Chapter 8.6, Problem 23P

(A)

Interpretation Introduction

Interpretation:

The best estimate of the fugacity in the liquid phase at temperature 400 K and pressure 100 bar for benzene.

Concept Introduction:

The expression of the vapor pressure using shortcut equation is,

$\begin{array}{c}{\log }_{10}{P}_r^{sat}=\frac{7}{3}\left(\omega +1\right)\left(1-\frac{1}{T_r}\right)\\ =\frac{7}{3}\left(\omega +1\right)\left(1-\frac{1}{T/T_c}\right)\end{array}$

Here, reduced vapor pressure is $P_r^{sat}$, reduced temperature is $T_r$, acentric factor is $\omega$, and temperature is $T$

The expression of the vapor pressure is,

$P^{sat}=P_r^{sat}\left(P_c\right)$

The expression of the reduced temperature $\left(T_r\right)$ is,

$T_r=\frac{T}{T_c}$

Here, system temperature is T.

The expression of the parameter used in calculation of a for Peng-Robinson EOS is,

$\kappa =0.37464+1.54226\omega -0.26993{\omega }^2$

Here, parameter is $\kappa$.

The Poynting equation to calculate the fugacity $\left(f\right)$ is,

$f=f^{sat}\exp \left\{\frac{{\underset{\_}{V}}^L\left(P-P^{sat}\right)}{RT}\right\}$

Here, molar volume in liquid phase is ${\underset{\_}{V}}^L$

Explanation of Solution

Refer Appendix C.1, “Critical point, enthalpy of phase change, and liquid molar volume”; obtain the properties for different components as in Table (1).

	Acentric factor $\left(\omega \right)$	Critical pressure $P_c\left(\text{bar}\right)$	Critical temperature $T_c\left(\text{K}\right)$	Molar volume in liquid phase ${\underset{\_}{V}}^L\left({\text{cm}}^{\text{3}}/\text{mol}\right)$
$\text{Benzene}$	0.210	48.95	562.05	89.41
$n\text{-hexane}$	0.3	30.25	507.60	131.59
Phenol	0.442	61.30	694.25	87.87

Write the expression of the vapor pressure using shortcut equation.

${\log }_{10}{P}_r^{sat}=\frac{7}{3}\left(\omega +1\right)\left(1-\frac{1}{T/T_c}\right)$        (1)

Substitute $\text{0}\text{.210}$ for $\omega$, $\text{562}\text{.05}\text{K}$ for $T_c$, and $\text{400}\text{K}$ for $T$ in Equation (1).

$\begin{array}{l}{\log }_{10}{P}_r^{sat}=\frac{7}{3}\left(0.210+1\right)\left(1-\frac{1}{\text{400}\text{K}/\text{562}\text{.05}\text{K}}\right)\\ {\log }_{10}{P}_r^{sat}=-1.143\\ P_r^{sat}={10}^{-1.143}\\ =0.0719\end{array}$

Write the expression of the vapor pressure.

$P^{sat}=P_r^{sat}\left(P_c\right)$        (2)

Substitute $0.0719$ for $P_r^{sat}$, and $\text{48}\text{.95}\text{bar}$ for $P_c$ in Equation (2).

$\begin{array}{c}{P}^{sat}=\left(0.0719\right)\left(\text{48}\text{.95}\text{bar}\right)\\ =3.52\text{bar}\end{array}$

Thus, the best estimate of vapor pressure at temperature of $\text{400}\text{K}$ is $\text{3}\text{.52}\text{bar}$.

Write the expression of the reduced temperature $\left(T_r\right)$.

$T_r=\frac{T}{T_c}$        (3)

Substitute $\text{400}\text{K}$ for T and $\text{562}\text{.05}\text{K}$ for $T_c$ in Equation (3).

$\begin{array}{c}{T}_r=\frac{\text{400}\text{K}}{\text{562}\text{.05}\text{K}}\\ =0.711\end{array}$

Write the expression of the parameter used in calculation of a for Peng-Robinson EOS.

