Chapter 3, Problem 3.44P

(a)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ethylene must calculate at ${25}^{\circ }\text{C}$ temperature and $12\text{bar}$ pressure by the truncated virial equation with given virial coefficients.

Concept Introduction:

The value of $Z$ is easily calculated by the given truncated virial equation and value of $V$ is find out by $PV=ZnRT$ .

Answer to Problem 3.44P

The $Z\text{and }V$ value for ethylene are $V=1894.065\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.928$ respectively.

Explanation of Solution

Given Information:

The virial equation is given as

  $Z=\frac{PV}{RT}=1+\frac{B}{V}+\frac{C}{V^2}$

  $\text{gas constant R in CGS unit is 82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}$

  $P=12\text{bar}$

  $T{\text{=25}}^{\text{0}}\text{C+273}\text{.15=298}\text{.15}\text{K}$

For ethylene, the second and third virial coefficients are: $B=-\text{140}{\text{cm}}^{\text{3}}{\text{.mol}}^{\text{-1}}$ and $C=7200{\text{cm}}^{\text{6}}{\text{.mol}}^{\text{-2}}$ .

From given virial equation

  $Z=\frac{PV}{RT}=1+\frac{B}{V}+\frac{C}{V^2}$

  $n\text{=1}\text{mol}$ is given, So

  $\frac{PV}{RT}=1+\frac{B}{V}+\frac{C}{V^2}$

Put the given values

  $\begin{array}{l}\frac{12\text{bar}\times V}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{298}\text{.15}\text{K}}=1+\frac{-140\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}{V}+\frac{7200\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ \text{4}\text{.9}\times {\text{10}}^{-4}V\frac{{\text{cm}}^3}{\text{mol}}=\frac{V^2-140\frac{{\text{cm}}^{\text{3}}}{\text{mol}}V+7200\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ \text{4}\text{.9}\times {\text{10}}^{-4}{V}^3\frac{{\text{cm}}^3}{\text{mol}}=V^2-140\frac{{\text{cm}}^{\text{3}}}{\text{mol}}V+7200\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}\end{array}$

On solving the equation

  $V=1894.065\frac{{\text{cm}}^3}{\text{mol}}$

So,

  $\begin{array}{l}Z=\frac{PV}{RT}=\frac{12\text{bar}\times 1894.065\frac{{\text{cm}}^3}{\text{mol}}}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{298}\text{.15}\text{K}}\\ Z=0.928\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ethylene must calculate at ${25}^{\circ }\text{C}$ temperature and $12\text{bar}$ pressure by the truncated virial equation with a value of $B$ virial coefficient from generalized Pitzer correlation.

Concept Introduction:

In order to find value of $Z$ from given truncated virial equation, first we calculate $B$ virial coefficient from generalized Pitzer correlation and then value of $V$ followed by value of $Z$ find out by $PV=ZnRT$ .

Answer to Problem 3.44P

The $Z\text{and }V$ value for ethylene are $V=1924.29\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.932$ respectively.

Explanation of Solution

Given Information:

The virial equation in reduced conditions is given as

  $\begin{array}{l}Z=1+\left[B^0+\omega B^1\right]\frac{T_r}{P_r}\\ \text{Or,}\\ V=\frac{RT}{P}+\frac{R{T}_C}{P_C}\left[B^0+\omega B^1\right]\end{array}$

Here $B^0=0.083-\frac{0.422}{T_r{}^{1.6}}$ and $B^1=0.139-\frac{0.172}{T_r{}^{4.2}}$

For ethylene at given temperature and pressure, the critical conditions are written from table B.1 in appendix B,

  $T_C=282.3\text{K}\text{and }{P}_C=50.4\text{bar, }\omega \text{=0}\text{.087}$

  $\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$

  $P=12\text{bar}$

  $T{\text{=25}}^{\text{0}}\text{C+273}\text{.15=298}\text{.15}\text{K}$

For calculation of $B$ virial coefficient from generalized Pitzer correlation,

  $\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{298.15\text{K}}{282.3\text{K}}\text{=1}\text{.056}\end{array}$

And

  $\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{12\text{bar}}{50.4\text{bar}}\text{=0}\text{.238}\end{array}$

So,

  $\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ B^0=0.083-\frac{0.422}{{1.056}^{1.6}}\\ B^0=-0.304\end{array}$

and

  $\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ B^1=0.139-\frac{0.172}{{1.056}^{4.2}}\\ B^1=2.183\times {10}^{-3}\end{array}$

