Kinetic Theory : Molecular Speeds

From the value of the molecular translational kinetic energy, ½MG², which is the same for all gases and equal (very approximately) to 34 x 10⁹ ergs at 0°, we can calculate the squares of the molecular speeds, G², by division by the molecular weight in grams, M, and multiplication by 2:

Thus, for oxygen, M = 32;.G² = 34 x 10⁹ x 2/32; the mean square speed G at 0° C = 46,000 cm. per sec., or 460 m. per sec. The mean speed, Ω, is G multiplied by 0.921, i.e., 425 m. per sec. In the case of hydrogen, the mean speed at 0° is 1694 m. per sec.

Mean Molecular Speed, Ω, at 0° in Metres per Second

(Velocities of sound in the gases are given in brackets.)

Hydrogen, 1694 (1286).
Helium, 1208.
Steam, 565 (401).
Nitrogen, 455 (337).
Oxygen, 425 (317).
Carbon dioxide, 362 (257).
Chlorine, 288 (206).
Mercury vapour, 170.

The speed of steam molecules (M = 18) is greater than that of oxygen molecules (M = 32); the speeds of hydrogen and helium are large relatively to those of the other gases. A speed of 1700 m. per sec. is 5500 ft. per sec., or more than a mile per sec., i.e., of the order of speed of a rifle bullet. Owing to these high speeds the kinetic energies of the minute fragments of matter which the molecules represent are high, and the pressure due to the molecular shower is thus explained. It is also seen that the molecular speeds are of the same order as, but greater than, the velocities of sound, u, in the gases. The formulae p = ⅓DG² and u = √(γp/D), where γ = c_p/c_v, give u = √(⅓)γ) * G.

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