Answer
$$
\begin{aligned}
V &=\frac{(5)(286.9)(353)}{401.3 \times 10^{3}} \\
&=1.26 \mathrm{~m}^{3}
\end{aligned}
$$
Therefore, the volume of the tank is $1.26 \mathrm{~m}^{3}$
Work Step by Step
Calculate the volume of the tank by using equation of state:
$$
p V=m R T \ldots \ldots \text { (1) }
$$
Here, Mass of air in air tank is $m$, the gas constant is $R$, the Absolute pressure is $p$, and Absolute temperature of compressed air in tank
Absolute temperature $=$ (gauge pressure + atmospheric pressure)
$$
\begin{aligned}
p &=300+101.3 \\
&=401.3 \mathrm{kPa}
\end{aligned}
$$
The gas constant of air is $R=286.9 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$
Substitute, $5 \mathrm{~kg}$ for $m, 401.3 \mathrm{kPa}$ for $p, 286.9 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ for $R, 401.3 \mathrm{kPa}$ for $p, 353 \mathrm{~K}$ for $T$ in in equation (1)
$$
\begin{aligned}
V &=\frac{(5)(286.9)(353)}{401.3 \times 10^{3}} \\
&=1.26 \mathrm{~m}^{3}
\end{aligned}
$$
Therefore, the volume of the tank is $1.26 \mathrm{~m}^{3}$
