All Chapters Covered
SOLUTION MANUAL
, @SOLUTIONSSTUDY
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:
f ( v)dv = 4;r(_!!!_) 312
2 2
exp{ -_!!! v }v dv
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
Jf ( v)vdv =
0 0
-=
V 0
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@....
) 2
, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
Therefore,
Va
2
mb
-2
Vb 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@....
) 3
, @SOLUTIONSSTUDY
1.4
We have the following two points that relate the Reamur temperature scale to the Celsius scale:
(o °C, 0 °Reamur) and (100 °C, 80 °Reamur)
Create an equation using the two points:
T (0 Reamur) = 0.8 T(° Celsius)
At 22 °C,
T = 17.6 °Reamur
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
, @SOLUTIONSSTUDY
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:
f ( v)dv = 4;r(_!!!_) 312
2 2
exp{ -_!!! v }v dv
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
Jf ( v)vdv =
0 0
-=
V 0
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@....
) 2
, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
Therefore,
Va
2
mb
-2
Vb 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@....
) 3
, @SOLUTIONSSTUDY
1.4
We have the following two points that relate the Reamur temperature scale to the Celsius scale:
(o °C, 0 °Reamur) and (100 °C, 80 °Reamur)
Create an equation using the two points:
T (0 Reamur) = 0.8 T(° Celsius)
At 22 °C,
T = 17.6 °Reamur
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
