., UA,1– UA,1,o = –5094.8 kJ

Similarly, in section B,

u – uo = –1056.8 kJ kmole–1, i.e., UB,1– UB,1,o = –5258.6 kJ

Applying the First law to the tank, Q – W = ∆U = 0, so that

Likewise, for isenthalpic throttling in sssf devices we can use the relation for (h – h)

Eight kmole of molecular nitrogen is stored in sections A and B of a rigid tank tion A corresponds to a pressure PA,1 = 119 bar and temperature TA,1 = 177 K In sec-tion B, PB,1 = 51 bar and TB,1 = 151 K The partition is suddenly ruptured Determinethe final equilibrium temperature T2 Assume that cv = cvo = 12.5 kJ kmole–1 K–1 (cvodoes not depend upon the temperature), and that the gas behavior can be described by

Sec-the RK equation of state P = RT/(v–b) – a/(T1/2 v(v+b)), where a = 15.59 bar m6

kmole–2, b = 0.02681 m3 kmole–1, Tc = 126.2 K, and Pc = 33.9 bar Assume that the

The entropy generated during adiabatic throttling can be determined using a similarprocedure.

Such calculations are useful in determining the final pressures and temperatures forshock tube experiments These experiments involve a pressurized gas in a section Athat is separated by a diaphragm from section B During the experiment, this dia-phragm is ruptured

7 Euken Coefficient – Throttling at Constant Volume

During the adiabatic expansion of pressurized gases, the following relation applies

du = cv dT +(T ∂P/∂T – P) dv

The Euken coefficient µE is related to a constant volume throttling process during which du =

0, i.e.,

µE = (∂T/∂v)u = –(T ∂P/∂T – P)/cv = – T2(∂/∂T(P/T))u/cv (165)For an RK fluid,

This coefficient is always negative, i.e., a specific volume increase is accompanied by a perature decrease during adiabatic irreversible expansion in a rigid system The correspondingentropy change is obtained by applying the relation

For an equation of state P = RT/(v-b) - a/Tn vm, Eq.(165) transforms to µE = (∂T/∂v)u

= - (n+1) a/(cv Tn vm) If cv has a constant value, one can integrate and obtain T(n+1) = (n+1)2a/(cv (m-1) v (m-1))+ C, m ≠ 1 If m=2, n=1, then for constant u, T2 =(4 a/(cv v))+ C

As per this model, at constant values of u, as v → 0, T→ ∞ and T → (Tig) (n+1) as v →

∞, since attractive forces become negligible as we approach ideal gas (ig) limit Hence, batic throttling of a closed system yields, T(n+1) - Tig (n+1) = (n+1)2 a/{cv (m-1) v (m-1)}

adia-Recall from Chapter 6 that a = RTc(n+1)(m+1)2 vc(m-1) /(4 m) Therefore, (TR(n+1) - Tig,R

(n+1))(cv/R)= (n+1)2 (m+1)2)((m2-1)/4m)(m-1)/(4 m (m-1) vR’ (m-1)) For m=2, n=1 (Berthelotequation), the reduced temperature change with pseudo-reduced volume change at constantenergy is provided by the expression (TR(n+1) - Tig,R (n+1))(cv/R) = (27/16) (1/ vR´)

a Physical Interpretation

Consider a container with two sections A and B Section A is filled with pressurizedgases and the second part B contains a vacuum If the partition in Section A is instantaneouslyremoved, the gas in Section A expands into Section B As a result of this process the overallinternal energy remains constant, but the intermolecular spacing increases (hence, the term(T∂P/∂T – P)dv > 0) Consequently, thermal portion of the energy cvdT must decrease (i.e., dT

< 0) in order to compensate for the increase in the intermolecular potential energy (In the case

of an ideal gas, there is a negligible change in the intermolecular potential energy, since thespecific volume is very large Therefore, (T∂P/∂T – P) dv = 0 and there is no change in tem-perature.) Note that no net boundary work is performed for a rigid system

L DEVELOPMENT OF THERMODYNAMIC TABLES

It is apparent from the information contained in the Chapters 6 and 7 thus far that aset of expressions can be developed for the thermodynamic properties of a fluid that exists inany phase if a state relation is known For example, properties for superheated vapors as H2Oand R134 (Tables A-4 and A-5) can be generated using the real gas state equations

