Lecture 5

Covers 11.6-11.9

Objectives: Students will be able to:

1. Compute $\phi_i$ from the Lee-Kesler correlations.
2. Use linear interpolation for 2-d tabular functions.
3. Define $\hat{f}_i$ in terms of $\mu_i$.
4. Derive the criterion for phase equilibrium in terms of $\hat{f}_i$.
5. Define the fugacity coefficient for a mixture in terms of the residual partial Gibbs energy.
6. Derive the fundamental residual property relation.
7. Compute mixture fugacity coefficients from volumetric properties.
8. Compute mixture fugacity coefficients for gases using the second virial EOS.
9. Compute mixture fugacity coefficients from pressure explicit EOS.
10. Compute mixture fugacity coefficients from Aspen.
11. Compare and contrast ideal solutions and ideal gases.
12. Define the ideal solution equations.
13. Identify the Lewis-Randall Rule for an ideal solution.
14. Define excess properties.
15. Define the activity coefficient.

Lecture Notes

1. Computing $\ln\phi_i$ from the Lee-Kesler correlations. The Lee-Kesler correlations are an empirical three-parameter corresponding states treatment for pure fluid properties. The correlations are typically valid over the entire range of the data listed in the tables (see Tables E.1-E.16 in Appendix E of the book). The basic idea of the Lee-Kesler correlation is that the properties of any nonpolar or slightly polar fluid can be represented in terms of the properties of two reference fluids, one that is a simple spherical reference fluid (like argon) and one that is a non-spherical molecule. The fugacity coefficient for a pure fluid can be computed from the Lee-Kesler tables from
   
   $$\phi_i = (\phi^0)(\phi^1)^{\omega}$$
   
   
   where $\phi^0$ and $\phi^1$ can be interpolated from Tables E.13-E.16 in terms of $T_r$ and $P_r$ and $\omega$ is the Pitzer acentric factor. You typically have to use 2-dimensional linear interpolation to find the correct values of $\phi^0$ and $\phi^1$ from the tables.

2. Linear interpolation in 2-dimensions.
   Assume that we have a function that depends on two variables, $f(x,y)$, that is tabulated at discrete values, $f(x_1,y_1)$, etc. You want to find the value of the function at $(x,y)$ through linear interpolation. You know how to do this with a single variable,
   
   $$f(x) = \frac{f(x_2)-f(x_1)}{x_2-x_1} (x - x_1) + f(x_1)$$
   
   
   You use exactly the same procedure for two variables, but now you have to interpolate in both $x$ and $y$. This gives a bit of an ugly equation:
   $$\begin{eqnarray*} f(x,y) &=& \frac{(x-x_1)}{(x_2-x_1)} \left\{\left[\frac{f(x_2,... ...rac{f(x_1,y_2)-f(x_1,y_1)}{(y_2-y_1)}\right](y-y_1) + f(x_1,y_1) \end{eqnarray*}$$
   
   
   
   
   

3. Example of computing $\phi_i$ from the Lee-Kesler correlations. Compute the fugacity of pure ethylbenzene at 250 bar and 420 K.
   1. Look up the critical properties and acentric factor from Table B.1. $T_c = 617.2$ K, $P_c = 36.06$ bar, $\omega = 0.303$
   2. Compute $T_r = T/T_c = 420/617.2 = 0.6805$ and $P_r = P/P_c = 250/36.06 = 6.933$
   3. Look up the values surrounding this state point in Tables E.13-E.16. We see that $0.65 < T_r < 0.7$ and $5.0 < P_r < 7.0$ so we use Tables E.15 and E.16. For $T_r = 0.65, P_r = 5.0$ we find $\phi^0 = 0.0244$, for $T_r = 0.65, P_r = 7.0$ we find $\phi^0 = 0.0245$, and $T_r = 0.7, P_r = 5.0, \phi^0 = 0.0406, T_r = 0.7, P_r = 7.0, \phi^0 = 0.0402$ from Table E.15. Linear interpolation of these values gives $\phi^0(0.6805, 6.933) = 0.034084$ Likewise, we find $\phi^1$ from interpolation of Table E.16, $\phi^1(0.6805, 6.933) = 0.05979$ The spreadsheet for computing these values can be found at

      http://puccini.che.pitt.edu/~karlj/Classes/CHE1007/interp.Example.xls

   4. Compute the final value
      
      $$\phi_i = (0.034084)(0.05979)^{0.303} = 0.0145$$
      
      