$\kappa =0.37464+1.54226\omega -0.26993{\omega }^2$        (4)

Substitute $0.210$ for $\omega$ in Equation (4).

$\begin{array}{c}\kappa =0.37464+1.54226\left(0.210\right)-0.26993{\left(0.210\right)}^2\\ =0.6866\end{array}$

Write the expression of the value of $\alpha$ expressed as a function of the reduced temperature.

$\alpha ={\left[1+\kappa \left(1-T_r^{0.5}\right)\right]}^2$        (5)

Substitute $0.6866$ for $\kappa$, and $0.711$ for $T_r$ in Equation (5).

$\begin{array}{c}\alpha ={\left[1+0.6866\left(1-{0.711}^{0.5}\right)\right]}^2\\ =1.22\end{array}$

Write the expression of the value of parameter a at the critical point.

$a_c=\frac{0.45724R^2{T}_c^2}{P_c}$        (6)

Here, gas constant is $R$.

Substitute $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for $R$, $\text{562}\text{.05}\text{K}$ for $T_c$, and $48.95\text{bar}$ for $P_c$ in Equation (6).

$\begin{array}{c}{a}_c=\frac{0.45724{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)}^2{\left(562.05\text{K}\right)}^2}{48.95\text{bar}}\\ =20.396\times {10}^6\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\end{array}$

Write the expression of the parameter $a$.

$a=a_c\alpha$        (7)

Substitute $1.22$ for $\alpha$ and $20.396\times {10}^6\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2$ for $a_c$ in Equation (7).

$\begin{array}{c}a=\left(20.396\times {10}^6\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\right)\left(1.22\right)\\ =2.50\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\end{array}$

Write the expression of the parameter $b$.

$b=\frac{0.07780R{T}_c}{P_c}$        (8)

Substitute $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for $R$, $\text{562}\text{.05}\text{K}$ for $T_c$, and $48.95\text{bar}$ for $P_c$ in Equation (8).

$\begin{array}{c}b=\frac{0.07780\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{562}\text{.05}\text{K}\right)}{48.95\text{bar}}\\ =74.3{\text{cm}}^{\text{3}}/\text{mol}\end{array}$

Write the expression of the molar volume $\left(\underset{\_}{V}\right)$ from the Peng-Robinson equation.

$P=\frac{RT}{\underset{\_}{V}-b}-\frac{a}{\underset{\_}{V}\left(\underset{\_}{V}+b\right)+b\left(\underset{\_}{V}-b\right)}$        (9)

Here, pressure is $P$.

Substitute $\text{3}\text{.52}\text{bar}$ for P, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, $\text{400}\text{K}$ for $T$, $2.50\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2$ for a, and $74.3{\text{cm}}^{\text{3}}/\text{mol}$ for b in Equation (9).

$\begin{array}{l}\text{3}\text{.52}\text{bar}=\left[\begin{array}{l}\frac{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{400}\text{K}\right)}{\underset{\_}{V}-74.3{\text{cm}}^{\text{3}}/\text{mol}}-\\ \frac{2.50\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2}{\underset{\_}{V}\left(\underset{\_}{V}+74.3{\text{cm}}^{\text{3}}/\text{mol}\right)+74.3{\text{cm}}^{\text{3}}/\text{mol}\left(\underset{\_}{V}-74.3{\text{cm}}^{\text{3}}/\text{mol}\right)}\end{array}\right]\\ \underset{\_}{V}=8728{\text{cm}}^{\text{3}}/\text{mol}\end{array}$

Write the expression of compressibility factor $\left(Z\right)$.

$Z=\frac{P\underset{\_}{V}}{RT}$        (10)

Substitute $\text{3}\text{.52}\text{bar}$ for P, $8728{\text{cm}}^{\text{3}}/\text{mol}$ for $\underset{\_}{V}$, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (10).

$\begin{array}{c}Z=\frac{3.52\text{bar}\left(8728{\text{cm}}^{\text{3}}/\text{mol}\right)}{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)400\text{K}}\\ =0.924\end{array}$

Write the expression of the dimensionless group that include the Peng-Robinson parameter $\left(A\right)$.

