Chapter 13.3, Problem 44P

(a)

To determine

The volume of the mixture using the ideal gas mixture.

Answer to Problem 44P

The volume of the mixture using the ideal gas mixture is $\underset{\_}{0.8{\text{m}}^{\text{3}}}$.

Explanation of Solution

Write the equation to calculate the mole number of oxygen and nitrogen gas using an ideal gas equation of state.

$N_{{\text{O}}_{\text{2}}}={\left(\frac{PV}{R_{u}T}\right)}_{{\text{O}}_{\text{2}}}$ (I)

$N_{{\text{N}}_2}={\left(\frac{PV}{R_{u}T}\right)}_{{\text{N}}_2}$ (II)

Here, pressure and temperature of oxygen and nitrogen gas is $P_{{\text{O}}_{\text{2}}},P_{{\text{N}}_2}$, $T_{{\text{O}}_{\text{2}}},T_{{\text{N}}_2}$, volume of oxygen and nitrogen gas is $V_{{\text{O}}_{\text{2}}},V_{{\text{N}}_2}$, universal gas constant of oxygen and nitrogen gas is $R_{u,{\text{O}}_{\text{2}}}\text{and}{R}_{u,{\text{N}}_2}$ respectively.

Calculate the mole number of the mixture.

$N_m=N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_2}$ (III)

Calculate the volume of the mixture.

$V_m=\frac{N_m{R}_u{T}_m}{P_m}$ (IV)

Here, pressure and temperature of the mixture is $P_m$ and $T_m$.

Conclusion:

Substitute $8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}$ for $R_u$, 200 K for $T_{{\text{O}}_{\text{2}}}$, $0.3{\text{m}}^{\text{3}}$ for $V_{{\text{O}}_{\text{2}}}$, and 8 MPa for $P_{{\text{O}}_{\text{2}}}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{\left(8\text{MPa}\right)\left(0.3{\text{m}}^{\text{3}}\right)}{\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =\frac{\left(8\text{MPa}\times \text{1000}\frac{\text{kPa}}{1\text{MPa}}\right)\left(0.3{\text{m}}^{\text{3}}\right)}{\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =1.443\text{kmol}\end{array}$

Substitute $8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}$ for $R_u$, 200 K for $T_{{\text{N}}_2}$, $0.5{\text{m}}^{\text{3}}$ for $V_{{\text{N}}_2}$, and 8 MPa for $P_{{\text{N}}_2}$ in Equation (II).

$\begin{array}{c}{N}_{{\text{N}}_2}=\frac{\left(8\text{MPa}\right)\left(0.5{\text{m}}^{\text{3}}\right)}{\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =\frac{\left(8\text{MPa}\times \text{1000}\frac{\text{kPa}}{1\text{MPa}}\right)\left(0.5{\text{m}}^{\text{3}}\right)}{\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =2.406\text{kmol}\end{array}$

Substitute 1.443 kmol for $N_{{\text{O}}_{\text{2}}}$ and 2.406 kmol for $N_{{\text{N}}_2}$ in Equation (III).

$\begin{array}{c}{N}_m=1.443\text{kmol}+2.406\text{kmol}\\ =3.849\text{kmol}\end{array}$

Substitute 3.849 kmol for $N_m$, $8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}$ for $R_u$, 200 K for $T_m$, and 8 MPa for $P_m$ in Equation (IV).

$\begin{array}{c}{V}_m=\frac{\left(3.849\text{kmol}\right)\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)\left(200\text{K}\right)}{8\text{MPa}}\\ =\frac{\left(3.849\text{kmol}\right)\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)\left(200\text{K}\right)}{8\text{MPa}\times \text{1000}\frac{\text{kPa}}{1\text{MPa}}}\\ =0.8{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of the mixture using the ideal gas mixture is $\underset{\_}{0.8{\text{m}}^{\text{3}}}$.

(b)

To determine

The volume of the mixture using Kay’s rule.

