Find the temperature of nitrogen with a mass of 4 g, which at a pressure of 0.1 MPa occupies a volume of 8.31 liters

Let’s translate all the values from given to the SI system:
m = 4g = 0.004 kg.
p = 0.1 MPa = 0.1 * 10 ^ 6 Pa.
V = 8.31 l. = 8.31 * 10 ^ -3 m³.
Mendeleev-Clapeyron equation
pV = (m / M) * R * T, where p is the gas pressure, V is the gas volume, m is the gas mass, M is the molar mass of the gas, R is the universal gas constant 8.31 J / (mol * K), T is the gas temperature.
Let us express the temperature from this expression:
T = p * V * M / (m * R)
The molar mass of nitrogen is M (N2) = 2 * 0.014 = 0.028 kg / mol.
Substitute the numerical values:
T = p * V * M / (m * R) = 0.1 * 10 ^ 6 * 8.31 * 10 ^ -3 * 0.028 / (0.004 * 8.31) = 700 K.
Answer: nitrogen temperature 700 K.