Table 2: Reference conditions and ideal gas properties for a few fluids

Property Steam Freon 22 R134A R152A Ammonia Nitrogen Carbon

1 Procedure for Determining Thermodynamic Properties

Thermodynamic properties can be determined, once the state equation, critical stants, and corresponding ideal gas properties are known Some useful formulas are listed be-low and some thermodynamic data is listed in Table 2 The reference conditions should bespecified For example for water, the reference condition is generally specified as that of thesaturated liquid at the triple point The choice of reference conditions is arbitrary Here,

Figure 23: Schematic illustration of a method

of determining the thermodynamic properties

of a material using a P–h diagram

range 0 K < T < 1000 K, the maximum error is less than 8%.

Solution

A schematic diagram of the procedure followed is illustrated in on a P–h diagram in

Figure 25 and Figure 24

First, the reference condition is selected at which hf = sf = 0 (e.g., the saturated liquidstate at the triple point of water at point A in the figure)

At the reference condition hg = 0 + hfg = 2501 kJ kg–1 (point B in Figure 25 and

Figure 24) Therefore,

sg = hfg/T = 2501.3÷273 = 9.17 kJ kg–1 K–1,

The entropy of the saturated vapor at 273.15 K and 0.0061 bar is 9.17 kJ kg– 1

K–1above the entropy of saturated liquid at same temperature and pressure

For the vapor at 273 K and 0.0061 bar (i.e., at PR,ref = 0.000028 and TR,ref = 0.422) thereduced correction factor (ho – h)/RTc≈ 0, since the pressure is low and the intermo-lecular attraction forces are weak Therefore,

Figure 24: Schematic illustration of the determination of enthalpy of a

vapor or a real gas with respect to the values at the reference

condi-tion

The correction or residual factors at 873 K and 250 bar can be obtained From charts

we see that at PR = 1.13, TR = 1.35, Z = 0.845 Therefore,

which is represented by point C

Other properties can be similarly obtained

2 Entropy

Applying the relation at a temperature of absolute zero, (∂v/∂T)P = –∂s/∂P = 0 fore, there is no entropy change with pressure at T = 0 K and s(Pref, 0 K) = 0 Two paths reachstate B from state “0”: path 0 FLMB, path 0 DEAB Consider the path 0DEAB illustrated in

There-Figure 25 The entropy s(T,P) equals the sum of the entropy change along the paths 0D andDE), the entropy addition at E due to the change from the solid to gas phase, entropy changedue to superheating EA and the entropy change along AB Therefore, at any state (T,P)

where ∂s/∂P = –(∂v/∂T)P Given a state equation, the above equation can be integrated to termine the conditions at point B in Figure 25 If Pref is a small pressure, the entropy at point Acorresponds to the ideal gas value and, hence, cp(Pref, T) = cp,o and sB(T,P) – sA(T,Pref) = s(T,P)– so(T,Pref) = s(T,P) – s0 (T,P) + R ln (P/Pref).

de-M SUMMARY

Ideal and real models of gas behavior are important, since they can be used to obtainthe thermodynamic properties of substances, which are used in conservation and balanceequations The ideal gas models are accurate at high temperatures and low pressures In states

in the vicinity of saturated vapor and at relatively higher pressures, the real gas models providemore accurate predictions of properties Using the real gas state equations we are able to de-termine differences in fluid behavior from its ideal state for properties, such as u, h, s, cv and

cp However, determination of µJT, inversion temperature, saturation temperature do not requireideal gas properties Further, the corresponding values of Z, hR,c, and sR,c and hfg along thesaturation lines can be obtained

Chapter 8

A PARTIAL MOLAL PROPERTY

In Chapter 7, we discussed how the thermodynamic properties of pure components of

a substance could be determined and used in the four important conservation and balanceequations for design of thermal systems (mass, energy, entropy and availability) Thermalsystems, however, generally involve mixtures of several substances, possibly in multiplephases In this chapter, we will discuss the state equations, generalized thermodynamic rela-tions, and thermo-physical-chemical properties for species k in a mixture as well as non-reacting mixtures

