**James Richard Fromm**

Since the number of molecules of gas present anywhere, *N*, is given by *nN*_{A},
it is possible to rewrite the overall kinetic energy expression as

*E*_{k} = *nN*_{A}*mv*^{2}/2 = *nMv*^{2}/2

In this expression *m* is the mass of a single molecule, so the molar mass *M*
is the product *mN*_{A}. Since the kinetic energy is also 3*nRT*/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)3*RT*/*M*

Example. Let us calculate the root-mean-square velocity of oxygen molecules at room
temperature, 25^{o}C. Using *v*(rms) = (the square root of)3*RT*/*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-m^{2}/s^{2},
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** *Z*_{1}
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)*d*^{2}
and length *v*(rms) each second, colliding with all molecules in the cylinder.

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 2*v*(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, *Z*_{1},
is:

*Z*_{1} = (pi)*d*^{2}*v*(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*/*N*_{A}; thus *N*/*V* = *pN*_{A}/*RT*.
This gives the mean free path of the gas molecules, *l*, as

(*v*(rms)/*Z*_{1})/(*N*/*V*) = *l* = *RT*/(pi)*d*^{2}*pN*_{A}(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, 25^{o}C, 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 vrms calculated in the previous example,

*l* = (8.3143 kg m^{2}/s^{2}K mol)(298.15 K)/3.14159(370 x 10^{-12}
m)^{2}(101325 kg/m s^{2}) (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 *Z*_{1}, which is the
root-mean-square velocity divided by the mean free path:

*Z*_{1} = (pi)*d*^{2}*pN*_{A}(the
square root of)2/*RT*

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 109 collisions per second per molecule.

For an ideal gas, the number of molecules per unit volume is given using *pV* = *nRT*
and *n* = *N*/*N*_{A} as

*N*/*V* = *N*_{A}*p*/*RT*

which for oxygen at 25^{o}C would be (6.0225 x 10^{+23} mol^{-1})(101325
kg/m s^{2})/(8.3143 kg m^{2}/s^{2} K mol)(298.15 K) or 2.46 x 10^{+25}
molecules/m^{3}. The number of collisions between **two** molecules
in a volume, *Z*_{11}, would then be the product of the number of collisions
each molecule makes times the number of molecules there are, *Z*_{1}*N*/*V*,
except that this would count each collision twice (since two molecules are involved in
each one collision). The correct equation must be

*Z*_{11} = (pi)*d*^{2}*p*^{2}*N*_{A}^{2}*v*(rms)
(the square root of)2/2*R*^{2}*T*^{2}

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.

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)(*M*_{2}/*M*_{1}) = (the square root of)(*d*_{2}/*d*_{1})

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 ^{238}UF_{6}
and ^{235}UF_{6}. 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 ^{238}UF_{6} and ^{235}UF_{6}
can be calculated as follows. The molar mass of ^{238}UF_{6} is 348.0343
and the molar mass of ^{238}UF_{6} 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 ^{235}U 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 uranium used in
the uranium atomic bomb dropped on Japan in 1945.

Copyright 1997 James R. Fromm