Hence, value of $V$ is

  $\begin{array}{l}V=\frac{RT}{P}+\frac{R{T}_C}{P_C}\left[B^0+\omega B^1\right]\\ V=\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 298.15\text{K}}{12\text{bar}}+\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 282.3\text{K}}{50.4\text{bar}}\left[-0.304+0.087\times 2.183\times {10}^{-3}\right]\\ V=1924.29\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

And value of compressibility factor,

  $\begin{array}{l}Z=\frac{PV}{RT}=\frac{12\text{bar}\times 1924.29\frac{{\text{cm}}^3}{\text{mol}}}{\text{83}\text{.144}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{298}\text{.15}\text{K}}\\ Z=0.932\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ethylene must calculate at ${25}^{\circ }\text{C}$ temperature and $12\text{bar}$ pressure by the Redlich/Kwong equation.

Concept Introduction:

The Redlich/Kwong equations is an iterative procedure. So, we will use hit and trial procedure and guess some values of $Z$ to achieve convergent value.

According to Redlich/Kwong equations, the molar volume of saturated vapor can be found using formula:

  $V=\frac{ZRT}{P}$

Answer to Problem 3.44P

From Redlich/Kwong equations, the molar volume of ethylene is:

$V=1917.046\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.928$ respectively.

Explanation of Solution

Given Information:

The Redlich/Kwong equation is

$Z=1+\beta -q\beta \frac{Z-\beta }{(Z+\epsilon \beta )(Z+\sigma \beta )}$

                 ....(1)

Here

$\beta =\Omega \frac{P_r}{T_r}$ and $q=\frac{\phi \alpha (T_r)}{\Omega T_r}$

From table 3.1 in the example based on Redlich/Kwong equation given in book, the values used to calculate the terms in equation (1) are:

For Redlich/Kwong:

  $\epsilon =0,\sigma =1,\psi =0.42748,\Omega =0.08664\text{and}\alpha {\text{(T}}_r)=T_r{}^{-1/2}$

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=12\text{bar}$

  $T{\text{=25}}^{\text{0}}\text{C+273}\text{.15=298}\text{.15}\text{K}$

  $\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{298.15\text{K}}{282.3\text{K}}\text{=1}\text{.056}\end{array}$

And,

  $\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{12\text{bar}}{50.4\text{bar}}\text{=0}\text{.238}\end{array}$

So,

  $\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.238}{1.056}\\ \beta =0.0195\end{array}$

  $\alpha {\text{(T}}_r)=T_r{}^{1/2}={1.056}^{-0.5}=0.973$

  $\begin{array}{l}q=\frac{\psi \alpha (T_r)}{\Omega T_r}=\frac{0.42748\times 0.973}{0.08664\times 1.056}\\ q=4.546\end{array}$

For ethylene, put values in equation (1)

  $Z=1+\beta -q\beta \frac{Z-\beta }{(Z+\epsilon \beta )(Z+\sigma \beta )}$

  $Z=1+0.0195-4.546\times 0.0195\times \frac{Z-0.0195}{(Z+0\times 0.0195)(Z+1\times 0.0195)}$

Using hit and trial method and compare both side of equation, the calculated value is:

  $Z=0.928$

Hence,

  $\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.928\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 298.15\text{K}}{12\text{bar}}\\ V=1917.046\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ethylene must calculate at ${25}^{\circ }\text{C}$ temperature and $12\text{bar}$ pressure by the Soave/Redlich/Kwong equation.

Concept Introduction:

The Soave/Redlich/Kwong equations is an iterative procedure. So, we will use hit and trial procedure and guess some values of $Z$ to achieve convergent value.

According to Redlich/Kwong equations, the molar volume of ethylene can be found using formula:

  $V=\frac{ZRT}{P}$

Answer to Problem 3.44P

From Soave/Redlich/Kwong equations, the molar volume of ethylene is:

$V=1918.285\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.9286$ respectively.