4. Mixture fugacity:
   1. Recall that for an ideal gas Eq. (10.28) gives
      

      $$\mu^{ig}_i = \mu_i^0(T) + RT\ln x_i P$$ (1)

      
      
      By analogy with pure fluids, we define the fugacity in a real fluid mixture as
      

      $$\mu_i = \mu_i^0(T) + RT\ln \hat{f}_i$$ (2)

      
      
      We denote $\hat{f}_i$ as the fugacity of $i$ in a mixture.
   2. Note that there are three different fugacities:
      $$\begin{eqnarray*} f &=& g(T,P,x_i) \mbox{ Total fugacity of a mixture}\\ f_i &=... ... \hat{f}_i &=& g(T,P,x_i) \mbox{ Fugacity of $i$\ in a mixture} \end{eqnarray*}$$
      
      
      
      
      

5. Phase equilibria and $\hat{f}_i$

   We know that phase equilibrium requires that
   

   $$\mu^\alpha_i = \mu_i^\beta = \cdots = \mu^\pi_i$$ (3)

   
   
   Substituting Eq. (2) into Eq. (3) gives
   

   $$\mu^0_i + RT\ln \hat{f}_i^\alpha = \mu_i^0 + RT\ln \hat{f}_i^\beta = \cdots = \mu^0_i + RT\ln \hat{f}_i^\pi$$ (4)

   
   
   since $\mu^0_i$ is only a function of temperature, it is the same in each phase. This leads directly to
   

   $$\hat{f}_i^\alpha = \hat{f}_i^\beta = \cdots = \hat{f}_i^\pi$$ (5)

   
   

6. The fugacity coefficient:

   If we subtract Eq. (1) from Eq. (2) we get

   $$\begin{eqnarray*} \mu_i - \mu_i^{ig} &=& \mu^R_i = \mu_i^0(T) - \mu_i^0(T) + RT\ln \hat{f}_i - RT\ln x_i P\\ &=& RT\ln\frac{\hat{f}_i}{x_iP} \end{eqnarray*}$$
   
   
   
   
   
   But recall that $\bar{G}_i = \mu_i$ so
   

   $$\mu_i - \mu_i^{ig} = \bar{G}_i - \bar{G}_i^{ig} = \bar{G}_i^R = RT\ln\frac{\hat{f}_i}{x_iP} = RT\ln\hat{\phi}_i$$ (6)

   
   
   This leads to the definition of the fugacity coefficient
   
   $$\hat{\phi}_i \equiv \frac{\hat{f}_i}{x_i P}$$
   
   
   What is the value of
   
   $$\lim_{P \rightarrow 0} \bar{G}^R_i = \mbox{ ?}$$
   
   
   Note that $\bar{G}^R_i = \bar{G}_i - \bar{G}_i^{ig}$ but $\lim_{P \rightarrow 0} \bar{G}_i = \bar{G}_i^{ig}$ so
   
   $$\lim_{P \rightarrow 0} \bar{G}^R_i = \bar{G}_i^{ig} - \bar{G}_i^{ig} = 0$$
   
   
   What is
   
   $$\lim_{P \rightarrow 0} \hat{\phi}_i = \mbox{?}$$
   
   
   Compare the limit of $\hat{f}_i$ with $f_i$ for $P \rightarrow 0$.
7. We have three separate notational cases:
   1. Total mixture properties: $f$, $\phi = f/P$.
   2. Pure fluids: $f_i$, $\phi_i = f_i/P$.
   3. Components in mixture (analogous to partial properties): $\hat{f}_i$, $\hat{\phi}_i = \hat{f}_i/(x_i P)$.
   Note that we use a circumflex ( $\hat{ }$ ) to represent the mixture properties, because they are like partial properties, but they have slightly different definitions. You can prove that it is $\ln\hat{\phi}_i$ that is a true partial property, not $\hat{f}_i$.

