Page 80 - Essentials of physical chemistry

42 Essentials of Physical Chemistry

ð 2 ð
mv 2 mv 2 m mv 2
1 1
4
4p e 2kT v dv 2kT v dv
2 4p e
mv 0 2 2 0
ð ð :
¼ ¼
2 1 mv 2 2 1 mv 2 2
4p e 2kT v dv 4p e 2kT v dv
0 0
At this point, we have to remind the reader of our agreement; the text will explain difficult topics in
simple ways, but the student still has to use scratch paper and write the notes over. Thus, we come to
two new integral forms, which need to be learned and used. These two formulas can be derived
using integration by parts but that is very time consuming in some cases, so it is acceptable to look
up various forms of integrals and use the results in tables. Some tables of integrals have thousands of
cases, but in this text, we really only need to learn three definite integrals, the two below and
1 1
ffiffiffiffi
ð 1 . 3 . 5 .. . (2p 1) p ð p!
p
2 2
2p a x
x e 2 2 dx ¼ , x (2pþ1) a x dx ¼
e
2 pþ1 (2pþ1) 2a (2pþ2)
a
0 0
ð
1 n!
n ax
one other, which will be useful in quantum chemistry, namely, the integral x e dx ¼ (nþ1) .
0 a
We will use these formulas many times in this text so that you will become accustomed to using
them. In the case at hand, we first note that the integrands have squared exponents to the base ‘‘e,’’
but the power of x in the integrand can be odd or even. Note also that in the exponent of the
Boltzmann factor, the value of a is squared relative to the value in the final answer. The formula for
the ‘‘odd case’’ (2p þ 1) can be checked if desired using integration-by-parts for a low value of the
ffiffiffiffi
p
exponent, but the ‘‘even case’’ (2p) is more difficult and one can wonder, where the factor of p
ð
1 2
e x dx. Then, we have for the square
comes from. Consider the basic integral I ¼
0
p=2
ð ð ð ð ð ð

2
2
1 1 1 1 1
2 x 2 y 2 (x þy ) r 2
I ¼ e dx e dy ¼ e dx dy ¼ e rdu dr, so that we can
0 0 0 0 0 0
integrate over the first quadrant in polar coordinates to obtain
" #
ð 1 2 1
r
2 p r 2 p e p 0 ( 1) p
I ¼ e rdr ¼ ¼ ¼
2 2 2 2 2 4
0 0
ffiffiffiffi
p=2. When the even case is reduced to simpler form using integration-by-parts, the
p
and then I ¼
ffiffiffiffi
final step will lead to the ( p=2) factor. There will be problems at the end of this chapter to build
p
your skill in applying these formulas, but now we can apply the formulas to obtain the average
kinetic energy in the spherical system. Note that the 3D form of this integral will have
2
2
v sin (u)dv du df and after integrating over (u, f), there will still be v dv as the differential over v.
!
5=2
m 2 2kT
5
Then, we have noting ¼ etc:
2kT m
!
ffiffiffiffi
p 5=2
p 2kT
1 . 3 .
2 3
mv 2 m 3m 2kT 3kT
m
:
2 2 p ffiffiffiffi 3=2 4 m 2
¼
! ¼
¼
p 2kT
1 .
2 2 m
This result is comforting in that the direct application of Boltzmann averaging process produces the
same result as our previous phenomenological derivation from the ideal gas law. Further, we could