All Chapters
Covered
SOLUTION MANUAL
, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
_!_ mJ7 2 = e;olecular
2
kT =e;olecular
2
.-. v l:
Assume the temperature is 22 °C. The mass of a single oxygen molecule is m = 5.14 x
10-26 kg . Substitute and solve:
V = 487.6 [mis]
The molecules are traveling really, fast (around the length of five football fields
every second). Comment:
We can get a better solution by using the Maxwell-Boltzmann distribution of speeds
that is sketched in Figure 1.4. Looking up the quantitative expression for this
expression, we have:
31
f ( v)dv = 2
exp -_!!! v 2 }v 2 dv
4;r(_!!!_) {
2;rkT 2kT
where.f(v) is the fraction of molecules within dv of the speed v. We can find the
average speed by integrating the expression above
J
00
-=
V 0
f ( v)vdv
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>
·- c
g> w
w
en ©
en u
Q) 0
(.) I....
0
a: 0@....
) 2
,
=
8kT =
449
[m/
s]
f (v)dv mn
J
00
0
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>
·- c
g> w
w
en ©
en u
Q) 0
(.) I....
0
a: 0@....
) 3
, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
Therefore,
2
Va mb
V-2b ma
Since mb is larger than ma , the molecules of species A move faster on average.
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>
·- c
g> w
w
en ©
en u
Q) 0
(.) I....
0
a: 0@....
) 4
Covered
SOLUTION MANUAL
, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
_!_ mJ7 2 = e;olecular
2
kT =e;olecular
2
.-. v l:
Assume the temperature is 22 °C. The mass of a single oxygen molecule is m = 5.14 x
10-26 kg . Substitute and solve:
V = 487.6 [mis]
The molecules are traveling really, fast (around the length of five football fields
every second). Comment:
We can get a better solution by using the Maxwell-Boltzmann distribution of speeds
that is sketched in Figure 1.4. Looking up the quantitative expression for this
expression, we have:
31
f ( v)dv = 2
exp -_!!! v 2 }v 2 dv
4;r(_!!!_) {
2;rkT 2kT
where.f(v) is the fraction of molecules within dv of the speed v. We can find the
average speed by integrating the expression above
J
00
-=
V 0
f ( v)vdv
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>
·- c
g> w
w
en ©
en u
Q) 0
(.) I....
0
a: 0@....
) 2
,
=
8kT =
449
[m/
s]
f (v)dv mn
J
00
0
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>
·- c
g> w
w
en ©
en u
Q) 0
(.) I....
0
a: 0@....
) 3
, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
Therefore,
2
Va mb
V-2b ma
Since mb is larger than ma , the molecules of species A move faster on average.
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>
·- c
g> w
w
en ©
en u
Q) 0
(.) I....
0
a: 0@....
) 4
