Chapter 3, Problem 3.85P

(a)

Interpretation Introduction

Interpretation:

The value of $\psi ,\Omega \text{and}{Z}_C$ for Redlich/Kwong equation of state should be calculated.

Concept Introduction :

To find the values of $\psi ,\Omega \text{and}{Z}_C$, first write the general cubic equation of state at critical conditions $P_C,V_C,T_C$ and write it in form of ${(V-V_C)}^3=V^3+3V{V}_C{}^2-3V^2{V}_C-V_C{}^3$ since at critical conditions $(V-V_C)=0$ and so ${(V-V_C)}^3=0$ and compare both equations to find three general derived equation in the terms of critical temperature, pressure and volume. Three equations and we will have five unknowns $P_C,V_C,T_C,a(T_C)\text{and}b$ . Replace them with $\psi ,\Omega \text{and}{Z}_C$ and find value of $\psi ,\Omega \text{and}{Z}_C$ using three equations. The general cubic equation of state is:

  $P=\frac{RT}{V-b}-\frac{a(T)}{(V+\epsilon b)(V+\sigma b)}$......(1)

For critical conditions,

  ${\frac{\partial P}{\partial V}}_{T,cr}={\frac{{\partial }^{2}P}{\partial V^2}}_{T,cr}=0$ and $P=P_C,V=V_C\text{and }T=T_C$

For Redlich/Kwong equation of state,

  $\epsilon =0,\sigma =1$

Answer to Problem 3.85P

  $\begin{array}{l}\psi =0.42748,\\ \Omega =0.08664\text{and}\\ Z_C=1/3\end{array}$

Explanation of Solution

Given information:

The general expressions for $\psi ,\Omega \text{and}{Z}_C$ are:

  $\psi =\frac{a{P}_C}{R^2{T}_C{}^2}$

  $\Omega =\frac{b{P}_C}{R{T}_C}$

And $Z_C=\frac{P_C{V}_C}{R{T}_C}$

The general cubic equation of state may itself be written for the critical conditions, from equation (1) at critical conditions,

  $P_C=\frac{R{T}_C}{V_C-b}-\frac{a(T_C)}{(V_C+\epsilon b)(V_C+\sigma b)}$

Now replace five unknowns $P_C,V_C,T_C,a(T_C)\text{and}b$ . Replace them with $\psi ,\Omega \text{and}{Z}_C$ using $\psi =\frac{a{P}_C}{R^2{T}_C{}^2}$, $\Omega =\frac{b{P}_C}{R{T}_C}$ and $Z_C=\frac{P_C{V}_C}{R{T}_C}$,

  $Z_C=\frac{P_C{V}_C}{R{T}_C}\Rightarrow \frac{R{T}_C}{P_C}=\frac{V_C}{Z_C}$

  $\begin{array}{l}\Omega =\frac{b{P}_C}{R{T}_C}\Rightarrow b=\frac{R{T}_C\Omega }{P_C}\Rightarrow b=\frac{V_C\Omega }{Z_C}\\ b=\frac{V_C\Omega }{Z_C}\end{array}$

  $\psi =\frac{a{P}_C}{R^2{T}_C{}^2}\Rightarrow a=\frac{\psi R^2{T}_C{}^2}{P_C}$

  $P_C=\frac{R{T}_C}{V_C-b}-\frac{a(T_C)}{(V_C+\epsilon b)(V_C+\sigma b)}$

  $P_C=\frac{R{T}_C}{V_C-b}-\frac{a(T_C)}{[V_C{}^2+V_C\sigma b+V_C\epsilon b+\epsilon \sigma b^2]}$

Taking L.C.M

  $P_C=\frac{R{T}_C[V_C{}^2+V_C\sigma b+V_C\epsilon b+\epsilon \sigma b^2]-a(T_C)\left(V_C-b\right)}{\left(V_C-b\right)[V_C{}^2+V_C\sigma b+V_C\epsilon b+\epsilon \sigma b^2]}$

  $\begin{array}{l}\Rightarrow P_C\left(V_C-b\right)[V_C{}^2+V_C\sigma b+V_C\epsilon b+\epsilon \sigma b^2]=\\ R{T}_C[V_C{}^2+V_C\sigma b+V_C\epsilon b+\epsilon \sigma b^2]-a(T_C)\left(V_C-b\right)\end{array}$

