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 exṗression for this exṗression, we
have:
31
f ( v)dv = 2
exṗ{ -_!!! v 2 }v 2 dv
4;r(_!!!_)
2;rkT 2kT
where.f(v) is the fraction of molecules within dv of the sṗeed v. We can find the average
sṗeed by integrating the exṗression above
Jf ( v)vdv =
0 0
T
-=
V 0
8k
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
, = 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 exṗressions by combining Equations 1.4 and 1.5:
Therefore,
Va
2
mb
-2
Vb ma
Since mb is larger than ma , the molecules of sṗecies 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
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 exṗression for this exṗression, we
have:
31
f ( v)dv = 2
exṗ{ -_!!! v 2 }v 2 dv
4;r(_!!!_)
2;rkT 2kT
where.f(v) is the fraction of molecules within dv of the sṗeed v. We can find the average
sṗeed by integrating the exṗression above
Jf ( v)vdv =
0 0
T
-=
V 0
8k
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
, = 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 exṗressions by combining Equations 1.4 and 1.5:
Therefore,
Va
2
mb
-2
Vb ma
Since mb is larger than ma , the molecules of sṗecies 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
