Answer
$\dot{m}=6.65\ \text{kg/min}$
Work Step by Step
The molar fraction of air is $$
y_{\text {air }}=1-y_{\mathrm{CH} 4}=1-0.15=0.85
$$ The molar mass of the mixture is determined from
$$
\begin{aligned}
M_m & =y_{\mathrm{CH} 4} M_{\mathrm{CH} 4}+y_{\text {air }} M_{\text {air }} \\
& =0.15 \times 16+0.85 \times 28.97 \\
& =27.02 \mathrm{~kg} / \mathrm{kmol}
\end{aligned}
$$ Given the engine displacement and speed and assuming that this is a 4-stroke engine flow rate is determined from ($2$ revolutions per cycle), the volume
flow rate is determined from $$
\dot{V}=\frac{\dot{n} V_d}{2}=\frac{(3000 \mathrm{rev} / \mathrm{min})\left(0.005 \mathrm{~m}^3\right)}{2 \mathrm{rev} / \mathrm{cycle}}=7.5 \mathrm{~m}^3 / \mathrm{min} $$ The specific volume of the mixture is $$
v=\frac{R_u T}{M_m P}=\frac{\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol} \cdot \mathrm{K}\right)(293 \mathrm{~K})}{(27.02 \mathrm{~kg} / \mathrm{kmol})(80 \mathrm{kPa})}=1.127 \mathrm{~m}^3 / \mathrm{kg}
$$ Hence the mass flow rate is $$
\dot{m}=\frac{\dot{v}}{v}=\frac{7.5 \mathrm{~m}^3 / \mathrm{min}}{1.127 \mathrm{~m}^3 / \mathrm{kg}}=6.65 \mathrm{~kg} / \mathrm{min}
$$
