Physical Chemistry Third Edition

396 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium

We have assumed that the probability distribution of each velocity component is

independent of the other two. The joint probability of three independent events is the

product of the probabilities of the three events, so that

g(v)g(|v|)g(v)f(vx)f(vy)f(vz) (9.3-7)

Since all directions are equivalent, we assert thatgwill depend only onv, the magnitude

of the vectorv. We now show that our assumptions determine the mathematical form

of the probability distributionf. We first differentiategwith respect tovx. Using the

formula for the speed, Eq. (9.2-11) and the chain rule, Eq. (B-9) of Appendix B, we

obtain

∂g

∂vx



dg

dv

∂v

∂vx



dg

dv

∂

∂vx

[

(v^2 x+v^2 y+v^2 z)^1 /^2

]



dg

dv

1

2

(v^2 x+v^2 y+v^2 z)−^1 /^2 (2vx)

dg

dv

vx

v

(9.3-8)

We divide this equation byvxto obtain

1

vx

∂g

∂vx



1

v

dg

dv

(9.3-9)

Sincevx,vy, andvzall occur in the expression for the speedvin the same way, the

corresponding equation for differentiation byvyorvzwould be the same except for

havingvyorvzin place ofvx. The right-hand side of each equation would be the same,

so that

1

vx

∂g

∂vx



1

vy

∂g

∂vy



1

vz

∂g

∂vz



1

v

dg

dv

(9.3-10)

We take the partial derivative ofgwith respect tovx:

(

∂g

∂vx

)

vy,vz

f(vy)f(vz)

df(vx)

dvx

(9.3-11)

Similar equations for (∂g/∂vy) and (∂g/∂vz) can be written, so that Eq. (9.3-10) becomes

1

vx

f(vy)f(vz)

df(vx)

dvx



1

vy

f(vx)f(vz)

df(vy)

dvy



1

vz

f(vx)f(vy)

df(vz)

dvz

(9.3-12)

Division of this equation byf(vx)f(vy)f(vz) gives

1

vxf(vx)

df(vx)

dvx



1

vyf(vy)

df(vy)

dvy



1

vzf(vz)

df(vz)

dvz

(9.3-13)

We haveseparated the variablesvx,vy, andvz. That is, each term contains only one

of the independent variables. Becausevx,vy, andvzare independent variables it is

possible to keepvyandvzconstant while allowingvxto range. The first term must

therefore be a constant function ofvx. The second term must be a constant function of

vy, and the third term must be a constant function ofvz. Each term equals the same

constant, which we denote byC. Setting the first term equal toCand multiplying by

vxf(vx), we obtain

df

dvx

Cvxf(vx) (9.3-14)