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Chapter 1

Review of Concepts Thermodynamics is used to describe processes that involve changes in temperature, transformation of energy, and the relationships between heat and work. The engineer has to deal with problems that require calculation of heat and work requirements for physical and chemical processes, determination of equilibrium conditions for chemical reactions and for the transfer of chemical species between phases. For various unit operations, Thermodynamics provides the equilibrium data and enthalpy data for design of distillation columns, absorption columns, evaporators, condensers and other units where heat exchange is involved. For unit processes, where chemical reactions are involved, thermodynamics provide information regarding heat effects, maximum conversion, and other parameters. Thermodynamics: the study of interrelation of various physical properties ( such as temperature, pressure, volume and entropy) and their relation to process properties ( sucj as heat and work) Chemical Engineering Thermodynamics: studies in particular the evaluation/ estimation of physical properties ( i.e energy, entropy) for a given setoff conditions (usually temperature, pressure, composition) and their use in process calculations (material & energy balances) Fundamental concepts and notation

1. Mass & Moles :

mass = molecular weight x Moles 2. Temperature:

Fahrenheit T > -459.67, Celsius T > -273.15 ToC = 5/9*(ToF-32)

Kelvin T>0 T9K) = ToC+273.15 Rankine T>0 T9oR) = 1.8 T(k) = T(of)+459.67

Note: Always we use absolute Temperature in thermodynamics. 3. Pressure

Simple definition is Force/area P has various units : 1Pa = 1N/m2; 1kPa = 1000N/m2 1 bar = 100 kPa = 105N/m2 1 atm = 1.01325 bar 1 atm = 14.696 psia

4. Volume and Density Actual volume is called the extensive or total volume Vt Most thermodynamic equations use the molar volume V V = Vt/n ; Vt = nV Related quantity is Specific volume V = Vt/m

Process Engineering Thermodynamics lecture notes – winter 2011-12 - Only for slow performing students

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Molar density is reciprocal of molar volume, and mass density is reciprocal of specific volume. ρ= 1/V

5. Energy Matter can posses 3 basic forms of energy: a) Kinetic energy – calculated by motion of mass relative to reference frame, E = ½

mV2. b) Potential energy – Calculated by position of center of mass relative to reference

frame. E = mgh c) Internal energy - Energy within the system, energies of the molecules in the

system relative to center of mass

There fore Total energy of the system

TE = (KE + PE) macroscopic + IE (due to translation, rotational and vibration of molecules)

6. Processes

Real processes occur due to imbalances in force ( driving force) Driving/ Imbalanced physical forces Motion ( Momentum transfer) Driving/ Imbalanced chemical forces Diffusion ( Mass transfer) Driving/ Imbalanced thermal forces heat flow ( heat transfer) Balanced forces no transfer (ie.) equilibrium Infinitesimal forces infinitely slow transfer ( also called quasi- static) Dissipative forces always work to retard process ( eg. friction) Reversible process if quasi-static and no dissipative forces. Represent an idealization that may/may not be approached.

7. Work a. Resulting from a change in system

From Physics work = force x distance

dW = Force x dL but pressure P = Force/Area Force = Pressure x Area There fore dW = Press x Area x dL But Volume = area x Length ( note Vt = nV) dVt = area x dL dW = Press x d(nV)

b. Work not associated with change in system Then it is refered as shaft work.

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There fore Total work W = Ws + System- Change-work

Sign convention : Work done by the system means Positive Work done on to the system means negative

If First law notation ∆U = Q- W

8. Heat The energy transfer that occurs when two objects at different temperatures come

in contact. All energy exchanges between objects can be classified as either Heat or Work

dQ = T d(nS)

1.1 System In order to deal with the subject in a rigorous manner, some of the basic concepts have to be reviewed. A thermodynamic system is any definite quantity of matter we are interested or prescribed region of space enclosed by a boundary (fig 1.1) to which thermodynamic analyses can be applied. The boundaries may be fixed or moveable. Work or heat can be transferred across the system boundary.

Figure 1.1. Piston (boundary) and gas (system)

In a closed system, matter cannot cross the boundary. Hence the principle of the conservation of mass is automatically satisfied whenever we employ a closed system analysis. When working with devices such as engines it is often useful to define the system to be an identifiable volume with flow in and out. This is termed a control volume.

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Fig 1.1a Sample control volume 1.2 Surroundings Beyond the system, another part of the universe is the surroundings. Everything outside the boundary is the surroundings. The surroundings contain all the ways by which one can observe the system and make manipulations. 1.3 Boundary The system and the surrounding may be separated by a boundary, which may be real like a wall of a holding tank or imaginary like that we draw on paper (fig 1.1). Different boundaries have different characteristics and it may influence the manipulations and observations that can be made on the system. 1.4 Interactions between system and surroundings To manipulate the system there has to be interactions between system and surrounding through the boundary. These interactions cause change in matter or energy. Change in amount of matter can be caused by bulk mode of transfer and diffusion mode of transfer. Change in energy can be caused by thermal mode and work mode of interaction.

1.5 The Concept of a ``State''

The thermodynamic state of a system is defined by specifying values of a set of measurable properties sufficient to determine all other properties. For fluid systems, typical properties are pressure, volume and temperature. More complex systems may require the specification of more unusual properties. As an example, the state of an electric battery requires the specification of the amount of electric charge it contains.

1.5 Equilibrium state of a system The condition of the system at a particular time is called state, such that if we change the temperature, we change the state of the system. This also includes phase changes e.g. changing liquid water at 100 deg C to steam at 100 deg C .In Thermodynamics focus is mainly on equilibrium states.

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Equilibrium means the state of a system in which properties have definite, unchanged values as long as external conditions are unchanged is called an equilibrium state.

[Mechanical Equilibrium]

[Thermal Equilibrium]

A system in thermodynamic equilibrium satisfies:

1. mechanical equilibrium (no unbalanced forces) 2. thermal equilibrium (no temperature differences) 3. chemical equilibrium.

1) mechanical equilibrium : there is absence of any net driving force for bulk mass transfer of for change in system volume this means here (ie.) balance of system pressure P with surrounding pressure or Pext P = Pext 2) Thermal equilibrium: That means there is no driving force for causing thermal interaction. The system temperature will be same as surrounding temperature. 3) Diffusional equilibrium: This means absence of net driving forces for causing diffusional mass transfer.

1.6 Intensive and extensive properties The thermodynamic state of a system is defined by specifying values of a set of measurable properties. For fluid systems, typical properties are pressure, volume and temperature. Properties may be extensive or intensive. Extensive properties depend on the quantity of matter in the system. Thus, if the system is divided into a number of sub-systems, the value of the property for the whole system is equal to the sum of the values for the parts, e.g. volume. Intensive properties do not depend on the quantity of matter present, e.g. temperature and pressure. Specific properties are extensive properties per unit mass, e.g. specific enthalpy, molar volume. Specific properties are intensive because they do not depend on the mass of the system. Properties describe states only when the system is in equilibrium. Changes in property values depend only on the initial and final

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value of the state and not on the path taken. Hence properties are also called state functions. 1.7 Process The state of a system can be changed from the initial state to the final state by applying a process through the available interactions fig (1.2).

Fig 1.2 Just as a system has a set of quantities for identifying its state, so does a process have a set of variables called process variables for monitoring the progress and characterizing its behavior, e.g. Pext, work, Heat. They are not state functions. The nature of the system’s boundary can limit the kind of processes that we can apply. Boundary Process Open Insulated Rigid Closed Isolated

Any mass transfer or energy transfer possible Adiabatic(no thermal interaction) Isometric( no volume change) No bulk or diffusional mass transfer No mass or energy interaction

The succession of states through which the system passes defines the path of the process. If, at the end of the process, the properties have returned to their original values, the system has undergone a cyclic process or a cycle. If a system and surroundings can be returned to its original state, then the process is a reversible process. If the state of the surroundings has changed then the process is an irreversible process. All real life processes are irreversible. 1.8 Changing the state through heat, work The state of a closed system can be changed by interaction with the surrounding through heat and work which are two different ways of energy transfer. Heat (denoted by Q) is energy transferred because of differences in temperature. Heat is identified as it comes across system boundaries. Any other means for changing the energy of a system is called work (denoted by W). There are push-pull work (e.g. in a piston-cylinder, lifting a weight), electric and magnetic work (e.g. an electric motor), chemical work, surface tension work, elastic work, etc. In defining sign convention for heat and work, we focus on the effects that the surroundings has on the system. Thus we define heat (Q) and work (W) as being positive when transferred to the system.

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1.9 State functions and path functions Consider at system at an initial state 1 undergoes a process & goes to a final state 2.(Fig 1.2) Since T, P are properties, changes will be the difference between the initial and final values. ∆T = T2 – T1

∆ P = P2 - P1

If the process proceeds by a sequence of differential steps, then we can write

TdTT

T

∆=∫2

1

Differentials following this are called exact differentials. Their integrals give changes, whose values depend only on the initial and final values and are independent of the integration path. dT refers to differential change in T. All properties are exact differentials. And hence they are also called state functions. A state function is one whose change on going from initial to final is independent of the route taken. In a cyclic process the change in the state function will be zero. In contrast process variables like work W do not form exact differentials. For a process variable dW represents a small amount of work, not change, because work is not a property of the system., If dW is integrated over the entire process, we get the total amount of work involved in the process.

WdW =∫2

1

If the change in a function is dependent on the route taken, then the function is known as a path function. Some familiar path functions are work, heat. 1.10 The phase rule For multi phase systems at equilibrium, the number of independent variables that must be specified to establish its intensive state is given by Gibb’s phase rule.

NF +−= π2

π Is the number of phases (a phase is a homogeneous region of matter e.g. gas, mixture of gases, liquid, crystalline solid etc), N is the number of chemical species and F is the degrees of freedom of the system. For a pure homogeneous fluid=1, π =1, therefore F=2. The state of the pure homogeneous fluid is fixed whenever two intensive thermodynamic properties are set. 1.11 Internal energy Internal energy (denoted by Ut) refers to the kinetic and potential energy changes associated with molecules and intermolecular forces. Because of their ceaseless motion, all molecules possess kinetic energy of rotation, vibration, potential energy resulting from intermolecular forces, bond energy, etc. This form of energy is named internal energy to distinguish it from the kinetic and potential energy of the system resulting from its

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macroscopic position or motion. Absolute values of internal energy cannot be measured, only changes from one state to another. Addition of heat or work to a substance increases the molecular activity, which in turn increases the internal energy. Specific internal

energy, represented as m

Uor

n

UU

tt

= where n and m are the number of moles or mass of

the system is an intensive property. 1.12 The first law of thermodynamics ( Conservation of Energy) One formal statement of the first law of thermodynamics is that although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form, it appears simultaneously in other forms.

∆(energy of system) + ∆(energy of surrounding) =0 ( Note: Energy (like mass) can be neither created nor destroyed) For a closed system, the total energy change equals the net energy transferred into it as heat and work from the surrounding.

∆(nU) = n(∆U)=Q + W ∆U is the change in the internal energy per unit mass or mole of the system. The quantity Q is defined as the heat absorbed by the system from the surroundings. The quantity W is defined as the amount of work done on the system by the surroundings. Ex 1.1 When a system is taken from state a to state b (ref fig) along path acb, 120 J of heat flows into the system and the system does 50 J of work.

