Chapter 13.3, Problem 32P

The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane. Determine the mole fractions of each constituent, the mixture’s apparent molecular weight, the partial pressure of each constituent when the mixture pressure is 1200 kPa, and the apparent specific heats of the mixture when the mixture is at the room temperature.

To determine

The mole fraction of ${\text{N}}_{\text{2}}$.

The mole fraction of $\text{He}$.

The mole fraction of ${\text{CH}}_{\text{4}}$.

The mole fraction of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

The apparent molecular weight of the mixture.

The partial pressure of ${\text{N}}_{\text{2}}$.

The partial pressure of $\text{He}$.

The partial pressure of ${\text{CH}}_{\text{4}}$.

The partial pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

The constant-pressure specific heat of the mixture, $c_p$.

The constant-volume specific heat, $c_v$.

Answer to Problem 32P

The mole fraction of ${\text{N}}_{\text{2}}$ is $\underset{\_}{0.08637}$.

The mole fraction of $\text{He}$ is $\underset{\_}{0.2015}$.

The mole fraction of ${\text{CH}}_{\text{4}}$ is $\underset{\_}{\text{0}\text{.6046}}$.

The mole fraction of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $\underset{\_}{0.1075}$.

The apparent molecular weight of the mixture is $\underset{\_}{16.12\text{kg/kmol}}$.

The partial pressure of ${\text{N}}_{\text{2}}$ is $\underset{\_}{\text{103}\text{.6}\text{kPa}}$.

The partial pressure of $\text{He}$ is $\underset{\_}{\text{241}\text{.8}\text{kPa}}$.

The partial pressure of ${\text{CH}}_{\text{4}}$ is $\underset{\_}{\text{725}\text{.5}\text{kPa}}$.

The partial pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $\underset{\_}{\text{129}\text{.0}\text{kPa}}$.

The constant-pressure specific heat of the mixture, $c_p$ is $\underset{\_}{2.121\text{kJ/kg}\cdot \text{K}}$.

The constant-volume specific heat, $c_v$ is $\underset{\_}{1.605\text{kJ/kg}\cdot \text{K}}$.

Explanation of Solution

Refer to Table A-1, Obtain the molar masses of ${\text{N}}_{\text{2}},\text{He},{\text{CH}}_{\text{4}}\text{,}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ as below:

$\begin{array}{l}{M}_{N_{\text{2}}}=28.0\text{kg/kmol}\\ M_{\text{He}}=4.0\text{kg/kmol}\\ M_{{\text{CH}}_4}=16.0\text{kg/kmol}\\ M_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}=0.6667\text{kmol}\end{array}$

Refer to Table A-2a, obtain the constant-pressure specific heats of the gases at room temperature.

$\begin{array}{l}{c}_{p,{\text{N}}_{\text{2}}}=1.039\text{kJ/kg}\cdot \text{K}\\ c_{p,\text{He}}=5.1926\text{kJ/kg}\cdot \text{K}\\ c_{p,{\text{CH}}_{\text{4}}}=2.2537\text{kJ/kg}\cdot \text{K}\\ c_{p,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}=1.7662\text{kJ/kg}\cdot \text{K}\end{array}$

Write the mole number of ${\text{N}}_{\text{2}}$.

$N_{{\text{N}}_{\text{2}}}=\frac{m_{{\text{N}}_{\text{2}}}}{M_{{\text{N}}_{\text{2}}}}$ (I)

Here, the mass of nitrogen gas is $m_{{\text{N}}_{\text{2}}}$.

Write the mole number of $\text{He}$.

$N_{\text{He}}=\frac{m_{\text{He}}}{M_{\text{He}}}$ (II)

Here, the mass of helium gas is $m_{{\text{N}}_{\text{2}}}$.

Write the mole number of ${\text{CH}}_4$.

$N_{{\text{CH}}_4}=\frac{m_{{\text{CH}}_4}}{M_{{\text{CH}}_4}}$ (III)

Here, the mass of methane gas is $m_{{\text{CH}}_4}$.

