Chapter 7: Equilibrium and Stability in One-Component Systems

Important Notation

Learning objectives Be able to: Identify the criterion for equilibrium subject to different constraints (specifications); Use the concept of stability to predict when vapor-liquid phase splitting occurs; Identify the conditions of phase equilibrium and of the critical point; Compute the fugacity of a pure liquid or vapor; Use fugacity in calculations of vapor-liquid equilibrium.

The criteria for equilibrium Consider, for simplicity, a simple thermodynamic system with known amounts of substances (on a mass or molar basis) . To define its equilibrium state it is necessary to specify two additional independent variables. Examples: Common in the design of steady-state processes Equilibrium in storage vessels of known T and V. Tank charges and discharges.

The criteria for equilibrium The criteria for equilibrium criteria for these specifications (and many more) can be derived in an organized way, starting from the equilibrium condition of a closed, isolated system: M, U, V are specified and constant

Equilibrium in an isolated system No mass in or out; neglect changes in kinetic and potential energies. The energy and entropy balances are: with because of the 2nd law of thermodynamics.

Equilibrium in an isolated system 0, no heat transfer to an isolated system 0, no change in volume in an isolated system 0, no heat transfer to an isolated system

Equilibrium in an isolated system for systems in which M, U, V are constant. Away from equilibrium, the system conditions change, increasing its entropy. After a long enough wait (from a fraction of a second to many years, depending on the system), the system attains a condition in which its state properties no longer change, including the entropy. At this state, the equilibrium state, the entropy has a maximum value.

Equilibrium in an isolated system Consider an isolated system composed of two subsystems containing a single chemical component:

Equilibrium in an isolated system Consider an isolated system composed of two subsystems containing a single chemical component: Find the equilibrium condition if the internal wall is diathermal, rigid, and impermeable.

Equilibrium in an isolated system Consider an isolated system composed of two subsystems containing a single chemical component: Find the equilibrium condition if the internal wall is diathermal, rigid, and impermeable.

Equilibrium in an isolated system

Equilibrium in an isolated system At equilibrium:

Equilibrium in an isolated system At equilibrium: This looks like a very abstract equilibrium condition. Based on your daily experience, what state property would be equal when equilibrium is reached in this problem?

Equilibrium in an isolated system From Chapter 4: In each subsystem, V and N are constant:

Equilibrium in an isolated system At equilibrium: Note, for example, that the pressures will in general be different, because the internal wall is rigid.

Equilibrium in an isolated system Consider an isolated system composed of two subsystems containing a single chemical component: Find the equilibrium condition if the internal wall is diathermal, moveable, and impermeable.

Equilibrium in an isolated system Consider an isolated system composed of two subsystems containing a single chemical component: Find the equilibrium condition if the internal wall is diathermal, moveable, and permeable.

Equilibrium in an isolated system The equilibrium state was found by setting: However, this mathematical condition is valid in a point of maximum, in a point of minimum, and in an inflection point. This is necessary condition for equilibrium, but is not sufficient. Additionally, to characterize a point of maximum, the following stability criterion must also be true:

Equilibrium in systems with constant T and V No mass in or out; neglect changes in kinetic and potential energies. The energy and entropy balances are: with because of the 2nd law of thermodynamics.

Equilibrium in systems with constant T and V for systems in which M, T, V are constant. Away from equilibrium, the system conditions change, decreasing its Helmholtz energy. After a long enough wait (from a fraction of a second to many years, depending on the system), the system attains a condition in which its state properties no longer change, including its Helmholtz energy. At this state, the equilibrium state, the Helmholtz energy has a minimum value.

Equilibrium in systems with constant T and P No mass in or out; neglect changes in kinetic and potential energies. The energy and entropy balances are: with because of the 2nd law of thermodynamics.

Equilibrium in systems with constant T and P for systems in which M, T, P are constant. Away from equilibrium, the system conditions change, decreasing its Gibbs energy. After a long enough waiting (from a fraction of a second to many years, depending on the system), the system attains a condition in which its state properties no longer change, including its Gibbs energy. At this state, the equilibrium state, the Gibbs energy has a minimum value.