For pure simple substances, a pair of properties can be used to determine the state of asystem Using those two properties, we can then determine the other system properties Morethan two properties are required for a corresponding situation involving a mixture to accountfor the mixture composition

1 Introduction

The mixture composition can be represented by the mole fraction XK or mass fraction

YK, or the molality Mo

The subscript k denotes the k–th component of a mixture The subscript “g” for the gas phase

is generally omitted herein

The molality is used to describe liquid solutions, i.e.,

Mo = 10–3× kmole of solute ÷ kg of solvent (1e)

d Molecular Weight of a Mixture

The mixture molecular weight M is the mixture per unit mole of the mixture, namely,

A solution is dilute if the mole fraction of the solute is much smaller than that of the solvent.More generally, a mixture is dilute when the value of the mole fraction of a particular compo-nent dominates the mole fractions of the other components

a Remarks

The nonmeasurable properties, e.g., A, G, and S, can be expressed in terms of themeasurable properties P, V, T, N1, N2, N3, …, etc The chemical potential, which gov-erns the direction of species transfer in a mixture, can be expressed in various forms.For a closed system with no chemical reactions dNk = 0

Based on the previous discussion, a thermodynamic property B (for instance, B = V)can be expressed as

The extensive property B is a partly homogeneous function of degree 1 with respect

to N1, N2 ,N3, …, i.e., if the temperature and pressure are held constant and the ber of moles of each species in the mixture is doubled (although the correspondingmole fractions remain unchanged), the value of B is also doubled

num-The corresponding partial molal property of the i–th species can be written in theform

3 Euler and Gibbs–Duhem Equations

The total differential of Eq (6a) is

Note that X1 + X2 + + XK = 1.

The number of independent variables in Eq (10) is ((K–1)+2) = K+1, while in Eq (6)

it is K+2 At constant temperature and pressure, Eq (9b) assumes the form

Consider partial molal property of species 1 Differentiating the partial molal property

ˆb1 = ∂B/∂N1 with respect to N2, we obtain the relation

∂ˆb1/∂N2 = ∂2B/∂N2∂N1 = ∂/∂N1(∂B/∂N2) = ∂ˆb2/∂N1, (15c)

which is a form of Maxwell’s relations Similarly, ∂ˆb1/∂N3 = ∂ˆb3/∂N1 Using Eq (15c) in Eq.(15a)

Dividing by N

Nk k j k

N, then dN1 = dX1 N + X1 dN If only N1 is altered, then the values of all Nj≠1 are constant Inthat case

dN1 = dX1 N + X1 dN 1, or (1 – X1) dN1 = dX1 N, i.e., dN1 = dX1 N/(1 – X1).Using this result in Eq (15e)

water molecules in the mixture is itspartial molal volume As more water is added to the mixture the total volume increases asshown by curve ADBC in Figure 1 The slope of the mixture volume with respect to the num-ber of moles of water provides a measure of the partial molal volume The point A in the figurerepresents a condition corresponding to trace amounts of water in the mixture, while point Brepresents alcohol in trace quantities An immiscible mixture is formed if two species do notmix at a molecular level and, in that case, the partial molal volume loses meaning.

Measured, 3.83 L

B C

(C ) A

Figure 1 a Mixing of two miscible species b Mixing of two

immiscible species c Determination of the partial molal

vol-ume from a plot of total mixture volvol-ume vs the number of

moles of water in a water/alcohol mixture

i Remarks

In Chapters 1 and 6 we have discussed the functional form for the intermolecular traction forces given by the Lennard– Jones empirical potential Φ(l) between a pair of mole-cules For like pairs of molecules,

For an unlike molecular pair consisting of species k and j,

where εkj = (εkεj)1/2, and the collision diameter σkj = (σk+σj)/2 Note that concentration effects

on σkj and εkj are not included in this model If Fk (= ∂φ/∂l) denotes the intermolecular tion forces between the molecules of species k, and Fkj denotes the corresponding forces be-tween dissimilar molecules of the two species k and j,

ho-If Fkj » Fkk at all concentrations, the mixture is miscible at a molecular level tively, if Fkj «Fkk, at all concentrations, the mixture is completely immiscible In some cases,the mixture is miscible up to a certain mole fraction beyond which Fkj « Fkk Such mixtures arecalled partially miscible mixtures

