Answer
(a) $0.215\text{ kJ/kg}\cdot\text{K}$
Work Step by Step
The formula for mixture gas constant: $$R_{mix}=\frac{R_u}{M_{mix}}$$ where $R_u=8.314\text{kJ/kmol}\cdot\text{K}$ and $M_{mix}$ is the apparent molar mass of the mixture.We have: $$\begin{aligned}
M_{N_2}&=2\cdot 28=56\text{ kg}\\
M_{CO_2}&=4\cdot 44=176\text{ kg}.
\end{aligned}$$ The total moles: $$n_{total}=2+4=6\text{ kmol.}$$ The total mass is $$m_{total}=56+176=232\text{ kg}.$$ We calculate the apparent molar mass: $$M_{mix}=\frac{m_{total}}{n_{total}}=\frac{232}{6}=38.67\text{ kg/kmol}.$$ Then the mixture gas constant is $$R_{mix}=\frac{8.314}{38.67}=0.215\text{ kJ/kg}\cdot\text{K}.$$ The correct answer is (a).