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Example 1 Use the information available in the steam tables of Appendix A.III of the textbook to show that the chemical potentials are equal in the vapor and liquid phases at 100oC and 0.10135 MPa. Solution: The target is to show that: The Gibbs energy is not directly available in the steam table but the enthalpy and entropy are:

Example 1 At 100oC and 0.10135 MPa:

Example 2 Find the normal boiling temperature of n-hexane according to the van der Waals and the Peng-Robinson equations of state, using the XSEOS package. Solution: The normal boiling temperature is the temperature at which there is vapor-liquid equilibrium (VLE) at the pressure of 1 atm (1.0132 bar). Departure (residual) functions

Example 2 Given an equation of state, this is, in general, a non-linear equation in temperature that needs numerical solution. Use XSEOS to solve it. How do the values you obtain compare to the experimental value of 342 K?

Example 3 Plot the vapor pressure of n-hexane as function of temperature according to the Peng-Robinson equation of state, using the XSEOS package. Compute the vapor pressure from 300 K to 450 K with 10 K intervals. Also, plot ln P vs 1/T. Solution: The phase equilibrium equation derived in Example 2 is valid: It has to be solved to find the pressure at each specified temperature along the interval.

Example 4 Plot the molar enthalpy of vaporization of n-hexane as function of temperature according to the Peng-Robinson equation of state, using the XSEOS package from 300 K to 450 K with 10 K intervals. Solution: The molar enthalpy of vaporization is the difference between the molar enthalpies of the vapor and liquid in equilibrium: Use the XSEOS solution of Example 3 as starting point.

Example 5 Plot the molar volume of the liquid and vapor phases in equilibrium as function of temperature according to the Peng-Robinson equation of state, using the XSEOS package from 300 K to 450 K with 10 K intervals. Solution: Use the XSEOS solution of Example 3 as starting point.

Stability of thermodynamic systems In, the previous examples the assumption was that the system had more than one phase. In many applications, it is important to find out whether a system splits into several phases. This is called stability analysis. There are two types of stability analysis: local and global. This chapter covers local stability analysis.

Stability of thermodynamic systems Consider an isolated system containing a single chemical component. Will the system split in two phases? ?

Stability of thermodynamic systems Entropy of the single-phase system Entropy of the two-phase system ? Key idea: If, for every attempt to split the system, , the original system is stable. In other words, splitting the system does not increase its entropy.

Stability of thermodynamic systems The analysis begins by assuming the system will split into two phases, I and II, with . Now, obtain the molar entropy in phases I and II as a Taylor series expansion around the conditions of the original system.

Stability of thermodynamic systems An analogous series expansion can be used for phase II.

Stability of thermodynamic systems

Stability of thermodynamic systems Pink boxes are equal to zero because of energy and volume conservation.

Stability of thermodynamic systems

Stability of thermodynamic systems The analysis of the other second derivatives is similar. Then: In matrix notation, we obtain what is called a quadratic form:

Stability of thermodynamic systems If the quadratic form is negative definite, the original system is stable, i.e.: If the quadratic form is negative definite, all the eigenvalues of the matrix are negative. Simplifying the notation: Your textbook has a slightly different derivation, with the same conclusions section 7.2)

Stability of thermodynamic systems Here, we will skip the formal solution using eigenvalues and use a simpler property of negative definitive matrices that says determinants of their submatrices along the diagonal are negative if of odd order, and positive if of even order, i.e.:

Stability of thermodynamic systems

Stability of thermodynamic systems The analysis of the other derivatives involves more algebraic steps. A summary of the implications of stability is:

Fugacity of a pure component The fugacity coefficient of a pure component is defined as: i.e., there is a simple and direct connection between the fugacity coefficient of a pure component and the departure (residual) molar Gibbs function at the same temperature and pressure:

Fugacity of a pure component For an ideal gas: The fugacity coefficient of a pure ideal gas is equal to 1.

Fugacity of a pure component To find the fugacity coefficient, consider its change in an isothermal process from zero pressure to the system pressure:

Fugacity of a pure component

Fugacity of a pure component What are the units of the fugacity coefficient?

Fugacity of a pure component Also, if you know how to evaluate departure (residual) functions, you know how to compute the fugacity coefficient.

Example 6 Find the expression of the fugacity coefficient of a pure component according to the van der Waals equation of state.

Example 6 Find the expression of the fugacity coefficient of a pure component according to the van der Waals equation of state. Solution: A practical difficulty: expressing the molar volume as function of pressure in the van der Waals equation of state is not easy.