(a) How much heat flows into the system along path aeb if the work done by the system is 30J?

(b) The system returns from b to a along path bda.If the work done on the system is 40 J, does the system absorb or liberate heat? How much?

Sol :

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Assume that the system changes only in the internal energy. (a) Applying 1st law,

acbacbtab WQU +=∆ = 120-50=70J

Since U is a state function tadb

taeb

tacb UUU ∆=∆=∆

aebaebtaeb WQU +=∆

3070 −=+= aebaebaeb QWQ

JQaeb 100=

(b) For path bda, 4070 +=+=−=∆−=∆ bdabdabda

tbda

tadb QWQUU

JQbda 1104070 −=−−=

Heat is therefore transferred to the surroundings from the system. 1.13 Enthalpy Enthalpy is equivalent to the sum of the internal energy of the system plus the product of its volume multiplied by the pressure exerted on it by its surroundings. So, enthalpy H = U + PV. Its value is determined by the temperature, pressure, and composition of the system at any given time. 1.14 Specific heat The specific heat of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1 °C. For gases and vapors, which expand when heated, the specific heat depends on how the substance is heated. The specific heat or molar heat capacity at constant volume Cv is found by keeping the volume of the system constant during heating.

Vv T

UC

∂∂= depending on whether U is the molar or specific internal energy.

For the constant volume process, the above can be written as dU = CvdT

Integrating, ∫=∆2

1

T

T

vdTCU

Applying the first law for a mechanically reversible constant volume process in a closed system,

dTCnUnQT

T

v∫=∆=2

1

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The specific heat or molar heat capacity at constant pressure Cp is found by keeping the pressure of the system constant during heating.

PP T

HC

∂∂=

For the constant pressure process, the above can be written as dH = CPdT

Integrating, ∫=∆2

1

T

T

P dTCH

Applying the first law for a mechanically reversible constant pressure process In a closed system

dTCnHnQT

T

P∫=∆=2

1

Ex 1.2 Liquid water at 180º C and 1002.7 kPa has an internal energy of 762 kJ/kg and a specific volume of 1.128 cm3/gm. What is its enthalpy? The water is brought to the vapor state at 300 º C and 1500 kPa where its internal energy is 2784.4kJ/kg and its specific volume is 169.7 cm3/gm. Calculate ∆U and ∆H for the process. Sol : (a) U1=762 kJ/kg, P1=1002.7kPa, v1=1.128 cm3/g 1 cm3 = 10-6m3 , 1 g = 10-3kg v1= 1.128 (10-6)/10-3m3)/kg = 1.128(10-3)m3 H1= U1 + P1v1 = 762 +(1002.7) (1.128x10-3) = 763.13kJ/kg (b) U2=2784.4 kJ/kg, P2=1500kPa, v2=169.7 cm3/g v2= 169.7 (10-6)/10-3m3)/kg = 169.7(10-3)m3 H2= U2 + P2v2 = 2784.4 +(1500)(169.7x10-3) = 3037.5 kJ/kg ∆U = U2-U1 = 2784.4 - 762 = 2022.4 kJ/kg ∆H = H2-H1 = 3037.5 - 763.13 = 2274.37 kJ/kg Tutorial 1 1) Define a thermodynamic system? 2) What is a process? 3) Classify the following properties as intensive and extensive: volume, pressure, temperature, specific volume, internal energy, and molar mass, mass. 4) How many degrees of freedom have each of the following systems? a) Liquid water in equilibrium with its vapor

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b) Liquid water in equilibrium with a mixture of water vapor and nitrogen c) A liquid solution of alcohol in water in equilibrium with its vapor d) A system comprising of chloroform, 1,4-dioxane and ethanol in a two phase system . 5) One mole of gas in a closed system undergoes a four step thermodynamic cycle. Use the data given in the following table to determine numerical values for the missing quantities.

6) In the following take Cv = 20.8 and Cp = 29.1 Jmol-1ºC-1 for nitrogen gas: a) Three moles of nitrogen at 30º C , contained in a rigid vessel, is heated to 250 º C. How much heat is required if the vessel has a negligible heat capacity? If the vessel weighs 100 kg and has a heat capacity of 0.5kJkg-1º C-1, how much heat is required? b) Four moles of nitrogen at 200º C is contains in a piston/cylinder arrangement. How much heat much be extracted from this system, which is kept at constant pressure to cool it to 40º C if the heat capacity of the piston and cylinder is neglected?

Step ∆Ut(J) Q(J) W(J)

1-2 2-3 3-4 4-1

-200 ? ? 4700

? -3800 -800 ?

-6000 ? 300 ?

1-2-3-4-1 ? ? -1400

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Chapter 2 PVT behavior of pure substances The heat and work flows on to a system depend on the thermodynamic properties such as enthalpy and internal energy. For example, the heat required to change the temperature of 1 kg water by 10 degrees will be different from that needed to change the temperature of one kg of air by 10 degrees. For fluids these properties are evaluated from measurements of molar volume as a function of temperature and pressure. The relationship between pressure (P), molar volume (V) and temperature (T) may be expressed mathematically as equations of state. E.g. PV=RT gives a simple realistic model of fluid behaviour. Equations of state are useful in metering of fluids, sizing of vessels and pipelines. The behaviour of a pure substance can be expressed in terms of PVT diagrams. Understanding of phase diagrams is essential to thermodynamics because there will be many occasions in which reference to a phase diagram (or a tabular representation of one) is essential in order to obtain needed property information. 2.1 Schematic representation of P T relation for a pure fluid PT diagram is actually the projection of the 3 d PVT diagram shown in the below figure.

Fig 2.1 On the PVT diagrams, one phase situations, solid (s), liquid (L), vapour (Va) appear as areas. Applying Gibb’s phase rule, two variables (P, V) or (T, V,) or (P, T) will be required to identify the state. In addition to solid, liquid and vapour phase, the graph also depicts fluid and gas which are liquid and vapour states extended to high temperatures.

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The two phase lines for solid - liquid, solid –vapour and liquid- vapour appear in pairs as shown in the figure. The lines for solid liquid equilibrium are the melting curves and are separated by the region marked S +L. The two melting curves are parallel and never intersect. The lines for solid vapour equilibrium are the sublimation curves and are separated by the region marked S + Va . The lines for vapour liquid equilibrium are branches of the vapour pressure curve and are separated by the region marked L + Va.

All three of the two-phase regions collapse to single lines or curves in the PT diagram (fig 2.2). This is because coexisting phases always exist at the same temperature and pressure. So, on a PT diagram the infinite number of mixtures of liquid and vapour that could coexist at a particular pressure and temperature will all collapse onto a single PT point along a curve.

Fig 2.2 Measurements of the vapour pressure of a pure substance both as a solid and liquid lead to pressure versus temperature curves shown by lines 1-2 and 2-C in the figure 2.2. The third line 2-3 gives the solid/liquid equilibrium relationship.Line1-2 the sublimation curve separates the solid and gas regions. Line 2-3 the fusion curve separates the solid and gas regions. Line 2-C, the vaporisation curve separates the liquid and gas regions. All three lines meet at the triple point where the three phases coexist in equilibrium .The vaporization curve 2-C terminates at point C ,the critical point, the coordinates being critical pressure Pc and the critical temperature, Tc the highest pressure and temperature at which a pure species can exist in vapour/liquid equilibrium. The vaporisation curve does not give any information about where inside the two phase region this state lies - it could be saturated liquid, saturated vapour or any one of the infinite number of mixtures of liquid and vapour between the two. Changes of state may be represented by lines on the PT diagram: an isothermal change by a vertical line; an isobaric change by a horizontal line. 2.2 PV diagram

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If a projection is made from the 3D surface onto a plot of pressure versus volume, a PV diagram is obtained (fig 2.3). One phase states occupies areas, two phase estates occupy lines and the solid - liquid - vapour equilibrium point projects as three points.

Fig 2.3 On the P V diagram, the two phase lines appear in pairs for although, the phases are in equilibrium at the same T and P, the two phases have different molar volumes. These two molar volumes can be connected by tie lines. Every tie line is at one particular temperature and pressure. An infinite number of tie lines exist. Each melting line extends without limit to high pressures. The two branches of the vapour pressure curve coincide at a maximum pressure which identifies the critical point. Three isotherms are also shown in the graph: a supercritical isotherm (T>Tc), critical isotherm (T=Tc), sub critical isotherm (T<Tc). The critical point passes through a point of inflection at the critical point, thus defining the critical point as

0=

∂∂

= cTTv

P, 0

2

2

=

∂∂

= cTTv

P

2.3 Mathematical representation of PVT behaviour The most convenient method of representing the PVT behaviour of a substance as a single phase, is through the mathematical expression called PVT equation of state.

0),,( =TVPf . An equation of state may be solved for any one of the three quantities P, V or T as a function of the other two. If V is considered as a function of T and P, That is, ),( PTVV = ,

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dPP

VdT

T

VdV

TP

∂∂+

∂∂=

The above partial derivatives have definite physical meanings and are related to two properties commonly tabulated for liquids and defined as follows.

Volume expansivity, PT

V

V

∂∂= 1β

Isothermal compressibility, TP

V

V

∂∂−= 1κ

As seen from fig 2.3, the isotherms for the liquid phase are very steep and closely spaced. Thus both β and κ are very small. For idealized case of incompressible fluid, β and κ are zero. Hence no equation of state can be written for an incompressible fluid as V is independent of T and P.

Hence, dPdTV

dV κβ −=

For small changes in T and P, κβ and may be taken constant.

Hence, )()(ln 12121

2 PPTTV

V−−−= κβ

Ex 2.1 For acetone at 200C and 1 bar, β =1.487x10-3 ºC-1,κ=62x10-6cm3g-1,V=1.287cm3g-1 . Find a) The pressure generated when acetone is heated at constant volume from 200C and 1 bar to 300C. b) The volume change when acetone is changed from 200C and 1 bar to 00C and 10 bars. Solution: a).

)()(ln 12121

2 PPTTV

V−−−= κβ

0= ( ) ( ) )tan(1212 VtconsPPTT −−− κβ

( ) ( )

barPP

barx

xTTPP

2412401240

240)10)(24()2030(102.6

10487.1

12

5

3

1212

=+=+=∴

==−

=−

=− −

−

κβ

b).

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0303.0)9)(1062()20)(10487.1(ln 63

1

2 −=−−=

−− xxV

V

gcmVVV

gcmV

V

V

/038.0287.1249.1

/249.1)287.1)(9702.0(

.9702.0

312

32

1

2

−=−=−=∆

==∴

=⇒

2.4 Virial equations of state Isotherms for gases and vapors lying on the right of fig 2.3 are simple curves for which V increases as P decreases. The product PV can be expressed as a power series in P: ......2 +++= cPbPaPV If '' , aCcaBb == ,

......)1( 3'2'' ++++= PDPCPBaPV

where etcCBa ,,, '' are constants for a given temperature and chemical species. It has been shown experimentally that a is the same function of temperature for all species and an be taken as a=RT The above equation can be written as ......)1( 3'2'' ++++= PDPCPBRTPV

Define RT

PVZ = called compressibility factor. With this definition and with a=RT, the

above equation reduces to ........1 2'' +++= PCPBZ An alternate expression for Z is also in common use:

......12

+++=V

C

V

BZ

Both the above equations are known as virial expansions and the parameters B’,C’,D’ and B,C,D are called virial coefficients.

