Solution Manual for Engineering and Chemical Thermodynamics, 2nd Edition by Michael Koretsky

https://www.stuvia.com/en-us/doc/8356075/solution-manual-for-engineering-and-chemical-thermodynamics-
2nd-edition-koretsky-all-9-chapters-covered

All Chapters Covered




SOLUTION MANUAL

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

1 emolecul
mV
2
2 ar k
3
kT emolecular
2 k

 3kT
V  m

26
Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m 5.14 10 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
V 
0 8kT
m 449 m/s



f (v)dv
0
Process Engineering
Channel




2

, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:

 2
V  2
3kT
 V 3kT
a b
ma mb

Therefore,
2
Va
 2  mb

Vb ma

Since mb is larger than m a , the molecules of species A move faster on average.
Process Engineering
Channel




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




4