CHAPTER 14: Kinetic Theory of Gases (3 Hours)

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CHAPTER 14 Kinetic Theory of Gases(3 Hours)
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Learning Outcome
14.1 Ideal gas equation (1 hour)

• At the end of this chapter, students should be
  able to
• Sketch
• P-V graph at constant temperature
• V-T graph at constant pressure
• P-T graph at constant volume of an ideal gas.
• Use the ideal gas equation

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14.1 Ideal Gas Equation

• 14.1.1 Boyles law
• states The pressure of a fixed mass of gas at
  constant temperature is inversely proportional
  to its volume.

if
OR
where
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• Graphs of the Boyles law.

a.
b.
T2 gt T1
d.
c.
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14.1.2 Charless law

• states The volume of a fixed mass of gas at
  constant pressure is directly proportional to its
  absolute temperature.

If
where
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• Graphs of the Charless law.

a.
b.
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14.1.3 Gay-lussacs (pressure) law

• states The pressure of a fixed mass of gas at
  constant volume is directly proportional to its
  absolute temperature.

If
where
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• Graphs of the Gay-lussacs (pressure) law.

a.
b.
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14.1.4 Equation of state for an ideal gas

• An ideal gas is defined as a perfect gas which
  obeys the three gas laws (Boyles, Charless and
  Gay-Lussacs) exactly.
• Consider an ideal gas in a container changes its
  pressure P, volume V and temperature T as shown
  in Figure 15.1.

2nd stage
1st stage
Figure 14.1
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• In 1st stage, temperature is kept at T1 ,
• Using Boyles law
• In 2nd stage, pressure is kept constant at P2 ,
• Using Charless law
• Equating eqs. (1) and (2), thus

OR
Final
Initial
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• Consider 1 mole of gas at standard temperature
  and pressure (S.T.P.), T 273.15 K, P 101.3
  kPa and Vm 0.0224 m3
• From equation (3),
• where R is called molar gas constant and its
  value is the same for all gases.
• Thus
• For n mole of an ideal gas, the equation of state
  is written as
• where n the number of mole gas

where
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where

• If the Boltzmann constant, k is defined as
• then the equation of state becomes

OR
where
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Example 14.1
The volume of vessel A is three times of the
volume vessel B. The vessels are filled with an
ideal gas and are at a steady state. The
temperature of vessel A and vessel B are at 300 K
and 500 K respectively as shown in Figure 15.2.
If the mass of the gas in the vessel A is
m, obtain the mass of the gas in the vessel B in
terms of m.
Figure 14.2
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Solution
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Solution
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Example 14.2
Refer to Figure 15.3. Initially A contains
3.00 m3 of an ideal gas at a temperature of 250 K
and a pressure of 5.00 ? 104 Pa, while B contains
7.20 m3 of the same gas at 400 K and 2.00 ? 104
Pa. Calculate the pressure after the connecting
tap has been opened and the system reached
equilibrium, assuming that A is kept at 250 K and
B is kept at 400 K.
Figure 14.3
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Solution
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Exercise 14.1
Given R 8.31 J mol?1 K?1 and NA 6.0 ? 1023
mol?1 1. Estimate the number of molecules
in a flask of volume 5.0 ? 10?4 m3 which
contains oxygen gas at a pressure of 2.0 ?
105 Pa and temperature of 300 K. ANS. 2.41 ?
1022 molecules 2. A cylinder contains a hydrogen
gas of volume 2.40 ? 10?3 m3 at 17 ?C and 2.32 ?
106 Pa. Calculate a. the number of molecules of
hydrogen in the cylinder, b. the mass of the
hydrogen, c. the density of hydrogen under these
conditions. (Given the molar mass of hydrogen
2 g mol?1) ANS. 1.39 ? 1024 molecules 4.62 g
1.93 kg m?3
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Learning Outcome
14.2 Kinetic theory of gases (1 hour)

• At the end of this chapter, students should be
  able to
• State the assumptions of kinetic theory of gases.
• Apply the equations of ideal gas,
• and pressure ,
• in related problems.
• Explain and use root mean square (rms) speed,
• of gas molecules.

