TDA 301T- 2015-10 - Thermodynamic Properties Real Substances

APPLIED THERMODYNAMICSTDA-301T 2015

Lecture 10a: Equilibrium and Stability in One-Component

SystemsBy

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

Course Contents1. Introduction2. Conservation of Mass3. Conservation of Energy4. Entropy – An additional Balance Equation5. 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 Component in a

Mixture10. Vapor-Liquid Equilibrium in Mixtures11. Chemical Equilibrium

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Course Books1. “Chemical, Biochemical & Engineering Thermodynamics”

4th 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 Chemical Thermodynamics” 2nd Edition, Milo D. Koretsky (2012)

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Assessment TDA 301TOFFERING TYPE NUMBER WEIGHT TOTAL1. Assignments 2 10% 20

2. Class Tests 2 20% 40

3. Practicals 1 10% 10

3. Semester Test 1 30% 30 Predicate 100% 100

(50)

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Objectives

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• This Chapter should enable the student : Identify the Criterion For Equilibrium in systems

subject to different constraints (Sec. 7.1) Use the concept of stability to identify when phase

splitting into vapour and liquid phases will occur Identify the conditions of phase equilibrium (Sec. 7.3) Identify the critical point of a fluid (Sec. 7.3) Calculate the fugacity of a pure substance that is a

gas or a liquid when a volumetric equation of state is available (Sec. 7.4)

Objectives II

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• At the end of this Chapter, the student must: Be able to calculate the fugacity of a pure liquid or

solid when a volumetric equation of state is not available (Sec. 7.4)

Use fugacity in the calculation of vapour-liquid equilibrium (Sec. 7.5)

Be able to determine the number degrees of freedom for a pure fluid (Sec. 7.6)

Criteria for Equilibrium

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• Means of mathematically identifying the state of equilibrium in a closed, isolated system.

i.e. system in which M, U, and V are constant):Entropy (Ch. 4)

Goal is to develop a means of identification of the equilibrium state for closed thermodynamic systems subject to other constraints, especially those of constant T and V and and constant T and P !

• Starting point: Energy and Entropy Balances:


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• Further: From the 2. Law of Thermodynamics we remember that for equilibrium or for reversible processes:

• the system was chosen in such a way that the only work term is that of the deformation of the system boundary ( dV/dt ).

• For a constant-volume system, exchanging no heat with its surroundings,

• Equation

reduces to: = 0 , i.e. U = constant

Ṡ gen


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• Since the total number of moles, or mass, is fixed in a closed, one-component system,

then: U = constant• and equation: = + Ṡ gen

becomes:

• entropy function can only increase in value during the approach to equilibrium!! Why?

= Ṡ gen 0


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• Answer: Ṡ gen 0 !• Equilibrium criterion for a closed, isolated system is:

• This equilibrium criterion can be used for identifying the final equilibrium state in a closed, isolated system

• that is initially non-uniform, or• in which several phases or components are present

Example for Equilibrium Criterion

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• Example:

• Single-phase, single-component fluid in an adiabatic, constant-volume container has been divided into two subsystems by an imaginary boundary.

• Each of these subsystems is assumed to contain the same chemical species of uniform thermodynamic properties.

• However, these subsystems are open to the flow of heat and mass across the imaginary internal boundary, and their temperature and pressure are not necessarily the same.


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• For the composite system consisting of the two subsystems, the total mass (or number of moles), internal energy, volume, and entropy, all of which are extensive variables,

are the sums of the respective quantities for the two subsystems, i.e.:

In addition:

We know that S = S ( U, V, N)


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• From S = S ( U, V, N) we can calculate the change in the entropy of system 1 due to changes in N1, U1, and V1 as follows:

• Then: dS1 =

In the same way: dS2 = dU2 + dV2 - dN2

Entropy change for the whole system is dS = dS1 + dS2


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• Therefore Entropy change is : dS = dU1 + dV1 - dN1 + dU2 + dV2- dN2

• Total number of moles, total volume and total internal energy were set to be constant

• Hence under the condition that:

dS = [ - ] dU1 + [ - ] dV1 - [ - ]dN1


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• Since S = maximum or dS = 0 for all system variations at constant N, U, and V we conclude by comparison of equations below :

dS = [ - ] dU1 + [ - ] dV1 - [ - ]dN1

That:


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• And since T1 = T2

• In other words, the equilibrium condition for the system illustrated above is satisfied if both subsystems have the same temperature, the same pressure, and the same molar Gibbs energy.

• For a single-component, single-phase system, this implies that the composite system should be uniform.


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Conclusion: • The condition dS = 0 may be used to identify a

possible equilibrium state of the closed, isolated system, that is, a state for which S is at a maximum

• However: dS = 0 necessary but not sufficient condition for a maximum; it could also describe a minimum!

• Therefore, an additional condition is required: The condition d2 S 0, together with dS = 0, assures us

that a maximum value of the entropy, and hence a true equilibrium state, has been identified rather than a metastable state (an inflection point) or an unstable state (minimum value of S).


