Chapter 19: The Kinetic Theory of Gases
Questions and Example Problems
N
n
p
NA

Msam
pV  nRT  Nk BT
Mmolar
2
nMv rms
3V
v rms 
Q V  nC V T
  CP / C V
3RT
M

3k B T
QP  nCP T
m
 Vf 

 Vo 
W  nRT ln 
K avg  32 k BT
Cp  C V  R
pV   constant

1
2d N / V
Eint  nC V T
2
TV  1  constant
Questions
a. Why does the boiling temperature of a liquid increase with pressure?
b. H2 is the most plentiful element in the universe, yet our atmosphere is mostly N2 and
O2. Why doesn’t it contain H2?
c. When two gases are mixed, if they are to be in thermal equilibrium they must have
the same average molecular speed. Is the statement correct? Why or why not?
d. If you fill a saucer with water at room temperature, the water, under normal
conditions, will evaporate completely. It is easy to believe that some of the more
energetic molecules can escape from the water surface, but how can all of them
eventually escape? Many of then – in fact the vast majority – do not have enough
energy to do so.
e. Suppose that you want to heat a gas so that its temperature will be as high as
possible. Would you heat it under conditions of constant pressure or constant
volume? Why?
f. The plunger of a bicycle tire pump is pushed down rapidly with the end of the pump
sealed so that no air escapes. And there is little time for heat to flow through the
cylinder wall. Explain why the cylinder of the pump becomes warm to the touch.
Ignore friction and assume that air behaves as in ideal gas
Example 19.1
Water has a molar mass of 18 g/mol.
a. How many water molecules are there in your body? (Assume that you are nearly all
water.)
b. How many drops of water are there in all the oceans of the world? (The mass of the
world’s oceans is about 1021 kg)
c. Which of these two numbers from (a) and (b) is the larger?
Example 19.2
A quantity of ideal gas at 10oC and 100 kPa occupies a volume of 2.5 m3.
a. How many moles of the gas are present?
b. If the pressure is now raised to 300 kPa and the temperature is raised to 30oC, how
much volume does the gas occupy?
19-1
Example 19.3
A sample of ideal gas expands from an initial pressure and volume of 32 atm and 1.0 L
to a final volume of 2.0 L. The initial temperature of the gas is 300 K. What are the final
pressure and temperature of the gas and how much work is done by the gas during the
expansion, if the expansion is isothermal?
Example 19.4
Determine the average value of the translational KE of the molecules of an ideal gas at
(a) 20oC (room temperature). (b) What are the rms values for (b) H2 and N2 at room
temperature (20oC), (c) H2 in outer space, and (d) free electrons in the sun atmosphere
(2×106 K)?
Example 19.5
One mole of an ideal diatomic gas goes from 1 to 3 along two different
paths as shown.
a. Rank the (i) change in internal energy and (ii) heat transfers for
transitions 1→3 and 1→2→3. Explain your reasoning.
b. During the transition along 1→3, what is the (iii) change in internal
energy of the gas and (iv) heating done on the gas?
c. Repeat part (b) for transition 1→2→3.
Example 19.6
The figure shows a thermodynamics process followed by 120 mg of helium.
a. Determine the pressure, temperature, and volume of the gas at points 1,
2, and 3. Put your results in a table for easy reading.
b. How much work is done on the gas during each of the three segments?
c. How much heat energy is transferred to or from the gas during each of the
three segments?
Solution
I need to figure out the signs and draw energy bar diagrams.
a. Helium gas is a monatomic gas and the given number of moles is
19-2
mHe 120  10 3 g

 0.030 mol  n
MHe
4 g/mol
The underlying structure is the 1st law but the details are given by the IGL. In other
words, to get the (p, V, T) at anyone state, is use the IGL equations with the IGL process
to setup the equations. Let’s setup a table and fill-in the values to make sure I get them
all.
State-1:
nRT1
p1V1  nRT1  p1 
mol
V1 nT0.030
1 133  273  406K
3
Transition 1→2 is
V1 10 cm3 103 m3
n
0.030  8.31 406
Pa  1.01 105 Pa  1 atm
103
isochoric (constant volume process) so that the IGL implies
T T
p
V  constant 
 1  2 
 T2  2 T1
 5T1  2030 K
p1 p2
p1 T1 406K

p2 /p1 5
Transition 2→3 is isothermal so
T  constant 
 p2 V2  p3 V3 
 V3 
p2
V1
 5V1  5  103 m3
p1 T1 406K
p2 /p1 5
I now summarize everything into my table
p(105Pa) T(K)
V(m3)
1
1
406
10-3
2
5
2030
10-3
3
1
2030
5×10-3
b. To determine the work done through each transition, we do the following:
 Transition 1→2 is isochoric: W 12 = 0
 Transition 2→3 is isothermic:
V 
5
W23  p2 V2 ln  3   5  105 Pa  103 m3  ln    815 J  W23
 1
 V2 

