AllChaptersCovered
f f
SOLUTION MANUAL
f
, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
f f f f f f f f f f f f
_!_ mJ7 2 =e;olecular
f f f f
2
kT=e;olecular f f
2
.-. vl: f
Assume the temperature is 22 °C. The mass of a single oxygen molecule ism = 5.14x 10-
f f f f f ff f f f f f f f f f f f f
26 f
kg . Substitute and solve:
f f f f
V=487.6 [mis]
f f f
The molecules are traveling really, fast (around the length of five football fields every second). Comm
f f f f f f f f f f f f f f f
ent:
We can get a better solution by using the Maxwell-
f f f f f f f f f
Boltzmann distribution of speeds that is sketched in Figure 1.4. Looking up the quantitative expression
f f f f f f f f f f f f f f
for this expression, we have:
f f f f f
f (v)dv =
f f f f 4;r(_!!!_) 312
exp{ -_!!! v }v dv ffff
2f
f
2f
f
2;rkT 2kT
where.f(v) is the fraction of molecules within dv of the speed v. We can find the average speed by integr
f f f f f f f f f f f f f f f f f f f
ating the expression above
f f f
Jf (v)vdv
0 0
f f
-= 0
V f
f = 8kT = 449 [m/s ] f f
0
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,f (v)dv
f mn
Q)
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, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
f f f f f f ff f f
Therefore,
Va 2
mb
V-2b ma
Since mb is larger than ma , the molecules of species A move faster on average.
f f f f f f f f f f f f f f
Q)
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Q)
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) 4
f f
SOLUTION MANUAL
f
, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
f f f f f f f f f f f f
_!_ mJ7 2 =e;olecular
f f f f
2
kT=e;olecular f f
2
.-. vl: f
Assume the temperature is 22 °C. The mass of a single oxygen molecule ism = 5.14x 10-
f f f f f ff f f f f f f f f f f f f
26 f
kg . Substitute and solve:
f f f f
V=487.6 [mis]
f f f
The molecules are traveling really, fast (around the length of five football fields every second). Comm
f f f f f f f f f f f f f f f
ent:
We can get a better solution by using the Maxwell-
f f f f f f f f f
Boltzmann distribution of speeds that is sketched in Figure 1.4. Looking up the quantitative expression
f f f f f f f f f f f f f f
for this expression, we have:
f f f f f
f (v)dv =
f f f f 4;r(_!!!_) 312
exp{ -_!!! v }v dv ffff
2f
f
2f
f
2;rkT 2kT
where.f(v) is the fraction of molecules within dv of the speed v. We can find the average speed by integr
f f f f f f f f f f f f f f f f f f f
ating the expression above
f f f
Jf (v)vdv
0 0
f f
-= 0
V f
f = 8kT = 449 [m/s ] f f
0
J
00
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>f
·- c
g>w
w f
en © ff
en u ff
Q)fff 0
(.)fff I....
0
a: 0@....
) 2
,f (v)dv
f mn
Q)
c
c
ro
..c
()
O>
c
·c
Q)
Q)
c O>f
·- c
g>w
w f
en © ff
en u ff
Q)fff 0
(.)fff I....
0
a: 0@....
) 3
, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
f f f f f f ff f f
Therefore,
Va 2
mb
V-2b ma
Since mb is larger than ma , the molecules of species A move faster on average.
f f f f f f f f f f f f f f
Q)
c
c
ro
..c
()
O>
·c
c
Q)
Q)
c O> f
·- c
g>w
w f
en © ff
en u ff
Q)fff 0
(.)fff I....
0
a: 0@....
) 4
