Physical Chemistry , 1st ed.

ticular gas particle; it could be anything. This does not imply that all possible
velocities will exist in any gas sample at equal proportions! For example, the
velocities of gas particles couldbe anywhere between zero and the speed of
light. The naive presumption would be that the average velocity is (0 c)/2 
^12 c(where crepresents the speed of light in a vacuum). This is clearly not the
case, as illustrated by the answer in Example 19.1 above.
In the following discussion, we sometimes use the terms “speed” and “ve-
locity” interchangeably. This is because in such cases we are interested more in
the magnitude of the quantity, not the direction. However, in cases where di-
rection is important to the discussion, it is pointed out explicitly.
There are several ways of defining an “average” speed, and there are also
ways of determining the distribution of speeds of gas particles in any sample.
First, let us consider the use of average speed from the previous section. We
have two expressions for the kinetic energy of a gas; they are

E^32 RT and NAEavgNA^12 mv^2 avg

(The second equation comes from multiplying equation 19.9 by Avogadro’s
number.) Equating the two expressions for the molar energy of the gas,

^32 RTNA^12 mv^2 avg^12 (NAm) v^2 avg

Since (NAm) is the molar mass of the gas, we will use Mto define the mo-
lar mass and rewrite the above equation as

^32 RT^12 Mv^2 avg

We can algebraically rearrange this to solve for vavg. Because we are going to be
taking the square root of the square of the average (or “mean”) speed, we de-
fine this average speed as the root-mean-square speed,or vrms.We get

vrms

3

M

RT

 (19.13)

This is one way to define an average speed.

Example 19.2

What is the temperature of the He sample in Example 19.1 if the answer is

considered to be vrms?

Solution

The velocity in Example 19.1 was 1282 m/s. Using standard units for all quan-

tities, we have

1282 

m

s





Notice that we have expressed the mass of the gas in kilogram units. This is

necessary for the units to work out properly. We need to decompose the unit

J into kgm^2 /s^2 ; when we do this, the expression above becomes

1282 m

s

6235.5T

s

m

^2 K

2



1.6435  106 

m

s^2

2

6235.5 T 

s

m

(^2) K
2

3  8.314 
mo

J

lK

 T



0.00400 

m

kg

ol



19.3 Definitions and Distributions of Velocities of Gas Particles 657