TDA 301T-8c - Thermodynamic Properties Real Substances

7/27/2019 TDA 301T-8c - Thermodynamic Properties Real Substances

1/127

APPLIED THERMODYNAMICS

TDA-301T 2012

Lecture 8: ThermodynamicProperties of

Real SubstancesBy

Prof Alex SofianosBsc Chem Eng, Msc, PhD Ind Chem (Germany), MBL (UNISA)


2/127

Course Contents

1. Introduction2. Conservation of Mass

3. Conservation of Energy

4. Entropy An additional Balance Equation

5. Liquefaction, Power Cycles, Explosions6. Thermodynamic Properties of Real Substances

7. Equilibrium and Stability One Component

8. Thermodynamics of Multi-Component Systems

9. Estimation of Gibbs Energy and Fugacity of a Componentin a Mixture

10. Vapor-Liquid Equilibrium in Mixtures

11. Chemical Equilibrium

2


3/127

Course Books

1. Chemical, Biochemical & Engineering Thermodynamics

4 Edition, Stanley I. Sandler , John Wiley and Sons (2006)

2. Introduction to Chemical Engineering Thermodynamics

(The McGraw-Hill Series in Civil and Environmental

Engineering) by J.M. Smith, Hendrik C. Van Ness, and

Michael Abbott (2004)

3. Introduction to Chemical Engineering Thermodynamics

by J.M. Smith (2001)4. Engineering and ChemicalThermodynamicsby Milo D.

Koretsky (2003)

3


4/127

Assessment TDA 301T

OFFERING TYPE NUMBER WEIGHT TOTAL

1. Assignments 4 5% 20

2. Class Tests 2 15% 30

3. Semester Test 1 50% 50

Predicate 100% 50

4


5/127

Objectives

5

At the end of this Chapter, the student must: Be able to evaluate partial derivatives of

thermodynamic variables with respect to one variable

(e.g. T), while holding a second variable (P) constant!

Be able to interrelate partial derivatives arising in

thermodynamics

Be able to obtain volumetric equation of state

parameters from critical properties

Be able to solve problems for real fluids using

volumetric equations of state (e.g. Van der-Waals or

Peng-Robinson)


6/127

Objectives II

6

At the end of this Chapter, the student must: Be able to construct tables and charts of

thermodynamic properties (Sec. 6.4).


7/127

Mathematical Preliminaries

7

Eight thermodynamic state variables thus far

discussed: P, T, V, S, U, H, A, G

If values of two of them were known: we could

calculate the values of the remaining six state

variables thermodynamic state of pure, single-phase system is

fixed!

Mathematically: we choose two variables as

independentvariables (for the pure, single-phasesystem, single component system)

Remainingsix variables are dependent variables


8/127


8

EXAMPLE: TandVindependent variables

Any other thermodynamic state variables are

dependent variables:

U: Internal Energy is a dependent variable!

U = U ( T, V)

This means that the Internal Energy is a function of

temperature and specific volume.

The change in internal energy dU, which is a result

of differential changes in T and V is therefore:


9/127


9

The subscript on each derivative denotes the variable

being held constant:

: Shows the differential change in molar

internal energy accompanying a diff. change

in temperature, whereby the molar volume is

constant.


10/127


10

Keep in mind that : (U /T)V

and (U /V)T

are

partial derivatives of the type

(

)Z

At the same time: X , Y , Z are used to denote the statevariables and are used here to represent the state

variables P, T, V, S, U, H, A and G

Goal of present chapter is to develop methods to

allow us to calculate the numerical values of thesederivatives

Needed for many thermodynamic applications


11/127


11

We can re-write

As

Above equation says that the Internal Energy is afunction of all three independent variables:

temperature, total volume andnumber of moles in

the system.


12/127


13/127


13

Note that in equation below

It is necessary to have two subscripts (i.e. two

variables are kept constant for each differential

change.

We remember that anyextensive propertycan be

considered to be theproductof the related intensive

property and its mass (or its mole number).

X = X N


14/127


14

Then the derivative of an extensive property with

respect to the intensive property related at a constant

mole number is:

Therefore:

We have already seen in Ch. 3 that CV is the constant

volume heat capacity

CV


15/127

Thermodynamic Partial

Derivatives

15


16/127

Extensive Extensive

Derivatives

16

Further, we can illustrate the derivation of an

extensive extensive derivative, i.e property withrespect to only an extensive property related at a

constant mole number is:

Similarly we get for CP: ??


