Figure 13.21Gas in a box exerts an outward pressure on its walls. A molecule colliding with a rigid wall has the direction of its velocity and momentum in thex-direction
reversed. This direction is perpendicular to the wall. The components of its velocity momentum in they- andz-directions are not changed, which means there is no
force parallel to the wall.
If the molecule’s velocity changes in thex-direction, its momentum changes from–mvxto+mvx. Thus, its change in momentum is
Δmv= +mvx–(–mvx)= 2mvx. The force exerted on the molecule is given by
(13.45)
F=
Δp
Δt
=
2 mvx
Δt
.
There is no force between the wall and the molecule until the molecule hits the wall. During the short time of the collision, the force between the
molecule and wall is relatively large. We are looking for an average force; we takeΔtto be the average time between collisions of the molecule
with this wall. It is the time it would take the molecule to go across the box and back (a distance 2 l)at a speed ofvx. ThusΔt= 2l/vx, and
the expression for the force becomes
(13.46)F=
2 mvx
2 l/vx
=
mvx^2
l
.
This force is due toonemolecule. We multiply by the number of moleculesNand use their average squared velocity to find the force
(13.47)
F=N
mvx^2
l
,
where the bar over a quantity means its average value. We would like to have the force in terms of the speedv, rather than thex-component
of the velocity. We note that the total velocity squared is the sum of the squares of its components, so that
(13.48)v^2 =vx^2 +vy^2 +vz^2.
Because the velocities are random, their average components in all directions are the same:
(13.49)vx^2 =vy^2 =vz^2.
Thus,
(13.50)v^2 = 3vx^2 ,
or
(13.51)v^2 x=^1
3
v^2.
Substituting^1
3
v^2 into the expression forFgives
CHAPTER 13 | TEMPERATURE, KINETIC THEORY, AND THE GAS LAWS 451