414
9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium
Let the total time-average force on A be denoted by F A . Each particle makes a
contribution to this force that is given by Eq. (9.5-5), so that the collection of particles
with velocities in the given range makes the contribution
⎛
⎝
contribution to F A due
to particles with velocities
in the given range dv x dv y dv z
⎞
⎠
2mv y
τ
NA v y τg(v) dv x dv y dv z
2mNA v
2
y g(v) dv x dv y dv z
(9.5-10)
Notice that τ cancels out of this equation. We now add up the contributions for molecules
with different velocities by integrating over the velocity components. Only molecules
with positive values of v y will strike the wall at the right end of the box, whereas the
components parallel to the wall can be positive or negative. The total time-average
force on A is
F A 2mNA
∞
−∞
∞
0
∞
−∞
v
2
y g(v) dv x dv y dv z
mNA
∞
−∞
∞
−∞
∞
−∞
v
2
y g(v)dv x dv y dv z
(9.5-11)
The integrand is an even function of v y , which means that if v y is replaced by −v y , the
integrand is unchanged. The value of the integral over v y from 0 to ∞ equals half of
that of the integral from −∞ to ∞, so we have replaced the lower limit by −∞ and
divided by 2. This integral is the same as the integral that gives the mean value of v 2
y ,
which equals
v 2
/3, so we can write
F A 2mNA
1
2
v
2
y
2mNA
1
6
v
2
1
3
mNA
3k B T
m
NA k B T
(9.5-12)
The pressure is the force per unit area so that
P
F A
A
N k B T
Nk B T
V
nRT
V
(9.5-13)
We have derived the ideal gas law for our model system, so we believe that it does
resemble a real dilute gas. It is also possible to assume that our system obeys the ideal
gas law and then to derive the expression for the energy in Eq. (9.3-24).
E X A M P L E 9.12
For an O 2 molecule moving with v y 444 m s −1 , find the average impulse (force multiplied
by the time over which the force is exerted) on a wall perpendicular to the y axis if the particle
collides with the wall.
F y
τ 2mv y (i)
(2)(0.0320 kg mol −1 )(444 m s −1 )
6.022 × 10 23 mol −1
4.72 × 10 −23 kg m s −1 4.72 × 10 −23 N s
9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium
Let the total time-average force on A be denoted by F A . Each particle makes a
contribution to this force that is given by Eq. (9.5-5), so that the collection of particles
with velocities in the given range makes the contribution
⎛
⎝
contribution to F A due
to particles with velocities
in the given range dv x dv y dv z
⎞
⎠
2mv y
τ
NA v y τg(v) dv x dv y dv z
2mNA v
2
y g(v) dv x dv y dv z
(9.5-10)
Notice that τ cancels out of this equation. We now add up the contributions for molecules
with different velocities by integrating over the velocity components. Only molecules
with positive values of v y will strike the wall at the right end of the box, whereas the
components parallel to the wall can be positive or negative. The total time-average
force on A is
F A 2mNA
∞
−∞
∞
0
∞
−∞
v
2
y g(v) dv x dv y dv z
mNA
∞
−∞
∞
−∞
∞
−∞
v
2
y g(v)dv x dv y dv z
(9.5-11)
The integrand is an even function of v y , which means that if v y is replaced by −v y , the
integrand is unchanged. The value of the integral over v y from 0 to ∞ equals half of
that of the integral from −∞ to ∞, so we have replaced the lower limit by −∞ and
divided by 2. This integral is the same as the integral that gives the mean value of v 2
y ,
which equals
v 2
/3, so we can write
F A 2mNA
1
2
v
2
y
2mNA
1
6
v
2
1
3
mNA
3k B T
m
NA k B T
(9.5-12)
The pressure is the force per unit area so that
P
F A
A
N k B T
Nk B T
V
nRT
V
(9.5-13)
We have derived the ideal gas law for our model system, so we believe that it does
resemble a real dilute gas. It is also possible to assume that our system obeys the ideal
gas law and then to derive the expression for the energy in Eq. (9.3-24).
E X A M P L E 9.12
For an O 2 molecule moving with v y 444 m s −1 , find the average impulse (force multiplied
by the time over which the force is exerted) on a wall perpendicular to the y axis if the particle
collides with the wall.
F y
τ 2mv y (i)
(2)(0.0320 kg mol −1 )(444 m s −1 )
6.022 × 10 23 mol −1
4.72 × 10 −23 kg m s −1 4.72 × 10 −23 N s