17/127

More Properties

Derivatives

17

Derivative: (

)Z

may be re-written as:

Order is not important:

First derivative is

change in X along a path

of constant X;

Change in X in response

to an imposed change in X


18/127

More Properties

Derivatives

18

Triple Product Rule:

If X is any dependent variable and Y and Z are two

independent variables, using the chain-rule for

differentiation, we may write:

X = X ( Y, Z ) Then we get:


19/127

More Properties

Derivatives

19

Triple Product Rule (continued):

The latter can then be simplified if we remember

that:

Then we get the Triple Product Rule:

Note the symmetric form of equation in which each

variable appears only once in each derivative position


20/127

More Properties

Derivatives

20

The Expansion Rule:

From equation:

which we saw above, we can derive the Expansion

Rule by introducing two additional variables, K and L:

Finally if we consider the special case, when L = K we

get the Chain Rule


21/127

More Properties

Derivatives

21

The Chain Rule:

From equation:

With L = Z :

Chain Rule

Hence we can introduce a new variable in evaluating

a partial derivative, as with Y in equation above


22/127

Thermodynamic Derivatives

22

Definitions:

Further Definitions:

Energy Balance + G dN


23/127


23

ForClosed Systems we can write *):

Then we have: V = T

And

Alternatively we can write:

*) NOTE: U = U (S, V) ; dU =


24/127


24

However we may remember that:

Then we have

And if we now divide by N we get:

Similarly we can derive:


25/127


25

Therefore from the First Law we can derive:


26/127


26

The Maxwell Relations:

Expressions for other derivatives:

Then we can get:


27/127


27

The Maxwell Relations:

Similar expressions for other derivatives:

These are the famous Maxwell Relations


28/127

F d t l Th d i


29/127

Fundamental Thermodynamic

Relation

29

So we can put these together to form an expression

for dU which only involves functions of state. For a hydrodynamic system, for instance:

dU = T dS - P dV

Thefundamental thermodynamic relation

It involves only functions of state, so it is true for all

changes, not just reversible ones

This suggests that the natural variable in which to

express U are S and V: U = U (S , V) That means thatinternal energywill be unchanged

for processes atconstantvolume and entropy--

F d t l Th d i


30/127

Fundamental Thermodynamic

Relation - ENTHALPY

30

notthe most common experimental conditions.

It is therefore useful to introduce other functions ofstate, called ``thermodynamic potentials'', which are

conserved, or whose change is easily calculated, in

common experimental conditions. These are:

Enthalpy H = U + P V; hence,

dH = d U +P dV + Vdp

and with: dU = T dS - P dV , the fundamental

thermodynamic relation, we obtain:dH = d U +p dV + V dP = T dS - P dV + P dV + V dP

dH = T dS + V dP


31/127

Helmholtz Free Energy

31

dH = T dS + V dP

At constant pressure (such as chemical reactions in atest tube), dP = 0 dH = T dS + V dP ; dH = T dS

and therefore: dH = dQrev so that

Heats of reaction are usually measured as enthalpies

of reaction (at constant pressure)

Helmholtz Free Energy (E)

Similarly we get: E = U T S

The derivative is: dE = dU TdS S dT We remember that: dU = TdS PdV

By substitution: dE = S dTPdV

E is constant at constant temperature and volume.


32/127

Gibbs Free Enthalpy Relation

32

dH = T dS + V dP

Gibbs Free Enthalpy (G)

Similarly we get: G = H T S

The derivative is: dG = dH TdS S dT

We remember that: dH = TdS + PdV By substitution: dG = TdS + PdVS dTPdV

Therefore: dG = VdPS dT

The Gibbs Free Enthalpy is constant at constant

temperature and pressure.

These are the conditions under which phase transitions

(melting, boiling) take place, and are also relevant to

chemical equilibrium


33/127

Maxwells Relations

33

We remember: dU = T dS P dV

fundamental thermodynamic relation

On purely mathematical grounds, we may re-write this

as: U = U (S, V) ; or

dU = (

)V dS + (

)S dV

dU = T dS P dV

By comparison with the fundamental relation we see

that:

TemperatureT = (

)V ; Pressure: P = - (

)S


34/127

Maxwells Relations

34

TemperatureT = (

)V ; Pressure: P = - (

)S

These definitions (especially the second) are interesting

in their own right.

But we can go further, by differentiating both sides of

the first equation by V and of the second by S :

(

)S = (

(

)V )S

(

)V = - (

(

)S ) V

the order of differentiation in the second derivation isirrelevant, righthand sides are equal, and thus soare the left hand sides, giving:

(

)S = - (

)V


35/127

Maxwells Relations

35

By starting with , E, H and G, we can get three more

relationsi. dU = TdS PdV> T = (

)V and P = - (

)S

ii. dE = -SdT PdV> S = (

)V and P = - (

)T

iii. dH = TdS + VdP > T = (

)P and V = - (

)S

iv. dG = -SdT + VdP > S = (

)V and V = - (

)T

The two equations involving derivatives of S areparticularly useful; as they provide a possibility to

determine S,


36/127

Heat Capacities

36

Heat capacities are related to changes of entropy with

temperature A heat capacity is the temperature change per unit

heat absorbed by a system during a reversible process

C dT = dQ rev

Specific heat capacity (C)'; the latter is the heatcapacity C per kg or per mole

The heat capacity is is different for different processes.