Answer to Problem 44P

The volume of the mixture using Kay’s rule is $\underset{\_}{0.79{\text{m}}^{\text{3}}}$.

Explanation of Solution

Write the equation of pseudo-reduced temperatures and pressures of ${\text{O}}_2\text{,}\text{and}{\text{N}}_{\text{2}}$.

$T_{R,{\text{O}}_{\text{2}}}=\frac{T_m}{T_{\text{cr},{\text{O}}_{\text{2}}}}$ (V)

$P_{R,{\text{O}}_{\text{2}}}=\frac{P_m}{P_{\text{cr},{\text{O}}_{\text{2}}}}$ (VI)

$T_{R,{\text{N}}_2}=\frac{T_m}{T_{\text{cr},{\text{N}}_2}}$ (VII)

$P_{R,{\text{N}}_2}=\frac{P_m}{P_{\text{cr},{\text{N}}_2}}$ (VIII)

Here, the critical temperature of ${\text{O}}_{\text{2}}$ is $T_{\text{cr},{\text{O}}_{\text{2}}}$, mixture temperature is $T_m$, mixture pressure is $P_m$, critical temperature of ${\text{N}}_2$ is $T_{\text{cr},{\text{N}}_2}$, critical pressure of ${\text{N}}_2$ is $P_{\text{cr},{\text{N}}_2}$, and critical pressure of ${\text{O}}_{\text{2}}$ is $P_{\text{cr},{\text{O}}_{\text{2}}}$.

Write the equation to calculate the mole number of oxygen and nitrogen gas using the kay’s rule.

$N_{{\text{O}}_{\text{2}}}={\left(\frac{PV}{Z{R}_{u}T}\right)}_{{\text{O}}_{\text{2}}}$ (IX)

$N_{{\text{N}}_2}={\left(\frac{PV}{Z{R}_{u}T}\right)}_{{\text{N}}_2}$ (X)

Here, compressibility factor of ${\text{O}}_{\text{2}}\text{and}{\text{N}}_2$ are $Z_{{\text{O}}_{\text{2}}}$ and $Z_{{\text{N}}_2}$.

Calculate the mole fraction of ${\text{O}}_{\text{2}}\text{and}{\text{N}}_2$.

$y_{{\text{O}}_{\text{2}}}=\frac{N_{{\text{O}}_{\text{2}}}}{N_m}$ (XI)

$y_{{\text{N}}_2}=\frac{N_{{\text{N}}_2}}{N_m}$ (XII)

Write the critical temperature of a gas mixture.

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{T}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{T}_{\text{cr},{\text{O}}_{\text{2}}}+y_{{\text{N}}_2}{T}_{\text{cr},{\text{N}}_2}\end{array}$ (XIII)

Write the critical pressure of a gas mixture.

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{P}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{P}_{\text{cr},{\text{O}}_{\text{2}}}+y_{{\text{N}}_2}{P}_{\text{cr},{\text{N}}_2}\end{array}$ (XIV)

Write the equation of reduced temperature and pressure.

$T_R=\frac{T_m}{{T^{\prime }}_{\text{cr},m}}$ (XV)

$P_R=\frac{P_m}{{P^{\prime }}_{\text{cr},m}}$ (XVI)

Calculate the volume of the mixture.

$V_m=\frac{Z_m{N}_m{R}_u{T}_m}{P_m}$ (XVII)

Conclusion:

Refer to Table A-1, obtain the critical temperatures and pressures of ${\text{O}}_2\text{,}\text{and}{\text{N}}_{\text{2}}$.

$\begin{array}{l}{T}_{\text{cr},{\text{O}}_{\text{2}}}=154.8\text{K}\\ P_{\text{cr},{\text{O}}_{\text{2}}}=5.08\text{MPa}\\ T_{\text{cr},{\text{N}}_2}=126.2\text{K}\\ P_{\text{cr},{\text{N}}_2}=3.39\text{MPa}\end{array}$

Substitute 200 K for $T_m$, 8 MPa for $P_m$ and obtained values from Table A-1 in Equations (V) to (VIII).