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Conclusion: • Thus, the sign of d 2S determines the stability of the

state found from the condition that dS = 0. Stability condition for equilibrium d2S < 0New Example• Possible to develop the equiIibrium and stability

conditions for systems subject to other constraints. • For a closed system at constant temperature and

volume, the energy and entropy balances are:

and


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• Eliminating Q between these two equations, and using the fact that: T dS = d(TS), since T is constant we obtain:

• This last part stems from the fact that T = constant and positive: T and Śgen

• Therefore: - T Śgen 0 • and as we saw with entropy we get for the Helmholtz

free energy


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New Example• For a closed system at constant temperature and

constant pressure , the energy and entropy balances are:

and

Combining the two above equations over ‘Q, we obtain:


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• This last part stems from the fact that T = constant and positive: T and Śgen

• Therefore again: - T Śgen 0 • and as we saw with entropy we get for the Gibs free

energy

• This is the most important of the equilibrium criteria and also shows the equality of the molar Gibbs energies as a condition for equilibrium.


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• Equality of Gibbs Free Energy as condition for equilibrium:

Equilibrium Condition in Problem Solving

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EXAMPLE: Two identical metal blocks of mass M with initial temperatures T1, i and T2, i , respectively, are in contact with each other in a well-insulated (adiabatic), constant-volume enclosure. Find the final equilibrium temperature of the blocks.

• Choose the two metal blocks as the system for writing the balance equations.


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• Balance equations: There is no exchange of mass between the blocks, so

the mass balance does not provide any useful information

The energy balance for the system is:

or by introducing of heat capacity:


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• Equation below can be simplified

• to:

• And by using the entropy balance for the system in which M, U, and V are constant, at equilibrium the entropy is a maximum.

• The final entropy of the system is


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• To calculate this final equilibrium temperature, we use the energy balance:

• The equilibrium equation can be satisfied if T1,f = T2,f = Tfinal ;

• that is, the final temperature of the two blocks must be equal:

• And

Proving the Equality of Gibbs Energies for Vapour-Liquid Equilibrium

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ILLUSTRATION 7.1-2 • Use the information in the steam tables of Appendix A

to show that Eq. 7.1-9c is satisfied at 100°C and 0.10135 MPa. The equilibrium equation can be satisfied if

• SOLUTION• From the saturated steam temperature .table, we

have at 100°C and 0.101 35 MPa


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ILLUSTRATION 7.1-2 Given:• 100°C and 0.101 35 Mpa

• We know that: Ḡ = Ḣ - T Ṧ • Then we have: ḠL = ḢL - T Ṧ L = 419.04 – 373.15 X 1.3069 = - 68.6 kJ/kg

ḠV = ḢV - T Ṧ V = 2676.1 – 373.15 X 7.3549 = - 68.4 kJ/kg


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ḠL = ḢL - T Ṧ L = 419.04 – 373.15 X 1.3069 = - 68.6 kJ/kg

ḠV = ḢV - T Ṧ V = 2676.1 – 373.15 X 7.3549 = - 68.4 kJ/kg

• These are close enough to justify ḠL = ḠV at this vapor-liquid phase equilibrium state and therefore:

ḠL = ḠV

Stability of Thermodynamic Systems

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• The result dS = 0 used to identify the equilibrium state of an initially non-uniform system constrained to remain at constant mass, internal energy, and volume.

• In this section we explore the information content of the stability criterion:

d2S 0 , at constant M, U, and V By studying the sign of the second differential of the

entropy, we are really considering the following question:


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• Suppose that a small fluctuation in a fluid property, say temperature or pressure, occurs in some region of a fluid that was initially at equilibrium;

is the character of the equilibrium state such that d2S < 0, and the fluctuation will dissipate, or such d2S > 0 , in which case the fluctuation grows until the

system evolves to a new equilibrium state of higher entropy

Since we know that fluids exist in thermodynamically stable state, we will accept that d2S < 0 for all real fluids at equilibrium, and rather establish the restrictions placed on the equations of state of fluids by this stability condition


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• Problem of the intrinsic stability of the equilibrium state in a pure single-phase fluid, and then

• Problem of the mutual stability of two interacting systems or phases

• Previous example of Figure 7.1-1 of the last section, equilibrium in a pure fluid at constant mass (actually, we will use number of moles), internal energy, and volume.

• Using the subdivision of the system into two subsystems, and writing the extensive properties N, U, V, and S as sums of these properties for each subsystem, we: were able to show that the condition:


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• Using the subdivision of the system into two subsystems, and writing the extensive properties N, U, V, and S as sums of these properties for each subsystem, we were able to show that the condition:

• for all system variations consistent with the constraints led to the requirements that at equilibrium:


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• Now we write an expression for the stability requirement d2S < 0 for this system and obtain:

• It now remains to evaluate the various entropy derivatives, so that the stability restrictions of Equation above and its derivatives can be put into a more usable form.


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• Starting with equation below:

• Which for an open system becomes:

• And:

Thermal Stability Criterion

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• Finally, since T is absolute temperature and positive, one condition for the existence of a stable equilibrium state of a fluid is that:

• This is the First or Thermal Stability Criterion• Which says that for an open system the constant-

volume heat capacity must be positive, so that internal energy increases as the fluid temperature increases

Cv > 0


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• Similarly we can derive

• And with quite a lot of Algebra we obtain:

• Which can be reduced to

Mechanical Stability Criterion

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• Finally we have the Second or mechanical stability criterion:

• Or as

• where kT : the isothermal compressibility of the fluid. • This result indicates that if a fluid is to be stable, its

volumetric equation of state must be such that the fluid volume decreases as the pressure increases at constant temperature.


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• The main conclusion from this exercise is that if a fluid is to exist in a stable equilibrium state, that is, an equilibrium state in which all small internal fluctuations will dissipate rather than grow, the fluid must be such that:

• and

or

Cv > 0


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• Since all real fluids exist in thermodynamically stable states, these equations must be satisfied for real fluids.

