Physical Chemistry Third Edition

414 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium

Let the total time-average force onAbe denoted by〈FA〉. 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〈FA〉due

to particles with velocities

in the given rangedvxdvydvz

⎞

⎠^2 mvy

τ

NAvyτg(v)dvxdvydvz

 2 mNAv^2 yg(v)dvxdvydvz (9.5-10)

Notice thatτcancels 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 ofvywill 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 onAis

〈FA〉 2 mNA

∫∞

−∞

∫∞

0

∫∞

−∞

v^2 yg(v)dvxdvydvz

mNA

∫∞

−∞

∫∞

−∞

∫∞

−∞

v^2 yg(v)dvxdvydvz (9.5-11)

The integrand is aneven functionofvy, which means that ifvyis replaced by−vy, the

integrand is unchanged. The value of the integral overvyfrom 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 ofv^2 y,

which equals

〈

v^2

〉

/3, so we can write

〈FA〉 2 mNA

1

2

〈

v^2 y

〉

 2 mNA

1

6

〈

v^2

〉



1

3

mNA

3 kBT

m

NAkBT (9.5-12)

The pressure is the force per unit area so that

P

〈FA〉

A

NkBT

NkBT

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).

EXAMPLE9.12

For an O 2 molecule moving withvy444ms−^1 , find the average impulse (force multiplied

by the time over which the force is exerted) on a wall perpendicular to theyaxis if the particle

collides with the wall.

〈

Fy

〉

τ 2 mvy(i)

( 2 )(0.0320 kg mol−^1 )(444 m s−^1 )

6. 022 × 1023 mol−^1

 4. 72 × 10 −^23 kg m s−^1  4. 72 × 10 −^23 Ns