Chapter 3, Problem 3.47P

(a)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ammonia must calculate at $320\text{K}$ 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 3.47P

The $Z\text{and }V$ value for ammonia are $V=1514.07\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.865$ respectively.

Explanation of Solution

Given Information:

The virial equation is given as

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

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

$P=15\text{bar}$

$T\text{=}320\text{K}$

For ammonia, the second and third virial coefficients are: $B=-\text{208}{\text{cm}}^{\text{3}}{\text{.mol}}^{\text{-1}}$ and $C=4378{\text{cm}}^{\text{6}}{\text{.mol}}^{\text{-2}}$ .

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

$\begin{array}{l}\frac{15\text{bar}\times V}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times 320\text{K}}=1+\frac{-\text{208}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}{V}+\frac{4378\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ 5.71\times {10}^{-4}V\frac{{\text{cm}}^3}{\text{mol}}=\frac{V^2-\text{208}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}V+4378\frac{{\text{cm}}^{\text{6}}}{{\text{mol}}^{\text{2}}}}{V^2}\\ \\ 5.71\times {10}^{-4}{V}^3=V^2-208V+4378\end{array}$

On solving the equation

$V=1514.07\frac{{\text{cm}}^3}{\text{mol}}$

So,

$\begin{array}{l}Z=\frac{PV}{RT}=\frac{15\text{bar}\times 1514.07\frac{{\text{cm}}^3}{\text{mol}}}{\text{82}\text{.05746}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{320}\text{K}}\\ Z=0.865\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ammonia must calculate at $320\text{K}$ 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 3.47P

The $Z\text{and }V$ value for ammonia are $V=1589.687\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.896$ 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 ammonia at given temperature and pressure, the critical conditions are given in table B.1 in Appendix B as,

$T_C=405.7\text{K}\text{and }{P}_C=112.8\text{bar, }\omega \text{=0}\text{.253}$

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

$P=15\text{bar}$

$T\text{=}320\text{K}$

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

$\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{320\text{K}}{405.7\text{K}}\text{=0}\text{.789}\end{array}$

And

$\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{15\text{bar}}{112.8\text{bar}}\text{=0}\text{.133}\end{array}$

So,

$\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ B^0=0.083-\frac{0.422}{\text{0}{\text{.789}}^{1.6}}\\ B^0=-0.533\end{array}$

and

$\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ B^1=0.139-\frac{0.172}{\text{0}{\text{.789}}^{4.2}}\\ B^1=-0.326\end{array}$

Hence, value of $V$ is

$\begin{array}{l}V=\frac{RT}{P}+\frac{R{T}_C}{P_C}\left[B^0+\omega B^1\right]\\ V=\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 320\text{K}}{15\text{bar}}+\frac{83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 405.7\text{K}}{112.8\text{bar}}\left[-0.533-0.253\times 0.326\right]\\ V=1589.687\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

And value of compressibility factor,

$\begin{array}{l}Z=\frac{PV}{RT}=\frac{15\text{bar}\times 1589.687\frac{{\text{cm}}^3}{\text{mol}}}{\text{83}\text{.144}\frac{{\text{cm}}^3\text{bar}}{\text{mol}\text{K}}\times \text{320}\text{K}}\\ Z=0.896\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ammonia must calculate at $320\text{K}$ 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 ammoniacan be found using formula:

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

Answer to Problem 3.47P

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

$V=1605.23\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.905$ 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:

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

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

$T\text{=320}\text{K}$

$\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{320\text{K}}{405.7\text{K}}\text{=0}\text{.789}\end{array}$

And

$\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{15\text{bar}}{112.8\text{bar}}\text{=0}\text{.133}\end{array}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.133}{0.789}\\ \beta =0.0146\end{array}$

$\alpha {\text{(T}}_r)=T_r{}^{1/2}={0.789}^{-0.5}=1.1258$

$\begin{array}{l}q=\frac{\phi \alpha (T_r)}{\Omega T_r}=\frac{0.42748\times 1.1258}{0.08664\times 0.789}\\ q=7.04\end{array}$

For ammonia, put values in equation (1)

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

$Z=1+0.0146-7.04\times 0.0146\times \frac{Z-0.0146}{(Z+0\times 0.0146)(Z+1\times 0.0146)}$

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

$Z=0.905$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.905\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 320\text{K}}{15\text{bar}}\\ V=1605.23\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ammonia must calculate at $320\text{K}$ 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 ammonia can be found using formula:

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

Answer to Problem 3.47P

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

$V=1589.27\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.896$ 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:

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

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

$T\text{=320}\text{K}$

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

$T_r=\text{0}\text{.789}$ and $P_r=\text{0}\text{.133}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.08664\times \frac{0.133}{0.789}\\ \beta =0.0146\end{array}$

$\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=1.203\end{array}$

$\begin{array}{l}q=\frac{\phi \alpha (T_r)}{\Omega T_r}=\frac{0.42748\times 1.203}{0.08664\times 0.789}\\ q=7.523\end{array}$

For ammonia, put values in equation (1)

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

$Z=1+0.0146-7.523\times 0.0146\times \frac{Z-0.0146}{(Z+0\times 0.0146)(Z+1\times 0.0146)}$

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

$Z=0.896$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.896\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 320\text{K}}{15\text{bar}}\\ V=1589.27\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The $Z\text{and }V$ value for ammonia must calculate at $320\text{K}$ 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 ammonia can be found using formula:

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

Answer to Problem 3.47P

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

$V=1578.63\frac{{\text{cm}}^3}{\text{mol}}$ and $Z=0.89$ 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:

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

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

$T\text{=320}\text{K}$

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

$T_r=\text{0}\text{.789}$ and $P_r=\text{0}\text{.133}$

So,

$\begin{array}{l}\beta =\Omega \frac{P_r}{T_r}=0.07779\times \frac{0.133}{\text{0}\text{.789}}\\ \beta =0.0131\end{array}$

$\begin{array}{l}\alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)={[}^1\\ \alpha {\text{(T}}_r)=1.174\end{array}$

$\begin{array}{l}q=\frac{\psi \alpha (T_r)}{\Omega T_r}=\frac{0.45724\times 1.174}{0.07779\times 0.789}\\ q=8.746\end{array}$

For ammonia, put values in equation (1)

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

$Z=1+0.0131-8.746\times 0.0131\times \frac{Z-0.0131}{(Z+(1-\sqrt{2})\times 0.0131)(Z+(1+\sqrt{2})\times 0.0131)}$

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

$Z=0.89$

Hence,

$\begin{array}{l}V=\frac{ZRT}{P}=\frac{0.89\times 83.144\frac{{\text{cm}}^3\text{bar}}{\text{K}\text{mol}}\times 320\text{K}}{15\text{bar}}\\ V=1578.63\frac{{\text{cm}}^3}{\text{mol}}\end{array}$