8. Fundamental residual property relation

   All the equations that apply to any thermodynamic property $M$ also apply to any residual property $M^R$. We can therefore write
   

   $$d(nG^R) = -nS^RdT + nV^RdP + \sum_{i=1}^C \mu^R_i dn_i$$
   
   
   How do you measure entropy? Entropy is a bit difficult to deal with so we will eliminate it by using a trick:
   $$\begin{eqnarray*} d\left(\frac{nG^R}{RT}\right) &=& \frac{1}{RT}d(nG^R) + nG^R ... ...- \frac{nH^R}{RT^2}dT + \sum_{i=1}^C \frac{\bar{G}^R_i}{RT} dn_i \end{eqnarray*}$$
   
   
   
   
   
   Recall from Eq. (6) that $\bar{G}^R_i = RT\ln\hat{\phi}_i$ so
   
   $$d\left(\frac{nG^R}{RT}\right) = \frac{nV^R}{RT}dP - \frac{nH^R}{RT^2}dT + \sum_{i=1}^C \ln\hat{\phi}_i dn_i$$
   
   
   This is the fundamental residual property relation. It is useful because we can compute other properties from it through taking partial derivatives of it. For example, if we hold the temperature and composition constant and take the derivative with respect to $P$ then we have
   
   $$\left(\frac{\partial (nG^R/RT)}{\partial P}\right)_{T,n_j} = \frac{nV^R}{RT}$$
   
   
   If we hold the pressure and composition constant and take the derivative with respect to $T$ then
   
   $$\left(\frac{\partial (nG^R/RT)}{\partial T}\right)_{P,n_j} = -\frac{nH^R}{RT^2}$$
   
   
   Finally, the partial molar $G^R$ is given by
   
   $$\left(\frac{\partial (nG^R/RT)}{\partial n_i}\right)_{T,P,n_{j \ne i}} = \ln \hat{\phi}_i$$
   
   
   We see that $RT\ln \hat{\phi}_i$ is the partial property to $G^R$, so
   
   $$G^R = RT\sum_i x_i \ln \hat{\phi}_i = RT\ln \phi$$
   
   

9. Example: Computing $\ln\hat{\phi}_i$ from $PVT$ data.

   The above equations give us a method to compute $\ln\hat{\phi}_i$
   

   $$\ln \hat{\phi}_i = \int_0^P(\bar{Z}_i - 1)\frac{dP}{P}$$ (7)

   
   
   where $\bar{Z}_i$ is the partial compressibility of $i$ in the mixture,
   $$\begin{eqnarray*} \bar{Z}_1 &=& Z + x_2 \frac{dZ}{dx_1}\\ \bar{Z}_1 &=& Z - x_1 \frac{dZ}{dx_1} \end{eqnarray*}$$
   
   
   
   
   
   The virial equation for a mixture truncated at the second virial coefficient is
   $$\begin{eqnarray*} Z &=& 1 + \sum_i\sum_j y_iy_j\frac{B_{ij}P}{RT} = 1 + \left(y... ...[y_1^2 B_{11} + (1-2y_1+y_1^2) B_{22} + 2(y_1-y_1^2)_{12}\right] \end{eqnarray*}$$
   
   
   
   
   
   Therefore,
   $$\begin{eqnarray*} \frac{dZ}{dy_1} &=& \left[2y_1 B_{11} + 2(y_1-1)B_{22} + 2(1-y... ...1-y_1^2)B_{22} - y_1^2 B_{11} + 2y_1^2B_{12}\right]\frac{P}{RT} \end{eqnarray*}$$
   
   
   
   
   