$A=\frac{aP}{R^2{T}^2}$        (11)

Substitute $2.50\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2$ for a, $\text{3}\text{.52}\text{bar}$ for P, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (11).

$\begin{array}{c}A=\frac{\left(2.50\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\right)\left(3.52\text{bar}\right)}{{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)}^2{\left(400\text{K}\right)}^2}\\ =0.080\end{array}$

Write the expression of the dimensionless group that include the Peng-Robinson parameter $\left(B\right)$.

$B=\frac{bP}{RT}$        (12)

Substitute $74.3{\text{cm}}^{\text{3}}/\text{mol}$ for b, $\text{3}\text{.52}\text{bar}$ for P, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (12).

$\begin{array}{c}B=\frac{\left(74.3{\text{cm}}^{\text{3}}/\text{mol}\right)\left(3.52\text{bar}\right)}{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(400\text{K}\right)}\\ =0.00786\end{array}$

Write the fugacity of a compound as described by the Peng-Robinson equation.

$\ln \left(\frac{f^{sat}}{P}\right)=\left(Z-1\right)-\ln \left(Z-B\right)-\frac{A}{2B\sqrt{2}}\ln \left[\frac{Z+\left(1+\sqrt{2}\right)B}{Z+\left(1-\sqrt{2}\right)B}\right]$        (13)

Here, the fugacity of a liquid or vapor that is at its vapor pressure is $f^{sat}$.

Substitute $0.924$ for $Z$, $\text{3}\text{.52}\text{bar}$ for $P$, $0.00786$ for $B$, and $0.080$ for $A$ in Equation (13).

$\begin{array}{l}\ln \left(\frac{f^{sat}}{3.52\text{bar}}\right)=\left[\begin{array}{l}\left(0.924-1\right)-\left[\ln \left(0.924-0.00786\right)\right]-\\ \frac{0.080}{2\left(0.00786\right)\sqrt{2}}\left\{\ln \left[\frac{0.924+\left(1+\sqrt{2}\right)\left(0.00786\right)}{0.924+\left(1-\sqrt{2}\right)\left(0.00786\right)}\right]\right\}\end{array}\right]\\ \frac{f^{sat}}{3.52\text{bar}}=e^{\left[\begin{array}{l}\left(0.924-1\right)-\left[\ln \left(0.924-0.00786\right)\right]-\\ \frac{0.080}{2\left(0.00786\right)\sqrt{2}}\left\{\ln \left[\frac{0.924+\left(1+\sqrt{2}\right)\left(0.00786\right)}{0.924+\left(1-\sqrt{2}\right)\left(0.00786\right)}\right]\right\}\end{array}\right]}\\ f^{sat}=\left(3.52\text{bar}\right)\left(0.929\right)\\ =3.27\text{bar}\end{array}$

Write the Poynting equation to calculate the fugacity $\left(f\right)$.

$f=f^{sat}\exp \left\{\frac{{\underset{\_}{V}}^L\left(P-P^{sat}\right)}{RT}\right\}$        (14)

Substitute $\text{3}\text{.27}\text{bar}$ for $f^{sat}$, $\text{89}\text{.41}{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}^L$, $\text{100}\text{bar}$ for $P$, $\text{3}\text{.52}\text{bar}$ for $P^{sat}$, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (14).

$\begin{array}{c}f=\left(\text{3}\text{.27}\text{bar}\right)\exp \left\{\frac{\text{89}\text{.41}{\text{cm}}^{\text{3}}/\text{mol}\left(\text{100}\text{bar}-\text{3}\text{.52}\text{bar}\right)}{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{400 K}\right)}\right\}\\ =\left(\text{3}\text{.27}\text{bar}\right)\left(1.297\right)\\ =4.24\text{bar}\end{array}$

Thus, the fugacity in the liquid phase at temperature of $\text{400}\text{K}$ and pressure of $\text{100}\text{bar}$ is $\text{4}\text{.24}\text{bar}$.