  $\begin{array}{l}\Rightarrow P_C[V_C{}^3+V_C{}^2\sigma b+V_C{}^2\epsilon b+\epsilon \sigma b^2{V}_C-b{V}_C{}^2-V_C\sigma b^2-V_C\epsilon b^2-\epsilon \sigma b^3]=\\ [R{T}_C{V}_C{}^2+R{T}_C{V}_C\sigma b+R{T}_C{V}_C\epsilon b+R{T}_C\epsilon \sigma b^2]-\left(a(T_C)V_C-a(T_C)b\right)\end{array}$

  $\begin{array}{l}\Rightarrow P_C{V}_C{}^3+P_C{V}_C{}^2\sigma b+P_C{V}_C{}^2\epsilon b+P_C\epsilon \sigma b^2{V}_C-P_{C}b{V}_C{}^2-P_C{V}_C\sigma b^2-P_C{V}_C\epsilon b^2-P_C\epsilon \sigma b^3=\\ [R{T}_C{V}_C{}^2+R{T}_C{V}_C\sigma b+R{T}_C{V}_C\epsilon b+R{T}_C\epsilon \sigma b^2]-a(T_C)V_C+a(T_C)b\end{array}$

Divide the whole equation by $P_C$

  $\begin{array}{l}{V}_C{}^3+V_C{}^2\sigma b+V_C{}^2\epsilon b+\epsilon \sigma b^2{V}_C-b{V}_C{}^2-V_C\sigma b^2-V_C\epsilon b^2-\epsilon \sigma b^3=\\ [\frac{R{T}_C{V}_C{}^2}{P_C}+\frac{R{T}_C{V}_C\sigma b}{P_C}+\frac{R{T}_C{V}_C\epsilon b}{P_C}+\frac{R{T}_C\epsilon \sigma b^2}{P_C}]-\frac{a(T_C)V_C}{P_C}+\frac{a(T_C)b}{P_C}\\ \end{array}$

Rearrange,

  $\begin{array}{l}\Rightarrow V_C{}^3+\left(\sigma b+\epsilon b-b-\frac{R{T}_C}{P_C}\right)V_C{}^2+\left(\epsilon \sigma b^2-\sigma b^2-\epsilon b^2-\frac{R{T}_C\sigma b}{P_C}-\frac{R{T}_C\epsilon b}{P_C}+\frac{a(T_C)}{P_C}\right)V_C-\\ \left(\epsilon \sigma b^3+\frac{R{T}_C\epsilon \sigma b^2}{P_C}+\frac{a(T_C)b}{P_C}\right)=0\end{array}$

In terms of non-critical condition

  $\begin{array}{l}\Rightarrow V^3+\left(\sigma b+\epsilon b-b-\frac{RT}{P}\right)V^2+\left(\epsilon \sigma b^2-\sigma b^2-\epsilon b^2-\frac{RT\sigma b}{P}-\frac{RT\epsilon b}{P}+\frac{a(T)}{P}\right)V-\\ \left(\epsilon \sigma b^3+\frac{RT\epsilon \sigma b^2}{P}+\frac{a(T)b}{P}\right)=0\end{array}$

The above equation is similar as ${(V-V_C)}^3=V^3+3V{V}_C{}^2-3V^2{V}_C-V_C{}^3$, Now compare both equations and at critical conditions

  $\sigma b+\epsilon b-b-\frac{R{T}_C}{P_C}=-3V_C$

  $\epsilon \sigma b^2-\sigma b^2-\epsilon b^2-\frac{R{T}_C\sigma b}{P_C}-\frac{R{T}_C\epsilon b}{P_C}+\frac{a(T_C)}{P_C}=3V_C{}^2$

And

  $\left(\epsilon \sigma b^3+\frac{R{T}_C\epsilon \sigma b^2}{P_C}+\frac{a(T_C)b}{P_C}\right)=V_C{}^3$

Now, the three general derived equations are the term of critical temperature, pressure and volume. Three equations and we will have five unknowns $P_C,V_C,T_C,a(T_C)\text{and}b$ . Replace them with $\psi ,\Omega \text{and}{Z}_C$ .