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14.2 Kinetic theory of gases

• The macroscopic behaviour of an ideal gas can be
  describe by using the equation of state but the
  microscopic behaviour only can be describe by
  kinetic theory of gases.
• 14.2.1 Assumption of kinetic theory of gases
• All gases are made up of identical atoms or
  molecules.
• All atoms or molecules move randomly and
  haphazardly.
• The volume of the atoms or molecules is
  negligible when compared with the volume occupied
  by the gas.
• The intermolecular forces are negligible except
  during collisions.
• Inter-atomic or molecular collisions are elastic.
• The duration of a collision is negligible
  compared with the time spent travelling between
  collisions.
• Atoms and molecules move with constant speed
  between collisions. Gravity has no effect on
  molecular motion.

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14.2.2 Force exerted by an ideal gas

• Consider an ideal gas of N molecules are
  contained in a cubical container of side d as
  shown in Figure 15.4.
• Let each molecule of the gas have the mass m and
  velocity v.
• The velocity, v of each molecule can be resolved
  into their components i.e. vx, vy and vz.

Figure 14.4
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• Consider, initially a single molecule moving with
  a velocity vx towards wall A and after colliding
  elastically , it moves in the opposite direction
  with a velocity ?vx as shown in Figure 15.5.
• Therefore the change in the linear momentum of
  the molecule is given by

Figure 14.5
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• The molecule has to travel a distance 2d (from A
  to B and back to A) before its next collision
  with wall A. The time taken for this movement is
• If Fx1 is the magnitude of the average force
  exerted by a molecule on the wall in the time ?t,
  thus by applying Newtons second law of motion
  gives
• For N molecules of the ideal gas,

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• where vx1 is the x component of velocity of
  molecule 1, vx2 is the x component of velocity of
  molecule 2 and so on.
• The mean (average ) value of the square of the
  velocity in the x direction for N molecules is
• Thus, the x component for the total force exerted
  on the wall of the cubical container is
• The magnitude of the velocity v is given by
• then

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• Since the velocities of the molecules in the
  ideal gas are completely random, there is no
  preference to one direction or another. Hence
• The total force exerted on the wall in all
  direction, F is given by

where
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14.2.3 Pressure of an ideal gas

• From the definition of pressure,

where
and
and
where
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• Since the density of the gas, ? is given by
• hence the equation (14.1) can be written as

where
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14.2.4 Root mean square velocity ( vrms)

• is defined as
• From the equation of state in terms of Boltzmann
  constant, k
• By equating the eqs. (14.4) and (14.2), thus
• Therefore

OR
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• Since
• therefore the equation of root mean square
  velocity of the gas molecules also can be written
  as

where
thus
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Example 14.3
Eight gas molecules chosen at random are found to
have speeds of 1,1,2,2,2,3,4 and 5 m s?1.
Determine a. the mean speed of the molecules, b.
the mean square speed of the molecules, c. the
root mean square speed of the molecules. Solution
a.
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Solution b.
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Example 14.4
A cylinder of volume 0.08 m3 contains oxygen gas
at a temperature of 280 K and pressure of 90 kPa.
Determine a. the mass of oxygen in the
cylinder, b. the number of oxygen molecules in
the cylinder, c. the root mean square speed of
the oxygen molecules in the
cylinder. (Given R 8.31 J mol?1 K?1, k 1.38 ?
10?23 J K?1, molar mass of oxygen, M 32 g
mol?1, NA 6.02 ? 1023 mol?1) Solution a.
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Solution .
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Exercise 14.2
Given R 8.31 J mol?1 K?1, Boltzmann constant, k
1.38?10?23 K?1 1. In a period of 1.00 s, 5.00 ?
1023 nitrogen molecules strike a wall with an
area of 8.00 cm2. If the molecules move with a
speed of 300 m s?1 and strike the wall head-on in
the elastic collisions, determine the pressure
exerted on the wall. (The mass of one N2
molecule is 4.68 ? 10?26 kg) ANS. 17.6
kPa 2. Initially, the r.m.s. speed of an atom of
a monatomic ideal gas is 250 m s?1. The pressure
and volume of the gas are each doubled while the
number of moles of the gas is kept constant.
Calculate the final translational r.m.s. speed of
the atoms. ANS. 500 m s?1 3. Given that the
r.m.s. of a helium atom at a certain temperature
is 1350 m s?1, determine the r.m.s. speed of an
oxygen (O2) molecule at this temperature. (The
molar mass of O2 is 32.0 g mol?1 and the molar
mass of He is 4.00 g mol?1) ANS. 477 m s?1
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Learning Outcome
14.3 Molecular kinetic energy and internal energy