Useful heat capacities are those at constant volume or

constant pressure (for a fluid). Since

dQ rev= T dS

we have at constant volume CVdT = T dS , so


37/127

Heat Capacities

37

we have at constant volume CVdT = T dS , so

i. CV= T()V Heat Capacity at constant volume

ii. C P= T(

)P Heat Capacity at constant pressure

At constant volume (reversible conditions!) no work is

done; therefore: dU = dQ rev+ PdV

dU = dQ rev

i. CV= (

)V expressed in terms of internal Energy

Also (usefully for chemists), at constant pressure dP=0

d H = dQ revwhere H is the enthalpy, so

ii. CV= (

)P expressed in terms of Enthalpy


38/127

Heat Capacities

38

The specific heat capacity is the heat capacity per unit

mass (or per mole). Heat capacities are notindependent of temperature (or

pressure) in general, butover a narrow temperature

range they are often treated as such, especially for a

solid. Together with two of Maxwell's relations, we now have

expressions for the partial derivatives of the entropy

with respect to all easily measurable variables (P, V , T).

These can be used to derive expressions for the entropychange in real processes as well! We can show how to

do that by using the Van der Waals Equation

P =

-

2


39/127

Heat Capacities

39

We can also derive a relation between , , and other

measurable properties of a substance which can bechecked experimentally: If

a is the isobaric thermal expansion coefficient and

kTis the isothermal compressibility

With

a =1

(

)P and kT= -

1

(

)T

We obtain

CPCV= VT2

kT

which is always greater than zero.


40/127

Heat Capacities

40

CPCV= VT2

kT

This relation is a firm prediction of thermal physicswithout any approximations whatsoever.

It has to be true! For real gases and compressible

liquids and solids it can be checked.

For relatively incompressible liquids and solids it is hard

to carry out processes at constant volume so CV may

not be well known and this equation can be used to

predict it.

For one mole of a van der Waals gas this gives

CPCV= R[ 1 -2 ()2

RTV 3]-1


41/127

Heat Capacities

41

For one mole of a van der Waals gas this gives

CPCV= R[ 1 - 2 ()2RTV 3 ]-1

In the idealgas case a, b 0 are zero (limit)

Above equation then reduces to

CPCV= R

Question

What is the entropy change during the expansion of a

van der Waals gas for which CV is a constant?

Equation of state for one mole: P =

-

2


42/127

Entropy Change Calculation

42

b represents the volume taken up by the finite size of

the molecules,

a/V2 is the reduction in pressure due to interactions

between the molecules (Van der Waal forces).

In this way the two most important corrections

neglected in the ideal gas are included.

However, the heat capacity at constant volume is still

independent of temperature and volume, as in an ideal

gas.

During an expansion, both the temperature and the

volume may change.

P =

-

2


43/127


43

To calculate the change in entropy, we need

()V=

from definition of CV= T()V

Equation can be modified:

Differentiation of V.-d. Waals equation:

(

)V=

(

)T= (

)V=

Then dS = (

)VdT + (

)TdV , using a Maxwell Relation If we now substitute:

(

)V=

and (

)V=

we obtain:


44/127


44

Finally dS =

dT +

Choose variables, either (V, T) or , (P, T) and

writing dS in terms of infinitesimal changes of these

variables

then using the definition of the heat capacities, anda Maxwell relation, we obtain something like the

second line.

Then we can calculate the total entropy change by

integrating, first at constant T and then at constantV :

S ( T1, V2) = S ( T1, V1) + [ (1,)

21

]T dV


45/127


45

Finally dS =

dT +

Choose variables, either (V, T) or , (P, T) and

writing dS in terms of infinitesimal changes of these

variables

then using the definition of the heat capacities, anda Maxwell relation, we obtain something like the

second line.



S ( T1, V2) = S ( T1, V1) + [ (1,)

21

]T dV


46/127

Heat Capacities

46



S ( T1, V2) = S ( T1, V1) + [ (1,)

21

]TdV

S ( T1, V2) = S ( T1, V1) + ()

21 )TdV

= S ( T1, V1) + (2 )

1)

Similarly:

S ( T1, V2) = S ( T1, V2) + [ (1,)

2

1]TdP

= S ( T1, V2) +

21

dT


47/127


47

S ( T2, V2) = S ( T1, V2) + [ (1,)

21

]TdP

= S (T1 , V2) +

21

dT

= S ( T1, V2) + CV lnT2T1

= S ( T1, V2) + CV 2 1+ CV ln T2T1

Therefore:

S ( T2, V2) - S ( T1, V2) = CV2

1

+ CV lnT2

T1

where in the second integration we used the fact that

CV is constant.


48/127


48

S ( T2, V2) - S ( T1, V2) = CV2

1+ CV ln

T2T1

We can check our result against those we have

already calculated for an ideal gas just by setting

b=0 and using the correct CV

for a monatomic gas CV= 3/2R for an isothermal expansion (T1 T2) we get:

S = R lnV2V1

for a reversible adiabatic expansion we can useT1 V1

-1 = T2 V2-1

We remember heat capacities at constant pressure and

at constant volume differ by a constant forideal gases:


49/127


49

We remember that heat capacities at constant

pressure and at constant volume differ by a constantforideal gases:

CPCV= n R

reversible adiabatic compression or expansion of an

ideal gas the pressure and volume change togetherin such a way that:

P V = constant, where =

Then from the Ideal Gas Law: p V = n R T it followsthat:

T V-1 = constant and T P(

1

1)

= constant


50/127


50

Expressions of S, U and H in terms of T and V or T and

P (which are the convenient choices (whymeasurement of S?)

