Question

Calculate the kinetic energy of 1 g molecule of gas at N.T.P. and what will be its value at 273°C?

Answer 1

Kinetic energy of 1 mol

E = \(\frac{3}{2}\) RT

At T = 273 K

E = \(\frac{3}{2}\) x 8.31 x 273 = 3402.9 J

or E = 3.40 x 10³ J

E₁ = 3.40 x 10³ J, T₁ = 273 K

E₂ = ?; T₂ = 273 + 273 = 2 x 273 K

\(\because\ \frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{2\times273}{273} =2\)

∴ E₂ = 2E₁ = 2 x 3.40 x 10³

or E₂ = 6.80 x 10₃ J