• In fact, no real fluid state for which either

> 0 or Cv < 0 has yet been found!

• Hence, for real fluids the criteria as set out below are valid:

Cv > 0

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Stability Criteria

• Next we consider the problems of identifying the equilibrium state for two interacting phases of the same molecular species but in different states of aggregation,• Determination of requirements for the stability

of this state. • An example of this is a vapor and a liquid in

equilibrium. To be general, we again consider u composite system isolated from its environment, except that here the boundary between the two subsystems is the real interface between the phases.

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Stability Criteria

• For this system we have:

Since N, U , and V are fixed, the equilibrium condition is that the entropy should attain a maximum value.

However. we allow for the fact that the states of aggregation in regions I and II are different, so that the fluids in these regions may follow different equations of state.

• Using the same analysis we find that at equilibrium (i.e., when dS = O), TI = TII , P I = P II and G I = G II

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Stability Criteria

• Using the same analysis we find that at equilibrium (i.e., when dS = O),

TI = TII and P I = P II are obvious conditions, but

G I = G II Not so obvious condition for equilibrium!

• Again, from the stability condition d2 S < 0, we obtain: (following the same analysis as before:

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Stability Criteria – More Phases

• Here, however, the two partial derivatives in each of the bracketed terms need not be equal, since the two phases are in different states of aggregation and thus obey different equations of state, or different roots of the same equation of state. • It is clear from a comparison with Eq. 7.2-2b that a

sufficient condition for Eq. 7.2-16 to be satisfied is that each phase to be intrinsically stable; that is, Eq. 7.2-16 is satisfied if, for each of the coexisting phases, the equations:

need to be satisfied!! Therefore, a condition for the mutual stability of two interacting subsystemsis that each subsystem be intrinsically stable.

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Stability Criteria – More Phases

• Therefore, a condition for the mutual stability of two interacting subsystems is that each subsystem be intrinsically stable.

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Phase Equilibria: Application of Equilibrium And Stability Criteria to an Equation Of State

• Isotherms computed using a the van der Waals equation of state (as an example of an EOS).

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Application of Equilibrium And Stability Criteria to an Equation Of State

• Isotherms are labeled so that T5 > T4 > T3 > T2 > T1

• The isotherm T3 has a single point, C, for which ()T = 0; at all other points on this isotherm ()T < 0. On the isotherms T4 and T5, ()T < 0 everywhere, whereas for the isotherms T1 and T2,

()T < 0 in some regions and ()T > 0 in other regions

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Application of Equilibrium And Stability Criteria to an Equation Of State

• The criterion for fluid stability requires that ()T < 0;• This is satisfied for the isotherms T4 and T5, but not in

the aforementioned regions of the T1 and T2 isotherms.• Thus we conclude that the regions A to B and A' to B' of

the isotherms T1 and T2 , respectively, are not physically realizable; that is,

they will not be observed in any experiment since a fluid in these states is not stable, and instead would go to an appropriate stable state

Not peculiarity of the van der Waals EOS; other EOS’s show very similar behaviour

Investigate this isotherm more thoroughly

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Equilibrium And Stability Criteria in EOS

• Take any constant-pressure line between PA and PB in this figure, such as Pa

• We see that it intersects the equation of state three times, corresponding to the fluid volumes V a V’ a , and V ‘’a:.

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• We see that it intersects the equation of state three times, corresponding to the fluid volumes V a ; V’ a , and V ‘’a

• One of these, V ‘’a, is on the part of the isotherm that is not possible by the stability criterion.

• However. the other two intersections, at V a and V’ a , are physically possible.

• This suggests that at a given pressure and temperature the system can have two different volumes;

• But:• This contradicts the axiom that two state variables

completely determine the state of a single-component, single-phase system!

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• This can only happen if equilibrium can exist between two phases of the same species that are in different states of aggregation (and hence density).

• The equilibrium between liquid water and steam at 100°C and 101.325 kPa (I atm) is one such example.

• Hence, there is a portion of the isotherm where the specific volume varies continuous at fixed temperature and pressure;

• Two-phase coexistence region, here the vapour-liquid coexistence region.

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• The variation of the overall (or two-phase) specific volume in this region arises from the fact that although the specific volumes of the vapour and liquid phases are fixed (since the temperature and pressure are fixed in each of the one-phase equilibrium subsystems),

the fraction of the mixture that is vapour, ω V, can vary continuously from 0 to 1.

In the two-phase region the specific volume of the two-phase mixture is given by:

where ω V and ω L are fractions of vapour and liquid, on a molar basis.

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where ω V and ω L are fractions of vapour and liquid, on a molar basis and

And ω V + ω L = 1

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Maxwell or Lever Rule


And ω V + ω L = 1

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Maxwell or Lever Rule


And ω V + ω L = 1

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Computing the Properties of a Two-Phase Mixture

Compute the total volume, total enthalpy, and total entropy of 1 kg of water at 100°C, half by weight of which is steam and the remainder liquid water.SOLUTIONFrom the saturated steam temperature table at 100°C, the equilibrium pressure is 0.101 35 MPa and

• Using equation:

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SOLUTIONWe obtain by substitution into :

• Using corresponding equation we get the enthalpy as well :

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SOLUTIONWe obtain by substitution into :

• Using corresponding equation we get the enthalpy as well :

• And the entropy:

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Two-Phase Mixture

• as the pressure is lowered along an isotherm on which a liquid-vapour phase transition occurs, the actual volume-pressure behaviour is as shown below (there is no sigmoid curve!):

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Two-Phase Mixture

• We can improve the representation of the two-phase region when using the van der Waals or other EOS by recognizing that all van der Waals loops, such as those in Fig. 7.3-2. should be replaced by horizontal lines (isobars), as shown in Fig. 7.3-3.