S = S (T, V) and

However accordingto the Maxwellrelationship


51/127


51

Then we obtain:

And by substituting:

We get:


52/127


52

To obtain complete expressions of dU and dH in

measurable entities we start again with equation:

And by substitution of the dS into previous equation:

We get:


53/127


53

Finally, we can express dH in a similar way as follows:

From equation above and the last one of the previous

slide:

d U

We get:

and:

Table of useful


54/127

Table of useful

Thermodynamic Functions

54

Table of Useful Thermodynamic


55/127


Functions (Coned)

55



56/127


Functions (Coned)

56

Ideal Gas And Absolute


57/127


Temperature Scale

57

If we consider that P, V and U are only functions of T

as in the equations:

P V = R T ; and P V = 1 ( T )

and U = 2

( T ); therefore

we obtain with

= CV (

)V + (

)V



58/127


Temperature Scale

58

Since:

CV (

)V = CV

Due to the fact that (

)V= 1

And : (

)V= 0

The last equation of previous slide

becomes:

Id l G S E i


59/127

Ideal Gas State Equations

59

Since: at constant volume can

be written as the derivative

And by integration we get the well known equation for

an ideal gas



60/127


Temperature Scale

60

From the same equation (below)

we can see that

CV=

and if we keep P constant we obtain:

= CV +

= CV (

)V + (

)V

Id l G St t E ti


61/127

Ideal Gas State Equations

61

By comparison of the two equations we can clearly

see that

CV=

STATEMENTS (Please read!) Similar partial derivatives with Different State Variables

held constant are NOT EQUAL (Illustration 6.2-1)

Similar partial derivatives with Different State Variables

held constant are NOT EQUAL (Illustration 6.2-2)

Changes in Thermodynamic


62/127


Properties of Real Substances

62

STATEMENTS By using partial derivatives develop expressions for

i. The coefficient of thermal expansion, a

ii. Isothermal compressibility factor ktiii. The Joule-Thomson Coefficient

iv. The difference CP - Cv

Using:

a. The Ideal Gas Equation: P V = R T

b. The Van der Waals Volumetric Equation of State

(P +

V2) (Vb) = R T

(Illustration 6.2.3)



63/127

g y

Properties of Real Substances

63

For real substances we need a volumetric equation of state

Ideal Gas Equation: P =

Van der Waals

Redlich Kwong

Peng Robinson

E ti f St t


64/127

Equations of State

64

For real substances we need a volumetric equation of state

Generalized Class

Virial

Benedict Webb

Rubin

H t C it


65/127

Heat Capacity

65

We start with:

And

Using the Commutative Property, which states that in amixed second derivative, the order of taking derivatives

is not important:

H t C it


66/127

Heat Capacity

66

We also use:

And

If we use the same method as with the heat capacity at

constant-volume for the constant-pressure heatcapacity

We convert above equation into the much more

versatile second equation

Integration of the above gives:

H t C it


67/127

Heat Capacity

67

And

By integration for P1= 0 and V1= we obtain:

Heat Capacity


68/127

Heat Capacity

68

The same results are expected if one considers the

constant-volume heat capacity:

And

By integration for P1= 0 and V1= we obtain for theconstant-volume heat capacity:

Change of State of a Real Gas


69/127


69

ILLUSTRATION 6.5-1

Compare solutions to Problem assuming the gas is

ideal, Using a Thermodynamic Properties Chart (Peng

Robinson) and Assuming Gas can be described by Van

der Waals Equation of State

Nitrogen gas is being withdrawn from a 0.15 m3 cylinder at 10

mol/min. Cylinder initially at 100 bar and 170 K. Cylinder well

insulated; no heat transfer between cylinder and gas.

i. How many moles of nitrogen will be in the cylinder at any

time?

ii. What will the temperature and the pressure of the

cylinder be after 50 minutes?



70/127


70

ILLUSTRATION 6.5-1

a. Assume that nitrogen IDEAL GAS

b. Use the nitrogen properties chart of Fig. 3.3-3.

c. Assume that nitrogen obeys the Van der Waals equation of

state.

Data: For a.- and b.- use :

CP* [J/mol K) = 27.2 + 4.2 X 10-3 (K)

SOLUTION We set up the mass and energy balances for the contents of

the cylinder (the System)



71/127


71

SOLUTION

Mass and Energy balances for the contents of the cylinder

I.

= or N(t) = N (t=0) + t

II.

= H with = -10 mol/min

We saw during Illustration 4.5.-2 that the entropy

balance for the portion of the gas remaining in the

cylinder is:

= 0 or S (t=0) = S(t)

Also from V = N V we get



72/127


72

SOLUTION

Also from V = N V we get:

N (t=0) =

V(=)=

.1

V(=)

These equations apply equally to ideal and non-ideal

gases

Calculation of N(t=0)

I. Ideal Gas Equation of State

P V = R T or V(t=0) = (=)

(=)

=8.141

1

1 = 1.4134 x10 -4 m3/mol



73/127


73


I. Ideal Gas Equation of State

P V = R T or V(t=0) = (=)

(=)

= 8.141

1 1

= 1.4134 x10 -4 m3/mol

And N (t=0) = 1061.3 mol

N (t) = (1061.3 10t) mol



74/127


74


ii. Using the Chart of 3.3.-3

(T=170K, P=100 bar) = (T=170K, P=10 MPa) 0.0035 m3/kg

V(T=170K, P=10 MPa) = N (T=170K, P=10 MPa) =

28.014 g/mol x 0.0035 m3/kg x1

1= 9.80 x 10 -5 m3/mol

And N (t=0) = .1

9.8 1 / = 1529.8 mol

N (t) = (1529.8 10t) mol



75/127


75


iii. Using the Van der Waals Equation

We use the constants given in Table 6.4.-1, whereby fornitrogen a value for a = 0.1368 [Pa 106/mol2] and for b =

3.864 [105m3/mol] are listed!