Explanation of Solution

Given Information:

The Soave/Redlich/Kwong equation is

$Z=1+\beta -q\beta \frac{Z-\beta }{(Z+\epsilon \beta )(Z+\sigma \beta )}$

                 ....(1)

Here

$\beta =\Omega \frac{P_r}{T_r}$ and $q=\frac{\phi \alpha (T_r)}{\Omega T_r}$

From table 3.1 in the example based on Soave/Redlich/Kwong equation given in book, the values used to calculate the terms in equation (1) are:

For Soave/Redlich/Kwong:

  $\begin{array}{l}\epsilon =0,\sigma =1,\psi =0.42748,\Omega =0.08664\text{and}\\ \alpha {\text{(T}}_r)={[}^1\end{array}$

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=12\text{bar}$

  $T{\text{=25}}^{\text{0}}\text{C+273}\text{.15=298}\text{.15}\text{K}$

The values of reduced temperature and pressure are same as found in subpart (c), From reference subpart (c),

$T_r=\text{1}\text{.056}$ and $P_r=\text{0}\text{.238}$

So,

  $\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.238}{1.056}\\ \beta =0.0195\end{array}$

  $\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=0.966\end{array}$

  $\begin{array}{l}q=\frac{\psi \alpha (T_r)}{\Omega T_r}=\frac{0.42748\times 0.966}{0.08664\times 1.056}\\ q=4.5135\end{array}$

For ethylene, put values in equation (1)

  $Z=1+\beta -q\beta \frac{Z-\beta }{(Z+\epsilon \beta )(Z+\sigma \beta )}$

  $Z=1+0.0195-4.5135\times 0.0195\times \frac{Z-0.0195}{(Z+0\times 0.0195)(Z+1\times 0.0195)}$

Using hit and trial method and compare both side of equation, the calculated value is:

  $Z=0.9286$

Hence,

  $\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.9286\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 298.15\text{K}}{12\text{bar}}\\ V=1918.285\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ethylene must calculate at ${25}^{\circ }\text{C}$ temperature and $12\text{bar}$ pressure by Peng/Robinson equation.

Concept Introduction:

The Peng/Robinson equation is an iterative procedure. So, we will use hit and trial procedure and guess some values of $Z$ to achieve convergent value.

According to Peng/Robinson equation, the molar volume of ethylene can be found using formula:

  $V=\frac{ZRT}{P}$

Answer to Problem 3.44P

From Soave/Redlich/Kwong equations, the molar volume of ethylene is:

$V=1900.52\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.92$ respectively.

Explanation of Solution

Given Information:

The Peng/Robinson equation is

$Z=1+\beta -q\beta \frac{Z-\beta }{(Z+\epsilon \beta )(Z+\sigma \beta )}$

                 ....(1)

Here

$\beta =\Omega \frac{P_r}{T_r}$ and $q=\frac{\phi \alpha (T_r)}{\Omega T_r}$

From table 3.1 in the example based on Peng/Robinson equation given in book, the values used to calculate the terms in equation (1) are:

For Peng/Robinson equation:

  $\begin{array}{l}\epsilon =1-\sqrt{2},\sigma =1+\sqrt{2},\psi =0.45724,\Omega =0.07779\text{and}{Z}_C=0.30740\\ \alpha {\text{(T}}_r)={[}^1\end{array}$

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=12\text{bar}$

  $T{\text{=25}}^{\text{0}}\text{C+273}\text{.15=298}\text{.15}\text{K}$

The values of reduced temperature and pressure are same as found in subpart (c), From reference subpart (c),

$T_r=\text{1}\text{.056}$ and $P_r=\text{0}\text{.238}$

So,

  $\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.07779\times \frac{0.238}{1.056}\\ \beta =0.0175\end{array}$

  $\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=0.9722\end{array}$

  $\begin{array}{l}q=\frac{\psi \alpha (T_r)}{\Omega T_r}=\frac{0.45724\times 0.9722}{0.07779\times 1.056}\\ q=5.411\end{array}$

For ethylene, put values in equation (1)

  $Z=1+\beta -q\beta \frac{Z-\beta }{(Z+\epsilon \beta )(Z+\sigma \beta )}$

  $Z=1+0.0175-5.411\times 0.0175\times \frac{Z-0.0175}{(Z+(1-\sqrt{2})\times 0.0175)(Z+(1+\sqrt{2})\times 0.0175)}$

Using hit and trial method and compare both side of equation, the calculated value is:

  $Z=0.92$

Hence,

  $\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.92\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 298.15\text{K}}{12\text{bar}}\\ V=1900.52\frac{{\text{cm}}^3}{\text{mol}}\end{array}$