Chapter 3, Problem 3.48P

(a)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for Boron trichloride must calculate at $300\text{K}$ temperature and $1.5\text{bar}$ 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 3.48P

The $Z\text{and }V$ value for Boron trichloride are $V=15667.9\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.9422$ respectively.

Explanation of Solution

Given Information:

The virial equation is given as

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

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

$P=1.5\text{bar}$

$T\text{=300}\text{K}$

For Boron trichloride, the second and third virial coefficients are: $B=-\text{724}{\text{cm}}^{\text{3}}{\text{.mol}}^{\text{-1}}$ and $C=93866{\text{cm}}^{\text{6}}{\text{.mol}}^{\text{-2}}$ .

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

$\begin{array}{l}\frac{1.5\text{bar}\times V}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{300}\text{K}}=1+\frac{-\text{724}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}{V}+\frac{93866\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ 6.09\times {10}^{-5}V\frac{{\text{cm}}^3}{\text{mol}}=\frac{V^2-724\frac{{\text{cm}}^{\text{3}}}{\text{mol}}V+93866\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ 6.09\times {10}^{-5}{V}^3=V^2-724V+93866\end{array}$

On solving the equation

$V=15667.9\frac{{\text{cm}}^3}{\text{mol}}$

So,

$\begin{array}{l}Z=\frac{PV}{RT}=\frac{1.5\text{bar}\times 15667.9\frac{{\text{cm}}^3}{\text{mol}}}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{300}\text{K}}\\ Z=0.9422\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for Boron trichloride must calculate at $300\text{K}$ temperature and $1.5\text{bar}$ 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 3.48P

The $Z\text{and }V$ value for Boron trichloride are $V=15989.4\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.961$ 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 Boron trichlorideat given temperature and pressure, the critical conditions are given as,

$T_C=452\text{K}\text{and }{P}_C=38.7\text{bar, }\omega \text{=0}\text{.086}$

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

$P=1.5\text{bar}$

$T\text{=300}\text{K}$

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

$\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{300\text{K}}{452\text{K}}\text{=0}\text{.664}\end{array}$

And

$\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{1.5\text{bar}}{38.7\text{bar}}\text{=0}\text{.0387}\end{array}$

So,

$\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ B^0=0.083-\frac{0.422}{\text{0}{\text{.664}}^{1.6}}\\ B^0=-0.729\end{array}$

and

$\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ B^1=0.139-\frac{0.172}{\text{0}{\text{.664}}^{4.2}}\\ B^1=-0.821\end{array}$

Hence, value of $V$ is

$\begin{array}{l}V=\frac{RT}{P}+\frac{R{T}_C}{P_C}\left[B^0+\omega B^1\right]\\ V=\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 300\text{K}}{1.5\text{bar}}+\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 452\text{K}}{38.7\text{bar}}\left[-0.729+0.086\times 0.821\right]\\ V=15989.4\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

And value of compressibility factor,

$\begin{array}{l}Z=\frac{PV}{RT}=\frac{1.5\text{bar}\times 15989.4\frac{{\text{cm}}^3}{\text{mol}}}{\text{83}\text{.144}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{300}\text{K}}\\ Z=0.961\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for Boron trichloridemust calculate at $300\text{K}$ temperature and $1.5\text{bar}$ 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 Boron trichloridecan be found using formula:

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

Answer to Problem 3.48P

From Redlich/Kwong equations, the molar volume of Boron trichlorideis:

$V=15930.4\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.958$ 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:

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

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=1.5\text{bar}$

$T\text{=}300\text{K}$

$\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{300\text{K}}{452\text{K}}\text{=0}\text{.664}\end{array}$

And

$\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{1.5\text{bar}}{38.7\text{bar}}\text{=0}\text{.0387}\end{array}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.0387}{0.664}\\ \beta =0.005\end{array}$

$\alpha {\text{(T}}_r)=T_r{}^{1/2}=\text{0}{\text{.664}}^{-0.5}=1.2272$

$\begin{array}{l}q=\frac{\phi \alpha (T_r)}{\Omega T_r}=\frac{0.42748\times 1.2272}{0.08664\times \text{0}\text{.664}}\\ q=9.119\end{array}$

For Boron trichloride, put values in equation (1)

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

$Z=1+0.005-9.119\times 0.005\times \frac{Z-0.005}{(Z+0\times 0.005)(Z+1\times 0.005)}$

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

$Z=0.958$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.958\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 300\text{K}}{1.5\text{bar}}\\ V=15930.4\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for Boron trichloridemust calculate at $300\text{K}$ 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 Boron trichloridecan be found using formula:

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

Answer to Problem 3.48P

From Soave/Redlich/Kwong equations, the molar volume of Boron trichlorideis:

$V=15913.76\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.957$ 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:

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

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=1.5\text{bar}$

$T\text{=}300\text{K}$

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

$T_r=\text{0}\text{.664}$ and $P_r=\text{0}\text{.0387}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.0387}{0.664}\\ \beta =0.005\end{array}$

$\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=1.2403\end{array}$

$\begin{array}{l}q=\frac{\phi \alpha (T_r)}{\Omega T_r}=\frac{0.42748\times 1.2403}{0.08664\times 0.664}\\ q=9.22\end{array}$

For Boron trichloride, put values in equation (1)

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

$Z=1+0.005-9.22\times 0.005\times \frac{Z-0.005}{(Z+0\times 0.005)(Z+1\times 0.005)}$

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

$Z=0.957$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.957\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 300\text{K}}{1.5\text{bar}}\\ V=15913.76\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for Boron trichloridemust calculate at $300\text{K}$ temperature and $1.5\text{bar}$ 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 Boron trichloridecan be found using formula:

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

Answer to Problem 3.48P

From Soave/Redlich/Kwong equations, the molar volume of Boron trichlorideis:

$V=15880.504\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.955$ 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:

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

$\text{gas constant R in CGS unit is }83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}$, $P=1.5\text{bar}$

$T\text{=}300\text{K}$

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

$T_r=\text{0}\text{.664}$ and $P_r=\text{0}\text{.0387}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.07779\times \frac{0.0387}{\text{0}\text{.664}}\\ \beta =0.0045\end{array}$

$\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=1.196\end{array}$

$\begin{array}{l}q=\frac{\psi \alpha (T_r)}{\Omega T_r}=\frac{0.45724\times 1.196}{0.07779\times 0.664}\\ q=10.587\end{array}$

For Boron trichloride, put values in equation (1)

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

$Z=1+0.0045-10.587\times 0.0045\times \frac{Z-0.0045}{(Z+(1-\sqrt{2})\times 0.0045)(Z+(1+\sqrt{2})\times 0.0045)}$

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

$Z=0.955$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.955\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 300\text{K}}{1.5\text{bar}}\\ V=15880.504\frac{{\text{cm}}^3}{\text{mol}}\end{array}$