Engineering and Chemical Thermodynamics 2nd Edition Koretsky Complete Solutions Manual and Worked Problems 2026

All Chapters Covered




SOLUTION MANUAL

, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:

1 2
mV  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
 



 
 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
 449 m/s
8kT
V  0


m
 f (v)dv
0
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2

, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
 2 3kT  2 3kT
Va  m Vb  m
a b


Therefore,

V a2 mb
2 
Vb ma

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