Then we have:

P = 100 bar = 107Pa=8.141

= .8641-

.8641 = 2

Cubic Equation: V(t=0) = 9.435x 10 -5 m3/mol



76/127


76

And N (t=0) =.1

9.4 1

/

= 1589.9 mol

N (t) = (1589.9 10t) mol

We can also compute the molar volume of the gas at anylater stage from the volume of cylinder (0.15 m3) and N(t)

V (t) =.1

()

Example: Van der Waals:

N (t) = (1589.9 10t) mol = 1589.9 10 x50 = 1089.9 mol

V (t) =.1

()= 1.376 x 10 -4 m3/mol

Ideal Gas


77/127

Ideal Gas

77

An ideal gas is an imaginary gas that satisfies the

following conditions:

Negligible interactions between the molecules,

Its molecules occupy no volume (negligible molecular

volume),

Collisions between molecules are perfectly elastic thisis, no energy is lost after colliding.

This fluid is imaginary because as we know there are no

ideal gases

Ideal Gas


78/127

Ideal Gas

78

The ideal gas EOS is inaccurate both qualitatively and

quantitatively

Qualitatively, ideal gas will not condense no matter what

pressure and temperature the system is subjected to.

Quantitatively,pressures and volumes used by the ideal

gas model are higher than the values that a real gas

would have.

These were the primary reasons that scientists have made

an effort to go beyond the ideal gas EOS: simply because it does not apply for most cases of interest

Ideal Gas


79/127

Ideal Gas

79

The ideal gas EOS is inaccurate both qualitatively and

quantitatively


pressure and temperature the system is subjected to.

P-v Isotherm for an Ideal Gas

1. volume of the gas becomes very largeat very low pressures (i.e., as , a

concept that agrees with what we

know from our experience in the

physical world).

2. And second, as (the volume of matter

just vanishes if the pressure is highenough: this concept we would not be

as willing to accept). These two

behaviors are a consequence of the

assumptions made in the ideal gas

model.

Ideal Gas


80/127

Ideal Gas

80


pressure and temperature the system is subjected to.Ideal EOS is therefore qualitatively wrong

P-v Isotherm for a Real Gas

This figure has an obvious discontinuity

(at the vapor+liquid transition).

For a real substance, as pressure

increases, there must be a point of

discontinuity that represents the phase

change.

Hence, we cannot hope to reproduce theP-V behavior (actual!) using the ideal

equation since no discontinuity is to be

found.

However, the real P-v isotherm can be

approximated by ideal behavior at low

pressures, as we can see from the plots.

Ideal Gas


81/127

Ideal Gas

81

Quantitatively, ideal gas pressures and volumes used by

the ideal gas model are higher than the values that a realgas would have.

Ideal EOS is therefore also quantitatively wrong

for most conditions of interest at a given volume and

temperature, the ideal gas model over-predicts the

pressure of the system:

P Ideal Gas > P Real Gas

real gas does have interaction forces between molecules ideal model assumes that the physical space that the

molecules themselves occupy is negligible. In reality

molecules are physical particles and they do occupy space

Real Gas


82/127

Real Gas

82

How can we adjust the ideal model to make it suitable for

real gases?

The answer: We said that once we have established

a base (ideal) model, we look at a real case by estimating

how close (or far) it performs with respect to the base

(ideal) case, and introducing the corresponding

corrections.

Such corrections will take into account all the

considerations that our original assumptions left out.

We introduce a correction factorZto account for the

discrepancies between experimental observations and

predictions from our ideal model.

Real Gas


83/127

Real Gas

83

We introduce a correction factorZ the compressibility

factor , which is defined as:Z = V Videal

V is the real volume occupied by the gas and Videal is the

volume that the ideal model predicts for the same

conditions. The ideal volume is given by:

Videal= n R T P

Hence, the equation of state for real gases:

P V = Z n R T for Z = 1, this equation collapses to the ideal gas model. In

fact, unity is the compressibility factor of any gas that behaves

ideally

Definition of Equation of State


84/127


84

Assuming an equilibrium state, the three properties

needed to completely define the state of a system arepressure (P), volume (V), and temperature (T).

Hence, we should be able to formulate an equation

relating these 3 variables, of the form f (P,T,V) = 0.

Equation of State (EOS) is a semi-empirical functionalrelationship between pressure, volume and temperature

of a pure substance.