Alterna-In miscible mixtures, energy must be initially utilized to overcome the intermolecularattraction forces between like molecules (e.g., k–k) Inserting a molecule of j and forming thej–k pairs alters the intermolecular attraction forces Consequently, the system may reject en-ergy (during exothermic mixing) or require it (endothermic mixing)

In an ideal solution Fkk = Fkj, and ˆvk(T, P, X1, X2, ) = vk(T, P) Recall that

dS – δQ/Tb = δσ,

the difference in the entropy after and before mixing is given by the expression

S – S – 0 = σ,

where σ > 0 Furthermore, since Sinitial = Σksk(T, P)Nk and Sfinal = Σkˆsk(T, P)Nk,

Σk(ˆsk(T, P) – sk(T, P)) Nk > 0

Even after ideal mixing, the entropy of a species k inside the mixture at T and P islarger than the entropy of the pure species at same temperature and pressure This is due to theincrease in the intermolecular spacing between the k and j molecules, which increases thenumber of quantum states for each species (Chapter 1)

4 Relationship Between Molal and Pure Properties

Figure 2 Assume that v1 = 0.04072 m3 kmole–1, v2 = 0.0181 m3 kmole–1

Determine the specific volume vid under ideal mixing

Plot ˆv1 and ˆv2 with respect to X1

a constant temperature bath held at 25oC and 1 atm The variation of X1 is shown in

A small amount of liquid water (species 2) is added to liquid methanol (species 1) in

Equation (D) indicates that a plot of vid with respect to X1 is linear However, urements indicate that this is not so.

meas-Using the measured results for v with respect to X1 shown in Figure 2, it is possible

to obtain ˆv1 and ˆv2 using Eqs (24d) and (24e) with b = v Since X1 + X2 = 1, dX1 =–dX2, and

B

S

X 1

Figure 2: Partial and molal volumes of methanol (1) and water (2) at 25 C, 1 atm (From

Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, 4th

Edi-tion, McGraw Hill Book Company, 1987, p 428 With permission.)

water (2) at 25ºC, 1 atm while Figure 3 illustrates a graphical method that can be used

to determine ˆv2 by applying Eq (F) For instance, a tangent to the v– X1 curve atpoint R yields the slope dv/dX1, and the intercept PS represents ˆv2 at that value of X1(i.e., at R) As X1→1, the mixture is virtually pure so that species 1 is mainly sur-rounded by like molecules, and

ˆv1,

x1→1 = v1 = 0.04072 m3 kmole–1 (Point D, Figure 2)

As X1→0, Eq (E) yields

We must, therefore, resort to experiments from which we find that

ˆv1,x

1 →0 = ˆv1 = 0.0375 m3 kmole–1 (point G, Figure 2)

At this condition, species 1 is in trace amounts and is surrounded mainly by unlikemolecules with force fields dominated by molecules of species 2 Likewise,

ˆv2,X1 →1 = v2 = 0.0181 m3 kmole–1, (point A) and

N D

A

ˆ

v

2ˆ

v

Figure 3: Determination of partial molal properties

Then for small values of X1, Eq (I) yields

Using a Taylor series expansion,

V(N1+dN1,N2+dN2,N3+dN3, ) = V(N1,N2,N3, ) + ∂V/∂N1dN1 + ∂V/∂N2dN2 + Suppose initially there is only species 1 and hence initial amounts of N2, N3 etc areequal to zero Assume that small amounts of species 2, 3, 4, … are being added tospecies 1 In that case, δN1 = 0, and δN2 = N2, δN3 = N3, δN4 = N4, … are small.Therefore,

V(N1, 0+N2, 0+N3) = V(N1) + ∂V/∂N1×(0) + ∂V/∂N2 N2 + ∂V/∂N3 N3 + , i.e.,V(N1, N2, N3, ) = V(N1) + ˆv2 N2 + ˆv3 N3 +

Dividing throughout by the total number of moles N (recall that N→ N1),

v (X1→1, X2→0) = v1 + ˆv2N2/N + ˆv3N3/N + = v1 + ˆv2X2 + ˆv3X3 + where X2, X3, … denote mole fractions of trace species In a salt (solute – species 2)and water (solvent – species 1) solution, an upper limit X2,upper≈ 0.3 exists at standardconditions, which is called the solubility limit In a mixing tank, the addition of saltbeyond a 30% salt mole fraction usually results in the settling of solid salt Therefore,one may not reach the limit X2→1 or X1→ 0 in a solution

c Example 3

and salt (solute – species 2) solution:

ˆv2

in terms of N2, and determine the value of ˆv1 Determine

kmole Obtain an expression for ˆv1 in terms of N2, and one in terms of X1

N1 = 1 kg/18.02 kg kmole = 0.0555 kmole,

Therefore, applying Eq (C),

ˆv1 = 18.02 – 505.4 N23/2 – 2151.6 N2 in units of l kmole–1 (D)The mole fraction X2 = (1 – X1) = N2/(N2 + N1) = N2/(N2 + 0.055), i.e.,

The actual volume V = ˆv1N1 + ˆv2N2 can differ from Vid, i.e., if the partition is moved in the rigid system, the final pressure may not be 100 bar

re-d Example 5

of N2 (species 2) at 160ºC and 100 kPa, and 25ºC and 100 kPa

Solution

Since the 433 K temperature is high, we expect each species to behave as though it is

an ideal gas in its pure state Therefore,

Vid = 0.3v1 = (433 K,100 kPa) + 9.7v2(433 K,100 kPa), where

Determine Vid for a gaseous mixture of 0.3 kmole of H2O (species 1) and 9.7 kmole

A large flexible tank is divided into two sections A and B by a partition Section Aconsists of 2 kmole of C2H2 (species 1) at 320 K and 100 bar, and section B consists

of 3 kmole of CO2 (species 2) at the same temperature and pressure The partition isremoved, but the temperature and pressure are maintained constant Determine VA,and VB using the RK equation of state for each component before mixing and the total

wa-Suppose we have 0.3 kmole of H2O(l) in compartment A at P =100 kPa and 9.7 kmole

of N2 at P= 100 kPa in compartment of B of a PCW assembly, which is immersed in

a bath at 25°C If the partition is removed and allowed to equilibrate at 25°C,100KPa, then the vapor will become H2O(g) Since the water changes phase aftermixing, it must be endothermic, i.e., heat must be supplied from the thermal bath toevaporate the water molecules Therefore, after mixing, Fkj « Fkk

5 Relations between Partial Molal and Pure Properties

We have discussed the partial molal volume and now focus on other partial molalproperties

a Partial Molal Enthalpy and Gibbs function

Since the enthalpy H(T, P, N1, N2, ) = U(T, P, N1, N2, ) + PV(T, P, N1, N2, ),the partial molal enthalpyˆhi = (∂H/∂Ni)T, P, N N

b Differentials of Partial Molal Properties

Applying the Gibbs–Duhem equation

Differentiating Eq (29) and using (31) to eliminate dgˆ ,k

Equations (37) to (39) are similar to the corresponding expressions for a pure substance Thus

if state equations are available for mixtures, ˆuk , ˆhk and ˆsk can be determined

i Remarks

Maxwell’s relations can be obtained using Eqs (31)–(34) These relations are similar

to those for pure components

Consider the derivative (∂/∂Ni(∂V/∂T)P, N N N N

j i K

1 , 2 , , ≠ , , ) T, P, N N N N

j i K

1 , 2 , , ≠ , , Switching der of differentiation, the expression equals the term

6 Ideal Gas Mixture

b Pressure

Another form of Eq (42) is

P = N1RT/V + N2RT/V + = p1(T,V,N1) + p2(T,V,N2) + … (44)

If each component alone occupies the whole volume, the pressure exerted by component k is

pk=Nk RT/V (which is called component pressure and is the same as the partial pressure forideal gases) Then the pressure exerted by the mixture

respec-Partially combusted products consist of propane (C3H8): 0.5, O2 : 2.5, CO2: 1.5, and

H2O: 2 kmoles at T = 298 K and P= 1bar Determine the partial molal volume of CO2

If a room contains the mixture at T and P, we can hypothesize that 0.5 kmole (i.e.,

3×1026 molecules) occupies 12.39 m3 while O2 occupies 61.95 m3

c Internal Energy

The internal energy of a system is the combined energy contained in all of the cules in the system In ideal gases the internal energy of a species in a mixture equals its en-ergy in a pure state at the temperature and total pressure of the mixture Therefore,

dS = NBcv,odT/T + NBR(dV/V)