• This construction ensures that the equilibrium phases will have the same temperature and pressure

• Question: at which pressure should the isobar be drawn, since any pressure such that

Pa < P < Pb

• will yield an isotherm like that in Fig. 7.3-3. • Answer: The pressure chosen must satisfy the last

condition for equilibrium, G I = G II

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Two-Phase Mixture

• To find out the equilibrium pressure we start from:

• We integrate between any two points along an isotherm of the equation of state,

• Then we have:

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Two-Phase Mixture

• To calculate the specific volume in each of the integrations we use the equation of state for the appropriate part of the van der Waals loop.

• Alternatively, we can find the equilibrium pressure graphically by noting that Eq. 7.3-2 requires that

• Areas I and II between the van der Waals loop and the constant pressure line in Fig. 7.3-2 be equal at the pressure at which the vapour and liquid exist in equilibrium.

• This vapour-liquid coexistence pressure, (a function of temperature!), is called the vapor pressure of the liquid and will be denoted by Pvap(T).

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Two-Phase Mixture

• of temperature!), is called the vapor pressure of the liquid and will be denoted by Pvap(T).

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Two-Phase Mixture

• We can continue in the manner described here to determine the phase behaviour of the fluid for all temperatures and pressures.

• For the van der Waals fluid, this result is the temperature!), is called the vapour pressure of the liquid and will be denoted by Pvap(T). An important feature of this figure is the dome-shaped. [two-phase co-existence region].

• The inflection point C of Fig. 7.3-1 is the peak of this dome, and therefore is the highest temperature at which the condensed phase (the liquid) can exist;

• this point is called the critical point of the liquid.

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Two-Phase Mixture

• It is worthwhile retracing the steps followed in identifying the existence and location of the two-phase region in the P-V plane:

1. The stability condition () T < 0 was used to identify the unstable region of an isotherm and thereby establish the existence of a two-phase region.

2. The conditions T' = T" and P' = P" were then used to establish the shape (but not the location) of the horizontal coexistence line in the P-V plane.

3. Finally. the equilibrium condition, G I = G II was used to locate the position of the co-existence line.

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Two-Phase Mixture• Three-dimensional PVT phase diagram is shown in

Fig. 7.3-5.

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Two-Phase Mixture• The P, V, T phase diagram for a substance with a

single solid phase diagram

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Two-Phase Mixture If we assume that it were possible to compute the

Gibbs energy as a function of temperature and pressure for any phase, either from an equation of state, experimental data, or statistical mechanics.

Then, at fixed pressure, we could plot G as a function of T for each phase, see Fig. for the vapor and liquid phases. equilibrium condition is G is a minimum! At phase transition temperature Tp. • the liquid is the equilibrium phase at temperature T<

Tp

• the vapor is the equilibrium phase above Tp

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Two-Phase Mixture• The angle of intersection θ between the liquid and

vapor Gibbs energy curves decreases as the pressure (and temperature) at which the intersection occurs

increases (provided P < Pc). • At the critical pressure, the two Gibbs energy curves intersect, with θ = 0: i.e., the two curves are collinear for some range of T around the critical temperature T.Thus, at the critical point,

Since: we find at the critical point: • And also

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Two-Phase Mixture

• Since the molar Gibbs energy, molar entropy, temperature, and pressure each have the same value in the vapor and the liquid phases, the values of all other state variables must be identical in the two equilibrium phases at the critical point.

• Therefore, the vapor and liquid phases become indistinguishable at the critical point.

This is exactly what is experimentally observed. At all temperatures higher than the critical

temperature, regardless of the pressure, only the vapor phase exists.

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Two-Phase Mixture• This is the reason for the abrupt termination of the

vapor-liquid coexistence line in the pressure-temperature plane at the critical point.

• The triple point, this is the intersection of the solid-liquid, liquid-vapor, and solid-vapor coexistence curves.

• It is the only point on the phase diagram where the solid, liquid, and vapor coexist at equilibrium.

Since the solid-liquid coexistence curve generally has a steep slope the triple-point temperature for most fluids is close to the normal melting temperature, i.e. the meltingtemperature at atmospheric pressure

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The Molar Gibbs Energy and Fugacity of a Pure Component

• We need to consider how one uses an equation of state to identify the states of vapor-liquid

equilibrium in a pure fluid.• The starting point is the equality of molar Gibbs

energies in the coexisting phases:

• Then using equation:

We obtain and

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To compute V as a function of T and P, only Eq. below will be considered further:

• Integration of equation above between any two pressures P1 and P2 (at constant temperature)

yields:

• If the fluid under consideration were an ideal gas, following equation would be valid:

We obtain then by substitution:

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• If we subtract the ideal gas equation from that of the real gas we obtain:

• Further, i. setting P1 equal to zero, ii. recognizing that at P = 0 all fluids are ideal gases

so that G(T1, P = 0 ) = G IG (T1 , P = 0), andiii. omitting all subscripts yields equation:

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Definition of Fugacity of a Pure Component

• If we define a new thermodynamic function, the fugacity, denoted by the symbol f .

