Chemical Sciences

The Structure of Matter

The Nature of Gases

Kinetic-Molecular Theory Explains the Gas Laws:

Velocities of Molecules

Since the number of molecules of gas present anywhere, N, is given by nNA, it is possible to rewrite the overall kinetic energy expression as

Ek = nNAmv2/2 = nMv2/2

In this expression m is the mass of a single molecule, so the molar mass M is the product mNA. Since the kinetic energy is also 3nRT/2, the square root of the square of the mean velocity, known as the root-mean-square velocity v(rms), of the molecules of the gas is proportional to the square root of its molar mass. The root-mean-square velocity, like the actual distribution of velocities embodied in the Maxwell law, is a function only of the absolute temperature.

v(rms) = (the square root of)3RT/M


Example. Let us calculate the root-mean-square velocity of oxygen molecules at room temperature, 25oC. Using v(rms) = (the square root of)3RT/M, the molar mass of molecular oxygen is 31.9998 g/mol; the molar gas constant has the value 8.3143 J/mol K, and the temperature is 298.15 K. Since the joule is the kg-m2/s2, the molar mass must be expressed as 0.0319998 kg/mol. The root-mean-square velocity is then given by:

v(rms) = (the square root of)3(8.3143)(298.15)/(0.0319998) = 482.1 m/s

A speed of 482.1 m/s is 1726 km/h, much faster than a jetliner can fly and faster than most rifle bullets.


The very high speed of gas molecules under normal room conditions would indicate that a gas molecule would travel across a room almost instantly. In fact, gas molecules do not do so. If a small sample of the very odorous (and poisonous!) gas hydrogen sulfide is released in one corner of a room, our noses will not detect it in another corner of the room for several minutes unless the air is vigorously stirred by a mechanical fan. The slow diffusion of gas molecules which are moving very quickly occurs because the gas molecules travel only short distances in straight lines before they are deflected in a new direction by collision with other gas molecules.

The distance any single molecule travels between collisions will vary from very short to very long distances, but the average distance that a molecule travels between collisions in a gas can be calculated. This distance is called the mean free path l of the gas molecules. If the root-mean-square velocity is divided by the mean free path of the gas molecules, the result will be the number of collisions one molecule undergoes per second. This number is called the collision frequency Z1 of the gas molecules.

The postulates of the kinetic-molecular theory of gases permit the calculation of the mean free path of gas molecules. The gas molecules are visualized as small hard spheres. A sphere of diameter d sweeps through a cylinder of cross-sectional area (pi)d2 and length v(rms) each second, colliding with all molecules in the cylinder, as shown in the Figure below.


Figure is not available.
The radius of the end of the cylinder is d because two molecules will collide if their diameters overlap at all. This description of collisions with stationary gas molecules is not quite accurate, however, because the gas molecules are all moving relative to each other. Those relative velocities range between zero for two molecules moving in the same direction and 2v(rms) for a head-on collision. The average relative velocity is that of a collision at right angles, which is v(rms) times the square root of 2. The total number of collisions per second per unit volume, Z1, is

Z1 = (pi)d2v(rms)(the square root of)2

This total number of collisions must now be divided by the number of molecules which are present per unit volume. The number of gas molecules present per unit volume is found by rearrangement of the ideal gas law to n/V = p/RT and use of Avogadro's number, n = N/NA; thus N/V = pNA/RT. This gives the mean free path of the gas molecules, l, as

(v(rms)/Z1)/(N/V) = l = RT/(pi)d2pNA(the square root of)2

According to this expression, the mean free path of the molecules should get longer as the temperature increases; as the pressure decreases; and as the size of the molecules decreases.


Example. Let us calculate the length of the mean free path of oxygen molecules at room temperature, 25oC, taking the molecular diameter of an oxygen molecule as 370 pm. Using the formula for mean free path given above and the value of the root-mean-square velocity v(rms) calculated in the previous example,

l = (8.3143 kg m2/s2K mol)(298.15 K)/3.14159(370 x 10-12 m)2(101325 kg/m s2) (6.0225 x 10+23 mol-1)((the square root of)2),

so l = 6.7 x 10-8 m = 67 nm. The utility of SI units and of the quantity calculus in this example should be obvious.