   We can now use Eq. (7) to compute the fugacity coefficients,
   $$\begin{eqnarray*} \ln \hat{\phi}_1 &=& \int_0^P(\bar{Z}_1 - 1)\frac{dP}{P} = \i... ...1-y_1^2)B_{22} - y_1^2 B_{11} + 2y_1^2B_{12}\right]\frac{P}{RT} \end{eqnarray*}$$
   
   
   
   
   

10. Computing $\hat{\phi}$ from the second virial EOS.
    We know that the virial EOS truncated at the second virial coefficient is
    
    $$Z = 1 + \sum_i\sum_j y_iy_j\frac{B_{ij}P}{RT}$$
    
    
    Note that this equation is only valid for gases as low to moderate pressure. It is not valid for liquids. For a binary mixture the fugacity coefficients are given by
    $$\begin{eqnarray*} \ln \hat{\phi}_1 &=& \left[(1-y_2^2)B_{11} - y_2^2 B_{22} + 2y... ...(1-y_1^2)B_{22} - y_1^2 B_{11} + 2y_1^2B_{12}\right]\frac{P}{RT} \end{eqnarray*}$$
    
    
    
    
    
    For a multicomponent mixture
    
    $$\ln \hat{\phi}_k = \frac{P}{RT}\left[B_{kk} + \frac{1}{2}\su... ...j\left(4B_{ik} - B_{ii} - 2B_{kk}-2B_{ij}+B_{jj}\right)\right]$$
    
    
    We have a correlation for $B_{ii}$ through Eq. (3.48), but we need a method for finding the cross coefficients, $B_{ij}$. These are called combining rules, and there are many of them. The cross coefficients analogous to Eq. (3.48) are given by
    
    $$B_{ij} = \frac{RT_{cij}}{P_{cij}}\left(B^0+\omega_{ij}B^1\right)$$
    
    
    with
    $$\begin{eqnarray*} B^0 &=& 0.083 - \frac{0.422}{T_r^{1.6}}\\ B^1 &=& 0.139 - \frac{0.172}{T_r^{4.2}} \end{eqnarray*}$$
    
    
    
    
    
    The combining rules proposed by Prausnitz are given by
    $$\begin{eqnarray*} \omega_{ij} &=& \frac{\omega_i + \omega_j}{2}\\ T_{cij} &=& \... ... V_{cij} &=& \left(\frac{V_{ci}^{1/3}+V_{cj}^{1/2}}{2}\right)^3 \end{eqnarray*}$$
    
    
    
    
    
    The $k_{ij}$ values are empirical binary interaction parameters. You have to look these up or use a correlation or assume they are zero.

11. Computing $f$, $\hat{f}_i$ from an equation of state.
    The fugacity coefficient of a mixture can be calculated from
    
    $$RT \ln \phi = \int_0^P \left(V - \frac{RT}{P}\right)dP$$
    
    
    We can then compute $\hat{\phi}_i$ or $\hat{f}_i$ from the partial property relations presented. But, most EOS are pressure explicit (i.e., $P$ as a function of $V,T$), so the above integral equation is not very useful. We can transform the above equation to the following form:
    
    $$\ln \phi = Z - 1 - \ln Z - \int_{\infty}^{V} \left( Z - 1\right)\frac{dV}{V}$$
    
    
    We can compute $\ln\hat{\phi}_i$ from the above equation and making use of the method of computing partial molar properties.
    
    $$\ln \hat{\phi}_i = \ln \phi - \sum_{k \ne i} x_k \left(\frac{\partial \ln \phi} {\partial x_k}\right)_{x_{j\ne i,k}}$$
    
    
    This is the approach that is taken in commercial software.

12. Notes are given on the web page on how to compute $\hat{\phi}_i$ and other properties from Aspen.

13. What makes an ideal solution ideal?
    1. Comparing ideal solutions with ideal gases:
       1. What would the energetic interactions of an ideal solution be?
          Compare this with ideal gas interactions.
       2. What would the size/shape/volume interactions of an ideal solution be like?
          Compare this with size/shape/volume interactions in an ideal gas.
       3. Is a mixture of ideal gases an ideal mixture?
    2. Which of the following ``solutions'' would you expect to be ideal?
       1. Red and blue marbles?
       2. Bowling balls and marbles?
       3. Velcro ``hooks'' and ``eyes''?
       4. Wooden blocks and stone blocks of the same size?
       5. Ortho-, meta-, and para-xylene?
       6. H$_2$O and H$_2$S?
       7. n-decane and n-octane?
       8. 2-hexanol and n-hexane?
    3. Discussion: When do you expect solutions to be ideal? Give some examples of binary mixtures that you think will be ideal.