$\begin{array}{l}{T}_{R,{\text{O}}_{\text{2}}}=1.292\\ P_{R,{\text{O}}_{\text{2}}}=1.575\\ T_{R,{\text{N}}_2}=1.585\\ P_{R,{\text{N}}_2}=2.360\end{array}$

Refer to Figure A-15, obtain the compressibility factor for ${\text{O}}_2\text{,}\text{and}{\text{N}}_{\text{2}}$ by reading the above values of reduced temperatures and reduced pressure.

$\begin{array}{l}{Z}_{{\text{O}}_{\text{2}}}=0.77\\ Z_{{\text{N}}_2}=0.863\end{array}$

Substitute 0.77 for $Z_{{\text{O}}_{\text{2}}}$, $8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}$ for $R_u$, 200 K for $T_{{\text{O}}_{\text{2}}}$, $0.3{\text{m}}^{\text{3}}$ for $V_{{\text{O}}_{\text{2}}}$, and 8 MPa for $P_{{\text{O}}_{\text{2}}}$ in Equation (IX).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{\left(8\text{MPa}\right)\left(0.3{\text{m}}^{\text{3}}\right)}{0.77\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =\frac{\left(8\text{MPa}\times \text{1000}\frac{\text{kPa}}{1\text{MPa}}\right)\left(0.3{\text{m}}^{\text{3}}\right)}{0.77\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =1.874\text{kmol}\end{array}$

Substitute 0.863 for $Z_{{\text{N}}_2}$, $8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}$ for $R_u$, 200 K for $T_{{\text{N}}_2}$, $0.5{\text{m}}^{\text{3}}$ for $V_{{\text{N}}_2}$, and 8 MPa for $P_{{\text{N}}_2}$ in Equation (X).

$\begin{array}{c}{N}_{{\text{N}}_2}=\frac{\left(8\text{MPa}\right)\left(0.5{\text{m}}^{\text{3}}\right)}{0.863\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =\frac{\left(8\text{MPa}\times \text{1000}\frac{\text{kPa}}{1\text{MPa}}\right)\left(0.5{\text{m}}^{\text{3}}\right)}{0.863\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)200\text{K}}\\ =2.787\text{kmol}\end{array}$

Substitute 1.874 kmol for $N_{{\text{O}}_{\text{2}}}$ and 2.787 kmol for $N_{{\text{N}}_2}$ in Equation (III).

$\begin{array}{c}{N}_m=1.874\text{kmol}+2.787\text{kmol}\\ =4.661\text{kmol}\end{array}$

Substitute 1.874 kmol for $N_{{\text{O}}_{\text{2}}}$, 4.661 kmol for $N_m$, and 2.787 kmol for $N_{{\text{N}}_2}$ in Equations (XI) and (XII).

$\begin{array}{l}{y}_{{\text{O}}_{\text{2}}}=0.402\\ y_{{\text{N}}_2}=0.598\end{array}$

Substitute 0.402 for $y_{{\text{O}}_{\text{2}}}$, 154.8 K for $T_{\text{cr},}{}_{{\text{O}}_{\text{2}}}$, 0.598 for $y_{{\text{N}}_2}$, and 126.2 K for $T_{\text{cr},}{}_{{\text{N}}_2}$ in Equation (XIII).

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}=\left(0.402\right)\left(154.8\text{K}\right)+\left(0.598\right)\left(126.2\text{K}\right)\\ =137.7\text{K}\end{array}$

Substitute 0.402 for $y_{{\text{N}}_2}$, 5.08 MPa for $P_{\text{cr},}{}_{{\text{O}}_{\text{2}}}$, 0.598 for $y_{{\text{N}}_2}$, and 3.39 MPa for $P_{\text{cr},}{}_{{\text{N}}_2}$ in Equation (XIV).

