What volume will neon with a mass of 5 g occupy, creating a pressure of 249 kPa at a temperature of 500 K?

Let us express the volume of gas V, which has mass m and is under pressure p at absolute temperature T, from the Mendeleev-Clapeyron equation: p ∙ V = R ∙ T ∙ m / M, where the universal gas constant R = 8.3 J / (mol ∙ K), and the mass of a mole M is determined through the mass of the molecule m₀ and Na by the formula M = m₀ ∙ Na. Avogadro’s number Na = 6.02 ∙ 10 ^ 23 mol ^ (- 1). Means: V = R ∙ T ∙ m / (M ∙ р). It is known that neon with a mass of m = 5 g = 0.005 kg, creating a pressure of p = 249 kPa = 249000 Pa at a temperature of T = 500 K, has a relative atomic mass of A = 20 amu, then m₀ = A ∙ 1.67 ∙ 10 ^ (- 27) kg / a.u. and M = A ∙ 1.67 ∙ 10 ^ (- 27) kg / a.u. ∙ Na. Hence:

V = R ∙ T ∙ m / (A ∙ 1.67 ∙ 10 ^ (- 27) kg / amu ∙ Na ∙ p).

We get:

V = (8.3 J / (mol ∙ K) ∙ 500 K ∙ 0.005 kg) / (20 amu ∙ 1.67 ∙ 10 ^ (- 27) kg / amu ∙ 6 , 02 ∙ 10 ^ 23 mol ^ (- 1) ∙ 249000 Pa);

V ≈ 4.1 ∙ 10 ^ (- 3) m ^ 3.

Answer: neon occupies a volume of ≈ 4.1 ∙ 10 ^ (- 3) m ^ 3.