From equation $\sigma b+\epsilon b-b-\frac{R{T}_C}{P_C}=-3V_C$

Divide by $b$

  $\sigma +\epsilon -1-\frac{R{T}_C}{b{P}_C}=-\frac{3V_C}{b}$

Since, $\Omega =\frac{b{P}_C}{R{T}_C}$ and $b=\frac{V_C\Omega }{Z_C}$

Gives

  $\begin{array}{l}\sigma +\epsilon -1-\frac{1}{\Omega }=-\frac{3Z_C}{\Omega }\\ \text{Or}\\ -\sigma -\epsilon +1+\frac{1}{\Omega }=\frac{3Z_C}{\Omega }\\ \frac{1+\Omega -\sigma \Omega -\epsilon \Omega }{\Omega }=\frac{3Z_C}{\Omega }\\ \text{Or,}\end{array}$

  $1+(1-\sigma -\epsilon )\Omega =3Z_C$ ......(2)

From equation $\epsilon \sigma b^2-\sigma b^2-\epsilon b^2-\frac{R{T}_C\sigma b}{P_C}-\frac{R{T}_C\epsilon b}{P_C}+\frac{a(T_C)}{P_C}=3V_C{}^2$

Divide by $b^2$

  $\begin{array}{l}\epsilon \sigma -\sigma -\epsilon -\frac{R{T}_C\sigma b}{P_C{b}^2}-\frac{R{T}_C\epsilon b}{P_C{b}^2}+\frac{a(T_C)}{P_C{b}^2}=\frac{3V_C{}^2}{b^2}\\ \text{Or}\\ \\ \epsilon \sigma -\sigma -\epsilon -\frac{R{T}_C\sigma }{P_{C}b}-\frac{R{T}_C\epsilon }{P_{C}b}+\frac{a(T_C)}{P_C{b}^2}=\frac{3V_C{}^2}{b^2}\end{array}$

Since, $\Omega =\frac{b{P}_C}{R{T}_C}$ and $b=\frac{V_C\Omega }{Z_C}$

  $a=\frac{\psi R^2{T}_C{}^2}{P_C}$

Gives

  $\epsilon \sigma -\sigma -\epsilon -\frac{\sigma }{\Omega }-\frac{\epsilon }{\Omega }+\frac{\frac{\psi R^2{T}_C{}^2}{P_C}}{P_C\frac{R^2{T}_C{}^2{\Omega }^2}{P_C{}^2}}=\frac{3V_C{}^2}{b^2}$

Or,

  $\epsilon \sigma -\sigma -\epsilon -\frac{\sigma }{\Omega }-\frac{\epsilon }{\Omega }+\frac{\psi }{{\Omega }^2}=\frac{3Z_C{}^2}{{\Omega }^2}$

Or,

  $\epsilon \sigma {\Omega }^2-\sigma {\Omega }^2-\epsilon {\Omega }^2-\sigma \Omega -\epsilon \Omega +\psi =3Z_C{}^2$

Or

  $\epsilon \sigma {\Omega }^2-\Omega (\epsilon +\sigma )(\Omega +1)+\psi =3Z_C{}^2$......(3)

From equation $\left(\epsilon \sigma b^3+\frac{R{T}_C\epsilon \sigma b^2}{P_C}+\frac{a(T_C)b}{P_C}\right)=V_C{}^3$

Divide by $b^3$

  $\left(\epsilon \sigma +\frac{R{T}_C\epsilon \sigma }{b{P}_C}+\frac{a(T_C)}{b^2{P}_C}\right)=\frac{V_C{}^3}{b^3}$

Since, $\Omega =\frac{b{P}_C}{R{T}_C}\Rightarrow b=\frac{\Omega R{T}_C}{P_C}$ and $b=\frac{V_C\Omega }{Z_C}$

  $a=\frac{\psi R^2{T}_C{}^2}{P_C}$

Gives

  $\begin{array}{l}\left(\epsilon \sigma +\frac{R{T}_C\epsilon \sigma }{\frac{\Omega R{T}_C}{P_C}{P}_C}+\frac{\frac{\psi R^2{T}_C{}^2}{P_C}}{{\frac{{\Omega }^2{R}^2{T}_C}{P_C{}^2}}^2{P}_C}\right)=\frac{Z_C{}^3}{{\Omega }^3}\\ \text{Or,}\\ \\ \left(\epsilon \sigma +\frac{\epsilon \sigma }{\Omega }+\frac{\psi }{{\Omega }^2}\right)=\frac{Z_C{}^3}{{\Omega }^3}\end{array}$

Or,

  $\begin{array}{l}\epsilon \sigma {\Omega }^3+\epsilon \sigma {\Omega }^2+\Omega \psi =Z_C{}^3\\ \text{Or,}\\ \end{array}$

Or

  $\epsilon \sigma {\Omega }^2-\Omega (\epsilon +\sigma )(\Omega +1)+\psi =3Z_C{}^2$