• and the related fugacity coefficient Ф by division with the pressure as:

• Fugacity is a well-defined function related to the exponential of the difference between the real

and ideal gas Gibbs energies.

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Definition of Fugacity

• In chemical thermodynamics, the fugacity (f) of a real gas is an effective pressure which replaces the true mechanical pressure in equilibrium calculations.

• It is equal to the pressure of an ideal gas which has the same chemical potential as the real gas.

EXAMPLE: Nitrogen gas (N2) at 0°C and a pressure of P=100 atm has a fugacity of f = 97.03 atm.[1] This means that the chemical potential of real nitrogen

at a pressure of 100 atm is less than if nitrogen were an ideal gas;

the value of the chemical potential is that which nitrogen as an ideal gas would have at a pressure of 97.03 atm.

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Definition of Chemical Potential

• In chemical thermodynamics, the chemical potential, μ (also known as partial molar free energy) is a form of potential energy that can be absorbed or released during a chemical reaction.

• The chemical potential of a species in the mixture can be defined as the derivative of the free energy of the system with respect to a change in the number of moles of just that species.

• Thus, it is the partial derivative of the free energy with respect to the amount of the species, all other species' concentrations in the mixture remaining constant, and at constant temperature.

• At constant pressure , the chemical potential is the partial molar Gibbs free energy.

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Definition of Chemical Potential

• At chemical equilibrium (or in phase equilibrium) the total sum of chemical potentials is zero, as the free energy is at a minimum.

• The fundamental equation of chemical thermodynamics for a system containing n constituent species, with the I - th species having Ni particles is, in terms of Gibbs energy:

• At constant temperature and pressure we get:

• The chemical potential is therefore the partial molar Gibbs free energy.

(at constant T and P)

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Definition of Fugacity• From this definition it is clear that the fugacity

has units of pressure, and that f → P, when P → 0; that is, the fugacity becomes equal to the pressure at

pressures low enough that the fluid approaches the ideal gas state.• The fugacity function has been introduced

because its relation to the Gibbs energy makes it useful for phase equilibrium calculations• The present criterion for equilibrium between

two phases was:

• with the restriction that: temperature and pressure be constant and equal in the two phases! • Now we can use the fugacities instead!

G I = G"

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Definition of Fugacity of a Pure Component

• Equation

• becomes by substitution

• and with

• equation above becomes:

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Equilibrium Criterion with Fugacity of a Pure Component

• Or in terms of the fugacity:

Equilibrium Condition

• and using the fugacity coefficients:

These equations follow directly from the equality of the molar Gibbs Free Energy in each of the two phases:

important forms of the equilibrium criteria ;

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• Important forms of the equilibrium criteria ; follow directly from the equality of the molar Gibbs

energy in each phase at phase equilibrium• In all equations of state pressure is an explicit

function of volume and temperature.• Therefore, it is useful to have an equation relating

the fugacity to an integral over volume (notpressure).

• From equation

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• From equation

• and

• we obtain by changing the variable of integration:

• with Z = P V / R T , the compressibility factor

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• In addition we know

• and for the temperature dependence of the fugacity given as the temperature dependence of the logarithm of the fugacity coefficient (f/P)

we get

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• The fugacity function is of central importance in phase equilibrium calculations, therefore we consider here the computation of the fugacity for pure gases, liquids, and solids

Fugacity of a Pure Gaseous Species

Using the above EOS, where the superscript V is used to designate the fugacity and Z the compressibility of the vapor phase

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Fugacity Pure Gaseous Species

• If we accept that EOS is applicable up to the pressure of interest, the fugacity of a pure gas can be computed by integration of this equation:

At very low pressures, where a gas can be described by the ideal gas equation of state

the following equations are valid, i.e. fugacity is equal to total pressure

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Fugacity of a Pure Gas

EXAMPLEFugacity CalculationCalculate the fugacity of steam at 300 oC and 8 MPa using the data given at the steam tables (superheated steam)SOLUTIONWe use following equation:

From the superheated vapour steam Tables at 300°C, we have at different pressures:

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Superheated Vapor Steam Tables at 300°C

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Fugacity of a Pure Gas

EXAMPLE We carry out numerical integration of the equation

using the data above and we find :

and finally we find the fugacity f:

Also, the fugacity coefficient, φ, in this case is

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Example: Calculation of Fugacity of a Pure Component

EXAMPLE 2Fugacity CalculationCalculate the fugacity of steam at 300 oC and 8 MPa using the data given at the steam tables (superheated steam)• At 300 oC and 0.01 MPa we have from the steam tables:

Ĥ = 3076.5 kJ/kg and Ŝ = 9.28 13 kJ/(kg K).

• Therefore, for the Gibbs free energy:

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EXAMPLEFugacity Calculation• By multiplying with the mole number (water: 18

g/mol) we get the molar Gibbs Free Energy:

• We have assumed a very low pressure P = 0.01MPa; steam behaves like an ideal gas:

Calculation of Fugacity of a Pure Component

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Fugacity Calculation• Then we obtain G IG [300 oC, 8 MPa] by substitution:


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Fugacity Calculation• For real steam at 300°C and 8 MPa, we have, from

the steam tables: Ĥ = 2785.0 kJkg and Ŝ = 5.7906 kJ/(kg K), so thatfrom: Ḡ = Ĥ - T Ŝ

Ḡ REAL [300 oC, 8 MPa] = 2785.0 kJkg – 573K x 5.7906 kJ/kg

K = - 533.88 KJ/Kg

And for the molar Gibbs free energy:

G REAL [300 oC, 8 MPa] = - 533.88 KJ/Kg x 18 g/mol = -9 617 J/mol


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Fugacity Calculation• Finally to calculate fugacity we use the equation

below:

• And obtain:

• Therefore fugacity, f [300 oC , 8 Mpa] = 6.402 MPa

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Calculation Of Fugacity Coefficient

Fugacity Coefficient • Finally we can calculate the fugacity coefficient:

And Ф = = 0.8003

These values are in excellent agreement with the results obtained in the previous Example

Ф =

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Calculation of Fugacity of Saturated Steam

EXAMPLE 3Compute the fugacity of saturated steam at 300 oC• First find the saturation pressure of steam (from the

steam tables) at 300°C: PSAT = 8.581 MPa • Following the previous illustration, we have

• Ḡ REAL [300 oC, 8.581 MPa] = -520.15 kJ/kg; and • G REAL [300 oC, 8.581 MPa] = -520.15 kJ/kg x 18 g/mol =

= - 9376 J/mol

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Calculation of Fugacity of Saturated Steam

EXAMPLECompute the fugacity of saturated steam at 300 oC

• Then we obtain for the fugacity:

Thus the fugacity of the vapour is: f V = 6.7337 MPa ; At equilibrium

f L = 6.7337 MPa ; therefore f V = f L

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Calculation of Fugacity using the virial equation of state

• At low to moderate pressures, the virial equation of state shortened after the second virial coefficient

if data for B(T) are available

• Then we obtain for the fugacity:

with Z

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EXAMPLECompute the fugacities of pure ethane and pure butane at 373.15 K and 1, 10, and 15 bar, assuming the virial equation of state can describe these eases at these conditionsGiven:

SOLUTION• We use the equations:

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Then we obtain for the fugacities:

Since the pressures are not very high, these results are reasonably accurate

However, the virial equation with only the second virial coefficient will be less accurate as the pressure increases.

In fact, at slightly above 15 bar and 373 K, n-butane condenses to form a liquid!

In this case the virial equation description is inappropriate

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Calculation of Fugacity of a Gasat Higher Pressures

• At higher pressures, a more complicated equation of state (or higher terms in the virial - expansion) must be used

• By using the van der Waals equation, (not very accurate! ) we obtain:

• This equation is sometimes applicable but:

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Calculation of Fugacity of a Gasusing the Peng Robinson EOS

• At higher pressures, a more complicated equation of state (or higher terms in the virial - expansion) must be used

• For hydrocarbons and simple gases, the Peng-Robinson equation provides a more accurate description

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Calculation of Fugacity of a Gasusing the Peng Robinson EOS

• Fugacities calculation with the Peng-Robinson equation

with A = αP/(RT)2 and B = Pb/RT

• Obviously, the mathematical calculations increase significantly, but there is software available that significantly facilitates matters

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Procedure for Calculation of Fugacity of a Gas Higher Order EOS

• To use either the virial, the van der Waals, Peng-Robinson, or other equation of state to calculate the fugacity of a gaseous species, the following procedure is used:

1. For a given value of T and P, use the chosen equation of state to calculate the molar volume V or, equivalently, the compressibility factor Z

2. When using cubic or other compIicated equations of state , the low-density (large V or Z) solution is used.

3. This value of V( or Z) is then used in the appropriate Fugacity equations to calculate the species fugacity coefficient, f/P, and then the fugacity.

4. The Peng-Robinson equation-of-state programs or MATHCAD worksheets are described in Appendix B

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Procedure for Hand Calculation of Fugacity of a Gas

• For hand calculations it is simpler, but less accurate, to compute the fugacity of a species using a specially prepared corresponding-states fugacity chart.

• To do this, we note that since, for simple gases and hydrocarbons, the compressibility factor Z obeys a corresponding-states relation (see Sec. 6.6), the fugacity coefficient f/P given by Eq. 7.4-6 can aIso be written in corresponding-states form as follows:

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Procedure for Hand Calculation of Fugacity of a Gas

• This can be simplified to:

• the fugacity coefficient can be tabulated in the corresponding-states manner.

• The corresponding-states correlation for the fugacity coefficient of nonpolar gases and liquids given in following Fig. 7.4-1 was obtained using Eq. above and the compressibility correlation

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Equilibrium State Of A System Of Several Phases - Gibbs Phase Rule

• GIBBS phase rule for a one-component System We know that to completely fix the equilibrium

thermodynamic state of a one-component, single-phase system, we must specify the values of two state variables

• EXAMPLE• to fix the equilibrium thermo-

dynamic state in either the vapour, liquid, or solid region both the temperature and pressure are needed

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Equilibrium State - System Of Several Phases - Gibbs Phase Rule

• Thus, we say that a one-component, single-phase system has two degrees of freedom.

• Further, to fix the total size or extent of the system we must also specify its mass or one of its extensive properties, such as total volume or total energy, from which the mass can be calculated!• equilibrium, single-component, multiphase

system.QUESTIONS• How many state variables must be specified to

completely fix the thermodynamic state of each phase when several phases are in equilibrium (i.e., how many degrees of freedom are there in a single-component, multiphase system)?