Example 6 Way around: first, substitute P in the formula. To do that, it is necesssary to express dP as function of dV. Then:

Example 6 It is now a calculus problem… When studying these slides, do all the steps to prove that: For most of the equations of state commonly used for chemical process design, the steps to derive expressions for fugacity coefficients are analogous, but generally more complicated because the equations of state are more complex.

Alternative expressions for fugacity coefficients

Fugacity of a pure component We showed that the phase equilibrium condition for a pure component is: Using fugacity coefficients:

Fugacity of a pure component At equilibrium, the vapor and liquid phases have equal temperature: These are just three different ways of expressing the VLE condition in terms of fugacity coefficients.

Example 7 Find the normal boiling temperature of n-hexane according to the Peng-Robinson equations of state, using the fugacity coefficient function of the XSEOS package. Solution: The normal boiling temperature is the temperature at which there is vapor-liquid equilibrium (VLE) at the pressure of 1 atm (1.0132 bar). The phase equilibrium condition is: Solve this non-linear equation to find the normal boiling temperature.

Fugacity of a pure component Starting from: Multiply both sides by pressure: The product in each side of this equation is the fugacity of the pure component:

Fugacity of a pure component What are the units of fugacity in the S.I.?

Fugacity of a pure component

Fugacity of a pure component Effect of changing pressure at constant temperature:

Fugacity of a pure component Effect of changing pressure at constant temperature:

Fugacity of a pure component Effect of changing temperature at constant pressure :

Fugacity of a pure component

Fugacity of a pure component Effect of changing temperature at constant pressure :

Fugacity of a pure liquid Besides integrating formulas that depend on an equation of state, there is another path to evaluating the fugacity of pure liquids, especially useful for approximate calculations. To prove it, consider the effect of pressure on the logarithm of fugacity at constant temperature (shown on a previous slide): Now compute the difference between ln f at the saturation pressure and an arbitrary pressure, at the same temperature:

Fugacity of a pure liquid Poynting correction The formula above is rigorous – it involves no approximation. In practice, it is common to use it with approximations. If the difference between the saturation pressure and the pressure is small, the Poynting correction is nearly 1. If the saturation pressure is small, the saturated vapor can be considered as an ideal gas and the fugacity coefficient is approximately equal to 1.

Fugacity of a pure liquid Poynting correction When making all these approximations:

Fugacity of a pure liquid Poynting correction A less drastic approximation is to keep all terms but assume the molar volume of the liquid is pressure-independent:

Example 8 Compute the fugacity of pure liquid water at 300oC and 25 MPa from steam table data. Solution: We will assume the molar volume is pressure-independent and equal to the saturated liquid molar volume.

Example 8 Saturated liquid molar volume at 300oC: The fugacity of the saturated liquid at 300oC is: Read illustration 7.4.3 in the book to see how to get this result.

Example 8 Saturation pressure at 300oC from steam table. Then:

Fugacity of a pure solid At a given temperature, a solid may have different crystalline structures (phases) at different pressures. Assuming the molar volume of all solid phases is equal: Here:

Example 9 Compute the fugacity of pure solid napthalene at 35oC and 50 bar. Data:

Example 9 Solution:

Clayperon equation Consider a pure component in vapor-liquid equilibrium. Then: Consider some perturbation is imposed, reaching a new state of equilibrium. Then: Subtracting these two expressions, we have:

Clayperon equation Assume the perturbation is arbitrarily small. Then: But:

Clayperon equation The Clayperon equation is also valid for phase changes between solid and liquid and between solid and vapor.

Clausius-Clayperon equation Starting from the Clayperon equation: Assume (this is why the Clausius-Clayperon equation is an approximation):

Clausius-Clayperon equation Then:

Example 10 Estimate the molar enthalpy of vaporization of n-hexane at its normal boiling point given the ln P vs. (1/T) plot. Use the Clausius-Clayperon equation.

Antoine equation Starting from the Clausius-Clayperon equation: Integrating this equation, we obtain: In this equation, A is an integration constant and B is defined as shown above. This a simple relation between the saturation pressure and temperature, but not accurate enough.

A, B, and C are obtained by fitting experimental vapor pressure data. Antoine equation The Antoine equation is an empirical modification: A, B, and C are obtained by fitting experimental vapor pressure data. Extensive parameter tables exist with parameters for hundreds of components.

Riedel equation Another empirical modification:

Harlecher-Braun equation Yet another empirical modification:

Recommendation Read chapter 7 in the textbook and review all examples.