(B)

Interpretation Introduction

Interpretation:

The best estimate of the fugacity in the liquid phase at temperature 400 K and pressure 100 bar for n-hexane.

Explanation of Solution

Substitute $\text{0}\text{.3}$ for $\omega$, $\text{507}\text{.60}\text{K}$ for $T_c$, and $\text{400}\text{K}$ for $T$ in Equation (1).

$\begin{array}{l}{\log }_{10}{P}_r^{sat}=\frac{7}{3}\left(\omega +1\right)\left(1-\frac{1}{T/T_c}\right)\\ {\log }_{10}{P}_r^{sat}=\frac{7}{3}\left(0.3+1\right)\left(1-\frac{1}{\text{400}\text{K}/\text{507}\text{.60}\text{K}}\right)\\ P_r^{sat}=0.153\end{array}$

Substitute $0.153$ for $P_r^{sat}$, and $\text{30}\text{.25}\text{bar}$ for $P_c$ in Equation (2).

$\begin{array}{c}{P}^{sat}=P_r^{sat}\left(P_c\right)\\ =\left(0.153\right)\left(\text{30}\text{.25}\text{bar}\right)\\ =4.62\text{bar}\end{array}$

Thus, the best estimate of vapor pressure at temperature of $\text{400}\text{K}$ is $\text{4}\text{.62}\text{bar}$.

Substitute $\text{400}\text{K}$ for T and $\text{507}\text{.60}\text{K}$ for $T_c$ in Equation (3).

$\begin{array}{c}{T}_r=\frac{T}{T_c}\\ =\frac{\text{400}\text{K}}{\text{507}\text{.60}\text{K}}\\ =0.788\end{array}$

Substitute $0.3$ for $\omega$ in Equation (4).

$\begin{array}{c}\kappa =0.37464+1.54226\omega -0.26993{\omega }^2\\ =0.37464+1.54226\left(0.3\right)-0.26993{\left(0.3\right)}^2\\ =0.8130\end{array}$

Substitute $0.8130$ for $\kappa$, and $0.788$ for $T_r$ in Equation (5).

$\begin{array}{c}\alpha ={\left[1+\kappa \left(1-T_r^{0.5}\right)\right]}^2\\ ={\left[1+0.8130\left(1-{0.788}^{0.5}\right)\right]}^2\\ =1.19\end{array}$

Substitute $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for $R$, $\text{507}\text{.60}\text{K}$ for $T_c$, and $30.25\text{bar}$ for $P_c$ in Equation (6).

$\begin{array}{c}{a}_c=\frac{0.45724R^2{T}_c^2}{P_c}\\ =\frac{0.45724{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)}^2{\left(507.60\text{K}\right)}^2}{30.25\text{bar}}\\ =26.92\times {10}^6\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\end{array}$

Substitute $1.19$ for $\alpha$ and $26.92\times {10}^6\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2$ for $a_c$ in Equation (7).

$\begin{array}{c}a=a_c\alpha \\ =\left(26.92\times {10}^6\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\right)\left(1.19\right)\\ =3.21\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\end{array}$

Substitute $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for $R$, $\text{507}\text{.60}\text{K}$ for $T_c$, and $30.25\text{bar}$ for $P_c$ in Equation (8).

$\begin{array}{c}b=\frac{0.07780R{T}_c}{P_c}\\ =\frac{0.07780\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{507}\text{.60}\text{K}\right)}{30.25\text{bar}}\\ =108.5{\text{cm}}^{\text{3}}/\text{mol}\end{array}$

Substitute $\text{4}\text{.62}\text{bar}$ for P, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, $\text{400}\text{K}$ for $T$, $3.21\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2$ for a, and $108.5{\text{cm}}^{\text{3}}/\text{mol}$ for b in Equation (9).

