Physical Chemistry Third Edition

394 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium

b. Construct a graph of the position of the particle as a

function of time form 1 .00 kg,f 1 .00 kg s−^1 ,

x 0 0, andv 0  10 .0ms−^1.

9.5An object is propelled upward at the surface of the earth with

an initial speed of 10.0 m s−^1. Ignoring air resistance, find

the maximum height of the object above its starting point.

9.6An object with mass 0.100 kg slides down a frictionless

surface of some arbitrary shape near the surface of the earth.

If it starts with zero speed, find its speed when it is at a

location 0.250 m lower in altitude than its initial position,

and show that this is independent of the shape of the

surface.

9.3 The Velocity Probability Distribution

If a system obeys classical mechanics, averaging over molecular states means averag-

ing over molecular coordinates and momenta. Coordinates and momenta vary continu-

ously, so we must integrate over their possible values instead of summing over discrete

energies as in Eq. (9.2-19). Consider a quantityuthat can take on any value between

u 1 andu 2. We define aprobability distributionf(u) such that

(

The probability thatulies

betweenu′andu′+du

)

f(u′)du (9.3-1)

whereduis an infinitesimal range of values ofuand whereu′is some value ofu.

The probability distributionf(u) is a probability per unit length on theuaxis, and we

call it aprobability density. Sincef(u) is a probability per unit length on the axis, the

probability thatulies betweenu′andu′+duis represented by an area under the curve

representingf(u) as shown schematically in Figure 9.5.

Shaded area equals f(u')du

f(

u)

u' u' 1 du

u

Figure 9.5 An Example Proba-
bility Density (Probability Distri-
bution).

The total probability is equal to the integral

(

total probability

)



∫u 2

u 1

f(u)du (9.3-2)

whereu 1 is the smallest possible value ofuandu 2 is the largest. This integral is equal to

the total area under the curve representingf(u) in Figure 9.5. We commonly multiply

the probability density by a constant so that the total probability is equal to 1, which is

callednormalizationof the probability distribution.

Themean valueofuis given by an equation that is analogous to Eq. (9.1-4) and

Eq. (9.2-19):

〈u〉

∫u 2

u 1

uf(u)du

∫u 2

u 1

f(u)du

(9.3-3a)

If the probability density is normalized, the denominator in this equation equals unity:

〈u〉

∫u 2

u 1

uf(u)du (fnormalized) (9.3-3b)