Write the mole number of ${\text{C}}_2{\text{H}}_6$.

$N_{{\text{C}}_2{\text{H}}_6}=\frac{m_{{\text{C}}_2{\text{H}}_6}}{M_{{\text{C}}_2{\text{H}}_6}}$ (IV)

Here, the mass of ethane gas is $m_{{\text{C}}_2{\text{H}}_6}$.

Write the equation to calculate the mole number of the mixture.

$N_m=N_{{\text{N}}_{\text{2}}}+N_{\text{He}}+N_{{\text{CH}}_{\text{4}}}+N_{{\text{C}}_2{\text{H}}_6}$ (V)

Write the formula to calculate the mole fraction of ${\text{N}}_{\text{2}}$.

$y_{{\text{N}}_{\text{2}}}=\frac{N_{{\text{N}}_{\text{2}}}}{N_m}$ (VI)

Write the formula to calculate the mole fraction of $\text{He}$.

$y_{\text{He}}=\frac{N_{\text{He}}}{N_m}$ (VII)

Write the formula to calculate the mole fraction of ${\text{CH}}_{\text{4}}$.

$y_{{\text{CH}}_{\text{4}}}=\frac{N_{{\text{CH}}_{\text{4}}}}{N_m}$ (VIII)

Write the formula to calculate the mole fraction of ${\text{C}}_2{\text{H}}_6$.

$y_{{\text{C}}_2{\text{H}}_6}=\frac{N_{{\text{C}}_2{\text{H}}_6}}{N_m}$ (IX)

Calculate the molar mass of the gas mixture.

$M_m=\frac{m_m}{N_m}$ (X)

Write the partial pressure of ${\text{N}}_{\text{2}}$.

$P_{{\text{N}}_{\text{2}}}=y_{{\text{N}}_{\text{2}}}{P}_m$ (XI)

Here, mixture pressure is $P_m$.

Write the partial pressure of $\text{He}$.

$P_{\text{He}}=y_{\text{He}}{P}_m$ (XII)

Write the partial pressure of ${\text{CH}}_{\text{4}}$.

$P_{{\text{CH}}_{\text{4}}}=y_{{\text{CH}}_{\text{4}}}{P}_m$ (XIII)

Write the partial pressure of ${\text{C}}_2{\text{H}}_6$.

$P_{{\text{C}}_2{\text{H}}_6}=y_{{\text{C}}_2{\text{H}}_6}{P}_m$ (XIV)

Write the equation to calculate the constant-pressure specific heat of the mixture.

$c_p={\text{mf}}_{{\text{N}}_{\text{2}}}{c}_{p,{\text{N}}_{\text{2}}}+{\text{mf}}_{\text{He}}{c}_{p,\text{He}}+{\text{mf}}_{{\text{CH}}_{\text{4}}}{c}_{p,{\text{CH}}_{\text{4}}}+{\text{mf}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{c}_{p,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ (XV)

Here, mass fraction of ${\text{N}}_{\text{2}}{\text{,He,CH}}_{\text{4}}\text{,and}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ are ${\text{mf}}_{{\text{N}}_{\text{2}}}$, ${\text{mf}}_{\text{He}}$, ${\text{mf}}_{{\text{CH}}_{\text{4}}}$, and ${\text{mf}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ respectively and constant pressure specific heat of ${\text{N}}_{\text{2}}{\text{,He,CH}}_{\text{4}}\text{,and}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ are $c_{p,{\text{N}}_{\text{2}}}$, $c_{p,\text{He}}$, $c_{p,{\text{CH}}_{\text{4}}}$, and $c_{p,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ respectively.

Calculate the gas constant of the mixture.

$R=\frac{R_u}{M_m}$ (XVI)

Here, the universal gas constant is $R_u$.

Calculate the constant volume specific heat.

$c_v=c_p-R$ (XVII)

Conclusion:

Substitute 15 kg for $m_{{\text{N}}_{\text{2}}}$ and $28\text{kg/kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{15\text{kg}}{\left(28\text{kg/kmol}\right)}\\ =0.5357\text{kmol}\end{array}$

Substitute 50 kg for $m_{\text{He}}$ and $44\text{kg/kmol}$ for $M_{\text{He}}$ in Equation (II).

