Physical Chemistry , 1st ed.

Since the mass of the particle is constant, it can be removed from the integral,

which is easily evaluated to get

Favg

tim

1

e

mvavg (19.4)

where vavgis the average change in the velocity of the gas particle. If there are

Ngas particles in the container, then the total average force would be

Favg,totalN

tim

1

e

mvavg (19.5)

where vavgrepresents the average change in velocity of the Nparticles.

(Again, we have not constrained the particles to have any particular velocity

yet.) This equation is a potential problem, because velocity is a vector in

three-dimensional space and so is the change in velocity,vavg. However, the

three-dimensional velocity can be separated into its one-dimensional compo-

nents, and the average force separated into three components that are equiv-

alent to each other. Thus, we can consider a one-dimensional problem, then

apply our conclusions to the other two dimensions. Let us assume that we are

considering the xdimension, corresponding to the dotted line in Figure 19.1.

The total force in the xdimension that is applied to the wall is

Favg,total,xN

tim

1

e

mvavg,x

But if we resolve the initial and final velocities of the gas particle in Figure 19.1,

we can see that vivf(that is, they have the same magnitude but opposite di-

rections). Therefore, in terms of the initial velocity,vavg,xvavg,x(vavg,x) 

2 vavg,x(where we are adding the xsubscript to the initial velocity also). The equa-

tion for the total average force in the xdimension becomes

Favg,total,x 2 N

tim

1

e

mvavg,x (19.6)

Finally, consider a box having dimensions abcin the x,y, and zdimen-

sions, as shown in Figure 19.2. What is the time amount in equation 19.6?

When the particle is not colliding with the wall, no force is being exerted on

it, nor by it on anything else (that’s one of the postulates). The time can be as

long as it takes for the particle to start at one wall, travel in the xdimension to

collide with the other wall, then travel all the way back to the opposite wall.

This means that the particle travels twice the xdimension, or a distance of 2a.

From the definition of velocity we have

vavg,x

ti

2

m

a

e



or, rewriting,

time 

va

2

v

a

g,x



If we substitute this expression for time in the denominator of equation 19.6,

we get

Favg,total,x 2 Nmvavg,x

Favg,total,xN

1

a

mv^2 avg,x

Nm

a

v^2 avg,x

 (19.7)

1





va

2

v

a

g,x



654 CHAPTER 19 The Kinetic Theory of Gases

c

a

b

Gas particle

x
z

y

Figure 19.2 A gas particle has a three-

dimensional velocity, so an understanding of pres-

sure must consider three dimensions. However, if

we assume that all three dimensions are equiva-

lent, the mathematical derivation of pressure is

much easier.