What is the volume of nitrogen weighing 1.05 kg at a pressure of 2.8 MPa and a temperature of 27 degrees?

Let’s translate the values of quantities in the SI system;

m = 1.05kg, p = 2.9 * 10 ^ 6Pa, T = 27⁰C + 273 = 300K, molar mass of nitrogen µ = 2.8 ∙ 10 ^ -3kg / mol;

Substitute the data into the formula V = m * RT / M * p; R = 8.31 J / mol ∙ K;

V = (1.05kg * 8.31 J / mol ∙ K ∙ 300K) / (2.8 ∙ 10 ^ -3kg / mol ∙ 2.9 ∙ 10 ^ 6Pa) = 26.1765 ∙ 10 ^ 2 / 8.12 ∙ 10 ^ 3m3 = 3.22 ∙ 10 ^ -3m3 = 0.322m3.