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}=\left(0.402\right)\left(5.08\text{MPa}\right)+\left(0.598\right)\left(3.39\text{MPa}\right)\\ =4.07\text{MPa}\end{array}$

Substitute 200 K for $T_m$, 8000 kPa for $P_m$, 137.7 K for ${T^{\prime }}_{\text{cr},m}$, and 4.07 MPa for ${P^{\prime }}_{\text{cr},m}$ in Equations (XV) and (XVI).

$\begin{array}{l}{T}_R=1.452\\ P_R=1.966\end{array}$

Refer to Figure A-15, obtain the compressibility factor of the mixture by reading the pseudo reduced temperature and pseudo reduced pressure of 1.452 and 1.966.

$Z_m=0.82$

Substitute 0.82 for $Z_m$, 4.661 kmol for $N_m$, $8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}$ for $R_u$, 200 K for $T_m$, and 8 MPa for P in Equation (XVII).

$\begin{array}{c}{V}_m=\frac{\left(0.82\right)\left(4.661\text{kmol}\right)\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)\left(200\text{K}\right)}{\left(8\text{MPa}\right)}\\ =\frac{\left(0.82\right)\left(4.661\text{kmol}\right)\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)\left(200\text{K}\right)}{\left(8\text{MPa}\times \text{1000}\frac{\text{kPa}}{1\text{MPa}}\right)}\\ =0.79{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of the mixture using Kay’s rule is $\underset{\_}{0.79{\text{m}}^{\text{3}}}$.

(c)

To determine

The volume of the mixture using the compressibility chart and Amagat’s law.

Answer to Problem 44P

The volume of the mixture using the compressibility chart and Amagat’s law is $\underset{\_}{0.80{\text{m}}^{\text{3}}}$.

Explanation of Solution

Write the equation of compressibility factor of the mixture.

$\begin{array}{c}{Z}_m={\displaystyle \sum y_i{Z}_i}\\ =y_{{\text{O}}_{\text{2}}}{Z}_{{\text{O}}_{\text{2}}}+y_{{\text{N}}_2}{Z}_{{\text{N}}_2}\end{array}$ (XVIII)

Here, the mole fraction of ${\text{O}}_2\text{and}{\text{N}}_{\text{2}}$ are $y_{{\text{O}}_{\text{2}}}\text{and}{y}_{{\text{N}}_{\text{2}}}$ respectively.

Calculate the volume of the mixture.

$V_m=\frac{Z_m{N}_m{R}_u{T}_m}{P_m}$ (XIX)

Conclusion:

Substitute 0.402 for $y_{{\text{O}}_{\text{2}}}$, 0.77 for $Z_{{\text{O}}_{\text{2}}}$, 0.598 for $y_{{\text{N}}_2}$, and 0.863 for $Z_{{\text{N}}_2}$, in Equation (XVIII).

$\begin{array}{c}{Z}_m=\left(0.402\right)\left(0.77\right)+\left(0.598\right)\left(0.863\right)\\ =0.83\end{array}$

Substitute 0.863 for $Z_m$, 4.661 kmol for $N_m$, $8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}$ for $R_u$, 200 K for $T_m$, and 8 MPa for P in Equation (XIX).

$\begin{array}{c}{V}_m=\frac{\left(0.863\right)\left(4.661\text{kmol}\right)\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)\left(200\text{K}\right)}{\left(8\text{MPa}\right)}\\ =\frac{\left(0.863\right)\left(4.661\text{kmol}\right)\left(8.314\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kmol}\cdot \text{K}}\right)\left(200\text{K}\right)}{\left(8\text{MPa}\times \text{1000}\frac{\text{kPa}}{1\text{MPa}}\right)}\\ =0.80{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of the mixture using the compressibility chart and Amagat’s law is $\underset{\_}{0.80{\text{m}}^{\text{3}}}$.