Or,

  $\epsilon \sigma {\Omega }^2(\Omega +1)+\Omega \psi =Z_C{}^3$......(4)

For Redlich/Kwong equation of state,

  $\epsilon =0,\sigma =1$

Put values in equation (2) (3) and (4),

  $\begin{array}{l}1+(1-1-0)\Omega =3Z_C\\ \\ \Rightarrow Z_C=\frac{1}{3}\end{array}$

Put values in equation (3),

  $\epsilon \sigma {\Omega }^2-\Omega (\epsilon +\sigma )(\Omega +1)+\psi =3Z_C{}^2$

  $\begin{array}{l}0\times 1\times {\Omega }^2-\Omega \times (0+1)(\Omega +1)+\psi =3\times {\frac{1}{3}}^2\\ \\ -\Omega (\Omega +1)+\psi =\frac{1}{3}\end{array}$

Put values in equation (4),

  $\begin{array}{l}\epsilon \sigma {\Omega }^2(\Omega +1)+\Omega \psi =Z_C{}^3\\ \\ 0\times 1\times {\Omega }^2\times (\Omega +1)+\Omega \psi ={\frac{1}{3}}^3\\ \text{Or,}\\ \\ \psi =\frac{1}{27\Omega }\end{array}$

From equation $-\Omega (\Omega +1)+\psi =\frac{1}{3}$, Put $\psi =\frac{1}{27\Omega }$

  $\begin{array}{l}-\Omega (\Omega +1)+\frac{1}{27\Omega }=\frac{1}{3}\\ \\ -{\Omega }^2-\Omega +\frac{1}{27\Omega }=\frac{1}{3}\\ \\ -27{\Omega }^3-27{\Omega }^2+1=9\Omega \\ \\ -27{\Omega }^3-27{\Omega }^2-9\Omega +1=0\end{array}$

After solving cubic equation, roots of $\Omega$ are

  $\Omega =0.08664,0.5433\pm 0.364\text{i}$

Since value of $\Omega$ cannot be complex number so, value of $\Omega$ is:

  $\Omega =0.08664$

And $\psi =\frac{1}{27\Omega }$

  $\begin{array}{l}\psi =\frac{1}{27\Omega }=\frac{1}{27\times 0.08664}=0.42748\\ \text{So,}\\ \\ \psi =0.42748\end{array}$

Hence proved.

(b)

Interpretation Introduction

Interpretation:

The value of $\psi ,\Omega \text{and}{Z}_C$ for Soave/Redlich/Kwong equation of state should be calculated.

Concept Introduction :

To find the values of $\psi ,\Omega \text{and}{Z}_C$, first write the general cubic equation of state at critical conditions $P_C,V_C,T_C$ and write it in form of ${(V-V_C)}^3=V^3+3V{V}_C{}^2-3V^2{V}_C-V_C{}^3$ since at critical conditions $(V-V_C)=0$ and so ${(V-V_C)}^3=0$ and compare both equations to find three general derived equation in the terms of critical temperature, pressure and volume. Three equations and we will have five unknowns $P_C,V_C,T_C,a(T_C)\text{and}b$ . Replace them with $\psi ,\Omega \text{and}{Z}_C$ and find value of $\psi ,\Omega \text{and}{Z}_C$ using three equations. The general cubic equation of state is:

  $P=\frac{RT}{V-b}-\frac{a(T)}{(V+\epsilon b)(V+\sigma b)}$......(1)

For critical conditions,

  ${\frac{\partial P}{\partial V}}_{T,cr}={\frac{{\partial }^{2}P}{\partial V^2}}_{T,cr}=0$ and $P=P_C,V=V_C\text{and }T=T_C$

For Soave/Redlich/Kwong equation of state,

  $\epsilon =0,\sigma =1$

Answer to Problem 3.85P

  $\begin{array}{l}\psi =0.42748,\\ \Omega =0.08664\text{and}\\ Z_C=1/3\end{array}$

Explanation of Solution

Given information:

The general expressions for $\psi ,\Omega \text{and}{Z}_C$ are:

  $\psi =\frac{a{P}_C}{R^2{T}_C{}^2}$

  $\Omega =\frac{b{P}_C}{R{T}_C}$

And $Z_C=\frac{P_C{V}_C}{R{T}_C}$

For Soave/Redlich/Kwong equation of state,

  $\epsilon =0,\sigma =1$

Put values in general equation (2) (3) and (4) found in subpart (1),

  $\begin{array}{l}1+(1-1-0)\Omega =3Z_C\\ \\ \Rightarrow Z_C=\frac{1}{3}\end{array}$

