Answer
$V_{m}=3.93$ m$^{3}$
Work Step by Step
The mole numbers of each component are $$
\begin{aligned}
N_{\mathrm{CH} 4} & =\frac{m_{\mathrm{CH} 4}}{M_{\mathrm{CH} 4}}=\frac{60 \mathrm{~kg}}{16 \mathrm{~kg} / \mathrm{kmol}}=3.75 \mathrm{kmol} \\
N_{\mathrm{C} 3 \mathrm{H} 8} & =\frac{m_{\mathrm{C} 3 \mathrm{H} 8}}{M_{\mathrm{C} 3 \mathrm{H} 8}}=\frac{25 \mathrm{~kg}}{44 \mathrm{~kg} / \mathrm{kmol}}=0.5682 \mathrm{kmol} \\
N_{\mathrm{C} 4 \mathrm{H} 10} & =\frac{m_{\mathrm{C} 4 \mathrm{H} 10}}{M_{\mathrm{C} 4 \mathrm{H} 10}}=\frac{15 \mathrm{~kg}}{58 \mathrm{~kg} / \mathrm{kmol}}=0.2586\ \mathrm{kmol}
\end{aligned}
$$ The mole number of the mixture is $$
N_m=N_{\mathrm{CH} 4}+N_{\mathrm{C} 3 \mathrm{H} 8}+N_{\mathrm{C} 4 \mathrm{Hl} 0}=3.75+0.5682+0.2586=4.5768\ \mathrm{kmol}
$$ The apparent molecular weight of the mixture is $$
M_m=\frac{m_m}{N_m}=\frac{100 \mathrm{~kg}}{4.5768 \mathrm{kmol}}=21.85 \mathrm{~kg} / \mathrm{kmol}
$$ Then the volume of this ideal gas mixture is $$
V_m=\frac{N_m R_u T}{P}=\frac{(4.5768 \mathrm{kmol})\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol} \cdot \mathrm{K}\right)(310 \mathrm{~K})}{3000 \ \mathrm{kPa}}=\mathbf{3 . 9 3\ \mathrm { m } ^ { 3 }}
$$