• At the end of this chapter, students should be
  able to
• Explain and use translational kinetic energy of
  gases,
• State the principle of equipartition of energy.
• Define degree of freedom.
• State the number of degree of freedom for
  monoatomic, diatomic and polyatomic gas
  molecules.
• Explain internal energy of gas and relate the
  internal energy to the number of degree of
  freedom.
• Explain and use internal energy of an ideal gas

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14.3.1 Molecular kinetic energy

• From equation (14.1), thus
• This equation shows that
• Rearrange equation (14.5), thus

P increases (?) When
and
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and
where
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• For N molecules of an ideal gas in the cubical
  container, the total average (mean) translational
  kinetic energy, E is given by

OR
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Principle of equipartition of energy

• States the mean (average) kinetic energy of
  every degrees of freedom of a molecule is
  
• Therefore

Mean (average) kinetic energy per molecule
OR
Mean (average) kinetic energy per mole
where
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Degree of freedom ( f )

• is defined as a number of independent ways in
  which an atom or molecule can absorb or release
  or store the energy.
• Monoatomic gas (e.g. He, Ne, Ar)
• The number of degrees of freedom is 3 i.e. three
  direction of translational motion where
  contribute translational kinetic energy as shown
  in Figure 15.6.

Figure 15.6
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• Diatomic gas (e.g. H2, O2, N2)
• The number of degrees of freedom is
• Polyatomic gas (e.g. H2O, CO2, NH3)
• The number of degrees of freedom is

Translational kinetic energy
Rotational kinetic energy
Figure 14.7
Figure 14.8
Translational kinetic energy
Rotational kinetic energy
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• Table 14.1 shows the degrees of freedom for
  various molecules.

Molecule Example Degrees of Freedom ( f ) Degrees of Freedom ( f ) Degrees of Freedom ( f ) Average kinetic energy per molecule,ltKgt
Molecule Example Translational Rotational Total Average kinetic energy per molecule,ltKgt



Monatomic
He
Diatomic
H2
Polyatomic
H2O
(At temperature of 300 K)
Table 14.1
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• Degrees of freedom depend on the absolute
  temperature of the gases.
• For example Diatomic gas (H2)
• Hydrogen gas have the vibrational kinetic energy
  (as shown in Figure 15.9) where contribute 2
  degrees of freedom which correspond to the
  kinetic energy and the potential energy
  associated with vibrations along the bond between
  the atoms.
• when the temperature,
• At 250 K
• At 250 750 K
• At gt750 K

Figure 15.9
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Example 14.5
A vessel contains hydrogen gas of 2.20 ? 1018
molecules per unit volume and the mean
square speed of the molecules is 4.50 km s?1
at a temperature of 50 ?C. Determine a. the
average translational kinetic energy of a
molecule for hydrogen gas, b. the pressure
of hydrogen gas. (Given the molar mass of
hydrogen gas 2 g mol?1, NA
6.02 ? 1023 mol?1 and k 1.38 ? 10?23 J
K?1) Solution
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Solution
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• 14.3.2 Internal energy, U
• is defined as the sum of total kinetic energy and
  total potential energy of the gas molecules.
• But in ideal gas, the intermolecular forces are
  assumed to be negligible thus the potential
  energy of the molecules can be neglected. Thus
  for N molecules,
• For N molecules of monoatomic gas ,

and
OR
OR
where
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Example 14.7
Neon is a monoatomic gas. Determine the internal
energy of 1 kg of neon gas at temperature of 293
K. Molar mass of neon is 20 g. Solution
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