All Chapters Covered
SOLUTION MANUAL
, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
1
mV 2 = emolecular
2 k
3
kT = emolecular
2 k
3kT
V
m
Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m = 5.14 10−26 kg .
Substitute and solve:
V = 487.6 m/s
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:
m m 2 2
f (v)dv = 4 exp − v v dv
2kT 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
f (v)vdv
= 449 m/s
8kT
V = 0
=
m
f (v)dv
0
Process Engineering Channel
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2
, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
2 3kT 2 3kT
Va = Vb =
ma mb
Therefore,
V2 mb
a =
Vb2 ma
Since mb is larger than ma, the molecules of species A move faster on average.
Process Engineering Channel
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3
, 1.4
We have the following two points that relate the Reamur temperature scale to the Celsius scale:
(0 º C, 0 º Reamur) and (100 º C, 80 º Reamur)
Create an equation using the two points:
T (º Reamur) = 0.8 T(º Celsius)
At 22 ºC,
T = 17.6 º Reamur
Process Engineering Channel
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4
SOLUTION MANUAL
, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
1
mV 2 = emolecular
2 k
3
kT = emolecular
2 k
3kT
V
m
Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m = 5.14 10−26 kg .
Substitute and solve:
V = 487.6 m/s
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:
m m 2 2
f (v)dv = 4 exp − v v dv
2kT 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
f (v)vdv
= 449 m/s
8kT
V = 0
=
m
f (v)dv
0
Process Engineering Channel
@ProcessEng
2
, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
2 3kT 2 3kT
Va = Vb =
ma mb
Therefore,
V2 mb
a =
Vb2 ma
Since mb is larger than ma, the molecules of species A move faster on average.
Process Engineering Channel
@ProcessEng
3
, 1.4
We have the following two points that relate the Reamur temperature scale to the Celsius scale:
(0 º C, 0 º Reamur) and (100 º C, 80 º Reamur)
Create an equation using the two points:
T (º Reamur) = 0.8 T(º Celsius)
At 22 ºC,
T = 17.6 º Reamur
Process Engineering Channel
@ProcessEng
4