A change in volume occupied by, say, species B from VB to V (e.g., if there are two adjacentrooms with O2 and N2 in each room and if the partition between them is removed, the N2 nowoccupies a larger volume making more quantum states available for energy storage) results in

an entropy change at same temperature Then,

Now we will generalize Eq (51) for several components in a mixture and add a subscript “0”

to denote that the gas is ideal, i.e.,

S0(T, P, N) = Σkˆsk0 Nk, and s = S0/N = Σkˆsk0Xk, or (54)

S0 (T, P, N) = Σkˆsk0(T, pk) Nk, and s = S0/N = Σkˆsk0 (T, Pk) Xk, and (55)

s = Σkˆsk Xk = Σkˆsk(T, Pk) Xk = Σk(ˆsk(T, P) – R lnXk) Xk (56)

f Gibbs Free Energy

The Gibbs free energy G = H– TS so that

A mixture contains 60% N2 and 40% O2 at 298 K and 2 bar The mixture is placed in

a cylinder A that is connected by a rigid semipermeable membrane to another der B that contains pure N2 at 298 K Cylinder B is maintained at constant pressurethrough a piston with adjustable weights When the pressure in cylinder B is rela-tively low, N2 is transferred from A to B As the pressure in cylinder B is raised (byplacing weights on the piston), the transfer of N2 from cylinder A to B ceases at 1.19

b Internal Energy and Enthalpy

In ideal mixture of liquids or real gases,

H2O in the solution is compressed from 200 to 250 kPa, with ˆvH O2 = 0.015 m3 kmole–1, dgˆH O

them-ˆsH O

2 (25ºC, 1 bar, 0.999) = sH O2 (25ºC, 1 bar) – R ln XH O

2 Pure water exists at 25ºC and 1 bar as a compressed liquid We will assume that theliquid is incompressible and that sH O2 (25ºC, Psat = 0.032 bar) = sH O2 (25ºC, 1 bar) =6.621 kJ kmole–1 K–1 Therefore,

ˆsH O

2 (25ºC, 1 bar, 0.999) = 6.621 – 8.314 × ln 0.999 = 6.629 kJ kmole–1

8 Fugacity

a Fugacity and Activity

As in Chapter 7 we define the fugacity for a component as

If ˆvk(P, T, Xk) is known, Eq (66) may be integrated at a given composition to obtain

ˆfk (T, P) Subtracting Eq (66) from Eq (65)

ˆgk(T, P,X1,X2 ) – gk (T, P) = RT ln αˆk + f(T) (70)Lake water at 25ºC and 1 bar absorbs air from the atmosphere If the air mole fraction

Since the above equation is applicable even when k exists in large amounts at which

gk→gk αk →1 0 then f(T) =0, and

ˆgk(T, P,X1,X2 ) – gk (T, P) = RT ln αˆk= ∫ (ˆvk (T, P,Xk) – vk(T, P))dP (71)Note that if ( ˆvk − ) is known as a function of P Then ( ˆgvk k− ) and hence gk αˆk can

be obtained as a function of T, P, Xk The partial molal Gibbs function of a species is veryimportant in determining the condition for its phase equilibrium (cf Chapter 9) in a multicom-ponent mixture and as well as at chemical equilibrium (cf Chapter 12) The fugacity is a usefulproperty For instance, since the chemical potentials µ k = ˆgk equal one another for the vaporand liquid in multiphase systems, i.e., ˆfk,liq = ˆfk,vapor The value of the Gibbs function is nor-mally a large negative number (g→(–∞) when Xk→0 and g→0 as Xk→1), particularly forchemically reacting species, but the corresponding fugacity values lie between zero and unity

b Approximate Solutions for ˆgk

In order to determine the value of ˆgk, we require the values of pure properties andactivities Here, we present some approximate schemes to evaluate ˆgk

i Ideal solution or the Lewis–Randall Model

For an ideal mixture,

ˆgkid(T, P, X1, X2, ) – gk(T, P) = RT ln αˆkid (72)Using Eqs (62c), (68) and (72),

The ideal solution model for component 1 of a binary mixture oftentimes fails when

