Chapter 13.3, Problem 41P

(a)

To determine

The mass of the gas treating it as an ideal gas mixture.

Answer to Problem 41P

The mass of the gas treating it as an ideal gas mixture is $4.441\times {10}^6\text{lbm}$.

Explanation of Solution

Write the expression to calculate the mole number of the substance $\left(N\right)$.

$N=\frac{m}{M}$ (I)

Here, the mass of the substance is $m$ and the molar mass of the substance is $M$.

Write the expression to calculate the mole fraction of the substance $\left(y\right)$.

$y=\frac{N}{N_m}$ (II)

Here, total mole number of the products is $N_m$.

Write the expression to calculate the individual gas constant $\left(R\right)$.

$R=\frac{R_u}{M}$ (III)

Here, universal gas constant is $R_u$.

Write the expression to calculate the mass of the gas from ideal gas expression.

$m=\frac{PV}{RT}$ (IV)

Here, pressure of the gas is $P$, volume of the gas is $V$, and the temperature is $T$.

Conclusion:

From the Table A-1 of “Molar mass, gas constant, and critical-point properties”, obtain the molar masses as follows:

$\begin{array}{l}{M}_{{\text{CH}}_4}=16.0\text{lbm}/\text{lbmol}\\ M_{{\text{C}}_2{\text{H}}_6}=30.0\text{lbm}/\text{lbmol}\end{array}$

Substitute $75\text{lbm}$ for $m_{{\text{CH}}_4}$ and $16.0\text{lbm}/\text{lbmol}$ for $M_{{\text{CH}}_4}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{CH}}_4}=\frac{75\text{lbm}}{16.0\text{lbm}/\text{lbmol}}\\ =4.6875\text{lbmol}\end{array}$

Substitute $25\text{lbm}$ for $m_{{\text{C}}_2{\text{H}}_6}$ and $30.0\text{lbm}/\text{lbmol}$ for $M_{{\text{C}}_2{\text{H}}_6}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{C}}_2{\text{H}}_6}=\frac{25\text{lbm}}{30.0\text{lbm}/\text{lbmol}}\\ =0.8333\text{lbmol}\end{array}$

Calculate the total number of moles $\left(N_m\right)$.

$\begin{array}{c}{N}_m=N_{{\text{CH}}_4}+N_{{\text{C}}_2{\text{H}}_6}\\ =4.6875\text{lbmol}+0.8333\text{lbmol}\\ =5.5208\text{lbmol}\end{array}$

Substitute $4.6875\text{lbmol}$ for $N_{{\text{CH}}_4}$ and $5.5208\text{lbmol}$ for $N_m$ in Equation (II).

$\begin{array}{c}{y}_{{\text{CH}}_4}=\frac{4.6875\text{lbmol}}{5.5208\text{lbmol}}\\ =0.8491\end{array}$

Substitute $0.8333\text{lbmol}$ for $N_{{\text{C}}_2{\text{H}}_6}$ and $5.5208\text{lbmol}$ for $N_m$ in Equation (II).

$\begin{array}{c}{y}_{{\text{C}}_2{\text{H}}_6}=\frac{0.8333\text{lbmol}}{5.5208\text{lbmol}}\\ =0.1509\end{array}$

Substitute $100\text{ lbm}$ for $m_m$ and $5.5208\text{lbmol}$ for $N_m$ in Equation (I).

$\begin{array}{c}{M}_m=\frac{100\text{ lbm}}{5.5208\text{lbmol}}\\ =18.11\text{lbm}/\text{lbmol}\end{array}$

Substitute $10.73\text{psia}\cdot {\text{ft}}^3/\text{lbmol}\cdot \text{R}$ for $R_u$ and $18.11\text{lbm}/\text{lbmol}$ for $M_m$ in Equation (III).

$\begin{array}{c}R=\frac{10.73\text{psia}\cdot {\text{ft}}^3/\text{lbmol}\cdot \text{R}}{18.11\text{lbm}/\text{lbmol}}\\ =0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\end{array}$

Substitute $2000\text{ psia}$ for $P$, $\left(1\times {10}^6{\text{ ft}}^3\right)$ for $V$, $0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, and $760\text{ R}$ for $T$ in Equation (IV).

$\begin{array}{c}m=\frac{2000\text{ psia}\times \left(1\times {10}^6{\text{ ft}}^3\right)}{0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times \left(760\text{ R}\right)}\\ =4.441\times {10}^6\text{lbm}\end{array}$

Thus, the mass of the gas treating it as an ideal gas mixture is $4.441\times {10}^6\text{lbm}$.

(b)

To determine

The mass of the gas using the compressibility chart.

Answer to Problem 41P

The mass of the gas using the compressibility chart is $4.684\times {10}^6\text{lbm}$.

Explanation of Solution

Write the expression to calculate the reduced pressure $\left(P_R\right)$.

$P_R=\frac{P}{P_{\text{cr}}}$ (V)

Here, pressure at the critical point is $P_{\text{cr}}$.

Write the expression to calculate the reduced temperature $\left(T_R\right)$.

