Chapter 1 Solutions
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
,1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
mV = ekmolecular
1 2
2
kT = ekmolecular
3
2
∴V ≈
3kT
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 = 4π exp− v v dv
3/ 2
2πkT 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
∞
V = = = 449 [m/s]
πm
8kT
∫ f (v)dv
∞
0
0
2
,1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
Va2 = Vb2 =
3kT 3kT
ma mb
Therefore,
=
Va2 mb
Vb2 ma
Since m b is larger than m a , the molecules of species A move faster on average.
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
4
Engineering and Chemical Thermodynamics
Wyatt Tenhaeff
Milo Koretsky
Department of Chemical Engineering
Oregon State University
,1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:
mV = ekmolecular
1 2
2
kT = ekmolecular
3
2
∴V ≈
3kT
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 = 4π exp− v v dv
3/ 2
2πkT 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
∞
V = = = 449 [m/s]
πm
8kT
∫ f (v)dv
∞
0
0
2
,1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
Va2 = Vb2 =
3kT 3kT
ma mb
Therefore,
=
Va2 mb
Vb2 ma
Since m b is larger than m a , the molecules of species A move faster on average.
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
4