It can be shown that ( )2

2

','RT

BCC

RT

BB

−==

The accuracy required determines the number of terms that are kept -- more terms makes the equation more accurate, but also more complicated to work with. Virial coefficients are different for each gas, but other than that are functions of temperature only. Coefficients are normally obtained by making measurements of P, V, and T, and fitting the equation. 2.5 The ideal gas

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The internal energy of a real gas is a function of pressure and temperature(phase rule). This pressure dependency arises as a result of forces between the molecules. If such forces do not exist, no energy would be required to alter the average intermolecular distance, and therefore no energy would be required to bring about volume and pressure changes in a gas at constant temperature. In the absence of molecular interactions, the internal energy of a gas depends on temperature only. These considerations of the behaviour of an ideal gas in which no intermolecular forces exist and of a real gas in the limit as pressure approaches zero lead to the definition of an ideal gas. Because the terms B/V, C/V2, etc of the virial expansion arise on account of molecular interactions, the virial coefficients B,C etc would be zero if there were no such interactions and the virial expansion would reduce to that for an ideal gas. Z=1 or PV =RT The ideal gas is only a model fluid that is useful because it is described by simple equations applicable as good approximation for actual gases. In engineering calculations, gases at very low pressures up to a few bars may be considered ideal. 2.6 Implied property relations for an ideal gas: The definition of heat capacity at constant volume leads for an ideal gas to the conclusion that Cv is a function of temperature only.

Cv

=

∂∂=

dT

dU

T

U

v

=Cv(T)

The defining equation for enthalpy, applied to an ideal gas leads to the conclusion that enthalpy H also is a function of temperature only. H= U+PV = U(T)+RT= H(T) The heat capacity at constant pressure Cp is a function of temperature only:

Cp( )

dT

TdH

T

H

p

=

∂∂= = Cp(T)

A useful relation between Cp and Cv for an ideal gas is obtained from above.

Cp RdT

dU

dT

dH +== = Cv + R

The above relation shows that Cp and Cv vary with temperature for an ideal gas in such a way that their difference is equal to R. For any change of state for an ideal gas, the above equations lead to ∆U=∫CvdT ∆H=∫CpdT 2.7 Equations for process calculations for ideal gases The processes considered are mechanically reversible closed system processes. a) The constant-volume process (isometric process)

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Change in internal energy between two states is given by:

dTCU V∫=∆ , dTCH P∫=∆

∫ =−= 0PdVW

By 1st law , dTCUQ v∫=∆=

Q is the quantity of heat added to the system. b) The constant-pressure process (isobaric process)

dTCU V∫=∆ dTCQH P∫==∆

dTCQ p∫= )( 12 TTRW −−=

By 1st law, dTCHQ p∫=∆=

c) The constant-temperature (isothermal process) The internal energy of an ideal gas does not change in an isothermal process.

0=∆=∆ HU

=

−=

1

2

1

2 lnlnV

VRT

P

PRTQ

=

−=

1

2

1

2 lnlnP

PRT

V

VRTW

d) The adiabatic process:

−

−=

−

−=

=

=

==⇒

=

−−

−−

γγ

γγ

γγγ

γγ

γ

γγ

1

1

21

1

1

211

1

1

2

1

2

1

1

2

2211

11

11

,

tan

tan

P

PRT

P

PVPW

bygivenisprocessadiabaticanofWork

P

P

V

V

T

THence

tconsVPVP

tconsPV

e) The irreversible process The above equations which give property changes – ∆U,∆ H are valid for ideal gases whether the process is reversible or irreversible., in both closed and open systems as the change in properties depend only on the initial and final states of the system. On the other hand, Q and W depend on the process. The work of an irreversible process is calculated by a two step procedure:

i) W is determined for a mechanically reversible process that accomplishes the same change of state as the actual irreversible process.

ii) The above result is multiplied or divided by the efficiency.

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Ex 2.2 Air is compressed from an initial condition of 1 bar and 250C to a final state of 5 bar and 250C by three different mechanically reversible processes. (a)Heating at constant volume followed by cooling at constant pressure; (b)Isothermal compression; (c)Adiabatic compression followed by cooling at constant volume. At these conditions, air may be considered an ideal gas with the constant heat capacities

CV=20.785 Kmol

J

.=(5/2) R CP=29.099

Kmol

J

.= (7/2) R.=8.314J/mol K

Calculate the work required, heat transferred, and the changes in internal energy and the enthalpy of air for each process. Solution: Basis : 1 mol of air Applying ideal gas law,

35

1

11 02479.0

101

15.298314.8m

x

x

P

RTV ===

Similarly, we get, 32 004958.0 mV =

As T is the same in the initial and final state. 0=∆=∆ HU in all the three cases

(a) For constant volume process, 2

2

1

1

T

P

T

P=

KP

TPT 75.1490)15.298)(5(

1

122 ===⇒

For constant volume process, )0,(sin =∆=∆= WceTnCUnQ V

=(20.785)(1490.75-298.15)= 24788 J. JTCH P 34703)15.29875.1490)(099.29( =−=∆=∆ /mole

W=0 For the second step at constant pressure,

JQUW

moleJTCU

JTCHQ

V

P

9915)34703(24788

/24788)75.149015.298)(785.20(

34703)75.149015.298)(099.29(

=−−−=−∆=−=−=∆=∆

−=−=∆=∆=

Hence for the entire process,

JW

JQ

H

U

991599150

99153470324788

03470334703

02478824788

=+=−=−==−=∆=−=∆

(b) for the isothermal compression of an ideal gas,

- 23 -

( ) JP

PnRTWQHence

HU

399015ln)15.298)(314.8(ln,

0

1

2 ==

=−=

=∆=∆

(c) Adiabatic compression: The initial step of adiabatic compression takes the air to its final volume of 0.004958 m3. The temperature T’ at this point is:

JW

moleJTCH

moleJTCU

UW

Hence

QstepthisFor

KV

VTT

P

V

5600

/7840)15.29857.567)(099.29(

/5600)15.29857.567)(785.20(

,

.0,

57.567004958.0

02479.0)15.298(

4.01

2

11

'

==−=∆=∆=−=∆=∆

∆=

=

=

=

=

−γ

For the second step, W=0 (constant volume).

moleJTCH

JTCUQ

P

V

/7840)57.56715.298)(099.29(

5600)57.56715.298)(785.20(

−=−=∆=∆−=−=∆=∆=

Hence for the entire process,

JW

JQHU

560005600

560056000078407840056005600

=+=−=−==−=∆=−=∆

The following figure shows these processes sketched on a PV diagram.

- 24 -

Ex 2.3 An ideal gas undergoes the following sequence of mechanically reversible processes in a closed system: (a)From an initial state of 70º C and 1 bar, it is compressed adiabatically to 150 ºC. (b)It is then cooled from 150 to 70 º C at constant pressure. (c)Finally it is expanded isothermally to the original state. Calculate W, Q, ∆U and ∆H for each of the three processes and for the entire cycle.

Take RCRC pv 2

5,

2

3 == . If the processes are carried out irreversibly, but so as to

accomplish exactly the same changes of state – the same changes of state in P, T, U and H, then different values of Q and W result. Calculate Q and W if each step is carried out with an efficiency of 80 %. R=8.314 J mol-1K-1 Sol: Basis: 1 mol of gas The cycle is shown on a PV diagram below

(a) For the adiabatic compression process, Q = 0;

( )( )( )( ) moleJTCH

moleJTCU

p

v

/166370150785.20

/99870150471.12

=−=∆=∆=−=∆=∆

- 25 -

Applying 1st law, W=∆ U=998 J

Pressure P2 is found 1

1

212

−

=

γγ

T

TPP = ( )

5.2

27370

2731501

++

=1.689 bar

(b) For the constant pressure process,

( )( )( )( ) moleJTCH

moleJTCU

p

v

/166315070785.20

/99815070471.12

−=−=∆=∆−=−=∆=∆

Q= ∆H= -1663 J W=∆U-Q = -998 – (- 1663) = 665 J (c) For ideal gases undergoing an isothermal process, ∆U and ∆H are zero;

=

=−=

1

2

1

2 lnlnP

PRT

P

PRTWQ = ( )( ) J1495

1

689.1ln15.343314.8 =

For the entire cycle, Q= 0-1663+1495=-168 J W=998+665-1495=165 J ∆U=998-998 +0=0 ∆H=1663-1663 +0 =0 a) If the property changes above are carried out irreversibly, for mechanically reversible adiabatic compression=998 J. If the process is 80 % efficient= 998/0.8=1248 J. This step cannot be adiabatic By the 1st law, Q= ∆U-W= 998-1248 = -250J b) The work for the mechanically reversible cooling process is 665 J. For the reversible process, W=665/0.8=831 J Q= ∆U-W = -998 - 831= -1839 J c) As work is done by the system in this step, the irreversible work in absolute value will be less than the reversible work of -1495 J W= (0.8)(-1495)= -1196 J Q= ∆U-W = 0 + 1196 = 1196 J For the entire cycle ∆U and ∆ H are zero with Q = -250 -1829+1196=-883 J W = 1248 +831 -1196 = 883 J 2.8 Applications of the virial equations At temperatures below critical temperatures and low to moderate pressures, the ideal gas equation ceases to apply. Under these conditions, you must deal with "real" gases. Real gases require more complex equations of state than do ideal gases , one of some of them being virial equations of state . The two forms of the virial equations are infinite series. Reasonably close values

- 26 -

are realized when two or three terms are used for gases and vapours at low to moderate pressures. As with the ideal gas equation, the temperatures and pressures used must be absolute. Ex 2.4 Reported values for the virial coefficients of methyl chloride at 125 º C are B = - 207.5 cm3 mol-1 and C = 18200 cm6mol-2 .Calculate V and Z for methyl chloride at 125ºC and 10 bar by a) The ideal gas equation

b) Virial equation RT

BP1

RT

PVZ +==

c) virial equation 2V

C

V

B1

RT

PVZ ++==

The absolute temperature is T = 398K and the appropriate value of the gas constant is R = 83.14 cm 3 bar mol-1K-1

Sol : a) By Ideal gas equation,

V ( )( )

10

39814.83

P

RT == = 3308.972 cm3.mol-1

Z=1

(b) Z RT

BP1

RT

PV +==

V

+=RT

BP1

P

RT= B

P

RT + = 3308.972 – 207.5= 3101 cm3mol-1

( )( )

( )( )39814.83

310110

RT

PVZ == =0.937

c) To facilitate iteration, write equation 2V

C

V

B1

RT

PVZ ++== as

++=+ 2

ii1i V

C

V

B1

P

RTV

For the first iteration, i=0,

++=

200

1 V

C

V

B1

P

RTV , take 3309.97V0 = cm3mol-1

V1 ( )( )

( )

+−=

297.3308

18200

97.3308

5.2071

10

39814.83

= 3106.97 cm3mol-1

- 27 -

V2

++=

211 V

C

V

B1

P

RT

( )( )

( )

+−=

297.3106

18200

97.3106

5.2071

10

39814.83

= 3094.22 cm3mol-1

Iteration continues until the difference Vi+1-V i is insignificant

++=

222

3 V

C

V

B1

P

RTV

( )( )

( )

+−=

222.3094

18200

22.3094

5.2071

10

39814.83

= 3093.36 cm3mol-1

++=

233

4 V

C

V

B1

P

RTV

( )( )

( )

+−=

236.3093

18200

36.3093

5.2071

10

39814.83

= 3093.30 cm3mol-1 V = 3093.30 cm3 mol-1 from which Z = 0.934 2.9 Cubic equations of State Virial equations cannot represent thermodynamic systems where both liquid and vapor are present. A "cubic" equation of state is needed to do this. One such is the Vander Waals equation and the Redlich Kwong equation. Van der Waals equation In real gases, the molecules occupy some space and so the actual volume available for the movement of the molecules is (V-b) where b is the correction for the volume occupied by the molecules. Similarly the velocity of a molecule about to strike the wall of the container is reduced due to attractive forces of the neighboring molecules. Since pressure is a direct consequence of the velocity with molecules strike the container walls, attractive forces decreases pressure.