$\begin{array}{c}{N}_{\text{He}}=\frac{5\text{kg}}{\left(4\text{kg/kmol}\right)}\\ =1.25\text{kmol}\end{array}$

Substitute 50 kg for $m_{{\text{CH}}_{\text{4}}}$ and $44\text{kg/kmol}$ for $M_{{\text{CH}}_{\text{4}}}$ in Equation (III).

$\begin{array}{c}{N}_{{\text{CH}}_{\text{4}}}=\frac{60\text{kg}}{\left(16\text{kg/kmol}\right)}\\ =3.75\text{kmol}\end{array}$

Substitute 20 kg for $m_{{\text{C}}_2{\text{H}}_6}$ and $30\text{kg/kmol}$ for $M_{{\text{C}}_2{\text{H}}_6}$ in Equation (IV).

$\begin{array}{c}{N}_{{\text{C}}_2{\text{H}}_6}=\frac{20\text{kg}}{\left(30\text{kg/kmol}\right)}\\ =0.6667\text{kmol}\end{array}$

Substitute $0.5357\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $1.25\text{kmol}$ for $N_{\text{He}}$, $3.75\text{kmol}$ for $N_{{\text{CH}}_{\text{4}}}$, and $0.6667\text{kmol}$ for $N_{{\text{C}}_2{\text{H}}_6}$ in Equation (V).

$\begin{array}{c}{N}_m=0.5357\text{kmol}+1.25\text{kmol}+3.75\text{kmol}+0.6667\text{kmol}\\ =6.2024\text{kmol}\end{array}$

Substitute $0.5357\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $6.2024\text{kmol}$ for $N_m$ in Equation (VI).

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{0.5357\text{kmol}}{6.2024\text{kmol}}\\ =0.08637\end{array}$

Thus, the mole fraction of ${\text{N}}_{\text{2}}$ is $\underset{\_}{0.08637}$.

Substitute $1.25\text{kmol}$ for $N_{\text{He}}$ and $6.2024\text{kmol}$ for $N_m$ in Equation (VII).

$\begin{array}{c}{y}_{\text{He}}=\frac{1.25\text{kmol}}{6.2024\text{kmol}}\\ =0.2015\end{array}$

Thus, the mole fraction of $\text{He}$ is $\underset{\_}{0.2015}$.

Substitute $3.75\text{kmol}$ for $N_{{\text{CH}}_{\text{4}}}$ and $6.2024\text{kmol}$ for $N_m$ in Equation (VIII).

$\begin{array}{c}{y}_{{\text{CH}}_{\text{4}}}=\frac{3.75\text{kmol}}{6.2024\text{kmol}}\\ =0.6046\end{array}$

Thus, the mole fraction of ${\text{CH}}_{\text{4}}$ is $\underset{\_}{\text{0}\text{.6046}}$.

Substitute $0.6667\text{kmol}$ for $N_{{\text{C}}_2{\text{H}}_6}$ and $6.2024\text{kmol}$ for $N_m$ in Equation (IX).

$\begin{array}{c}{y}_{{\text{C}}_2{\text{H}}_6}=\frac{0.6667\text{kmol}}{6.2024\text{kmol}}\\ =0.1075\end{array}$

Thus, the mole fraction of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $\underset{\_}{0.1075}$.

Substitute 100 kg for $m_m$ and $6.2024\text{kmol}$ for $N_m$ in Equation (X).

$\begin{array}{c}{M}_m=\frac{100\text{kg}}{6.2024\text{kmol}}\\ =16.12\text{kg/kmol}\end{array}$

Thus, the apparent molecular weight of the mixture is $\underset{\_}{16.12\text{kg/kmol}}$.

Substitute 0.08637 for $y_{{\text{N}}_{\text{2}}}$ and 1200 kPa for $P_m$ in Equation (XI).

