Question

Calculate for hydrogen at 27°C:

(i) Kinetic energy of 1 g molecule of gas.

(ii) Kinetic energy of 1 g of gas.

(Molecular weight of hydrogen = 2, R = 8.31 J mol^-1 K^-1)

Answer 1

T = 27°C = 27 + 273 = 300 K; R = 8.31 J mol^-1 K^-1

(i) Kinetic energy of 1 g molecule or 1 mol of hydrogen

E = \(\frac{3}{2}\) RT = \(\frac{3}{2}\) x 8.31 x 300

= 3 x 8.31 x 150

= 3739.5 J

= 3.74 x 10³ J

(ii) Molecular weight of hydrogen M = 2

∵ 2 g kinetic energy of hydrogen = 3.74 x 10³ J

∴ 1 g kinetic energy of hydrogen = \(\frac{3.74\times10^3}{2}\)

= 1.87 x 10³ J