Engineering and Chemical Thermodynamics 2nd Edition by Koretsky | Complete Solutions Manual with Step-by-Step Answers

1.2


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2

, 1.3


An approximate solution can be found if we combine Equations 1.4 and 1.5:

1 V emolecular
m
2
2 k
molecular
3 kT 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:

 m v 2 v2 dv
3/ 2
 
f (v)dv 4 m  exp 
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 (v)vdv 8kT
V  0
 449 m/s

m
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f (v)dv
0
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3

, 1.4
Derive the following expressions by combining Equations 1.4 and 1.5:

 3kT  3kT
Va2  Vb2 
ma mb

Therefore,

Va2 m
2  b
Vb ma

Since mb is larger than ma , the molecules of species A move faster on average.
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4

, 1.5
We have the following two points that relate the Reamur temperature scale to the Celsius scale:

0 ºC, 0 º Reamur and 100 º C, 80 º Reamur 
Create an equation using the two points:

T ºReamur0.8 Tº Celsius

At 22 ºC,

T 17.6 º Reamur
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5