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QUESTIONS• How many additional variables need be specified,

(of what type?), to fix the distribution of mass (or number of moles) between the phases, therefore fixing the overall molar properties of the

composite, multiphase system? Finally, • what additional information is needed to fix the

total size of the multiphase system?• It might appear that if P phases are present, the

system would have 2P degrees of freedom.; However, actual number of degrees of freedom is

fewer, since the requirement that the phases be in equilibrium puts constraints on the values of certain state variables in each phase

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QUESTIONS• It might appear that if P phases are present, the

system would have 2P degrees of freedom.; Equilibrium constraint: TI = TII

P- 1 relations to be satisfied• Similarly: at equilibrium the pressure in each phase

must be the same: PI = PII

so that there are an additional P- 1 restrictions• Finally, at equilibrium, the molar Gibbs energies

must be the same in each phase:GI (T, P) = GII (T, P)

additional P - 1 restrictions

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• A total of 3(P- 1) restrictions on the 2P state variables needed to fix the thermodynamic state of each of the P phases, the number of degrees of freedom for the single-component, multiphase system is:

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Gibbs Phase Rule - Discussion

• We have already indicated that a single-phase (P = 1) system has two degrees of freedom;

Gibbs Phase Rule; F = 3 – P = 2• To specify the thermodynamic state of each phase in a

two-phase system (i.e., vapor-liquid, vapor-solid, or solid-liquid coexistence regions), it is clear from Fig.that we need specify only the temperature or the pressure of the system; the value of the othervariable can then be obtained from the appropriate coexistence Curve!

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• Setting P = 2 confirms that a two-phase system has only a single degree of freedom.

• Finally, since the solid, liquid, and vapor phases coexist at only a single point, the triple point, a single-component, three-phase system has no degrees of freedom :

Phases Number: P = 3F = 3 - P = 3 – 3 = 0

Hence, to fix the thermo-dynamic state of each of the Pphases in equilibrium, we must specify 3 - properties of the individual phases.

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• If there are P phases, there are P values of ω i the mass fraction in phase i, which must be determined.

• Sum of mass fractions must be unity

• Then we need P - 1 additional equations of the form:

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• For example, if we know that steam and water are in equilibrium at some temperature T (which fixes the single-degree freedom of this two-phase system: F= 3 – P = 3 – 2 = 1) ),

the equation of state or the steam tables can be used to obtain the equilibrium pressure, specific enthalpy, entropy, and volume of each of the phases,

• but not the mass distribution between the phases. • However, if the volume [or enthalpy or entropy, etc.)

per unit mass of the two-phase mixture were known, this would be sufficient to determine the distribution of mass between the two phases, and then all the other overall thermodynamic properties

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Gibbs Phase Rule - Use

EXAMPLEUse of the Gibbs Phase Rule a. Show, using the steam tables, that fixing the

equilibrium pressure of a steam-water mixture at 1.0135 bar is sufficient to completely fix the thermodynamic states of each phase

b. Show that additionally fixing the specific volume of the two-phase system Ṽ at 1 m3/kg is sufficient to determine the distribution of mass between the two phases.

c. What is the total enthalpy of 3.2 m3 of this steam-water mixture

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SOLUTIONa. Using the steam tables we show that fixing the

pressure at 1.0135 bar is sufficient to determine the temperature and the thermodynamic properties of each phase:

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SOLUTIONb. To determine the relative amounts of each of the

phases, we need information from whichωV and ωL can be determined. Here the specific volume of the two-phase mixture is given so we can use Eqs.

to determine the mass distribution between the two phases:

and ωV = 0.5975 and ωL = 0.4025

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SOLUTIONc. The total mass is given

• From the results in parts (a) and (b) we can compute the enthalpy per unit mass of the two-phase mixture:

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Vapour Pressure of a Fluid

Definition• Vapor pressure or equilibrium vapor pressure is

the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or l iquid) at a given temperature in a closed system. • The equilibrium vapor pressure is an indication of

a liquid's evaporation rate. • It relates to the tendency of particles to escape

from the liquid (or a solid). • A substance with a high vapor pressure at normal

temperatures is often referred to as volatile

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• Vapor pressure of any substance increases non-linearly with the temperature according to the Clausius–Clapeyron relation. • The atmospheric pressure boiling point of a

liquid (also known as the normal boiling point) is the temperature at which the vapor pressure equals the ambient atmospheric pressure. • With any incremental increase in that

temperature, the vapor pressure becomes sufficient to overcome atmospheric pressure and lift the liquid to form vapor bubbles inside the bulk of the substance.

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• One phase transition property of high importance is the slope of the coexistence curves in the P-T

pIane; • the slope of the vapor-liquid equilibrium line

gives the rate of change of the vapor pressure of the liquid with temperature,

• the slope of the vapor-solid coexistence line is equal to the change with temperature of the vapour

pressure of the solid (called the sublimation pressure), and • the inverse of the slope of the liquid-solid

coexistence line gives the change of the melting temperature of the solid with pressure.

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• The slope of any of these phase equilibrium coexistence curves can be found by starting with Eq .

• For a small change we can write:

• which the becomes:

• Since, at equilibrium, T1 = T2 and P1 = P2

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• Finally we can write:

• and

• or

• which the becomes:

The Clapeyron Equation • the application of Eq. above to the vapor-liquid

coexistence curve is of great interest, because this gives the change in vapor pressure

with temperature.

