Answer
Calculate the characteristics gas constant of nitrogen using the formula,
$\mathrm{R}_{n}=\frac{\mathrm{R}}{\mathrm{M}}$
Here, $\mathrm{R}_{n}$ is the characteristics gas constant of nitrogen, $\mathrm{R}$ is universal gas constant and $\mathrm{M}$ is
molecular weight of nitrogen.
Substitute $8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$ for $\mathrm{R}, 28 \mathrm{~kg} / \mathrm{kmol}$ for $\mathrm{M}$.
$\mathrm{R}_{n}=\frac{8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}{28 \mathrm{~kg} / \mathrm{kmol}}$
$\mathrm{R}_{n}=\frac{8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}{28 \mathrm{~kg} / \mathrm{mol}} \times 10^{3}$
$\mathrm{R}_{n}=0.297 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$
Calculate the temperature of the gas using the Ideal gas equation.
$P=\rho R_{n} T$
Here, $P$ is the pressure, $\rho$ is the density and $T$ is the temperature.
Substitute $400 \mathrm{kPa}$ for $\mathrm{P}, 0.297 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $\mathrm{R}_{n}$ and $4 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho$.
$400=4 \times 0.297 \times T$
$T=336.7 \mathrm{~K}$
Convert the temperature into degrees Celsius.
$T=336.7-273$
$T=63.7^{\circ} \mathrm{C}$
Therefore, the temperature of nitrogen is $63.7{ }^{\circ} \mathrm{C}$.
Work Step by Step
Calculate the characteristics gas constant of nitrogen using the formula,
$\mathrm{R}_{n}=\frac{\mathrm{R}}{\mathrm{M}}$
Here, $\mathrm{R}_{n}$ is the characteristics gas constant of nitrogen, $\mathrm{R}$ is universal gas constant and $\mathrm{M}$ is
molecular weight of nitrogen.
Substitute $8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$ for $\mathrm{R}, 28 \mathrm{~kg} / \mathrm{kmol}$ for $\mathrm{M}$.
$\mathrm{R}_{n}=\frac{8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}{28 \mathrm{~kg} / \mathrm{kmol}}$
$\mathrm{R}_{n}=\frac{8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}{28 \mathrm{~kg} / \mathrm{mol}} \times 10^{3}$
$\mathrm{R}_{n}=0.297 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$
Calculate the temperature of the gas using the Ideal gas equation.
$P=\rho R_{n} T$
Here, $P$ is the pressure, $\rho$ is the density and $T$ is the temperature.
Substitute $400 \mathrm{kPa}$ for $\mathrm{P}, 0.297 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $\mathrm{R}_{n}$ and $4 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho$.
$400=4 \times 0.297 \times T$
$T=336.7 \mathrm{~K}$
Convert the temperature into degrees Celsius.
$T=336.7-273$
$T=63.7^{\circ} \mathrm{C}$
Therefore, the temperature of nitrogen is $63.7{ }^{\circ} \mathrm{C}$.