The apparently slow diffusion of gas molecules takes place because the molecules travel only a very short distance before colliding. At room temperature and atmospheric pressure, oxygen molecules travel only (6.7 x 10-8 m)/(370 x 10-12 m) = 180 molecular diameters between collisions. The same thing can be pointed out using the collision frequency for a single molecule Z1, which is the root-mean-square velocity divided by the mean free path:

Z1 = (pi)d2pNA(the square root of)2/RT = v(rms)/l

For oxygen at room temperature, each gas molecule collides with another every 0.13 nanoseconds (one nanosecond is 1.0 x 10-9 s), since the collision frequency is 7.2 x 10+9 collisions per second per molecule.

For an ideal gas, the number of molecules per unit volume is given using pV = nRT and n = N/NA as

N/V = NAp/RT

which for oxygen at 25oC would be (6.0225 x 10+23 mol-1)(101325 kg/m s2)/(8.3143 kg m2/s2 K mol)(298.15 K) or 2.46 x 10+25 molecules/m3. The number of collisions between two molecules in a volume, Z11, would then be the product of the number of collisions each molecule makes times the number of molecules there are, Z1N/V, except that this would count each collision twice (since two molecules are involved in each one collision). The correct equation must be

Z11 = (pi)d2p2NA2v(rms) (the square root of)2/2R2T2

If the molecules present in the gas had different masses they would also have different speeds, so an average value of v(rms) would be using a weighted average of the molar masses; the partial pressures of the different gases in the mixture would also be required. Although such calculations involve no new principles, they are beyond our scope. However, the number of collisions which occur per second in gases and in liquids are extremely important in chemical kinetics, so we shall return to this topic in other sections.

Graham's Law of Effusion and Diffusion
Root-mean-square velocities of gas molecules are sometimes directly useful, but the comparison of velocities explains the results of, and is useful in, studies of effusion of molecules through a small hole in a container or diffusion of molecules through porous barriers. The comparison between two gases is most conveniently expressed as:

v(rms)1/v(rms)2 = (the square root of)(M2/M1) = (the square root of)(d2/d1)

This equation gives the velocity ratio in terms of either the molar mass ratio or the ratio of densities d. The ratio of root-mean-square velocities is also the ratio of the rates of effusion, the process by which gases escape from containers through small holes, and the ratio of the rates of diffusion of gases.

This equation is called Graham's law of diffusion and effusion because it was observed by Thomas Graham (1805-1869) well before the kinetic-molecular theory of gases was developed. As an empirical law, it simply stated that the rates of diffusion and of effusion of gases varied as the square root of the densities of the gases. Graham's law is the basis of many separations of gases. The most significant is the separation of the isotopes of uranium as the gases 238UF6 and 235UF6. Fluorine has only one isotope, so the separation on the basis of molar mass is really a separation on the basis of isotopic mass.


Example. The ratio of root-mean-square velocities of 238UF6 and 235UF6 can be calculated as follows. The molar mass of 238UF6 is 348.0343 and the molar mass of 238UF6 is 352.0412. The mass ratio is 1.011513 and the ratio of root-mean-square velocities is 1.00574. Although the difference is small, many kilograms of 235U have been separated using this difference in the gas-diffusion separation plant at Oak Ridge, Tennessee, U. S. A. This plant prepared the uranium for the Manhattan Project of the Second World War and produced the uranum used in the uranium atomic bomb dropped on Japan in 1945.
[NEXT: Equations of State for Gases]
[PREVIOUS: Kinetic-Molecular Theory Explains the Gas Laws: Partial Pressure]
Copyright 1995 James A. Plambeck (Jim.Plambeck@ualberta.ca). Updated September 26, 1995.