14. Equations for ideal solutions.
    
    1. Partial properties of ideal soluations.
       The partial molar Gibbs free energy of an ideal solution is especially important for mixtures. It is defined as
       

       $$\bar{G}_i^{id} = \mu_i^{id} = G_i + RT \ln x_i$$ (8)

       
       
       Other properties are given by
       

       $$\bar{S}^{id}_i = S_i - R\ln x_i$$ (9)

       $$\bar{A}^{id}_i = A_i + RT\ln x_i$$ (10)

       $$\bar{U}^{id}_i = U_i$$ (11)

       $$\bar{H}^{id}_i = H_i$$ (12)

       $$\bar{V}^{id}_i = V_i$$ (13)

       
       

    2. Mixture properties of ideal solutions.
       
       

       $$G^{id} = \sum_i x_i G_i + RT\sum_i x_i \ln x_i$$ (14)

       $$S^{id} = \sum_i x_i S_i - R\sum_i x_i \ln x_i$$ (15)

       $$A^{id} = \sum_i x_i A_i + RT\sum_i x_i \ln x_i$$ (16)

       $$H^{id} = \sum_i x_i H_i$$ (17)

       $$U^{id} = \sum_i x_i U_i$$ (18)

       $$V^{id} = \sum_i x_i V_i$$ (19)

       
       

15. The Lewis-Randall Rule.
    
    1. Derivation of the Lewis-Randall Rule.
       Can we find a simple expression for the fugacity of an ideal solution? Recall that for a pure fluid
       

       $$G_i = \mu_i = \mu^0_i(T) + RT \ln f_i$$ (20)

       
       
       For a mixture the fugacity is given by
       

       $$\mu_i = \bar{G}_i = \mu^0_i(T) + RT \ln \hat{f}_i$$ (21)

       
       
       We can write Eq. (21) for an ideal solution, and this must be equal to Eq. (8), which gives
       

       $$\mu_i^{id} = \bar{G}_i^{id} = \mu^0_i(T) + RT \ln \hat{f}^{id}_i = \mu_i^{id} = G_i + RT \ln x_i$$ (22)

       
       
       Substituting Eq. (20) for $G_i$ we get
       
       $$\bar{G}_i^{id} =\mu^0_i(T) + RT \ln \hat{f}^{id}_i = \mu^0_i(T) + RT \ln f_i + RT \ln x_i$$
       
       
       Simplifying,
       
       $$RT \ln \hat{f}^{id}_i = RT \ln f_i + RT \ln x_i = RT \ln(x_i f_i)$$
       
       
       The definition of the ideal solution fugacity is therefore,
       

       $$\hat{f}^{id}_i \equiv x_i f_i$$ (23)

       
       
       The ideal solution fugacity coefficient can be obtained by dividing both sides of Eq. (23) by $x_iP$ to get
       

       $$\hat{\phi}^{id}_i \equiv \phi_i$$ (24)

       
       

    2. Example: Compare pure and mixture fugacities computed from Aspen. Test the Lewis-Randall rule by computing mixture and pure fluid fugacities from Aspen for two equimolar mixtures:
       1. n-hexane + n-heptane at 40 bar and 300 K.
          See http://puccini.che.pitt.edu/~karlj/Classes/CHE1007/mixture3.apw

       2. benzene + water at 40 bar and 300 K.

16. Excess properties.
    Recall that residual properties measured the difference between real and ideal gas behavior. It turns out that there are some very good (deep) reasons for defining an analogous function to measure the difference between real and ideal solution behavior. Take a good course in Statistical Mechanics to understand the reasons for this.