We can also apply an EOS to a mixture by invoking

appropriate mixing rules.The functional form of an EOS

can be expressed as:

f( P, T, V, ak, k=1, nP) = 0 , where ak= EOS parameters.



85/127


85

Most applicable EOS today are semi-empirical, in the sense

that they have some theoretical basis but their parameters(ak) must be adjusted.

The number of parameters (np) determines the

category/complexity of the EOS. For instance, 1-parameter

EOS are those for which np = 1, and 2-parameter EOS arethose for which np= 2. The higher np is, the more complexis the EOS.

Also, in general terms, the more complex the EOS, the more

accurate it is. However, this is not always the case; in somecases a rather simple EOS can do a very good job.



86/127


86

A great number of equations of state have been proposed

to describe real gas behavior. Only few have persisted through the years, this because of

their relative simplicity.

In the petroleum industry, the most common modern EOS

are the Peng Robinson EOS and Soave-Redlich-Kwong EOS

Both of these are cubic EOS and hence derivations of the

van der Waals EOS. Other more complex EOS, although

they have not yet found widespread application in our field:

Lee Kesler EOS

Benedict-Webb-Rubin

Benedict-Webb-Rubin-Starling



87/127


87

We want to use EOS as the basis for generating data:

volumetric data, thermophysical data, and to help usperform vapor/liquid equilibrium (VLE) calculations.

Not other area of thermodynamics to which so many hours

of research have been devoted as that of the topic of EOS.

Among the properties derived from an EOS include:

Densities (vapor and liquid),

Vapor pressures of pure components,

Critical pressures and temperatures for the mixture,

Vapor-Liquid equilibrium (VLE) information,

Thermodynamic properties (H, S, G, A).

Equation of State


88/127

Equation of State

88

1873: With van der Waals, a quantitative approach was taken

for the first time. He was an experimentalist and proposed thecontinuity of gases and liquid that won for him a Nobel Prize.

He has provided the most important contribution to EOS

development.

1927: Ursell proposed a series solution (polynomial functionalVirial form) for EOS: P = 1 + b/V + c/V2 + d/V3 +

1940: Benedit, Webb, & Rubbin proposed what can be called the

Cadillac of EOS, i.e., the most sophisticated and most accurate

for some systems. However, the price to pay is that it is toocomplicated and not easy to use.

1949: Redlich & Kwong introduced a temperature dependency to

the attraction parameter a of the vdW EOS.

Equation of State


89/127

Equation of State

89

1955: Pitzerintroduced the idea of the acentricfactor to

quantify the non-sphericity of molecules and was able to relateit to vapor pressure data.

1972: Soave modified the RK EOS by introducing Pitzers acentricfactor.

1976: Peng and Robinson proposed their EOS as a result of astudy sponsored by the Canadian Gas Commission, in which the

main goal was finding the EOS best applicable to natural gas

systems.

Since then, there has not been any radical improvement to SRK

and PR EOS

Equation of State


90/127

Equation of State

90

1955: Pitzerintroduced the idea of the acentricfactor to

quantify the non-sphericity of molecules and was able to relateit to vapor pressure data.

1972: Soave modified the RK EOS by introducing Pitzers acentricfactor.

1976: Peng and Robinson proposed their EOS as a result of astudy sponsored by the Canadian Gas Commission, in which the

main goal was finding the EOS best applicable to natural gas

systems.

Since then, there has not been any radical improvement to SRK

and PR EOS

van der Waals EOS


91/127

van der Waals EOS

91

van der Waals EOS (vdW EOS) has been around for more than

one hundred years, we still recognize its crucial contribution inrevolutionizing our thinking about EOS.

We talk about vdW EOS mainly because of pedagogical reasons,

not because it finds any practical application in todays world.

In fact, vdW EOS is not used for any practical design purposes.However, most of the EOS being used widely today for practical

design purposes have been derived from vdW EOS.

The contributions of vdW EOS can be summarized as follows:

It radically improved predictive capability over ideal gas EOS,

First to predict continuity of matter between gas and liquid,

It formulated the Principle of Corresponding States (PCS),

It laid foundations for modern cubic EOS.

van der Waals EOS


92/127

van der Waals EOS

92

van der Waals EOS (vdW EOS) accounted for the non-zero

molecular volume and non-zero force of attraction of a realsubstance.

Van der Waals realized that there is a point at which the volume

occupied by the molecules cannot be neglected.

He recognized that molecules must have a finite volume, andthat volume must be subtracted from the volume of the

container.

At the same time, he modified the pressure term to

acknowledge the fact that molecules do interact with each other

though cohesive forces. These are the two main valuable

recognitions that he introduced.

van der Waals EOS


93/127

93

van der Waals EOS (vdW EOS) focused his attention on

modifying the terms P and V in the original ideal gaslaw by introducing an appropriate correction.

Looking back at the inequality of equation Pideal PrealvdW proposed that the difference of both pressures is the

result of the attraction forces cohesive forces neglected in the ideal model and then he postulated the

term P attraction, an inverse function of the meandistance between molecules. Pideal= Preal + P

Then he introduced a as a constant of proportionality,Pideal= Preal + a / V

2

van der Waals EOS


94/127

94

Next van der Waals took care of the inequality with regard

to the volume. Any particle occupies a physical space;hence, the space available to the gas is less than the total

volume of its container.