This reduction in pressure is equal to2 v

a.

- 28 -

( )bVV

aP −

+2

=RT

Imposing mathematical conditions at the critical point, parameters a and b can be determined.

Pc =

c

c

V

RT

8

3

a c

2c

2

64P

T27R=

b =c

c

8P

RT

c

ccc RT

VPZ =

Redlich/Kwong equation : Another two parameter real gas equation is the Redlich-Kwong equation. It is almost always more accurate than the van der Waals equation and often more accurate than some equations with more than two parameters. The Redlich-Kwong equation is

P ( ) ( )( )( )bVVT

a

bV

RT

+−

−=

( )

c

c

P

TRa

2242748.0=

( )

c

c

P

RTb

08664.0=

2.10 Theorem of Corresponding States : Acentric Factor Experimental observation shows that compressibility factors Z for different fluids exhibit similar behavior when correlated as a function of reduced temperature Tr and reduced pressure Pr;

By definition c

r T

TT = and

cr P

PP =

- 29 -

These dimensionless thermodynamic co-ordinates provide the basis for the simplest form of the theorem of corresponding states:

All fluids, when compared at the same reduced temperature and the reduced pressure, have approximately the same compressibility factor (Z) , and all deviate from ideal gas behavior to about the same degree.

Corresponding-states correlations of Z based on this theorem are called two-parameter correlations, because they require use of the two reducing parameters Tc and Pc. Although these correlations are very nearly exact for the simple fluids (argon, krypton and xenon) systematic deviations are observed for more complex fluids. Appreciable improvement results from introduction of a third corresponding states parameter (in addition to Tc and Pc), characteristic of molecular structure; the most popular such parameter is the acentric factor ω, introduced by K.S.Pitzer and coworkers. The basic premise of the three parameter theorem of corresponding states can be written as

All fluids, having the same value of ω, when compared at the same reduced temperature and the reduced pressure, have approximately the same compressibility factor (Z) , and all deviate from ideal gas behavior to about the same degree.

2.11 Generalised correlation for gases: Equations of state that express Z in terms of Tr and Pr are said to be generalised because of their general applicability to all gases and liquids. Generalised correlations find widespread use. Most popular are the correlations developed by Pitzer and co-workers for the compressibility factor Z and for the second and third virial coefficients B, C. Pitzer correlation for the compressibility factor Z:

Z = Z0 + ω Z1 Z0 and Z1 are functions of both Tr and Pr. When ω=0, as is the case for simple fluids, the second term disappears, and Z0 becomes identical with Z. Because the second term of above equation is relatively a small correction to this correlation, its omission does not introduce large errors, and a correlation for Z0 may be used alone for quick but less accurate estimates of Z than are obtained from a three-parameter correlation. The above is a simple linear relation between Z and ω for given values of Tr and Pr. Lee and Kesler have developed tables, which present values of Z0 and Z1 as functions of Tr and Pr Pitzer correlation for the second virial coefficient:

- 30 -

Analytical expressions were given to Z0 and Z1 using the simplest form of the virial equation.

r

r

T

PB

RT

BPZ

^

11 +=+= where ^

B is a reduced second virial

coefficient given by :

c

c

RT

BPB =^

Pitzer and coworkers proposed the correlation.

10^

BBB ω+=

Therefore r

r

r

r

T

PB

T

PBZ 101 ω++=

r

r

T

PBZ 00 1+=

r

r

T

PBZ 11 =

0B and 1B are functions of reduced temperature only represented by the following equations:

6.1

0 422.0083.0

rTB −=

2.4

1 172.0139.0

rTB −=

Ex 2.5 Determine the molar volume of n-butane at 510 K and 25 bar by each of the following :

(a) The ideal gas equation (b) The generalized compressibility factor equation

(c) Equation r

r

T

PB

RT

BPZ

^

11 +=+= with the generalized correlation for ^

B

(a) By the ideal gas equation

V( )( )

25

51014.83==P

RT= 1696.1cm3mol-1

(b) Tc = 425.2K Pc = 38 bar ω= 0.200

- 31 -

2.11.425

510 ==rT 659.038

25 ==rP

From graphs or Lee Kesler tables , Z0=0.865, Z1=0.038 Z= 0.865 + (0.200)(0.038)=0.873

V( )( )( )

25

51014.83873.0

P

ZRT == = 1480.7 cm3mol-1

(c) 6.1

0 422.0083.0

rTB −=

6.12.1

422.0083.0 −= = -0.232

2.4

1 172.0139.0

rTB −=

2.42.1

172.0139.0 −= = 0.059

10^

BBB ω+= = -0.232+ (0.2)(0.059)=-0.220

r

r

T

PB

RT

BPZ

^

11 +=+= = ( ) 879.02.1

659.0220.01 =−+ ,

V=( )( )( )

25

51014.83879.0=P

ZRT= 1489.1 cm3mol-1

Ex 2.6 A mass of 500 g of gaseous ammonia is contained in a 30000 cm3 vessel immersed in a constant temperature bath at 65 º C. Calculate the pressure of the gas by: (a) The ideal gas equation (b) A generalized correlation Sol : The molar volume of ammonia in the vessel is

02.17500

30000===Mm

V

n

VV

tt

= 1021.2cm3mol-1

(a) By the ideal gas equation,

( )( )

2.1021

15.2736514.83 +==V

RTP = 27.53 bar

(b) For ammonia , 253.0,8.112,7.405 === ωbarPKT cc ,

- 32 -

because the reduced pressure is low,( 244.08.11253.27 =≈rP ), the generalized virial coefficient should suffice.

834.07.405

15.338 ==rT

6.1

0 422.0083.0

rTB −=

6.1834.0

422.0083.0 −= = -0.481

2.41 172.0

139.0rT

B −= 2.4834.0

172.0139.0 −= = -0.229

10

^

BBB ω+= = -0.481 + (0.253)(-0.229) = - 0.539

( )( )( )17.161

8.112

7.40514.83539.0 −=−==∧

c

c

P

RTBB cm3 mol-1

RT

BP

RT

PVZ +== 1

( )

( )( )( )

( ) barBV

RTP

RTBVP

BVRT

P

RT

BP

RT

PV

77.2317.1612.1021

15.33814.83

1

1

=−−

=−

=

=−

=−

=−

- 33 -

Tutorial 2

1. For water at 500C and 1 bar, .10418.4 15 −−= barxκ To what pressure must water be compressed at 500C to change its density by 1 %? Assume that κ is independent of P. 2. Five kilograms of liquid carbon tetrachloride undergoes a mechanically reversible, isobaric change of state at 1 bar during which the temperature changes from 273 K to 293 K . Determine ∆ Vt,W,Q, ∆Ht,∆Ut . The properties for liquid carbon tetra chloride at 1 bar and 273 K may be assumed independent of temperature: β = 1.2 x 10-3 K-1, Cp = 0.84kJkg-1K-1 and ρ = 1590 kgm-3. 3. One mole of an ideal gas CP=(7/2) R and CV=(5/2) R, expands from P1=10 bar and T1=600k to P2=1 bar by each of the following paths:

a. Constant volume; b. Constant temperature; c. Adiabatically.

Assuming mechanical reversibility, calculate W, Q, HandU ∆∆ for each process. Sketch each path on a single PV diagram. 4. One mole of an ideal gas initially at 30 º C and 1 bar is changed to 130º C and 10 bar by three different mechanically reversible processes:

a) The gas is first heated at constant volume until its temperature is 130 º C: then it is compressed isothermally until its pressure is 10 bar.

b) The gas is first heated at constant pressure until its temperature is 130 º C; then it is compressed isothermally to 10 bar.

c) The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130 º C

Calculate Q, W, ∆ H and ∆ U in each case. Take RCp 2

7= and RCv 2

5=

5. Estimate the following: a) The volume occupied by 18 kg of ethylene at 55 º C and 35 bar b) The mass of ethylene contained in a 0.25 m3 cylinder at 50º C and 115 bar

- 44 -

CHAPTER 3

THERMOCHEMISTRY

Thermo chemistry is the study of heat produced or consumed in a chemical reaction

3.1 Heat capacity (specific heat) It is the quantity of heat required in J to raise the temperature of 1 kg mass of a substance by 1 K.

Heat capacity of water is 4186 Kkg

J

−.

3.2 Sensible heat It is the quantity of heat effect (enthalpy change) associated with temperature change but no phase change when a known quantity of substance is heated or cooled.

TmCatSensiblehe P ∆=

Where, m is the mass of substance in kg, CP is the heat capacity of substance in Kkg

J

−

and T∆ is the temperature change. 3.3 Enthalpy of vaporization or Latent heat It is the quantity of heat effect (enthalpy change) associated with phase change but no temperature change when a known quantity of substance is heated or cooled. Estimates of the latent heat of vaporization of a pure liquid at any temperature from the known value at a single temperature may be based on the empirical equation given by Watson :

38.0

1

2

1

2

1

1

−−

=r

r

T

T

λλ

where 21,λλ are the latent heats of vaporization at

temperatures T1,T2 in Jg-1 3.4 Temperature Dependence of Heat Capacity Temperature dependence of the heat capacity for a substance in the ideal gas state is given by the empirical equation:

- 45 -

22 −+++= DTCTBTAR

C igP

A, B, C and D are constants characteristic of the particular substance.