Put values in equation (3),

  $\epsilon \sigma {\Omega }^2-\Omega (\epsilon +\sigma )(\Omega +1)+\psi =3Z_C{}^2$

  $\begin{array}{l}0\times 1\times {\Omega }^2-\Omega \times (0+1)(\Omega +1)+\psi =3\times {\frac{1}{3}}^2\\ \\ -\Omega (\Omega +1)+\psi =\frac{1}{3}\end{array}$

Put values in equation (4),

  $\begin{array}{l}\epsilon \sigma {\Omega }^2(\Omega +1)+\Omega \psi =Z_C{}^3\\ \\ 0\times 1\times {\Omega }^2\times (\Omega +1)+\Omega \psi ={\frac{1}{3}}^3\\ \text{Or,}\\ \\ \psi =\frac{1}{27\Omega }\end{array}$

From equation $-\Omega (\Omega +1)+\psi =\frac{1}{3}$, Put $\psi =\frac{1}{27\Omega }$

  $\begin{array}{l}-\Omega (\Omega +1)+\frac{1}{27\Omega }=\frac{1}{3}\\ \\ -{\Omega }^2-\Omega +\frac{1}{27\Omega }=\frac{1}{3}\\ \\ -27{\Omega }^3-27{\Omega }^2+1=9\Omega \\ \\ -27{\Omega }^3-27{\Omega }^2-9\Omega +1=0\end{array}$

After solving cubic equation, roots of $\Omega$ are

  $\Omega =0.08664,0.5433\pm 0.364\text{i}$

Since value of $\Omega$ cannot be complex number so, value of $\Omega$ is:

  $\Omega =0.08664$

And $\psi =\frac{1}{27\Omega }$

  $\begin{array}{l}\psi =\frac{1}{27\Omega }=\frac{1}{27\times 0.08664}=0.42748\\ \text{So,}\\ \\ \psi =0.42748\end{array}$

Hence proved.

(c)

Interpretation Introduction

Interpretation:

The value of $\psi ,\Omega \text{and}{Z}_C$ for Peng/Robinson equation of state should be calculated.

Concept Introduction :

To find the values of $\psi ,\Omega \text{and}{Z}_C$, first write the general cubic equation of state at critical conditions $P_C,V_C,T_C$ and write it in form of ${(V-V_C)}^3=V^3+3V{V}_C{}^2-3V^2{V}_C-V_C{}^3$ since at critical conditions $(V-V_C)=0$ and so ${(V-V_C)}^3=0$ and compare both equations to find three general derived equation in the terms of critical temperature, pressure and volume. Three equations and we will have five unknowns $P_C,V_C,T_C,a(T_C)\text{and}b$ . Replace them with $\psi ,\Omega \text{and}{Z}_C$ and find value of $\psi ,\Omega \text{and}{Z}_C$ using three equations. The general cubic equation of state is:

  $P=\frac{RT}{V-b}-\frac{a(T)}{(V+\epsilon b)(V+\sigma b)}$......(1)

For critical conditions,

  ${\frac{\partial P}{\partial V}}_{T,cr}={\frac{{\partial }^{2}P}{\partial V^2}}_{T,cr}=0$ and $P=P_C,V=V_C\text{and }T=T_C$

For Peng/Robinson equation of state,

  $\epsilon =1-\sqrt{2},\sigma =1+\sqrt{2}$

Answer to Problem 3.85P

  $\begin{array}{l}\psi =0.45724,\\ \Omega =0.07779\text{and}\\ Z_C=0.30740\end{array}$

Explanation of Solution

Given information:

The general expressions for $\psi ,\Omega \text{and}{Z}_C$ are:

  $\psi =\frac{a{P}_C}{R^2{T}_C{}^2}$

  $\Omega =\frac{b{P}_C}{R{T}_C}$

And $Z_C=\frac{P_C{V}_C}{R{T}_C}$

For Peng/Robinson equation of state,

  $\epsilon =1-\sqrt{2},\sigma =1+\sqrt{2}$

Put values in general equation (2) (3) and (4) found in subpart (1),

  $1+(1-\sigma -\epsilon )\Omega =3Z_C$

  $\begin{array}{l}1+(1-1-\sqrt{2}-1+\sqrt{2})\Omega =3Z_C\\ \\ \Rightarrow 1-\Omega =3Z_C\end{array}$

