Question 24.1: A gas mixture from a hydrocarbon reforming process contains ......

Let A={\mathrm{H_2}},B={\mathrm{CO_{2}}},{\mathrm{~and~}}C={\mathrm{CH}}_{4}. Assuming ideal gas behavior, the total molar concentration is

c=\frac{P}{R T}=\frac{1.5\;\mathrm{atm}}{(0.08206\;\mathrm{m^{3}}\cdot\mathrm{atm/kgmole}\cdot{ K})(673\;{ K})}=2.72\times10^{-2}\;\mathrm{kgmole}/\mathrm{m}^{3}

For an ideal gas, volume percent composition is equivalent to mole percent composition. The molar concentration of species A is

c_{A}=y_{A}\,c=(0.50)(2.72\times10^{-2}\,\mathrm{kgmole/m^{3}})=1.36\times10^{-2}\,\mathrm{kgmole}/\mathrm{m^{3}}

Similarly, c_{B}=1.10\times10^{-2}\,\mathrm{kgmole}/\mathrm{m}^{3}, and c_{C}=2.72\times10^{-3}\,\mathrm{kgmole}/\mathrm{m}^{3}. The mass fraction of each species is determined by the mole fraction and molecular weight of each species:

w_{A}=\frac{{y_{A}}M_{A}}{{y_{A}M_{A}}+{y_{B}}M_{B}+{y_{C}}M_{C}}=\frac{(0.50)(2)}{(0.50)(2)+(0.40)(44)+(0.10)(16)}=0.0495\frac{\mathrm{g~H}_{2}}{\mathrm{total~g}}

Likewise, w_{B}=0.871,\;\mathrm{and}\;w_{C}=0.0793. The mass density of the gas mixture is

\rho=\rho_{A}+\rho_{B}+\rho_{C}=c_{A}M_{A}+c_{B}M_{B}+c_{C} M_{C}

\quad=\left(1.36\times10^{-2}~{\frac{\mathrm{kgmole}}{\mathrm{m^{3}}}}\right)\left({\frac{2\,{\mathrm{g}}}{\mathrm{gmole}}}\right)+\left(1.10\times10^{-2}~{\frac{\mathrm{kgmole}}{\mathrm{m^{3}}}}\right)\left({\frac{44\,{\mathrm{g}}}{\mathrm{gmole}}}\right)

\quad \quad+\left(2.72\times10^{-3}\frac{\mathrm{kgmole}}{\mathrm{m^{3}}}\right)\left(\frac{16\,{\mathrm{g}}}{{\mathrm{gmole}}}\right)=0.555\,\frac{\mathrm{kg}}{\mathrm{m^{3}}}