Transition 2→3 is isobaric:
W31  p3 (V1  V3 )  1.01 105  (1  5)  103  405 J  W31
c. To determine the heating through each transition, we do the following:
 Transition 1→2 is isochoric heating:
Q12  Eint  nCV T  32 nR(T2  T1 )  32 (p2 V2  p1V1 )

3
2
5  10
5

 103  1 105  103 Pa  m3  609 J  Q12

Transition 2→3 is isothermic heating:
Q23  W23  815 J  Q23

Transition 2→3 is isobaric heating:
Q31  nCP T  52 nR(T1  T3 )  32 (p1V1  p3 V3 )

5
2


 1 105  103  5  105  103 J  1013 J  Q31
19-3
Example 19.7
In a bottle of champagne, the pocket of gas (primarily carbon dioxide) between the liquid
and the cork is at pressure of p1 = 5.00 atm and temperature of 5oC. When the cork is
pulled from the bottle, the gas undergoes an adiabatic expansion until its pressure
matches the ambient air pressure of 1.00 atm.
a. Draw a pV-diagram for the situation and what is the ratio of the molar specific heats?
Explain.
b. What is its temperature at the end of the adiabatic expansion?
Solution
Let p1, V1, and T1 represent the pressure, volume, and temperature of the initial state of
the gas, and let p2, V2, and T2 be the pressure, volume, and temperature of the final
state. Since the process is adiabatic p1V1 = p2V2. Combining with the ideal gas law, pV
= nRT, we obtain
p1V1  p1(T1 / p1 )  p11 T1  constant 
 p11 T1  p12 T2
With  = 4/3 which gives (1−)/ =−1/4, the temperature at the end of the adiabatic
expansion is
p 
T2   1 
 p2 
1

 5.00 atm 
T1  

 1.00 atm 
1/4
(278 K)  186 K  87C  T2 .
19-4
Example A
The temperature of 3.00 mol of a gas with CV = 6.00 cal/mol∙K is to be raised 50.0 K.
a. If the process is at constant volume, draw a pV-diagram of the situation and
determine Q, W, ∆Eint, and ∆K (total translational kinetic energy) of the gas?
b. Repeat (a) if instead the process is at constant pressure.
c. Repeat (a) if instead the process is adiabatic.
Solution
I would expect that because the internal energy and the translation kinetic energy are
temperature dependent, these values will be the same for all three processes.
Furthermore, since the question specifically states translational kinetic energy, it means
monatomic: CV =3R/2. The given quantities are T = 50 K, n = 3.0 mol, and converting
joules to calories in the ideal gas constant value gives R ≈ 2.0 cal/mol·K.

a. Constant volume process: since the work done by the gas is W = 0, and the
change in the internal energy is
Eint  nCV T  3.0  6.0  50  900 cal  Eint
The first law gives
Eint  Q  W  Q  900 cal
The change in the total translational kinetic energy is
K  32 nRT  32  3.0  2.0  50  450cal  K
b. Constant pressure process: since the change in internal energy is the same as
above (∆Eint = 900 cal), the work done at constant pressure is
W  pV pV nRT  nRT  3.0  2.0  50  300 cal  W
The first law gives
Q  Eint  W  900  300  1200cal  Q
The change in the translational kinetic energy is
K  32 nRT  32  3.0  2.0  50  450cal  K
c. Adiabiatic process: once again the change in internal energy is the same as above
(∆Eint = 900 cal) but by definition of adiabatic Q =0. The first law leads to
Eint  Q  W 
 W  900 cal
The change in the translational kinetic energy is
K  32 nRT  32  3.0  2.0  50  450cal  K
Example B
An ideal gas undergoes an adiabatic compression from (p, V, T) = (1.0 atm, 1.0×106 L,
0.0oC) to (p, V) = (1.0×105 atm, 1.0×103 L). (a) Is the gas monatomic, diatomic, or
polyatomic? (b) What is its final temperature? (c) How many moles of gas are present?
(d) What is the total translational kinetic energy per mole before and after the
compression?
Solution
a. We use p1V1 = p2V2 to compute :


ln  V V  ln 1.0  10
ln p1 p2 
2
1

  5  monatomic
L 3
ln 1.0atm 1.0  105 atm
3
L 1.0  106
b. Using the gas law in ratio form, the final temperature is
19-5
T2  T1
p2 V2
p1V1
1.0  10
  273K 
5

atm 1.0  10 3 L
1.0 atm  1.0  10
6
L

  2.7  10
4
K
c. The number of moles of gas present is
1.01 105 Pa 1.0  103 cm3
p1V1
n

 4.5  104 mol
RT1
 8.31 J/mol  K  273K 
d. The total translational energy per mole before the compression is
K1  32 RT1  32  8.31 J/mol  K  273K   3.4  103 J



and after the compression,
K 2  32 RT2  32  8.31 J/mol  K  2.7  10 4 K  3.4  10 5 J


19-6