X1→0, i.e., when species 1 is surrounded by a large amount of species 2 Consider a mixture of

3 mole percent of H2O (component 1) and 97 % (mole %) of N2 (component 2) at 1 bar and

300 K Under these conditions, water (component 1) exists as a vapor in the mixture The tion ˆf1id = X1 f1(300 K, 1 bar) = f1(300 K, Psat) + vf(1 – Psat)/(RT) yields liquid–like fugacities(see Chapter 7 for fk) This is clearly inappropriate for the H2O which exists as vapor in themixture Since there are very few water molecules in the mixture, the N2 molecules impose theforce fields in the H2O–N2 mixture which are negligible attractive force fields Therefore, thewater exists in vapor form in the mixture At low values of X1, the actual fugacity of compo-nent 1 in the mixture corresponds to that of gaseous H2O

rela-We can obtain (dˆf1/dX1) at low values of X1 (e.g., slope at A for curve AEC in Figure

4) assuming that this gradient is constant and determine the value of ˆf1, a method known asHenry’s Law (HL), i.e.,

ˆf1id(HL) = X1(dˆf1/dX1)x

Line ADB represents ˆfH OHL

2 in a H2O–N2 mixture, which is valid when XH O

2 →0 The lated fugacity of H2O at B is called a hypothetical fugacity at XH O

extrapo-2 =1 and is obtained from

HL Note that when water exists at low concentrations at 25ºC and 1 bar, the water in themixture may exist as a gas while at high concentrations it may exist as liquid with N dissolved

in liquid Similarly we can use the slope at point C to determine ˆfH O2 when XN

2→0 Henry’sLaw accurately predicts the fugacity of a component k when Xk→0, while the Lewis-RandallRule predicts ˆfk reasonably well when Xk→1

c Standard States

Instead of expressing ˆgk in terms of gk(T, P), we can express ˆgk in terms of g(T,

Po), where Po denotes a reference pressure If we add and subtract gk(T, Po) to the LHS of Eq.(71), then

We have discussed four possible ways to express ˆgk(T, P) in terms of the Gibbsfunction of (1) a pure component at specified T and P (cf Eq (71)); (2) for a pure component

at (T, Po) (cf Eq (75)); (3) for a pure component at (T, 1 bar) (cf Eq (76)); and (4) for thek–th species in mixture at (T, Po) (cf Eq (79))

HL id O H

fˆ 2 ,

B

LR id O H

fˆ 2 ,

Figure 4: Illustration of Lewis Randall Rule and Henry’s Law for the

estimation of fugacity

i Gas Mixtures

In the context of Eq.(76), if the mixture consists of real gases and the component k is

a gas in its pure state at any temperature and pressure,

ˆgk(T, P, Xk) – gk,o(T, 1) = RT ln (fk(T, P) ˆak/1)

For an ideal mixture of real gases ˆak = Xk, i.e.,

ˆgk(T, P, Xk) – gk,o(T, 1) = RT ln (fk(T, P) Xk /1) (80)For an ideal gas mixture at fk(T, P) = P, i.e.,

ˆgk(T, P, Xk) – gk,o(T, 1) = RT ln (P Xk/1) = RT ln(pk/1) (81)

For a liquid mixture, the standard state can be chosen for the pure liquid component atany temperature, but at atmospheric pressure Applying Eq (76),

ˆgk(T, P, Xk(l)) – gk(T, 1) = RT ln (fk(l) (T, P) αˆk( l)/fk(T,1)) (82)The Poynting correction can be used thereafter to simplify this expression for fk(T,1) (cf.Chapter 7) in terms of saturation properties The procedure for solid mixtures is similar to thatfor liquid mixtures

d Evaluation of the Activity of a Component in a Mixture.

We have discussed how to determine the value of ˆgkid for ideal solutions (Eqs (72)and (62c)) For non ideal mixtures, if a relation for ˆvk in terms of Xk, T, and P is available, then

a relation for ˆgk(T, P,Xk) can be obtained (Eqs (62a) and (71) Recall from Eq (62c)

dis-ˆgk – ˆgkid = R T ln (γk) = R T ln ( ˆφk/ φk) (87)

If the ideal state is selected as that for an ideal gas,

ˆgk – ˆgk,o = ˆgk = R T ln ( ˆφk), (88)since φk = 1 for an ideal gas.

