Physical Chemistry Third Edition

9.3 The Velocity Probability Distribution 397

Equation (9.3-14) is adifferential equation. Solving it means finding a functionfthat

satisfies the equation. It can be solved by a different version of separation of variables,

in which we separate a dependent variable from an independent variable. We multiply

Eq. (9.3-14) bydvxand divide it byf(vx):

1

f

df

dvx

dvxCvxdvx (9.3-15)

We recognize (df/d vx)dvxasdfand write

1

f

dfCvxf(vx) (9.3-16)

An indefinite integration of both sides of Eq. (9.3-16) gives

ln(f)

Cv^2 x

2

+A (9.3-17)

whereAis a constant of integration. We take the exponential (antilogarithm) of each

side of this equation to obtain

f(vx)eAeCv

(^2) x/ 2
(9.3-18)
We require thatfbe normalized:
∫∞
−∞
f(vx)dvx

∫∞

−∞

eAeCv

(^2) x/ 2
dvx 1 (9.3-19)
Because we are using nonrelativistic classical mechanics, speeds greater than the speed
of light are not excluded and we must take infinite limits for the integrations. The
constantCmust be negative, because otherwise the integrand in Eq. (9.3-19) would
grow without bound for large magnitudes ofvxand the integral would diverge (become
infinite). We letb−Cso thatbis positive. We eliminate the parameterAby the
normalization condition
1 eA

∫∞

−∞

e−bv

(^2) x/ 2
dvxeA

(

2 π

b

) 1 / 2

(9.3-20)

where we have looked up the definite integral in Appendix C. Equation (9.3-18) now

becomes

f(vx)

(

b

2 π

) 1 / 2

e−bv

(^2) x/ 2
(9.3-21)
The probability distribution for all three components becomes
g(v)f(vx)f(vy)f(vz)

(

b

2 π

) 3 / 2

e−bv

(^2) x/ 2
e−bv
(^2) y/ 2
e−bv
(^2) z/ 2



(

b

2 π

) 3 / 2

e−b(v

(^2) x+v (^2) y+v (^2) z)/ 2


(

b

2 π

) 3 / 2

e−bv

(^2) / 2
(9.3-22)
Becausev^2 x+v^2 y+v^2 zv^2 , the square of the speed,gdepends only on the speed,
as we anticipated. It can also be written as a function of the kinetic energy of the
particle,ε:
g(v)

(

b

2 π

) 3 / 2

e−bε/m (9.3-23)