What is the temperature of 1.6 g of oxygen under a pressure of 0.1 MPa and occupying a volume of 1.6 liters?

m = 1.6 g = 1.6 * 10 ^ -3 kg.
P = 0.1 MPa = 0.1 * 10 ^ 6 Pa.
V = 1.6 l = 1.6 * 10 ^ -3 m ^ 3.
M = 0.032 kg / mol.
R = 8.31 J / mol * K.
T -?
For gas, the Mendeleev-Cliperone equation is valid: P * V = m * R * T / M, where P is the gas pressure, V is the gas volume, m is the gas mass, R is the gas constant, T is the absolute temperature, M is the molar mass …
T = M * P * V / m * R.
T = 0.032 kg / mol * 0.1 * 10 ^ 6 Pa * 1.6 * 10 ^ -3 m ^ 3 / 1.6 * 10 ^ -3 kg * 8.31 J / mol * K = 393 K.
Answer: oxygen temperature T = 393 K.