We will now relate the fugacity coefficient ˆφk in terms of Z Multiplying Eq (88) by

Nk,

G- G0 = ΣR T Nk ln ( ˆφk) or g - g0 = ΣR T Xk ln ( ˆφk) (89)Define

k, since Zˆ

k = (∂/∂Nk)(NZ)=(∂/∂Nk)(PV/R T)= ˆvk/ (R T/P) = ˆvk/ vko

=ˆvk/ vo = specific volume of k-th component in the mixture /ideal gas specific volume of thesame component

f Fugacity Coefficient Relation in Terms of State Equation for P

Recall that G = A+ PV, i.e.,

Using Eq (96) in the Gibbs–Duhem Equation (95), one obtains the Duhem–Margules Relation

at given T and P , i.e.,

In a mixture containing two components,

We will see later that for ideal gas mixtures ˆfk = pk Thus if we know the experimental value

of the partial pressure p1 (or ˆf1) and its variation with X1, then the variation of p2 (or ˆf2) withrespect to X1 is provided by Eq (99)

h Ideal Mixture of Real Gases

In an ideal mixture of real gases,

i Mixture of Ideal Gases

In an ideal gas, φk = 1, and fk = P Therefore,

The value of gO2 of a real gas is lower than in an ideal gas

At 300 K and 100 bar, the activity

Since the pressure is low, we expect

each gas to behave as an ideal gas in its

pure state Therefore, at 160ºC and 100

Determine the enthalpy Hid for a mixture

containing 3 kmole of H2O (species 1)

and 97 kmole of N2 (species 2) at 160ºC

ˆ

αO id

2 in an 80 % N2

h1 = 2796.2 kJ kg × 18.02 kg kmole = 50387 kJ kmole (Steam Tables A-4 C).

At 298 K and100 bar, water exists as a liquid when in its pure state, but in the mixture

it exists as a vapor We will specify

Hid = 3 h1(298 K, 100 bar, liquid) + 97 h2(298 K, 100 bar)

= 3 × 105 × 18.02 + 97 × 298 × 28.97 = 843000 kJ

If we use a hypothetical ideal gas state for pure water at 25ºC and 100 kPa, then

Hid = 3 h1(298 K, 100 bar, ideal gas) + 97 h2(298 K, 100 bar)

= 3 × 18.02 × 2547 + 97 × 28.97 × 298 = 975000 kJ

Remarks

It is more accurate to determine Hid using enthalpies of pure components in the samestate as they exist in a mixture We have defined a hypothetical state to determine thepure substance property to account for the phase change from the “natural phase” ofthe pure state This process is schematically illustrated in Figure 5 The fugacity can

ψsw = Xw(ψw(T, P) + RTo ln Xw) + Xs (ψs (T, P) + RTo ln Xs) (F)Substituting Eq (F) in Eq (C) ,we obtain the relation

Since Xs = 0.011, Xw= 0.989,

wopt = 8.314×298×(0.989×ln 0.989 0.011×ln 0.011)

= -150 kJ per kmole of sea water or -152 kJ/kmole of pure water

Note that there is complete separation of water and salt here

s,e = N˙s,i, then,

w,,pure ψˆw,e–dN˙

Furthermore, since the composition change is negligible, ψˆw,e ≈ ψˆw,i, ψˆs,e≈ ψˆs,i ;Eq.(I) assumes the form,

δW˙

opt = dN˙

w,pure (ψˆw,e–ψw,e) = dN˙

w,pure (ψw,e + RT ln Xw,e –ψw,e) = dN˙

wRT ln Xw,e, i.e.,

δW˙

opt/dN˙

w,pure =RT ln Xw,e = 8.314 × 298 × ln(0.989) ≈ –27.40 kJ kmole–1 of pure water

Note that there is no complete separation of salt from sea water Thus, the salt centration in sea water leaving the system is higher

con-j Relation between Gibbs Function and Enthalpy

Recall from Chapter 7 that

gk – gk,o = RT ln fk/P, i.e., ∂((gk – gk,o)/RT)/∂T = ∂/∂T(ln fk/P) (103)Furthermore,