$T_R=\frac{T}{T_{\text{cr}}}$ (VI)

Here, temperature at the critical point is $T_{\text{cr}}$.

Write the expression to calculate the compressibility factor of the mixture $\left(Z_m\right)$.

$Z_m=y_{{\text{CH}}_4}{Z}_{{\text{CH}}_4}+y_{{\text{C}}_2{\text{H}}_6}{Z}_{{\text{C}}_2{\text{H}}_6}$ (VII)

Here, compressibility factor for ${\text{CH}}_4$ is $Z_{{\text{CH}}_4}$ and compressibility factor for ${\text{C}}_2{\text{H}}_6$ is $Z_{{\text{C}}_2{\text{H}}_6}$.

Write the expression to calculate the mass from using the compressibility factor.

$m=\frac{PV}{Z_{m}RT}$ (VIII)

Conclusion:

Substitute $760\text{R}$ for $T_m$ and $343.9\text{R}$ for $T_{{\text{cr,CH}}_4}$ in Equation (VI).

$\begin{array}{c}{T}_{R,{\text{CH}}_4}=\frac{760\text{R}}{343.9\text{R}}\\ =2.210\end{array}$

Substitute $2000\text{psia}$ for $P_{{\text{CH}}_4}$ and $673\text{psia}$ for $P_{{\text{cr,CH}}_4}$ in Equation (V).

$\begin{array}{c}{P}_{R,{\text{CH}}_4}=\frac{2000\text{psia}}{673\text{psia}}\\ =2.972\end{array}$

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced pressure of 2.972 and reduced temperature of 2.210 as 0.98

Substitute $760\text{R}$ for $T_m$ and $549.8\text{R}$ for $T_{{\text{cr,C}}_2{\text{H}}_6}$ in Equation (VI).

$\begin{array}{c}{T}_{R,{\text{C}}_2{\text{H}}_6}=\frac{760\text{R}}{549.8\text{R}}\\ =1.382\end{array}$

Substitute $1500\text{psia}$ for $P_{{\text{C}}_2{\text{H}}_6}$ and $708\text{psia}$ for $P_{{\text{cr,C}}_2{\text{H}}_6}$ in Equation (V).

$\begin{array}{c}{P}_{R,{\text{C}}_2{\text{H}}_6}=\frac{1500\text{psia}}{708\text{psia}}\\ =2.119\end{array}$

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced pressure of 2.119 and reduced temperature of 1.382 as 0.77

Substitute $0.8491$ for $y_{{\text{CH}}_4}$, 0.98 for $Z_{{\text{CH}}_4}$, 0.1509 for $y_{{\text{C}}_2{\text{H}}_6}$, and 0.77 for $Z_{{\text{C}}_2{\text{H}}_6}$ in Equation (VII).

$\begin{array}{c}{Z}_m=0.8491\left(0.98\right)+0.1509\left(0.77\right)\\ =0.9483\end{array}$

Substitute $2000\text{ psia}$ for $P$, $\left(1\times {10}^6{\text{ ft}}^3\right)$ for $V$, 0.9483 for $Z_m$, $0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, and $760\text{ R}$ for $T$ in Equation (VIII).

$\begin{array}{c}m=\frac{2000\text{ psia}\times \left(1\times {10}^6{\text{ ft}}^3\right)}{0.9483\times 0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times \left(760\text{ R}\right)}\\ =4.684\times {10}^6\text{lbm}\end{array}$

Thus, the mass of the gas using the compressibility chart is $4.684\times {10}^6\text{lbm}$.

(c)

To determine

The mass of the gas using Dalton’ law.

Answer to Problem 41P

The mass of the gas using the Daltonls law is $4.575\times {10}^6\text{lbm}$.

Explanation of Solution

Write the expression to calculate the reduced volume specific volume $\left(v_R\right)$.

$v_R=\frac{V/\text{m}}{R{T}_{\text{cr}}/P_{\text{cr}}}$ (IX)

Conclusion:

From the Table A-2E of “Ideal gas specific heats of various common gases”, obtain the gas constants as follows:

$\begin{array}{l}{R}_{{\text{CH}}_4}=0.6688\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\\ R_{{\text{C}}_2{\text{H}}_6}=0.3574\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\end{array}$

Substitute $\left(1\times {10}^6{\text{ ft}}^3\right)$ for $V$, $4.441\times {10}^6\text{lbm}$ for $m$, $0.6688\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R_{{\text{CH}}_4}$, $343.9\text{R}$ for $T_{{\text{cr,CH}}_4}$, and $673\text{psia}$ for $P_{{\text{cr,CH}}_4}$ in Equation (IX).

$\begin{array}{c}{v}_{R,{\text{CH}}_4}=\frac{\left(1\times {10}^6{\text{ ft}}^3\right)/4.441\times {10}^6\text{lbm}}{0.6688\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times 343.9\text{R}/673\text{psia}}\\ =0.8782\end{array}$

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced specific volume of 0.8782 and reduced temperature of 2.210 as 0.98

Substitute $\left(1\times {10}^6{\text{ ft}}^3\right)$ for $V$, $4.441\times {10}^6\text{lbm}$ for $m$, $0.3574\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R_{{\text{C}}_2{\text{H}}_6}$, $549.8\text{R}$ for $T_{{\text{cr,C}}_2{\text{H}}_6}$, and $708\text{psia}$ for $P_{{\text{cr,C}}_2{\text{H}}_6}$ in Equation (IX).

