Answer
$M_{m}=19.03$ lbm/lbmol
Work Step by Step
The mole numbers of gases are $$
\begin{aligned}
& m_{\mathrm{CO}_2}=1 \mathrm{lbm} \longrightarrow N_{\mathrm{CO}_2}=\frac{m_{\mathrm{CO}_2}}{M_{\mathrm{CO}_2}}=\frac{1 \mathrm{lbm}}{44 \mathrm{lbm} / \mathrm{lbmol}}=0.0227 \mathrm{lbmol} \\
& m_{\mathrm{CH}_4}=3 \mathrm{lbm} \longrightarrow N_{\mathrm{CH}_4}=\frac{m_{\mathrm{CH}_4}}{M_{\mathrm{CH}_4}}=\frac{3 \mathrm{lbm}}{16 \mathrm{lbm} / \mathrm{lbmol}}=0.1875 \mathrm{lbmol} \\
& N_m=N_{\mathrm{CO}_2}+N_{\mathrm{CH}_4}=0.0227 \mathrm{lbmol}+0.1875 \mathrm{lbmol}=0.2102 \mathrm{lbmol} \\
& y_{\mathrm{CO}_2}=\frac{N_{\mathrm{CO}_2}}{N_m}=\frac{0.0227 \mathrm{lbmol}}{0.2102 \mathrm{lbmol}}=0.108 \\
& y_{\mathrm{CH}_4}=\frac{N_{\mathrm{CH}_4}}{N_m}=\frac{0.1875 \mathrm{lbmol}}{0.2102 \mathrm{lbmol}}=0.892
\end{aligned}
$$ Then the partial pressures become
$$
\begin{aligned}
& P_{\mathrm{CO}_2}=y_{\mathrm{CO}_2} P_m=(0.108)(20 \text { psia })=\mathbf{2 . 1 6} \text { psia } \\
& P_{\mathrm{CH}_4}=y_{\mathrm{CH}_4} P_m=(0.892)(20 \text { psia })=\mathbf{1 7 . 8 4} \text { psia }
\end{aligned}
$$ The apparent molar mass of the mixture is
$$
M_m=\frac{m_m}{N_m}=\frac{4 \mathrm{lbm}}{0.2102 \mathrm{lbmol}}=19.03 \mathrm{lbm} / \mathrm{lbmol}
$$