$\begin{array}{l}P=\frac{RT}{\underset{\_}{V}-b}-\frac{a}{\underset{\_}{V}\left(\underset{\_}{V}+b\right)+b\left(\underset{\_}{V}-b\right)}\\ \text{4}\text{.62}\text{bar}=\left[\begin{array}{l}\frac{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{400}\text{K}\right)}{\underset{\_}{V}-108.5{\text{cm}}^{\text{3}}/\text{mol}}-\\ \frac{3.21\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2}{\underset{\_}{V}\left(\underset{\_}{V}+108.5{\text{cm}}^{\text{3}}/\text{mol}\right)+108.5{\text{cm}}^{\text{3}}/\text{mol}\left(\underset{\_}{V}-108.5{\text{cm}}^{\text{3}}/\text{mol}\right)}\end{array}\right]\\ \underset{\_}{V}=6252{\text{cm}}^{\text{3}}/\text{mol}\end{array}$

Substitute $\text{4}\text{.62}\text{bar}$ for P, $6252{\text{cm}}^{\text{3}}/\text{mol}$ for $\underset{\_}{V}$, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (10).

$\begin{array}{c}Z=\frac{P\underset{\_}{V}}{RT}\\ =\frac{4.62\text{bar}\left(6252{\text{cm}}^{\text{3}}/\text{mol}\right)}{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)400\text{K}}\\ =0.869\end{array}$

Substitute $3.21\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2$ for a, $\text{4}\text{.62}\text{bar}$ for P, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (11).

$\begin{array}{c}A=\frac{aP}{R^2{T}^2}\\ =\frac{\left(3.21\times {10}^7\text{bar}\cdot {\text{cm}}^6/{\text{mol}}^2\right)\left(4.62\text{bar}\right)}{{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)}^2{\left(400\text{K}\right)}^2}\\ =0.134\end{array}$

Substitute $108.5{\text{cm}}^{\text{3}}/\text{mol}$ for b, $\text{4}\text{.62}\text{bar}$ for P, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (12).

$\begin{array}{c}B=\frac{bP}{RT}\\ =\frac{\left(108.5{\text{cm}}^{\text{3}}/\text{mol}\right)\left(4.62\text{bar}\right)}{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(400\text{K}\right)}\\ =0.015\end{array}$

Substitute $0.869$ for $Z$, $\text{4}\text{.62}\text{bar}$ for $P$, $0.015$ for $B$, and $0.134$ for $A$ in Equation (13).

$\begin{array}{l}\ln \left(\frac{f^{sat}}{P}\right)=\left(Z-1\right)-\ln \left(Z-B\right)-\frac{A}{2B\sqrt{2}}\ln \left[\frac{Z+\left(1+\sqrt{2}\right)B}{Z+\left(1-\sqrt{2}\right)B}\right]\\ \ln \left(\frac{f^{sat}}{4.62\text{bar}}\right)=\left[\begin{array}{l}\left(0.869-1\right)-\left[\ln \left(0.869-0.015\right)\right]-\\ \frac{0.134}{2\left(0.015\right)\sqrt{2}}\left\{\ln \left[\frac{0.869+\left(1+\sqrt{2}\right)\left(0.015\right)}{0.869+\left(1-\sqrt{2}\right)\left(0.015\right)}\right]\right\}\end{array}\right]\\ \frac{f^{sat}}{4.62\text{bar}}=e^{\left[\begin{array}{l}\left(0.869-1\right)-\left[\ln \left(0.869-0.015\right)\right]-\\ \frac{0.134}{2\left(0.015\right)\sqrt{2}}\left\{\ln \left[\frac{0.869+\left(1+\sqrt{2}\right)\left(0.015\right)}{0.869+\left(1-\sqrt{2}\right)\left(0.015\right)}\right]\right\}\end{array}\right]}\\ f^{sat}=\left(4.62\text{bar}\right)\left(0.883\right)\\ =4.08\text{bar}\end{array}$

Substitute $\text{4}\text{.08}\text{bar}$ for $f^{sat}$, $\text{131}\text{.59}{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}^L$, $\text{100}\text{bar}$ for $P$, $\text{4}\text{.62}\text{bar}$ for $P^{sat}$, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, and $\text{400 K}$ for T in Equation (14).

