Physical Chemistry , 1st ed.

dimensions, we would get the same conclusions for yand z, and so we also have

two other relationships with the same constant K:



v

1

y

g

g

y

y

(

(

v

v

y

y

)

)K and 

v

1

z

g

g

z

z

(

(

v

v

z

z

)

)K (19.22)

Solving any one expression therefore gives us an understanding of all four

(including ). Let us confine ourselves to the xdimension. We want to know

what functions satisfy the expression



v

1

x

g

g

x

x

(

(

v

v

x

x

)

)K

where Kis some constant. To simplify the understanding of the derivation, let

us rewrite gx explicitly as the derivative with respect to vx:



v

1

x

K

We can rearrange this expression by collecting all terms in gx, including the dif-

ferential, on one side and all terms in vxitself, including itsdifferential, on an-

other side. We get, leaving the variable labels off the function gx:



d

g

g

x

xKv

xdvx

We integrate both sides of this equation, remembering to include an integra-

tion constant (because we are not integrating between specific limits). The in-

tegral on the left side is simply ln gx, the natural logarithm of the function gx.

The integral on the right side is a simple power function. Because we are try-

ing to isolate a form for gx, we will put the integration constant on the right

side of the equation with the vxterms. We get

ln gx^12 Kvx^2 C

where Cis some arbitrary integration constant (whose value will be deter-

mined later). Taking the inverse logarithm of both sides, we have a preliminary

form for gx:

gxe(1/2)Kvx

(^2) C
e(1/2)Kvx
2
eC (19.23)
That is, the distribution function gxis an exponential function involving the
square of the velocity. Since eCis just some constant, we can define eCas the
constant Aand write equation 19.23 as
gxAe(1/2)Kvx
2
(19.24)
All that remains are two issues. First, we point out that the above derivation
is also applicable to the yand zdimensions, so we can also say that
gyAe(1/2)Kvy
2
gzAe(1/2)Kvz
2
The constants Aand Kare the same for all three equations, since we are as-
suming that each dimension is equivalent. Second, we need to determine what
Aand Kare. It turns out that they are related. For starters, remember that
equation 19.15 requires that


vx
gx(vx) dvx 1

g
x
v
(v
x
x)

gx(vx)
660 CHAPTER 19 The Kinetic Theory of Gases