398 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium
Identification of the Parameterb
We need some additional information before we can identify the parameterb. We accept
as an experimental fact that the energy of a monatomic dilute gas is given byE(T)−E 0
3
2
nRT3
2
NkBT (9.3-24)wherekBis Boltzmann’s constant, equal toR/NAv, whereTis the absolute temperature,
wherenis the amount of gas in moles, and whereE 0 is a constant. We show in Chapter 2
that this assertion can be deduced from heat capacity data on monatomic gases. Because
a constant can be added to the energy without physical effect, we can choose to set the
constantE 0 equal to zero.
Viewed microscopically, the energy of the model gas isNtimes the mean molecular
energy,〈ε〉:EN〈ε〉N〈κ〉N1
2
m〈
v^2〉
(9.3-25)
where〈
v^2〉
is the mean of the square of the speed and where we have set the potential
energy equal to zero. The mean molecular kinetic energy is〈κ〉m
2〈
v^2 x+v^2 y+v^2 z〉
m
2(〈
v^2 x〉
+
〈
v^2 y〉
+
〈
v^2 z〉)
(9.3-26)
Because thex,y, andzvelocity component probability distributions are the same
function, the three terms in Eq. (9.3-26) will be equal to each other after averaging,
and we can write〈κ〉3 m
2〈
v^2 x〉
(
3 m
2)(
b
2 π) 3 / 2 ∫∞
−∞v^2 xe−bv(^2) x/ 2
dvx
∫∞
−∞v^2 xe−bv(^2) y/ 2
dvy
∫∞
−∞v^2 xe−bv(^2) z/ 2
dvz
(
3 m
2)(
b
2 π) 1 / 2 ∫∞
−∞v^2 xe−bv(^2) x/ 2
dvx (9.3-27)
The last two integrals in the first version of this equation cancel two of the factors of
(b/ 2 π)^1 /^2. We look up the remaining integral in Appendix C and obtain
〈ε〉
(
3 m
2)(
b
2 π) 1 / 2 ∫∞
−∞v^2 xe−bv(^2) x/ 2
dvx
2
(
3 m
2)(
b
2 π) 1 / 2 ∫∞
0v^2 xe−bv(^2) x/ 2
dvx
(
3 m
2)(
b
2 π) 1 / 2
1
4
(
8 π
b^3) 1 / 2
3 m
2 b(9.3-28)
where we have used the fact that this integral from−∞to∞is twice the integral from
0to∞, because the integrand is an even function (has the same value for−vxas for
vx). We now haveEN
3 m
2 b