$\begin{array}{c}{v}_{R,{\text{C}}_2{\text{H}}_6}=\frac{\left(1\times {10}^6{\text{ ft}}^3\right)/4.441\times {10}^6\text{lbm}}{0.3574\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times 549.8\text{R}/708\text{psia}}\\ =3.244\end{array}$

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced specific volume of 3.244 and reduced temperature of 1.382 as 0.92

Substitute $0.8491$ for $y_{{\text{CH}}_4}$, 0.98 for $Z_{{\text{CH}}_4}$, 0.1509 for $y_{{\text{C}}_2{\text{H}}_6}$, and 0.92 for $Z_{{\text{C}}_2{\text{H}}_6}$ in Equation (VII).

$\begin{array}{c}{Z}_m=0.8491\left(0.98\right)+0.1509\left(0.92\right)\\ =0.9709\end{array}$

Substitute $2000\text{ psia}$ for $P$, $\left(1\times {10}^6{\text{ ft}}^3\right)$ for $V$, 0.9709 for $Z_m$, $0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, and $760\text{ R}$ for $T$ in Equation (VIII).

$\begin{array}{c}m=\frac{2000\text{ psia}\times \left(1\times {10}^6{\text{ ft}}^3\right)}{0.9709\times 0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times \left(760\text{ R}\right)}\\ =4.575\times {10}^6\text{lbm}\end{array}$

Thus, the mass of the gas using the Daltonls law is $4.575\times {10}^6\text{lbm}$.

(d)

To determine

The mass of the gas using Kay’s rule.

Answer to Problem 41P

The mass of the gas using Kay’s rule is $4.579\times {10}^6\text{lbm}$.

Explanation of Solution

Write the formula to calculate the pseudo critical temperature $\left(T_{\text{cr,}m}\right)$.

$T_{\text{cr,}m}=y_{{\text{CH}}_4}{T}_{{\text{cr,CH}}_4}+y_{{\text{C}}_2{\text{H}}_6}{T}_{{\text{cr,C}}_2{\text{H}}_6}$ (X)

Write the formula to calculate the pseudo critical pressure $\left(P_{\text{cr,}m}\right)$.

$P_{\text{cr,}m}=y_{{\text{CH}}_4}{P}_{{\text{cr,CH}}_4}+y_{{\text{C}}_2{\text{H}}_6}{P}_{{\text{cr,C}}_2{\text{H}}_6}$ (XI)

Conclusion:

Substitute $0.8491$ for $y_{{\text{CH}}_4}$, 0.1509 for $y_{{\text{C}}_2{\text{H}}_6}$, $343.9\text{R}$ for $T_{{\text{cr,CH}}_4}$, and $549.8\text{R}$ for $T_{{\text{cr,C}}_2{\text{H}}_6}$ in Equation (X).

$\begin{array}{c}{T}_{\text{cr,}m}=0.8491\left(343.9\text{R}\right)+0.1509\left(549.8\text{R}\right)\\ =375\text{R}\end{array}$

Substitute $0.8491$ for $y_{{\text{CH}}_4}$, 0.1509 for $y_{{\text{C}}_2{\text{H}}_6}$, $673\text{psia}$ for $P_{{\text{cr,CH}}_4}$, and $708\text{psia}$ for $P_{{\text{cr,C}}_2{\text{H}}_6}$ in Equation (XI).

$\begin{array}{c}{P}_{\text{cr,}m}=0.8491\left(673\text{psia}\right)+0.1509\left(708\text{psia}\right)\\ =678.3\text{psia}\end{array}$

Substitute $760\text{R}$ for $T_m$ and $375\text{R}$ for $T_{\text{cr,}m}$ in Equation (VI).

$\begin{array}{c}{T}_R=\frac{760\text{R}}{375\text{R}}\\ =2.027\end{array}$

Substitute $2000\text{psia}$ for $P_{{\text{CH}}_4}$ and $678.3\text{psia}$ for $P_{\text{cr,}m}$ in Equation (V).

$\begin{array}{c}{P}_R=\frac{2000\text{psia}}{678.3\text{psia}}\\ =2.949\end{array}$

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced pressure of 2.949 and reduced temperature of 2.027 as 0.97

Substitute $2000\text{ psia}$ for $P$, $\left(1\times {10}^6{\text{ ft}}^3\right)$ for $V$, 0.97 for $Z_m$, $0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for $R$, and $760\text{ R}$ for $T$ in Equation (VIII).

$\begin{array}{c}m=\frac{2000\text{ psia}\times \left(1\times {10}^6{\text{ ft}}^3\right)}{0.97\times 0.5925\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\times \left(760\text{ R}\right)}\\ =4.579\times {10}^6\text{lbm}\end{array}$

Thus, the mass of the gas using Kay’s rule is $4.579\times {10}^6\text{lbm}$.