All Chapters Covered
SOLUTION MANUAL
,
An approximate solution can be found if we combine Equations 1.4 and 1.5:
1 V emolecul
m
2 ar k
2
3
kT molecular
e
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:
3/ 2
m
f (v)dv 4 m exp v 2 v2 dv
2 kT 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
8kT
449 m/s
(v)vdv m
V 0
Process Engineering Channel
f
(v)dv
0
@ProcessEng
2
,
Derive the following expressions by combining Equations 1.4 and 1.5:
3kT 3kT
Va2 Vb2
ma mb
Therefore,
Va2 mb
2
Vb
ma
Since mb is larger than m a , the molecules of species A move faster on average.
Process Engineering Channel
@ProcessEng
3
,
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
SOLUTION MANUAL
,
An approximate solution can be found if we combine Equations 1.4 and 1.5:
1 V emolecul
m
2 ar k
2
3
kT molecular
e
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:
3/ 2
m
f (v)dv 4 m exp v 2 v2 dv
2 kT 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
8kT
449 m/s
(v)vdv m
V 0
Process Engineering Channel
f
(v)dv
0
@ProcessEng
2
,
Derive the following expressions by combining Equations 1.4 and 1.5:
3kT 3kT
Va2 Vb2
ma mb
Therefore,
Va2 mb
2
Vb
ma
Since mb is larger than m a , the molecules of species A move faster on average.
Process Engineering Channel
@ProcessEng
3
,
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