Let assume we can experimentally determine the physical

space that all the molecules in the container occupy, andthat we call it b, or the co-volume.

vdW then proposed: Vavailable = Vtotalb


95/127

van der Waals EOS


96/127

96

where:P = absolute pressure

V = molar volume

T = absolute temperature

R = universal gas constant

How do we calculate a and b for each substance?

As can be inferred from their definitions, a and b aredifferent for different substances; i.e.,;

a CH4 aCH3-CH3 a C3H8. andb CH4 bCH3-CH3 b C3H8 How do we relate a and b to well-known and easily-

obtainable physical properties of substances?

Criticality Conditions

van der Waals EOS


97/127

97

How do we relate a and b to well-known and easily-

obtainable physical properties of substances? Criticality Conditions

Indeed, in the previous chapter we recognized that the

critical isotherm of a pure substance has a point of inflexion

(change of curvature) at the critical point

A plot of the isotherms of the Van der Waals equation of

state is shown in the next slide

van der Waals EOS


98/127

98

The red and blue

isotherms similarto

those ofan ideal gas Black isotherm

exhibits an unusual

feature; virtually no

change in pressure as

the volume changes Green isotherm has a

region where the

pressure decreases with

decreasing volume

dramatic jump involume (doted line)

signifies aphase

transition; from a

gaseous to a liquid

state

van der Waals EOS


99/127

99

- One region where the vander Waals equation works well

is for temperatures that are

slightly above the critical

temperature Tc of a substance

- Observe that inert gases like

Helium have a low value

of a as one would expect since

such gases do not interact very

strongly, and that large

molecules like Freon have

large values of b.

van der Waals EOS


100/127

100

Further, we see that the black

isotherm represents a boundarybetween those isotherms along

which no such phase transition

occurs and those that exhibit

phase transitions in form ofdiscontinuous changes in the

volume.

The black isotherm is called the

critical isotherm; thepointatwhich the isotherm is flatand

haszero curvature is called a

critical point.

van der Waals EOS


101/127

101

This condition of horizontal inflexion of the critical isotherm

at the critical point is mathematically imposed by theexpression:

These conditions are called the criticality conditions.

It turns out that when one imposes these conditions on vd

Waals equation, one is able to derive expressions for the

parameters a and b as a function of critical properties

= 0 and

2

2 0

van der Waals EOS


102/127

102

Using these two conditions, we can solve for the critical

volume Vcandcritical temperature Tcand correlate thesewith the Van der Waals constants:

Starting with

At the critical point we have;

(

)T|at Tc, Vc, Pc = 0 or

(

2

2 )T|atTc,Vc,Pc= 0

We get a and b as a function of the critical properties

van der Waals EOS


103/127

103

Problem is pure algebra, let's rewrite the simultaneous

equations that must be solved for the two unknowns VandT(which solutions we will call VCand TC):

(1) (2)

One way is to multiply Equation (1) by:2

()

Then we get: (3)

By addition of equations (2) & (3) :

we obtain eq. (4)

van der Waals EOS


104/127

104

Divide Equation (4) by 2an2 and multiply it by V3 (and bring

the negative term to the other side of the equal sign) toget,

(5)

This we now solve with re-arrangement to obtain the

critical volume: (6)

We substitute Vc in equation (1)

Then we get: (7)

Which can be simplified: or

(8)

van der Waals EOS


105/127

105

The critical pressure is obtained by substituting

VCand TC into the van der Waals equations ofstate as solved for p:

Then we get:

(9)

And finally: (10)

van der Waals EOS


106/127

106

From Eq. 6 we can now calculate b:

b = anda = 98

We can further calculate the compressibilityZat the critical

point as follows:

P V = Z nRT at the critical point we have:

Zc =

and Zc =

8 0.375

We can express a and b in terms of the critical temperatureand critical pressure:

a =222

64 b =

van der Waals EOS


107/127

107

Or we can express a and b in terms of the critical

volume and critical pressure:

a = 3Pc V2

c b =

If we substitute the above values into the vd

Waals Equation we get:

And finally:

van der Waals EOS


108/127

108

If we define the:

reduced temperature: Tr =

reduced pressure: Pr =

reduced volume: Vr =

Equation:

becomes:

[Pr +

] [3Vr 1] = 8 Tr

van der Waals EOS


109/127

109

Equation:

is extremely interesting: It suggests that at given

values of the reduced temperatures andreduced

pressures all van der Waals fluids will have thesame value of reduced volume

Further conclusion:the same reduced form

equation of state applies, no matter whata and

b may be for the particular fluid. If two fluids have the same reduced pressure,

reduced volume, and reduced temperature, we

say that their states are in corresponding states

[Pr +

] [3Vr 1] = 8 Tr

Parameters for van der Waals EOS


110/127

110

Substance a (L2 atm/mol2) b (L/mol)

He 0.0341 0.0237

H2 0.244 0.0266

O2 1.36 0.0318

H2O 5.46 0.0305

CCl4 20.4 0.1383

The parameterb is related to the size of each molecule. The

volume that the molecules have to move around in is not just

the volume of the container V, but is reduced to ( V - nb ). The parametera is related to intermolecular attractive force

between the molecules, and n/V is the density of molecules.