Put values in equation (3),

  $\epsilon \sigma {\Omega }^2-\Omega (\epsilon +\sigma )(\Omega +1)+\psi =3Z_C{}^2$

  $\begin{array}{l}(1-\sqrt{2})\times (1+\sqrt{2})\times {\Omega }^2-\Omega \times (1-\sqrt{2}+1+\sqrt{2})(\Omega +1)+\psi =3\times Z_C{}^2\\ \end{array}$

Put $1-\Omega =3Z_C$ in above equation,

  $\begin{array}{l}(1-\sqrt{2})\times (1+\sqrt{2})\times {\Omega }^2-\Omega \times (1-\sqrt{2}+1+\sqrt{2})(\Omega +1)+\psi =3\times Z_C{}^2\\ \\ -{\Omega }^2-2\Omega (\Omega +1)+\psi =3\times {\left(\frac{1-\Omega }{3}\right)}^2\\ \\ -{\Omega }^2-2{\Omega }^2-2\Omega +\psi =\frac{{\left(1-\Omega \right)}^2}{3}\\ \\ -3{\Omega }^2-2\Omega +\psi =\frac{{\left(1-\Omega \right)}^2}{3}\\ \\ -9{\Omega }^2-6\Omega +3\psi =1+{\Omega }^2-2\Omega \\ \end{array}$

  $-10{\Omega }^2-4\Omega +(3\psi -1)=0$......(4)

Put values in equation (4),

  $\begin{array}{l}\epsilon \sigma {\Omega }^2(\Omega +1)+\Omega \psi =Z_C{}^3\\ \\ (1-\sqrt{2})\times (1-\sqrt{2})\times {\Omega }^2\times (\Omega +1)+\Omega \psi ={\left(\frac{1-\Omega }{3}\right)}^3\\ \text{Or,}\\ \\ -{\Omega }^2\times (\Omega +1)+\Omega \psi ={\left(\frac{1-\Omega }{3}\right)}^3\\ \\ -27{\Omega }^3-27{\Omega }^2+27\Omega \psi =1-{\Omega }^3+3{\Omega }^2-3\Omega \\ \end{array}$

  $-26{\Omega }^3-30{\Omega }^2+(27\Omega \psi +3\Omega -1)=0$......(5)

From equation 4

  $\begin{array}{l}-10{\Omega }^2-4\Omega +(3\psi -1)=0\\ \text{multiply by 3}\\ \\ -30{\Omega }^2-12\Omega +(9\psi -3)=0\\ \end{array}$

  $\begin{array}{l}-30{\Omega }^2-12\Omega +(9\psi -3)=0\\ \\ 30{\Omega }^2+12\Omega +3=9\psi \\ \\ \psi =\frac{30{\Omega }^2+12\Omega +3}{9}\end{array}$

Put value in equation 5

  $\begin{array}{l}-26{\Omega }^3-30{\Omega }^2+(27\Omega \psi +3\Omega -1)=0\\ \\ -26{\Omega }^3-30{\Omega }^2+\left\{27\Omega \times \left(\frac{30{\Omega }^2+12\Omega +3}{9}\right)+3\Omega -1\right\}=0\\ \\ -26{\Omega }^3-30{\Omega }^2+\left\{3\Omega \times (30{\Omega }^2+12\Omega +3)+3\Omega -1\right\}=0\\ \\ -26{\Omega }^3-30{\Omega }^2+\left\{90{\Omega }^3+36{\Omega }^2+9\Omega +3\Omega -1\right\}=0\\ \\ 64{\Omega }^3+6{\Omega }^2+12\Omega -1=0\\ \end{array}$

After solving cubic equation, roots of $\Omega$ are

  $\Omega =0.07779,0.08577\pm 0.43988\text{i}$

Since value of $\Omega$ cannot be complex number so, value of $\Omega$ is:

  $\Omega =0.07779$

And $\psi =\frac{30{\Omega }^2+12\Omega +3}{9}$

  $\begin{array}{l}\psi =\frac{30{\Omega }^2+12\Omega +3}{9}=\frac{30\times {0.07779}^2+12\times 0.07779+3}{9}\\ \text{So,}\\ \\ \psi =0.45724\end{array}$

And

  $\begin{array}{l}1-\Omega =3Z_C\\ \\ 1-0.07779=3Z_C\\ \\ Z_C=0.30740\end{array}$

Hence proved.