∫ ∫==∆=1

0

1

0

T

T

T

T

pp dT

R

CRdTCHQ for a mechanically reversible constant pressure closed

system process Evaluation of the integral is accomplished by substitution of Cp as a function of T

( )

)]()1(3

)1(2

[

1])1(3

)1(2

[

)1

()1(3

)1(2

)1(

020

2200

0

230

200

0

0

330

2200

0

0

0

TTT

DT

CT

BAdT

R

C

T

DT

CT

BATdT

R

C

T

Twhere

T

DT

CT

BATdT

R

C

T

T

P

T

T

P

T

T

P

−++++++=

−++++++=

=

−+−+−+−=

∫

∫

∫

ττττ

ττ

τττ

τ

τττττ

Ex 3.1 Calculate the heat required to raise the temperature of 1 mol of methane from 533.15 to 873.15 K in a steady-flow process at a pressure sufficiently low that methane may be considered an ideal gas. Data: A=1.702 B=9.081x10-3 C=-2.164x10-6 Solution:

- 46 -

JTx

Tx

T

dTR

CRHQ

KT

KT

igP

19778)]1(3

10164.2

)1(2

10081.9)1(702.1[314.8

6377.115.533

15.873

15.873

15.533

330

6

220

3

0

15.873

15.533

0

=−−

−+−=

=∆=

==

==

−

−

∫

τ

ττ

τ

Ex 3.2: Given that the latent heat of vaporization of water at 100º C is 2257Jg-1, estimate the latent heat at 300º C. Let 1λ = latent heat at 100 º C = 2257 J g-1 and 2λ = latent heat at 300 º C

577.01.647

3731 ==rT 886.0

1.647

5732 ==rT

( ) 1371577.01

886.012257

38.0

2 =

−−=λ J g-1

Tutorial 3

- 47 -

1. For steady flow in a heat exchanger at atmospheric pressure, calculate the heat transferred:

(a) When 10 mol of SO2 is heated from 200 to 1100 º C (b) When 12 mol of propane is heated from 250 to 1200 º C 2. How much heat is required when 10,000 kg of CaCO3 is heated at atmospheric

pressure from 50 º C to 880 º C.? 3 Handbook values for the latent heats of vaporization in Jg-1 is given in the table

below for the following pure liquids at 0º C. Calculate the latent heat of vaporization at the normal boiling point (Appendix 1)

CHAPTER – 4

THERMOCHEMISTRY (CONTINUATION) 4.1 Standard Enthalpy of Reaction

∆H at 0º C Chloroform Methanol Tetrachloromethane

270.9 1189.5 217.8

- 48 -

mMlLbBaA +→+ The standard enthalpy of reaction is defined as the enthalpy change when a moles of A and b moles of B in their standard states at temperature T react to form l moles of L and m moles of M in their standard states at the same temperature T. 4.2 Standard Conditions Gas: The standard state is in the ideal gas state at 1 bar Liquid/solid: The standard state is the real pure liquid or solid at 1 bar 4.3 Standard Enthalpy of Formation Standard enthalpy of formation is defined as the heat change when 1 mole of compound is formed from its constituent elements at standard conditions of temperature and pressure.The usual temperature is 298.15K or 25 º C The standard heat of formation of a compound at this temperature is represented by

0298fH∆

Ex 4.1 Calculate the standard heat at 298.15 K for the following reaction:

)(22)()(4 222 gClOHgOgHCl +→+ Data: Standard enthalpy of formation HCl(g)= 923070

298 −=∆ fH Jmole-1

Standard enthalpy of formation of H2O(g)= 2418180298 −=∆ fH jmole-1

Writing the formation equations so that their sum yields the desired reaction equations so that , the he following combination gives the desired result : 4HCl(g)→2H2(g) + 2Cl2 (g) ( )( )307,9240

298 =∆H

2H2(g) + O2(g)→2H2O(g) ( )( )818,24120298 −=∆H

______________________________

)

sin(

)Re(22

1:

42

4232

322

elementsitsfromformednotisSOH

cereactionformationanotSOHSOOH

actionFormationOHCHHOCExample

→+

→++

- 49 -

4HCl(g) + O2(g)→2H2O(g) + 2Cl2(g) 408,1140298 −=∆H J

4.4 Standard enthalpy of Combustion It is the Quantity of heat released when 1 mole of compound reacts with oxygen under standard conditions of temperature and pressure. 4.5 Temperature dependence of 0H∆ The general chemical reaction may be written as:

.

........... 41312111

formulachemicaltheisAtcoefficientricstoichiometheiswhere

AAAA

iiννννν +→++

The species on the left hand side are reactants and the sign convention is negative and on the right hand side are products for which the sign convention is positive.

HH ifi

i ∆∑∆ = 00 ν

Ex. :

3,2,1,

23

232

322

−=+=−=→+

HNHNthen

NHHN

ννν

For standard reactions, products and reactants are always at the standard-state pressure of 1 bar. Standard state enthalpies are therefore functions of temperature only. The fundamental equation relating heats of reaction to temperature is as follows

dTR

RT

T

0

P0

298

0

0

∆C∆H∆H ∫+=

HH and ∆∆ 0

298

0 are heats of reaction at temperature T and at reference temperature

298 K respectively.

- 50 -

DCBforsdefinitionousanawith

AA

T

DT

CT

BTAdT

R

ii

i

T

T

PC

∆∆∆

=∆

−∆+−∆+−∆+−∆=

∑

∫∆

,,log

)1

()1(3

)1(2

)1()(0

330

2200

0

0

ν

τττττ

Ex 4.2 Calculate the standard heat of the methanol-synthesis reaction at 1073.15 K

)()(2)( 32 gOHCHgHgCO →+

Data:

molkJH /135.90)525.110(66.2000

15.298−=−−−=∆

i νi A 103B 106C 10-5D CH3OH 1 2.211 12.216 -3.450 0.000 CO -1 3.376 0.557 0.000 -0.031 H2 -2 3.249 0.422 0.000 0.083 ∆A = (1)(2.211) + (-1)(3.376) + (-2)(3.249) = -7.663 ∆B = (1)(12.216x10-3) + (-1) (0.557x 10-3) + (-2)(0.422 x 10-3)= 10.809x10-3

∆C = (1)(-3.45x10-6) + (-1)(0) + (-2)(0) = -3.45x10-6 ∆D = (1)(0) + (-1)(-0.031x105) + (-2)(0.083 x 105) = -0.135 x 105 Substituting in the above equation,

( )( ) ( )( )2

15.29815.107310809.1015.29815.1073663.7

22315.1073

15.298

0 −×+−−=∆ −

∫ R

CP

( )( ) ( )

−×−−−×−+−

15.298

1

15.1073

110135.0

3

15.29815.10731045.3 5336

= -1618.653

- 51 -

dTR

RT

T

PCHH ∫∆∆∆ +=

0

00

298

0

( )( )653.1618314.890135015.1073 −+−=∆H = -103593 J

Tutorial 4 1) If the enthalpy of combustion of urea, )()( 22 sCONH , at 29.15 K is 631660 J/mol

when the products are CO2(g), H2O(l) and N2(g), what is 0298,fH∆ for urea at 298.15 K?

Data:

Standard enthalpy of formation of CO2(g) = -393509mol

J

Standard enthalpy of formation of H2O(l) = --285830mol

J

- 52 -

2) Calculate the standard enthalpy of the following reaction (oxidation of ethylene to ethylene oxide) at 873.15 K by choosing 1 mol of ethylene as basis.

( ) )()(2

1)( 22242 gOCHgOgHC →+

I νi A 103B 106C 10-5D CH3OH 1 2.211 12.216 -3.450 0.000 CO -1 3.376 0.557 0.000 -0.031 H2 -2 3.249 0.422 0.000 0.083 3) Determine the standard heat of each of the following reactions at 25 º C: (a) ( ) ( )gNHgHgN 322 2)(3 →+

(b) ( ) ( ) ( ) ( )gOHgNOgOgNH 223 6454 +→+

(c) ( ) ( ) ( ) ( )gNOlHNOlOHgNO +→+ 322 23

(d) ( ) ( ) ( ) ( )sCaOgHClOHsCaC +→+ 2222

CHAPTER – 5

Thermodynamic properties of fluids Numerical values for thermodynamic properties are essential to the calculation of heat and work quantities for industrial processes. Fundamental property relations can be developed using the first and second law of thermodynamics. The only requirement being that the system be closed and that the changes occur between equilibrium states. 5.1 Internal energy: Internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale. The first law of thermodynamics is the application of conservation of energy principle.

- 53 -

The change in internal energy of a system is equal to the heat added to the system minus the work done by the system ∆U=Q-W 5.2 Enthalpy H=U+PV P and V are the pressure and volume and U is the internal energy. Enthalpy (H) is exactly measurable state variable since it is defined in terms of three other precisely defined state variables. It is parallel to first law of thermodynamics for a constant pressure system. Q=∆U=P∆V, in this case, Q=∆H. Enthalpy is a useful quantity for tracking chemical reactions. If as a result of an exothermic reaction, some energy is released to a system, it has to show up in some measurable form in terms of the state variables. An increase in enthalpy H=U+PV might be associated with an increase in internal energy which could be measured by calorimetry or with work done by the system or a combination of two. The internal energy U might be thought of as the energy required to create a system in the absence of changes in temperature or volume. But if the process changes with volume, as in a chemical reaction, which produces a gaseous product, then work must be done to produce the change in volume. For a constant pressure process, the work you must do to produce a volume change ∆V is P∆V. Then P∆V term can be interpreted as the work you must do to create room for the system if you presume it started at zero volume. 5.3 Helmholtz free energy: F=U-TS If the system is created in an environment of temperature T, then some of the energy can be obtained by spontaneous heat transfer from the environment to the system. The amount of this spontaneous energy transfer is TS where S is the final entropy of the system. 5.4 Gibbs free energy: Gibbs free energy, G=U-TS+PV = H-TS The internal energy might be thought of as the energy required to create a system in the absence of changes in temperature or volume. But as discussed in defining enthalpy, an additional amount of work PV must be done if the system is created from a very small volume in order to create room for the system. As discussed in defining Helmholtz free

- 54 -

energy, an environment at constant temperature T will contribute an amount TS to the system, reducing the overall investment necessary for creating the system. This net energy contribution for a system created in environment temperature from a negligible initial volume is the Gibbs free energy. Relation between the properties can be shown graphically below:

These fundamental property relations are general equations for a homogeneous fluid of constant composition. 5.5 Thermodynamics relations:

H = ENTHALPY

TS G = FREE ENERGY

TS PV A=TOTAL USEFUL WORK

PV U= INTERNAL ENERGY

UNAVAIL-ABLE ENERGY

EXTERNAL WORK

- 55 -

In the design of chemical plants and equipments, two types of thermodynamic properties are needed.

a. Directly measurable properties like pressure, temperature, volume etc. b. Properties which can not be measured directly. Examples are enthalpy, entropy,

free energy etc. The thermodynamic relations establish a relationship between directly measurable properties and properties which are not directly measurable. Thus the properties under the second category may be evaluated from their relation with directly measurable properties or from thermodynamic relations. 5.6 Theorem of exact differential:

yx

F

x

Nand

xy

F

y

M

obtainweationdifferentifurtherBy

y

FN

x

FM

NdyMdxdFor

dyy

Fdx

x

FdF

Fofaldifferentitotalthethen

yxFFIf

yx

xy

xy

∂∂∂=

∂∂

∂∂∂=

∂∂

∂∂=

∂∂=

+=

∂∂+

∂∂=

=

22

,,

),(

Since the order of differentiation in mixed second derivatives is immaterial, these equations gives,

xy

M

∂∂

=yx

N

∂∂

When F is a function of x and y, Mdx+Ndy is an exact differential expression. The functions, dU, dH, dA and dG are exact differential expressions.