$\begin{array}{c}f=f^{sat}\exp \left\{\frac{{\underset{\_}{V}}^L\left(P-P^{sat}\right)}{RT}\right\}\\ =\left(\text{4}\text{.08}\text{bar}\right)\exp \left\{\frac{\text{131}\text{.59}{\text{cm}}^{\text{3}}/\text{mol}\left(\text{100}\text{bar}-\text{4}\text{.62}\text{bar}\right)}{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{400 K}\right)}\right\}\\ =\left(\text{4}\text{.08}\text{bar}\right)\left(1.46\right)\\ =5.95\text{bar}\end{array}$

Thus, the fugacity in the liquid phase at temperature of $\text{400}\text{K}$ and pressure of $\text{100}\text{bar}$ is $\text{5}\text{.95}\text{bar}$.

(C)

Interpretation Introduction

Interpretation:

The best estimate of the fugacity in the liquid phase at temperature 400 K and pressure 100 bar for phenol.

Concept Introduction:

The expression of the vapor pressure using Antoine equation is,

$P^{sat}={10}^{\left[A-\frac{B}{C+T\left(°\text{C}\right)}\right]}$

Here, vapor pressure is $P^{sat}$, temperature in degree Celsius is $T\left(°\text{C}\right)$, and Antoine coefficients are $A,B,\text{and}C$.

Explanation of Solution

Convert the unit of temperature from Kelvin to degree Celsius.

$\begin{array}{c}T=400\text{K}\\ =\left(400-273.15\right)°\text{C}\\ =126.85°\text{C}\end{array}$

Write the expression of the vapor pressure using Antoine equation.

$P^{sat}={10}^{\left[A-\frac{B}{C+T\left(°\text{C}\right)}\right]}$        (15)

Refer Appendix E, “Antoine Equations”; obtain the following properties for component 1 and 2 as in Table (1).

Parameters	Phenol
A	7.1330
B	1516.79
C	174.95

Substitute $126.85°\text{C}$ for T, $7.1330$ for A, $1516.79$ for B, and $174.95$ for C in Equation (15).

$\begin{array}{c}{P}^{sat}={10}^{\left[7.1330-\frac{1516.79}{174.95+126.85}\right]}\\ =128\text{mm}\cdot \text{Hg}\left(\frac{0.00133\text{bar}}{1\text{mm}\cdot \text{Hg}}\right)\\ =0.171\text{bar}\end{array}$

Thus, the best estimate of vapor pressure at temperature of $\text{400}\text{K}$ is $\text{0}\text{.171}\text{bar}$.

Assume ideal gas behavior, as the obtained value of pressure is less,

$f^{sat}=P^{sat}$

Substitute $\text{0}\text{.171}\text{bar}$ for $P^{sat}$.

$f^{sat}=\text{0}\text{.171}\text{bar}$

Substitute $\text{0}\text{.171}\text{bar}$ for $f^{sat}$, $\text{87}\text{.87}{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}^L$, $\text{100}\text{bar}$ for $P$, $\text{0}\text{.171}\text{bar}$ for $P^{sat}$, $83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R and $\text{400 K}$ for T in Equation (14).

$\begin{array}{c}f=f^{sat}\exp \left\{\frac{{\underset{\_}{V}}^L\left(P-P^{sat}\right)}{RT}\right\}\\ =\left(\text{0}\text{.171}\text{bar}\right)\exp \left\{\frac{\text{87}\text{.87}{\text{cm}}^{\text{3}}/\text{mol}\left(\text{100}\text{bar}-\text{0}\text{.171}\text{bar}\right)}{\left(83.14\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{400 K}\right)}\right\}\\ =\left(0.171\text{bar}\right)\left(1.302\right)\\ =0.223\text{bar}\end{array}$

Thus, the fugacity in the liquid phase at temperature of $\text{400}\text{K}$ and pressure of $\text{100}\text{bar}$ is $\text{0}\text{.223}\text{bar}$.