Chapter 3, Problem 3.46P

(a)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for sulfur hexafluoride must calculate at ${75}^{\circ }\text{C}$ temperature and $15\text{bar}$ 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 3.46P

The $Z\text{and }V$ value for sulfur hexafluoride are $V=1557.91\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.818$ respectively.

Explanation of Solution

Given Information:

The virial equation is given as

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

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

$P=15\text{bar}$

$T{\text{=75}}^{\text{0}}\text{C+273}\text{.15=348}\text{.15}\text{K}$

For sulfur hexafluoride, the second and third virial coefficients are: $B=-\text{194}{\text{cm}}^{\text{3}}{\text{.mol}}^{\text{-1}}$ and $C=15300{\text{cm}}^{\text{6}}{\text{.mol}}^{\text{-2}}$ .

From given virial equation

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

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

Put the given values

$\begin{array}{l}\frac{15\text{bar}\times V}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{348}\text{.15}\text{K}}=1+\frac{-\text{194}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}{V}+\frac{15300\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ 5.25\times {10}^{-4}V\frac{{\text{cm}}^3}{\text{mol}}=\frac{V^2-194\frac{{\text{cm}}^{\text{3}}}{\text{mol}}V+15300\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ 5.66\times {10}^{-4}{V}^3=V^2-194V+15300\end{array}$

On solving the equation

$V=1557.91\frac{{\text{cm}}^3}{\text{mol}}$

So,

$\begin{array}{l}Z=\frac{PV}{RT}=\frac{15\text{bar}\times 1557.91\frac{{\text{cm}}^3}{\text{mol}}}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{348}\text{.15}\text{K}}\\ Z=0.818\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for sulfur hexafluoride must calculate at ${75}^{\circ }\text{C}$ temperature and $15\text{bar}$ 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 3.46P

The $Z\text{and }V$ value for sulfur hexafluoride are $V=1735.029\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.899$ 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 sulfur hexafluoride at given temperature and pressure, the critical conditions are given as,

$T_C=318.7\text{K}\text{and }{P}_C=37.6\text{bar, }\omega \text{=0}\text{.286}$

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

$P=15\text{bar}$

$T{\text{=75}}^{\text{0}}\text{C+273}\text{.15=348}\text{.15}\text{K}$

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

$\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{348.15\text{K}}{318.7\text{K}}\text{=1}\text{.0924}\end{array}$

And

$\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{15\text{bar}}{37.6\text{bar}}\text{=0}\text{.399}\end{array}$

So,

$\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ B^0=0.083-\frac{0.422}{\text{1}{\text{.0924}}^{1.6}}\\ B^0=-0.283\end{array}$

and

$\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ B^1=0.139-\frac{0.172}{\text{1}{\text{.0924}}^{4.2}}\\ B^1=0.02033\end{array}$

Hence, value of $V$ is

$\begin{array}{l}V=\frac{RT}{P}+\frac{R{T}_C}{P_C}\left[B^0+\omega B^1\right]\\ V=\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 348.15\text{K}}{15\text{bar}}+\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 318.7\text{K}}{37.6\text{bar}}\left[-0.283+0.286\times 0.02033\right]\\ V=1735.029\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

And value of compressibility factor,

$\begin{array}{l}Z=\frac{PV}{RT}=\frac{15\text{bar}\times 1735.029\frac{{\text{cm}}^3}{\text{mol}}}{\text{83}\text{.144}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{348}\text{.15}\text{K}}\\ Z=0.899\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for sulfur hexafluoride must calculate at ${75}^{\circ }\text{C}$ temperature and $15\text{bar}$ 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 sulfur hexafluoridecan be found using formula:

$V=\frac{ZRT}{P}$

Answer to Problem 3.46P

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

$V=1714.44\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.888417$ 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:

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

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=15\text{bar}$

$T{\text{=75}}^{\text{0}}\text{C+273}\text{.15=348}\text{.15}\text{K}$

$\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{348.15\text{K}}{318.7\text{K}}\text{=1}\text{.0924}\end{array}$

And

$\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{15\text{bar}}{37.6\text{bar}}\text{=0}\text{.399}\end{array}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.399}{1.0924}\\ \beta =0.0316\end{array}$

$\alpha {\text{(T}}_r)=T_r{}^{1/2}={1.0924}^{-0.5}=0.957$

$\begin{array}{l}q=\frac{0.42748\times 0.957}{0.08664\times 1.0924}\\ q=4.32243\end{array}$

For sulfur hexafluoride, put values in equation (1)

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

$Z=1+0.0316-4.32243\times 0.0316\times \frac{Z-0.0316}{(Z+0\times 0.0316)(Z+1\times 0.0316)}$

Using hit and trial method and compare both side of equation, the calculated value from scientific calculator 991ES-PLUS or 991MS is:

$Z=0.888417$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.888417\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 348.15\text{K}}{15\text{bar}}\\ V=1714.44\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for sulfur hexafluoride must calculate at ${75}^{\circ }\text{C}$ temperature and $15\text{bar}$ 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 sulfur hexafluoride can be found using formula:

$V=\frac{ZRT}{P}$

Answer to Problem 3.46P

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

$V=1727.15\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.895$ 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:

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

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=15\text{bar}$

$T{\text{=75}}^{\text{0}}\text{C+273}\text{.15=348}\text{.15}\text{K}$

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

$T_r=\text{1}\text{.0924}$ and $P_r=\text{0}\text{.399}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.399}{1.0924}\\ \beta =0.0316\end{array}$

$\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=0.919\end{array}$

$\begin{array}{l}q=\frac{\phi \alpha (T_r)}{\Omega T_r}=\frac{0.42748\times 0.919}{0.08664\times 1.0924}\\ q=4.151\end{array}$

For sulfur hexafluoride, put values in equation (1)

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

$Z=1+0.0316-4.151\times 0.0316\times \frac{Z-0.0316}{(Z+0\times 0.0316)(Z+1\times 0.0316)}$

Using hit and trial method and compare both side of equation, the calculated value from scientific calculator 991ES-PLUS or 991MS is:

$Z=0.895$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.895\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 348.15\text{K}}{15\text{bar}}\\ V=1727.15\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for sulfur hexafluoride must calculate at ${75}^{\circ }\text{C}$ temperature and $15\text{bar}$ 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 sulfur hexafluoride can be found using formula:

$V=\frac{ZRT}{P}$

Answer to Problem 3.46P

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

$V=1702.06\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.882$ 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:

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

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=15\text{bar}$

$T{\text{=75}}^{\text{0}}\text{C+273}\text{.15=348}\text{.15}\text{K}$

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

$T_r=\text{1}\text{.0924}$ and $P_r=\text{0}\text{.399}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.07779\times \frac{0.399}{\text{1}\text{.0924}}\\ \beta =0.0284\end{array}$

$\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=0.9296\end{array}$

$\begin{array}{l}q=\frac{\psi \alpha (T_r)}{\Omega T_r}=\frac{0.45724\times 0.9296}{0.07779\times 1.0924}\\ q=5.0019\end{array}$

For sulfur hexafluoride, put values in equation (1)

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

$Z=1+0.0284-5.0019\times 0.0284\times \frac{Z-0.0284}{(Z+(1-\sqrt{2})\times 0.0284)(Z+(1+\sqrt{2})\times 0.0284)}$

Using hit and trial method and compare both side of equation, the calculated value from scientific calculator 991ES-PLUS or 991MS is:

$Z=0.882$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.882\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 348.15\text{K}}{15\text{bar}}\\ V=1702.06\frac{{\text{cm}}^3}{\text{mol}}\end{array}$