- 56 -

5.7 Maxwell relations: We can write the following relations, called Maxwell relations using the criterion of theorem of exactness.

TP

TV

PS

VS

P

S

T

V

V

S

T

P

S

V

P

T

S

P

V

T

∂∂−=

∂∂

∂∂=

∂∂

∂∂=

∂∂

∂∂−=

∂∂

Tutorial 5 1. Show that:

dPT

V

T

dTCdS

PP

∂∂−=

2. Show that:

- 57 -

dVT

P

T

dTCdS

VV

∂∂+=

3. Prove that, VV

CV

U =

∂∂

4. Prove that, PP

CT

H =

∂∂

5. Show that for an ideal gas,

0.

0.

=

∂∂

=

∂∂

T

T

P

Hb

V

Ua

6. Show that for a PVT system,

dVV

dTdPκκ

β 1−=

7. Prove that, κβ 22

TV

V

P

T

VTCC

T

PVP =

∂∂

∂∂−=−

- 59 -

CHAPTER 6

CHEMICAL REACTION EQUILIBRIA

The rate of a chemical reaction and the maximum possible (or equilibrium) conversion of a chemical reaction are of primary concern and depend on the temperature, pressure and composition of the reactants. Ex: Consider the reaction, oxidation of SO2 to SO3. A catalyst is required if a reasonable reaction rate is to be attained. With V2O5 catalyst, the rate becomes appreciable at about 573.15 K and continues to increase at higher temperatures. On the basis of rate alone, one would operate the reactor at the highest practical temperature. However, the equilibrium conversion to SO3 falls as temperature rises decreasing from about 90 % at 793.15 K to 50 % at about 953.15 K. These values represent maximum possible conversion regardless of catalyst or reaction rate. Hence both equilibrium and rate has to be considered in the process design .Equilibrium conversions can be obtained by thermodynamic calculations. 6.1 Reaction coordinate: The general chemical reaction can be written as:

.............. 44332211 ++→++ AAAA νννν

where iν is the stoichiometric coefficient and Ai stands for chemical formula. iν s are

called as stoichiometric numbers (positive for products and negative for reactants). Ex:

224 3HCOOHCH +→+

3

1

1

1

2

2

4

==

−=

−=

H

CO

OH

CH

νννν

Stoichiometric number of an inert species is zero.

- 60 -

The changes in the numbers of moles of the species are in direct proportion to the stoichiometric numbers. If 0.5 mol of CH4 disappears by reaction, 0.5 mol of H2O must also disappear. Simultaneously 0.5 mol of CO and 1.5 mol of H2 are formed.

εεν

εννν

νν

νν

),....2,1(,.

....3

3

2

2

1

1

1

1

3

3

1

1

2

2

Niddnei

ddndndn

dndn

dndn

ii ==

====

=

=

is called reaction coordinate which characterizes the extent or degree to which a reaction has taken place.

νεεν

νε

νεεν

ενε

++

==

+=⇒

+==

=+=⇒

=

∑∑∑

∫∫

0

0

0

0

0

0

,

),....3,2,1(0

n

n

n

ny

isispeciesoffractionmole

nn

nnn

Ninn

ddn

iiii

ii

ii

ii

iii

i

n

n

i

i

i

Ex. 1: For a system in which the following reaction occurs

224 3HCOOHCH +→+ Assume that there are initially 2 mol CH4, 1 mol H2O, 1 mole CO and 4 mol H2. Determine the mole fractions yi as functions of ε. sol: For the given reaction, 23111 =++−−==∑ iνν

For the given numbers of moles of species initially present,

- 61 -

84112 =+++==∑ ioo nn

εε28

24 +

−=CHy ε

ε28

12 +

−=OHy

ε

ε28

1

++=COy

εε

28

342 +

+=Hy

6.2 Multi reaction stoichiometry: When two or more independent reactions proceed simultaneously, the stoichiometric numbers are doubly spaced to identify their association with both species and the reaction. Thus νi,j refers to stoichiometric number of species i in reaction j.

Ex. 3: Consider a system in which the following reactions occur:

)2.......(42

)1.......(3

2224

224

HCOOHCH

HCOOHCH

+→++→+

Where (1) and (2) represent the values of j, the reaction index. If there are initially 2 mol CH4 and 3 mol H2O, determine expressions for yi as functions of ε1 and ε2. Sol : The stochiometric numbers ji,ν can be arrayed as follows

__________________________________________________ j CH4 H2O CO CO2 H2 עj

__________________________________________________

∑

∑

∑

∑

∑

∑ ∑∑∑∑

∑

∑

+

+=∴

+=∴

=

=

+=+=

=+=⇒

==

jjj

jj

jii

i

jj

j

ijij

ii

j ijjij

i jji

ii

jj

jiii

jj

jii

n

n

y

nn

nnn

Ninn

Niddn

εν

εν

εν

νν

νν

ενεν

εν

εν

0

,0

0

,

,0,0

,0

,

)(

),.....2,1(

),.....2,1(

- 62 -

1 -1 -1 1 0 3 2 2 -1 -2 0 1 4 2 __________________________________________________

21

21

4 225

2

εεεε

++−−

=CHy 21

21

2 225

23

εεεε

++−−

=OHy

21

1

225 εεε

++=COy

21

2

2 225 εεε

++=COy

22

21

2 225

43

εεεε

+++

=Hy

Tutorial 6

1. Develop expressions for the mole fractions of reacting species as functions of the reaction co-ordinate for:

- 63 -

(a). A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction:

)(6)(4)(5)(4 223 gOHgNOgOgNH +→+

(b). A system initially containing 3 mol H2S and 5 mol O2 and undergoing the reaction:

)(2)(2)(3)(2 2222 gSOgOHgOgSH +→+

(c). A system initially containing 3 mol NO2, 4 mol NH3 and 1 mol N2 and undergoing the reaction:

)(12)(7)(8)(6 2232 gOHgNgNHgNO +→+

Consider a vessel which initially contains only n0 moles of H2O vapor. If decomposition occurs according to the reaction,

222 2

1OHOH +→

Find expressions which relate the number of moles and the mole fraction of each chemical species to the reaction coordinate ε. 2. A system initially containing 2 mol C2H4 and 3 mol O2 undergoes the reactions:

)(2)(2)(3)(

)()()(2

1)(

22242

22242

gOHgCOgOgHC

gOCHgOgHC

+→+

→+

Develop expressions for the mole fractions of the reacting species as functions of the reaction coordinates for the two reactions.

3. A system formed initially of 2 mol CO2, 5 mol H2 and 1 mol CO undergoes the

reactions:

)()()()(

)()()(3)(

222

2322

gOHgCOgHgCO

gOHgOHCHgHgCO

+→++→+

Develop expressions for the mole fractions of the reacting species

as functions of the reaction coordinates for the two reactions.

CHAPTER – 7

CHEMICAL REACTION EQUILIBRIA (CONTINUATION)

- 64 -

7.1 Equilibrium criteria to chemical reactions: The total Gibbs energy, Gt of a closed system at constant T and P must decrease during an irreversible process and that the condition of equilibrium is reached when Gt attains a minimum value. At equilibrium state, .0)( , =PT

tdG

When Gt is plotted against ,ε the reaction coordinate, the reaction coordinate has the

minimum value eε at the minimum of the curve.

7.2 Relationship between Standard Gibbs energy change and the equilibrium constant:

∆−=RT

GK

0

exp

KelvininetemperaturtheisT

tconsgasuniversalKkmol

JR

changeenergyGibbsdardstheisG

tconsmequilibriutheisK

tan.

8314

tan

tan0

==

∆

Effect of temperature on the equilibrium constant: We know that the differential change in Gibbs energy,

SdTVdPdG −= ------(1) G=H-TS---------------- (2) where G is Gibbs energy, S is the entropy and V is the volume. G=G(T,P), that is Gibbs energy is a function of temperature and pressure.

⇒

==

−−−=−=

)tan(0

,,vartan

)()(

1122

tconsPdP

thenetemperaturonlywithyreactionofchangespropertydardsSince

dTRT

TSHSdTVdP

RTdT

RT

GdG

RTRT

Gd

- 65 -

( )

+

∆+

∆+

∆+∆−=∆⇒

+∆+∆+∆+∆+−=

−−−−−−∆−∆+∆+∆+=∆∴

−−−−−−∆+=∆

−−−−−+∆=⇒

∆=⇒

∆−=∆

∫

∫

IT

DT

CT

BTARTJG

IT

DT

CT

BTA

RT

JK

resultfollowingthegivesAequationinCequationofonSubstituti

CT

DT

CT

BTA

R

J

R

H

egrationoftconsanotherisJwhere

BdTCJH

reactionofchangeenthalpydardSthatknowWe

egrationoftconstheisIwhere

AIdTRT

HK

RT

H

dT

Kd

RT

H

dTRT

Gd

P

220

22

320

00

2

0

2

0

2

00

1

262ln)(

262ln)(ln

:)()(

)(32

)(

.inttan

)(

,tan

.inttan

)(ln

ln

Standard Gibbs energy change 0G∆ and standard enthalpy change 0H∆ values can be found from literature and these values when substituted in the above equations give the integration constants I and J. Then, the values of I and J can be substituted in the equation for ln K. This gives the relationship between equilibrium constant K and temperature T. Ex 7.1: Calculate the equilibrium constant for the vapor phase hydration of ethylene at 145 and 3200C. The pertinent chemical reaction can be represented by the following equation:

)()()( 52242 gOHHCgOHgHC →+

Solution:

55

66

33

10121.010)121.000(

1061.110)0392.4002.6(

10157.410)45.1394.14001.20(

376.147.3424.1518.3

xxD

xxC

xxB

A

−=−−=∆−=−+−=∆

=−−=∆

−=−−=∆

−−

−−

- 66 -

Standard enthalpy change of reaction at 298 K is calculated as follows:

7.5308

15.298

12100)15.298)(105367.0()15.298)(100785.2()15.298)(376.1(

314.8

45792

3677.3)15.298)(314.8(

8348ln

8348)228572(68430168490

45792)241818(52510235100

3623

0

0298

0298

−=⇒

+−+−=−

==∆−=

−=−−−−=∆

−=−−−−=∆

−−

R

J

xxR

J

RT

GK

mol

JG

mol

JH

125.7

)15.298)(2(

12100

)15.298)(102683.0()15.298)(100785.2()15.298ln(376.115.298

7.53083677.3

2

263

−=⇒

+−

−+−= −−

I

I

xx

The general expression for Kln is as follows:

300

200

2263

1091.2,15.593320

1026.14,15.418145

125.72

12100102683.0100785.2ln376.1

7.5308ln

−

−

−−

======∴

−−−+−=

xKKCTwhen

xKKCTwhen

TTxTxT

TK

Tutorial 7 1) Calculate the equilibrium constant for the following reaction at 1500C:

)()(2)( 32 gOHCHgHgCO →+

- 67 -

2) For the ammonia synthesis reaction written:

)()(2

3)(

2

1322 gNHgHgN →+

with 0.5 mol N2 and 1.5 mol H2 as the initial amounts of reactants and with the assumption that the equilibrium mixture is an ideal gas, show that:

( ) 5.0299.111 −+−= KPeε

3) Three members A, B and C are asked to find the equilibrium composition at a particular T and P and for given initial amounts of reactants for the following gas phase reaction:

223 2

5332 NOHNONH +→+ -------------(1)

Each solves the problem correctly in a different way. A bases her solution on reaction (1) as written. B, who prefers whole numbers, multiples reaction (1) by 2:

223 5664 NOHNONH +→+ -------------(2)

C, who usually does things backward, deals with the reaction:

NONHNOH 322

53 322 +→+ ------------(3)

Write the chemical equilibrium equations for the three reactions, indicate how the equilibrium constants are related, and show why A,B and C all obtain the same result. 4) The following reaction reaches equilibrium at 773.15 K and 2 bar:

)(2)(2)()(4 222 gClgOHgOgHCl +→+ If the system initially contains 5 mol HCl for each mole of oxygen, find the composition of the system at equilibrium. Assume ideal gases. 5) The following reaction reaches equilibrium at 923.15 K and atmospheric pressure:

)(2)()( 222 gHCNgHCgN →+

- 68 -

If the system initially is an equimolar mixture of nitrogen and acetylene, find the composition of the system at equilibrium. What would be the effect of doubling the pressure? Assume ideal gases. 6) The following reaction reaches equilibrium at 623.15 K and 3 bar:

)()()( 5223 gOHHCgHgCHOCH →+

If the system initially contains 1.5 mol H2 for each mole of acetaldehyde, find the composition of the system at equilibrium. What would be the effect in reducing the pressure to 1 bar? Assume ideal gases.