$\begin{array}{c}{P}_{{\text{N}}_{\text{2}}}=\left(0.08637\right)1200\text{kPa}\\ =103.6\text{kPa}\end{array}$

Thus, the partial pressure of ${\text{N}}_{\text{2}}$ is $\underset{\_}{\text{103}\text{.6}\text{kPa}}$.

Substitute 0.08637 for $y_{\text{He}}$ and 1200 kPa for $P_m$ in Equation (XII).

$\begin{array}{c}{P}_{\text{He}}=\left(0.2015\right)1200\text{kPa}\\ =241.8\text{kPa}\end{array}$

Thus, the partial pressure of $\text{He}$ is $\underset{\_}{\text{241}\text{.8}\text{kPa}}$.

Substitute 0.6046 for $y_{{\text{CH}}_{\text{4}}}$ and 1200 kPa for $P_m$ in Equation (XIII).

$\begin{array}{c}{P}_{{\text{CH}}_{\text{4}}}=\left(0.6046\right)1200\text{kPa}\\ =725.5\text{kPa}\end{array}$

Thus, the partial pressure of ${\text{CH}}_{\text{4}}$ is $\underset{\_}{\text{725}\text{.5}\text{kPa}}$.

Substitute 0.1075 for $y_{{\text{C}}_2{\text{H}}_6}$ and 1200 kPa for $P_m$ in Equation (XIV).

$\begin{array}{c}{P}_{{\text{C}}_2{\text{H}}_6}=\left(0.1075\right)1200\text{kPa}\\ =129.0\text{kPa}\end{array}$

Thus, the partial pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $\underset{\_}{\text{129}\text{.0}\text{kPa}}$.

Substitute 0.15 for ${\text{mf}}_{{\text{N}}_{\text{2}}}$, 0.05 for ${\text{mf}}_{\text{He}}$, 0.60 for ${\text{mf}}_{{\text{CH}}_{\text{4}}}$, 0.20 for ${\text{mf}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$, $1.039\text{kJ/kg}\cdot \text{K}$ for $c_{p,{\text{N}}_{\text{2}}}$, $5.1926\text{kJ/kg}\cdot \text{K}$ for $c_{p,\text{He}}$, $2.2537\text{kJ/kg}\cdot \text{K}$ for $c_{p,{\text{CH}}_{\text{4}}}$, and $1.7662\text{kJ/kg}\cdot \text{K}$ for $c_{p,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ in Equation (XV).

$\begin{array}{c}{c}_p=\left[\begin{array}{l}\left(0.15\right)\left(1.039\text{kJ/kg}\cdot \text{K}\right)+\left(0.05\right)\left(5.1926\text{kJ/kg}\cdot \text{K}\right)+\\ \left(0.60\right)\left(2.2537\text{kJ/kg}\cdot \text{K}\right)+\left(\text{0}\text{.20}\right)\left(1.7662\text{kJ/kg}\cdot \text{K}\right)\end{array}\right]\\ =2.121\text{kJ/kg}\cdot \text{K}\end{array}$

Thus, the constant-pressure specific heat of the mixture, $c_p$ is $\underset{\_}{2.121\text{kJ/kg}\cdot \text{K}}$.

Substitute $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$ and 16.12 kg/kmol for $M_m$ in Equation (XVI).

$\begin{array}{c}R=\frac{8.314\text{kJ/kmol}\cdot \text{K}}{16.12\text{kg/kmol}}\\ =0.5158\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $0.5158\text{kJ/kg}\cdot \text{K}$ for R and $2.121\text{kJ/kg}\cdot \text{K}$ for $c_p$ in Equation (XVII).

$\begin{array}{c}{c}_v=2.121\text{kJ/kg}\cdot \text{K}-0.5158\text{kJ/kg}\cdot \text{K}\\ =1.605\text{kJ/kg}\cdot \text{K}\end{array}$

Thus, the constant-volume specific heat, $c_v$ is $\underset{\_}{1.605\text{kJ/kg}\cdot \text{K}}$.