The net effect of the intermolecular attractive force is to

reduce the pressure for a given volume and temperature.

Example: van der Waals EOS


111/127

111

Use the van der Waals equation to calculate the pressure

of a sample of 1.0000 mol of oxygen gas in a 22.415 L

vessel at 0.0000C (Note: For an ideal gas, thistemperature and volume would lead to conditions of STP,

i.e. a pressure of exactly 1.0000 atm)

Solution:

V = 22.4 L T = (0.000 + 273.15) = 273.15K

a (O2) = 1.36 L2 atm/mol2 (From Table)

b (O2) = 0.0318 L /mol (From Table)

(nRT)/(V-nb) = 1.0014 atm - a (n/V)2 = -0.0027 atm

p = (nRT)/(V-nb) - a (n/V)2 = 0.9987 atm

If O2 was perfectly ideal, the pressure would be 1.0000 atm.

van der Waals EOS


112/127

112

The first term on the right-hand-side of the van der Waalsequation, (nRT)/(V-nb), represents the ideal gas pressure

corrected for finite molecular volume. In other words the

volume is slightly less than 22.415 L due to the mole of O2 in it.

the molecules collide a bit more frequently with the walls ofthe container, because they have less room to fly around in in

the middle of the container.

van der Waals Solver


113/127

113

http://www.webqc.org/van_der_waals_gas_law.html

The value of 1.0014 atm for this term represents this increase

in pressure.

The second term in the equation, - a (n/V)2, represents the

reduction in pressure due to molecular stickiness.

This correction to ideal gas behavior dominates the finite

volume correction, leading to a compressibility factor,

Z=pV/nRT of less than 1 (0.9987, in fact).

Van der Waals equation calculatoris a powerful online tool

for solving problems using Van der Waals equation equation.

Select a quantity to solve for and one of the Van der Waals

equation equations to use.

van der Waals Solver
http://www.webqc.org/van_der_waals_gas_law.htmlhttp://www.webqc.org/van_der_waals_gas_law.html


114/127

114

http://www.webqc.org/van_der_waals_gas_law.html

Solve for Equations

P

T

n

V

Input values and select units

Temperature (T)

Volume (V)

Gas constant (R) (std) J/(mol*K)

Number of moles (n)Parameters (gas)

Measure of the attraction between the particles (a) J*m3/mol2

Volume excluded by a mole of particles (b) m3/mol

Solve for Pressure (P)

Principle of Corresponding States
http://www.webqc.org/van_der_waals_gas_law.htmlhttp://www.webqc.org/van_der_waals_gas_law.htmlhttp://www.webqc.org/van_der_waals_gas_law.html


115/127

115

Accuracy of Van der Waals equation not very

good

Critical volume of a fluid is known with less

experimental accuracy than either the critical

temperature or critical pressure.

Compressibility Factor comparison

We saw that Zc|Van der Waals = Pc Vc / R Tc = 0.375

In reality critical compressibility for most fluids is

in the range of 0.23 0.31 Ideal gas Zc|Ideal gas = 1 !!! Totally out!



116/127

116


experimental accuracy than either the critical

temperature or critical pressure.

Critical volume discrepancy can be calculated:

Zc|

Van der Waals/Z

c|

real= 3 /8 . Z

c|

real

with Zc|real real fluid critical compressibility

Attempt to express compressibility factor as a

function of reduced pressure and temperature



117/127

117


experimental accuracy than either the criticaltemperature or critical pressure.

Critical volume discrepancy can be calculated:

Zc|Van der Waals /Zc|real = 3 /8 . Zc|real

with Zc|real real fluid critical compressibility

Attempt to express compressibility factor as a

function of reduced pressure and temperature

functional relationship between Tr, Prand Z is

determined from experimental data or another

EOS, which should be more accurate


118/127

Critical and other Constants

for selected Fluids (con ed)


119/127

( )

119



120/127

120



121/127

121

Graph is pretty accurate but there are

systematic deviations from the simple

corresponding-states relation above.

In particular, the compressibility factors for the

inorganic fluids are almost always below those

for the hydrocarbons.

Furthermore, if equation above were universally

valid, all fluids would have :he same value

of the critical compressibility Zc = Z(Pr= 1, Tr= I);

however: from Table , it is clear that Z, for most

real fluids ranges from 0.23 to 0.31.


122/127


123/127

123



124/127

124

Here P vap(Tr= 0.7) is the vapor pressure of the

fluid at Tr= 0.7, a temperature near the normalboiling point.

In this case the corresponding-states relation

would be of the form:

However, things become too complex!

Generalized Equation of State


125/127

125

van der Waals equation and corresponding-

states charts for both the compressibility factor Z and the thermodynamic departure

functions

modern application of the corresponding states

idea is to use generalized equations of state. Concept already studied with van der Waals Eq.

With the constants a and b obtained fromcritical properties:

PENG ROBINSON

Generalized Equation of State


126/127

q

126

With


127/127

THANK YOU

TDA 301T-8c - Thermodynamic Properties Real Substances

Documents