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Clausius - Clapeyron Equation

• At temperatures for which the vapor pressure is not very high, it is found that VV >> VL and

ΔV ≈ VV . • If, in addition, the vapor phase is ideal, we have

• And the Clausius Clapeyron Equation:

• After integration this becomes:

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Clausius - Clapeyron Equation

129

• Plots of vapour pressure, P vs T are nonlinear curves (see examples below)

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• However, the Clausius-Clapeyron equation relates P, T and the enthalpy of vaporization, Hvap in a straight line relationship as discussed below:

• Plots of vapor pressure, P vs T are nonlinear curves (see examples below):

• With ΔHvap the enthalpy of vaporization

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• This form of the equation a linear equation,

y = mx + b ; where

y = lnP ; x = 1/T,

• m = - ΔHvap / R is the slope, and • b = C (a constant) is the intercept.

• Taking the following data , we can determine Hvap for C6F6 (hexafluorobenzene):

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• Taking the following data , we can determine ΔHvap for C6F6 (hexafluorobenzene):

Temperature (K)Vapor Pressure (torr)

1/T ln P

280.0 32.42 0.003571 3.478

300.0 92.47 0.003333 4.527

320.0 225.1 0.003125 5.417

330.0 334.4 0.003030 5.812

340.0 482.9 0.002941 6.170

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• Taking the plot of ln P vs 1/T , we can determine ΔHvap for C6F6 (hexafluorobenzene):

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• Slope = -4288.1 (K) = -ΔHvap/R

or ΔHvap = R x 4288.1 (J/mol-K)x(K)

ΔHvap = 8.3145 x 4288.1 (J/mol) = 35.65 x103 J/mol

Finally, ΔHvap = 35.65 kJ/mol

• The quality of the "fit" of the line, correlation coefficient=0.9999 testifies both to the

• quality of the data and the • correctness of the equation.

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• The second form of the Clausius-Clapeyron equationln(P2 / P1) = ΔHvap/R(1/T1-1/T2)

• is useful to predict the vapour pressure at one temperature if it is known at another.

• Taking the natural antilog of (or exponentiating) both sides, we get:

P2/P1 = exp[ΔH/R(1/T1-1/T2) ] or

P2 = P1exp[ΔH/R(1/T1-1/T2) ]• This last expression gives the shape of the vapour

pressure curves shown above.

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This formula allows us to calculate the pressure, P2, of a pure gas as a function of temperature, T, if we know:

P2 = P1 e [ΔH/R(1/T1-1/T2) ]

P1 the pressure of the gas at temperature T1

• ΔHvap the heat of vaporization of the liquid usually given in kJ/mol

• BUT don't forget to CONVERT to J/mol to be consistent with gas constant value commonly used

• R the gas constant in J/mol: R = 8.3145 J/mol

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EXAMPLE• Water, has a heat of vaporisation, ΔHvap= 44.0 kJ/mol

and its vapour pressure is 1 atm at 100 °C. Find the vapour pressure of water at T2 = 25 °C and at

T2 ‘= 125 °C?SOLUTION:• Convert ΔHvap to J/mol:• ΔHvap =44.0 kJ/mol x 1000 J/kJ = 44 000 J/mol• Use the gas constant value, R=8.3145 J mol-1 K-1• Convert temperatures to K:• T1=373.15 K, T2=298.15 K or T2 ‘= 398.15 K Plug into equation P2 = P1 e [ΔH/R(1/T1-1/T2) ]

and solve for P2

Answer: P2 = 0.028 atm and P2 ‘= 2.44 atm

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ILLUSTRATION 7.4-7Calculation of the Fugacity of Liquid Water from Density Data• Compute the fugacity of liquid water at 300°C and

25 MPa SOLUTION:

• We start from equation above describing the fugacity of a liquid at a certain temperature and pressure

Fugacity of Liquids

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SOLUTION:• The exponential term in this equation, known as the

Poynting pressure correction, accounts for the increase in fugacity due to the fact that the system pressure is higher than the vapour pressure of the liquid.

• We can assume that VL = VL Sat since water being a liquid, is not compressible

Fugacity of Liquids

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SOLUTION:• We saw from the previous example that: f sat (300°C) = 6.7337 MPa and from the steam

tables at 300°C : Pvap = 8.581 MPa and ṼL = 0.001 404 m3/kg,

so that

• By substitution into equation of the previous slide we obtain:

Fugacity of Liquids

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SOLUTION:• For the fugacity:

• Therefore the fugacity of liquid water at 300°C and 25 MPa is 7.347 MPa:

Fugacity of Liquids

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ILLUSTRATION 7.4-8• Calculation of the Fugacity of Ice from Density Data

Compute the fugacity of ice at -5°C and pressures of 0.1 MPa, 10 MPa, and 100 MPa . We know that at -5°C, the sublimation pressure of ice is 0.4 kPa and its specific gravity is 0.915.

SOLUTION• We start from equation below describing the

fugacity of a solid at a certain temperature and pressure

• We assume that ice is incompressible over the range from 0.4 kPa to 100 MPa, so that we can calculate its molar volume over this pressure range as:

Fugacity of Solids

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SOLUTION• Molar volume of ice over this pressure range is:

• Equation below gives the fugacity of a solid:

• By substitution we obtain:

Fugacity of Solids

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• By substitution we obtain for 0.1 MPa:

• We shall see that change in fugacity of a condensed species (L or S) with small pressure changes is small.

• Here we find that the fugacity of ice increases by 10 percent for an increase in pressure from the sublimation pressure to 100 bar, and by a factor of 2.5 as the pressure on ice increases to 1000 bar.

Fugacity of Solids

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SOLUTION• Similarly we obtain for 10 MPa:

• And for 100 MPa

Fugacity of Solids

THANK YOU

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TDA 301T- 2015-10 - Thermodynamic Properties Real Substances

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