- 59 -

CHAPTER 8

FUNDAMENTALS IN SOLUTION THERMODYNAMICS AND FLUID P HASE EQUILIBRIA

8.1 Definitions Residual Property The residual property is the difference between the fluid property and ideal gas property. Excess Property The excess property is the difference between the solution property and ideal solution property. Equilibrium Equilibrium is a static condition in which no changes occur in the properties of a system with time. Chemical Potential The term chemical potential is equivalent to free energy for one mole. Raoult’s Law The partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

Partial Pressure = Vapor pressure × Mole fraction ( Py P xi isat

i= ) Partial Pressure In any mixture of gases, each gas exerts its own pressure known as partial pressure. This partial pressure is independent of the other gas present in the mixture. Ideal Mixture An ideal mixture is the one which obeys Raoult’s law. Example: Hexane and Heptane

- 60 -

Benzene and Methylbenzene Azeotrope Azeotrope is defined as a mixture of liquids that has a constant boiling point because the vapor has the same composition as the liquid mixture. For azeotropic mixture Mole fraction in Vapor phase = Mole fraction in Liquid phase 8.2 General Phase Equilibrium Relations Phase equilibrium variables: T, P, moles of each component n Consider a closed system with no chemical reactions. The fundamental property relation with T and P as independent variables involved the Gibbs free energy:

( ) ( ) ( )( )

( )P

T

T

nGnS

P

nGnV

dTnSdPnVnGd

=−

=

−=

∂∂∂

∂

The 1st and 2nd laws and definition of G (G=H-TS) are involved in deriving this property relation. If the system can exchange mass with its surroundings (another phase), then it becomes an open system, and the fundamental property relationship must be modified:

( ) ( ) ( )

( )i

snPTii

i

N

ii

Gn

nG

dndTnSdPnVnGd

j

=

∂=

+−= ∑=

',,

1

∂µ

µ

This new defined property µi is called the chemical potential for component i in a mixture. At equilibrium, the temperature and pressure must be the same in all phases, for thermal and mechanical equilibrium considerations (i.e., no driving forces to cause fluxes of heat or fluid movement). Considering a system considering two phases at temperature T and pressure P In the vapor(α) and liquid phase(β) we can write the following fundamental property relations

- 61 -

( ) ( ) ( )( ) ( ) ( ) βββββ

ααααα

µ

µ

ii

ii

dndTnSdPnVnGd

dndTnSdPnVnGd

∑

∑

+−=

+−=

The total change in Gibbs free energy for the entire system is the sum of the above equations

( ) ( ) ( ) ∑∑ ++−= ββαα µµ iiii dndndTnSdPnVnGd

But, since this is a closed system, the last two terms must sum to zero. Also, any mass leaving one phase must enter the other.

( ) 0

0

=−

−=

=+

∑

∑∑

αβα

βα

ββαα

µµ

µµ

iii

ii

iiii

dn

dndn

dndn

Since the dn's are arbitrary independent variables, we conclude that

Niii ,....,3,2,1== βα µµ

This can be expanded to any number of phases. Thus, in addition to thermal and mechanical equilibrium, the chemical potential for any given component must be the same in all phases at equilibrium. This is phase equilibrium (i.e., no driving forces for diffusion). At equilibrium in a system of π phases and N components,

µ µ µα β πi i i i N= = = =.... , ,....,1 2

8.3 Calculations Using Raoult's Law Bubble-point pressure problem -- T,x given -- P,y unknown.

- 62 -

P

xPy

xPPPy

isat

i

isat

ii

=

== ∑∑

For a binary system,

satsat PxPxP 2211 += ( )( )

P

Pxy

sat11

1 =

Bubble point temperature problem -- P,x given -- T,y unknown. A trial and error procedure must be followed, where T is assumed, the vapor pressures calculated, and then see if the correct total pressure is obtained (or the vapor mole fractions sum to unity).

∑∑ ==P

xPy i

sati

i 1

Dew-point pressure problem -- T,y given -- P,x unknown. No trial and error is needed, as P can be directly calculated.

∑

∑

=

=

=

sati

i

i

sati

ii

P

yP

x

P

Pyx

1

1

- 63 -

Dew-point temperature problem -- P,y given -- T,x unknown. Guess a T, find the vapor pressures, and see if the liquid mole fractions sum to unity.

∑∑

==

sati

ii P

Pyx 1

Ex 8.1 The binary system acetonitrile(1),nitromethane(2) confirms closely to Raoult’s law. Vapor pressures for the pure species are given by the following Antoine equations:

( )

( )209

644.29722043.14ln

224

47.29452724.14ln

2

1

+−=

+−=

tP

tP

sat

sat

. P in kPa, T in ºC a) Generate data for P versus x1 and P versus y1 for a temperature of 75ºC and prepare graph. b) Generate data for t versus x1 and t versus y1 for a pressure of 70 kPa and prepare graph.

Sol : At 75º C vapor pressures calculated are 21.831 =satP kPa, 98.412 =satP kPa a) Writing Raoult’s law for each species,

sat

sat

PxPy

PxPy

222

111

=

=

Since 121 =+ yy , adding the above gives

satsat PxPxP 2211 +=

putting 12 1 xx −= ,the above equation reduces to

( ) 1212 xPPPP satsatsat −+=

- 64 -

For 01 =x to 11 =x , calculate P , and

( )( )P

Pxy

sat11

1 =

1x 1y P

0.0 0.2 0.4 0.6 0.8 1.0

0.0 0.33 0.57 0.75 0.89 1.0

41.98 50.23 58.47 66.72 74.96 83.21

b) For a given pressure, the temperature range is bounded by the saturation temperatures

satt1 and satt2

At kPaPP sati 70==

58.89

84.69

2

1

=

=sat

sat

t

t

in º C When x1 is known along with P

+=+= 2

2

1122211 x

P

PxPPxPxP

sat

satsatsatsat

or

+

=

22

11

2

xP

Px

PP

sat

sat

sat (1)

209

64.2972

224

47.29450681.0.ln

2

1

++

+−=

ttP

Psat

sat

(2)

a) choose a value of sat

sat

P

P

2

1 calculated at some intermediate temperature

b) calculate satP2 from (1)

- 65 -

c) calculate t from the Antoine equation for species 2 :

209ln2043.14

64.2972

2

−−

=satP

t

d) determine a new value of sat

sat

P

P

2

1 by equation (2) and a new value of satP2 by equation

(1) e) return to step (c) and continue iteration till convergence

When 1y is known along with P, use satsat

sat

satsat PPyy

P

PyPyP

2121

1

2211

1

+=

+=

( )satsatsat PPyyPP 21211 +=

224ln2724.14

47.2945

1

−−

=satP

t

For purpose of generating data to prepare txy diagram, select values of t between

satt1 and satt2 ,calculate satP1 and satP2 and evaluate x1 by satsat

sat

PP

PPx

21

21 −

−=

for example at 86º C, 1181 =satP and 03.622 =satP kPa

142.003.62118

03.62701 =

−−=x

( )( )

24.070

118142.0111 ===

P

Pxy

sat

values generated as shown below

1x 1y t º C

0.0 0.1424 0.3184 0.5156 0.7378

0.0 0.2401 0.4742 0.6759 0.8484

89.58 86 82 78 74

- 66 -

1.0 1.0 69.84 8.4 Fugacity The chemical potential can also be defined as

( )TfRT iii θµ += ˆln

Where iµ is the chemical potential of species i

if̂ is the fugacity of species i in solution

iθ is a function of temperature.

. In terms of fugacities, the phase equilibrium relations do not change:

$ $ ...... $ , ,....,f f f i Ni i i

α β π= = = = 1 2 Fugacity is equal to the pressure when the pressure approaches zero (Ideal gas condition). Fugacity Coefficient : The fugacity coefficient is defined as the ratio of the fugacity to pressure.

essure

FugacitytCoefficienFugacity

Pr=φ

The fugacity coefficient takes into account the departure of gas phase from ideal gas behavior. If the value of φ > 1 then it shows the deviation from ideal gas behavior. Fugacities in Mixtures - Ideal Behavior

- 67 -

Gas Phase -- Ideal Gases

$f Pyi

Vi=

Liquid Phase -- Ideal Solutions

$f f xi

Li

Li=

The coefficient on the RHS is the fugacity of the pure liquid at the same T,P as the actual liquid mixture. This relation is called the Lewis-Randall rule. Note: Ideal solutions exhibit no volume change on mixing and no heat of mixing.

8.5 Activity and activity coefficient The activity of any component is defined as ratio of partial fugacity of the component to fugacity of pure component.