    The excess properties are analogous to residual properties. An excess property for any thermodynamic property $M$ is defined as
    

    $$M^E \equiv M - M^{id}$$
    
    
    Excess properties are defined for mixtures only. They have no meaning for pure fluids. As before, partial properties exist for any valid thermodynamic function for a mixture:
    
    $$\bar{M}^E_i = \bar{M}_i - \bar{M}^{id}_i$$
    
    

17. Defining the activity coefficient.
    The excess Gibbs free energy and excess partial molar Gibbs free energies are extremely important for computing solution properties. The Gibbs free energies determine which phases are stable. The excess properties define the activity coefficient, which is used to compute phase and chemical equilibria for highly nonideal mixtures. It can be shown that
    $$\begin{eqnarray*} \bar{G}^E_i &=& \bar{G}_i - \bar{G}^{id}_i = RT \ln\frac{\hat{... ... G^E &=& G - G^{id} = RT\sum_i x_i \ln \frac{\hat{f}_i}{x_i f_i} \end{eqnarray*}$$
    
    
    
    
    
    These equations define the activity coefficient, $\gamma_i$.
    
    $$\gamma_i = \frac{\hat{f}_i}{x_i f_i} = \frac{\hat{\phi}_i}{\phi_i}$$
    
    
    Thus,
    $$\begin{eqnarray*} \bar{G}^E_i &=& RT \ln\gamma_i\\ G^E &=& RT\sum_i x_i \ln \gamma_i \end{eqnarray*}$$
    
    
    
    
    
    The standard definition for the fundamental property relations holds:
    
    $$d\left(\frac{nG^E}{RT}\right) = \frac{nV^E}{RT}dP - \frac{nH^E}{RT^2}dT + \sum_{i=1}^C \ln \gamma_i dn_i$$
    
    
    The Gibbs-Duhem equation for $G^E$ is
    
    $$\sum_i x_i d\ln \gamma_i = 0$$
    
    

18. Example: The Margules equation is often used to correlate excess Gibbs free energies. For a binary mixture the Margules expression is
    
    $$\frac{G^E}{RT} = x_1x_2(Ax_2+Bx_1)$$
    
    
    1. Find $\ln \gamma_1$ and $\ln \gamma_2$.
    2. Check to see that $G^E = RT\sum_i x_i \ln \gamma_i$.
    3. Check to see if the Gibbs-Duhem equation is satisfied.
    1. Find $\ln \gamma_1$ and $\ln \gamma_2$.
       
       $$\begin{eqnarray*} \ln \gamma_1 &=& \frac{\bar{G}^E_1}{RT} = \frac{G^E}{RT} + x_2... ..._1^3\\ \ln \gamma_2 &=& 2Ax_1^2 - Bx_1^2 - 2Ax_1^3 + 2Bx_1^3\\ \end{eqnarray*}$$
       
       
       
       
       

    2. Check to see that $G^E = RT\sum_i x_i \ln \gamma_i$.
       $$\begin{eqnarray*} \frac{G^E}{RT}&=& x_1\ln \gamma_1 + x_2 \ln \gamma_2\\ &=& A... ...\ &=& Ax_1 + Bx_1^2 - 2Ax_1^2 + Ax_1^3 - Bx_1^3 \mbox{ Q.E.D.} \end{eqnarray*}$$
       
       
       
       
       

    3. Check to see if the Gibbs-Duhem equation is satisfied.
       The Gibbs-Duhem equation is
       
       $$x_1 d\ln\gamma_1 + x_2 d\ln \gamma_2 = 0$$
       
       
       $$\begin{eqnarray*} x_1\frac{d\ln\gamma_1}{dx_1} &=& -4Ax_1 + 2Bx_1 + 10Ax_1^2 -8B... ...\\ & & + 2Bx_1- 2Bx_1 + 6Ax_1^3 -6Ax_1^3 + 6Bx_1^3- 6Bx_1^3 = 0 \end{eqnarray*}$$