( )0

ˆ

fponent f pure comFugacity o

fent the compongacity of Partial Fu

Activity a

=

Activity Coefficient : The activity coefficient is defined as the ratio of the activity to the mole fraction.

ionMole fract

ActivitytCoefficienActivity =γ

The activity coefficient takes into account the departure of liquid phase from ideal solution behavior. If the value of γ > 1 then it shows the deviation from ideal solution behavior. Ex 8.2

- 68 -

A mixture of ethylene and oxygen (67%C2H4 and 33%O2) exists at 25 atm and 0°C at which partial fugacities of components are reported to be

Partial fugacity of C2H4 42

ˆHCf = 14.38 atm

Partial fugacity of O2 2

ˆOf = 8.12 atm

The pure component fugacities are given by

Pure component fugacity of C2H4 atm 6.20042

=HCf

Pure component fugacity of O2 atm 5.2402

=Of

Calculate

1. Activity of C2H4 and O2 2. Activity Coefficient of C2H4 and O2

Solution

( )0

ˆ

fponent f pure comFugacity o

fent the compongacity of Partial Fu

a)Activity (

=

Activity of C2H4 is given by

69.06.20

28.14ˆ

0

42

42

42===

HC

HC

HC f

fa

Activity of O2 is given by

33.05.24

12.8ˆ02

2

2===

OO

f

Ofa

- 69 -

( )ionMole fract

ActivitytCoefficienActivity =γ

Activity Coefficient of C2H4 is given by

03.1

67.0

69.0

42

4242

===Hion of Cmole fract

a HCHCγ

Activity of Coefficient O2 is given by

00.1

33.0

33.0

2

22

===ion of Omole fract

aOOγ

8.6 Lewis Randall Rule Lewis Randall rule states that Partial Fugacity of any component = Pure component Fugacity × Mole fraction

iii xff ×= 0ˆ

Ex 8.3 A mixture of ethylene and oxygen (67%C2H4 and 33%O2) exists at 25 atm and 0°C. The pure component fugacities are given by

Pure component fugacity of C2H4 atm 6.20042

=HCf

Pure component fugacity of O2 atm 5.2402

=Of

Calculate

1. Partial Fugacity of C2H4 and O2 using Lewis Randall Rule

- 70 -

2. Activity of C2H4 and O2 3. Activity Coefficient of C2H4 and O2

Solution Partial Fugacity of any component = Pure component Fugacity × Mole fraction

Partial Fugacity of C2H4 onmolefractiHity of Cnent fugacPure compof HC ×= 4242

ˆ

atm 8.1367.06.20ˆ42

=×=HCf

Partial Fugacity of O2 onmolefractiity of Onent fugacPure compof O ×= 22

ˆ

atm 08.833.05.24ˆ2

=×=Of

Activity of C2H4 is given by

67.06699.06.20

8.13ˆ

0

42

42

42====

HC

HC

HC f

fa

Activity of O2 is given by

33.03298.05.24

08.8ˆ

0

2

22

====O

o

o f

fa

Activity Coefficient of C2H4 is given by

9998.0

67.0

6699.0

42

4242

===Hion of Cmole fract

a HCHCγ

Activity of Coefficient O2 is given by

9993.0

33.0

3298.0

2

22

===ion of Omole fract

aOOγ

- 71 -

8.7 Models for Excess Gibbs Energy The excess property is the difference between the solution property and ideal solution property.

idE MMM −=

The excess Gibbs Energy is given by

idE GGG −= Activity Coefficients have traditionally been calculated from correlating equations for

RTG E equations, hence the models of excess Gibbs energy is important in chemical engineering thermodynamics The excess Gibbs energy is a function of temperature, pressure and compositions, but for liquids at low to moderate pressures it is a weak function of pressure. Hence the pressure dependence of activity coefficient is usually neglected, thus Excess Gibbs energy is a function of composition at constant temperature.

( )N

Exxxg

RT

G,......,, 21= (Const T)

Redlich Kister Equation

The excess Gibbs energy is given by

( ) ( )2212121

xxCxxBARTxx

G E−+−+=

Margules Equation

If B=C=0, then

ARTxx

G E=

21 or

21xAxRT

G E=

The activity coefficients are calculated by

221ln Ax=γ 212ln Ax=γ

- 72 -

This is the simplest expression for the excess Gibbs energy.

Van Laar Equation The excess Gibbs energy is given by

2121 BxAx

AB

RTxx

G E

+=

The activity coefficient are calculated by

2

2

1

1

1

ln

+

=

Bx

Ax

Aγ

2

1

2

2

1

ln

+

=

Ax

Bx

Bγ

8.8 Deviation from Ideal Behavior

If the excess Gibbs free energy RT

GE is positive, then 1lnγ and 2lnγ are positive, ten

these systems are said to exhibit positive deviation from Ideality. Example: Furan and Carbon tetrachloride at 30°C

If the excess Gibbs free energy RT

GE is negative, then 1lnγ and 2lnγ are negative, ten

these systems are said to exhibit negative deviation from Ideality. Example: Tetrahydrofuran and Carbon tetrachloride at 30°C

8.9 Vapor- Liquid Equilibrium in mixtures non idea l behaviour Let us consider a closed system considsting of coexisting vapor and liquid phases. The system is maintained at constant temperature (T) and Pressure (P). The condition of equilibrium between the two phases is given by Chemical potential of one phase = Chemical Potential of another phase

- 73 -

Chemical potential of Vapor phase = Chemical Potential of Liquid phase (1) We know that the chemical potential is defined as

θµ += fRT ln (2) Hence the condition for equilibrium is given by

Fugacity of Vapor phase

v

if__

= Fugacity of Liquid phase

l

if__

(3)

Where v represent vapor phase and l represent liquid phase.

Fugacity in Vapor phase Pyf ii

v

i φ=

__

(4) Where y represent vapor phase composition φ is the fugacity coefficient in vapor phase P represent the pressure

Fugacity in Liquid phase satii

oii

l

i Pxfxf γγ ==

__

(5) Where x represent liquid phase composition γ is the activity coefficient

satP represents the saturation pressure Substituting equation (4) and (5) in equation (3) we get

satiiii PxPy γφ =

(6) At low pressure (upto atleast 1 bar), the vapor phase can be assumed to behave like an ideal gas and hence 1=iφ , hence the equation (6) reduces to

satiii PxPy γ=

(7) This equation is the modified Raoult’s law. If the vapor is in equilibrium with the ideal solution, 1=iγ , then equation (6) reduces to

satii PxPy =

(8) This reduces to the Raoult’s law.

- 74 -

8.10 Relative Volatility and Margules Constant For a binary system the excess Gibbs energy of the liquid phase is given by

21xBxRTG E = Where B is a weak function of temperature only. Making usual assumptions for low pressure vapor liquid equilibrium, show that relative volatility of species 1 to species 2 at infinite dilutions of species 1 is given by

( ) ( )BP

Px

sat

Sat

exp02

1112 ==α .

Proof Relative Volatility of species 1 to 2 is given by

22

1112 xy

xy=α

(1) For azeotrope mixture 0.112 =α We know from the modified Raoult’s law

satiiii PxPy γ= (i=1,2,….., N)

(2)

P

P

x

y satii

i

i γ=

(3) For Component 1

P

P

x

ysat

1

1

1 1γ

=

(4) For component 2

P

P

x

ysat22

2

2 γ=

(5) Substituting Equations (4) and (5) in equation (1), we get

- 75 -

sat

sat

sat

sat

P

P

P

P

P

P

22

11

22

1

12

1

γ

γ

γ

γ

α ==

(6)

If 21xBxRTG E = Margules Equation

Then 221ln Bx=γ and 2

12ln Bx=γ

At infinite dilutions of species 1 x1=0 and x2=1.0, B== ∞∞21 lnln γγ and 12 =γ

Then equation 6 becomes

sat

sat

P

P

2

1112

∞=

γα

(7)

Sub. B== ∞∞21 lnln γγ in the equation 7 we get

)exp(2

112 B

P

Psat

sat

=α

(8) Hence it is proved that relative volatility of species 1 to species 2 at infinite dilutions of

species 1 is given by ( ) ( )BP

Px

sat

Sat

exp02

1112 ==α .

8.11 Azeotropic Composition and Azeotropic Pressure

The excess Gibbs energy for a particular system is 21xBxRTG E = , where B is a function of temperature only. Assuming the validity of modified Rauolt’s law show that

at every temperature for which an azeotrope exists. The azeotropic composition azx and

azeotropic pressure azp are related by ( )( )

2/1

2

1

1 ln

ln1

1

+=sataz

sataz

az PP

PP

x

Proof:

- 76 -

Let us consider a azeotropic system with the azeotropic composition azx and azeotropic

pressure azp

The excess Gibbs energy of the system is represented by 21xBxRTG E =

Then 221ln Bx=γ and 2

12ln Bx=γ For azeotropic system the modified Rauolt’s law is assumed to be valid We know from the modified Rauolt’s law for the azeotropic system

satiii

azi PxPy γ= (i=1,2,….., N)

(1) At azeotropic point ii xy =

Therefore sub. This in equation we get

satP

P

11 =γ

(2)

satP

P

22 =γ

(3)

sat

az

P

P

11 lnln =γ

(4)

sat

az

P

P

22 lnln =γ

(5) Divide equation (4) by equation (5)

( )( )

=sataz

sataz

PP

PP

2

1

2

1

ln

ln

ln

ln

γγ

(6)

( )( )

=sataz

sataz

PP

PP

Bx

Bx

2

1

21

22

ln

ln

(7) Replacing 12 1 xx −= in equation (7) and taking square root on both sides and replacing

the composition as aezotropic composition azxx =

- 77 -

( )( )

2/1

2

11

ln

ln1

1

=−

sataz

sataz

az

az

PP

PP

x

x

(8)

( )( )

2/1

2

1

1 ln

ln1

1

=−sataz

sataz

az PP

PP

x

(9)

( )( )

2/1

2

1

1 ln

ln1

1

+=sataz

sataz

az PP

PP

x

(10) Hence it is proved that At every temperature for which an azeotrope exists. The

azeotropic composition azx and azeotropic pressure azp are related by

( )( )

2/1

2

1

1 ln

ln1

1

+=

sataz

sataz

az PP

PP

x

Ex 8.4

For the system methanol (1) / methyl acetate(2) , the following equations provide a reasonable correlation for the activity coefficients:

2

21ln Ax=γ 212ln Ax=γ where A = 2.771 – 0.00523T

Using Antoinne’s equations and assuming validity of Modified Raoult’s law, calculate P

and {yi } for T = 318.15K and x1=0.25 . Sol :

- 78 -

( )

( )424.53

54.266525326.14ln

424.33

31.364359158.16ln

2

1

−−=

−−=

TP

TP

sat

sat

kPaP

kPaPsat

sat

64.65

51.44

2

1

=

=

A = 2.771-(0.00523)(318)=1.107

( )( )21 75.0107.1exp=γ = 1.864

( )( )22 25.0107.1exp=γ = 1.072

P = (0.25)(1.864)(44.51) +(0.75)(1.072)(65.64) = 73.50 kPa

718.0

282.0

2

1111

=

==

yP

Pxy

satγ

Tutorial 8 1) Assuming Raoult’s law to be valid for the system

Benzene(1)/ethylbenzene (2),

- 79 -

a)Prepare a Pxy diagram for a temperature of 100º C b) prepare a txy diagram for a pressure of 101.33 kPa Vapor pressures of the pure species are given by the following Antoine’s

equations

20.213

47.32790045.14ln

79.220

51.27888858.13ln

2

1

+−=

+−=

tP

tP

sat

sat

where t is in º C and vapor pressure in kPa

2) The system acetone (1)/acetonitrile(2),the vapor pressures of the pure

species are given by,

224

47.29452724.14ln

22.237

46.29405463.14ln

2

1

+−=

+−=

tP

tP

sat

sat

where t is in º C and vapor pressure in kPa

Assuming Raoult’s law to describe the vapor liquid equilibrium states of the system,determine,

a) x1 and y1 for the equilibrium phases at 54º C and 65 kPa b) t and y1 for for P = 65 kPa and x1 = 0.4 c) P and y1 for t= 54º C and x1= 0.4 d) t and x1 for P= 65 kPa and y1 = 0.4 e) P and x1 for t= 54ºC and y1=0.4