The Kinetic Theory of Gases

1
Chapter 21

• The Kinetic Theory of Gases

2

• The description of the behavior of a gas in
  terms of the macroscopic state variables P, V,
  and T can be related to simple averages of
  microscopic quantities such as the mass and speed
  of the molecules in the gas. The resulting theory
  is called the Kinetic Theory of Gases.
• From the point of view of kinetic theory, a gas
  consist of a large number of molecules making
  elastic collisions with each other and with the
  walls of container.

3

• In the absence of external forces (we may
  neglect gravity), there no preferred position of
  the molecule in the container, and no preferred
  directions for its velocity vector.
• The molecules are separated, on the average, by
  distances that are large compared with their
  diameters, and they exert no forces on each other
  except when they collide.

4

• This final assumption is equivalent to assuming
  a very low gas density, which is the same as
  assuming that the gas is an ideal gas.
• Because momentum is conserved, the collisions
  the molecules make with each other have no effect
  on the total momentum in any directions thus
  such collisions may be neglected.

5
Molecular Model of an Ideal Gas

• The model shows that the pressure that a gas
  exerts on the walls of its container is a
  consequence of the collisions of the gas
  molecules with the walls
• It is consistent with the macroscopic description
  developed earlier

6
Assumptions for Ideal Gas Theory

• The gas consist of a very large number of
  identical molecules, each with mass m but with
  negligible size (this assumption is approximately
  true when the distance between the molecules is
  large, compared to the size).
  The consequence ? for negligible size
  molecules we can neglect the intermolecular
  collisions.
• The molecules dont exert any
  action-at-distance forces on each other. This
  means there are no potential energy changes to be
  considered, so each molecules kinetic energy
  remains unchanged. This assumption is fundamental
  to the nature of an ideal gas.

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Assumptions for Ideal Gas Theory

• The molecules are moving in random
  directions with a distribution of speeds that is
  independent of direction.
• Collisions with the container walls are
  elastic, conserving the molecules energy and
  momentum.

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Pressure and Kinetic Energy

• Assume a container is a cube
• Edges are length d
• Look at the motion of the molecule in terms of
  its velocity components
• Look at its momentum and the average force

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Pressure and Kinetic Energy

• Since the collision is elastic, the y - component
  of molecules velocity remains unchanged, while
  the x - component reverses sign. Thus the
  molecule undergoes the momentum change of
  magnitude 2mvx.
• After colliding with the right hand wall, the x -
  component of molecules velocity will not change
  until it hits the left-hand wall and its x -
  velocity will again reverses.
• ?t 2d / vx

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Pressure and Kinetic Energy

• The average force, due to the each molecule on
  the wall
• To get the total force on the wall, we sum over
  all N molecules. Dividing by the wall area A then
  gives the force per unit area, or pressure

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Pressure and Kinetic Energy
is just the average of the squares of x
components of velocities
Since,
12
Pressure and Kinetic Energy
Since, the molecules are moving in random
directions the average quantities ,
, and must be equal and the
average of the molecular speed
and v2 3vx2, or vx2 v2/3 . Then the
expression for pressure
13
Pressure and Kinetic Energy

• The relationship can be
  written
• This tells us that pressure is proportional to
  the number of molecules per unit volume (N/V) and
  to the average translational kinetic energy of
  the molecules

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Pressure and Kinetic Energy

• This equation also relates the macroscopic
  quantity of pressure with a microscopic quantity
  of the average value of the square of the
  molecular speed
• One way to increase the pressure is to increase
  the number of molecules per unit volume
• The pressure can also be increased by increasing
  the speed (kinetic energy) of the molecules

15

• A 2.00-mol sample of oxygen gas is confined to a
  5.00-L vessel at a pressure of 8.00 atm. Find the
  average translational kinetic energy of an oxygen
  molecule under these conditions.

16

• A 2.00-mol sample of oxygen gas is confined to a
  5.00-L vessel at a pressure of 8.00 atm. Find the
  average translational kinetic energy of an oxygen
  molecule under these conditions.

17
Molecular Interpretation of Temperature

• We can take the pressure as it relates to the
  kinetic energy and compare it to the pressure
  from the equation of state for an ideal gas
• Therefore, the temperature is a direct measure of
  the average molecular kinetic energy

18
Molecular Interpretation of Temperature

• Simplifying the equation relating temperature and
  kinetic energy gives
• This can be applied to each direction,
• with similar expressions for vy and vz

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A Microscopic Description of Temperature

• Each translational degree of freedom contributes
  an equal amount to the energy of the gas
• A generalization of this result is called the
  theorem of equipartition of energy

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Theorem of Equipartition of Energy

• Each degree of freedom contributes ½kBT to the
  energy of a system, where possible degrees of
  freedom in addition to those associated with
  translation arise from rotation and vibration of
  molecules

21
Total Kinetic Energy

• The total kinetic energy is just N times the
  kinetic energy of each molecule
• If we have a gas with only translational energy,
  this is the internal energy of the gas
• This tells us that the internal energy of an
  ideal gas depends only on the temperature

22
Molar Specific Heat

• Several processes can change the temperature of
  an ideal gas
• Since ?T is the same for each process, ?Eint is
  also the same
• The heat is different for the different paths
• The heat associated with a particular change in
  temperature is not unique

23
Molar Specific Heat

• We define specific heats for two processes that
  frequently occur
• Changes with constant volume
• Changes with constant pressure
• Using the number of moles, n, we can define molar
  specific heats for these processes

24
Molar Specific Heat

• Molar specific heats
• Q nCV ?T for constant-volume processes
• Q nCP ?T for constant-pressure processes
• Q (under constant pressure) must account for both
  the increase in internal energy and the transfer
  of energy out of the system by work
• Q (under constant volume) account just for change
  the internal energy
• Qconstant P gt Qconstant V for given values of n
  and ?T

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Ideal Monatomic Gas

• A monatomic gas contains one atom per molecule
• When energy is added to a monatomic gas in a
  container with a fixed volume, all of the energy
  goes into increasing the translational kinetic
  energy of the gas
• There is no other way to store energy in such a
  gas

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Ideal Monatomic Gas

• Therefore,
• ?Eint is a function of T only
• At constant volume,
• Q ?Eint nCV ?T
• This applies to all ideal gases, not just
  monatomic ones

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Monatomic Gases

• Solving
• for CV gives CV 3/2 R 12.5 J/mol . K for
  all monatomic gases
• This is in good agreement with experimental
  results for monatomic gases

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Monatomic Gases

• In a constant-pressure process, ?Eint Q W and
  
• Change in internal energy depends only on
  temperature for an ideal gas and therefore are
  the same for the constant volume process and for
  constant pressure process
• CP CV R

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Monatomic Gases

• CP CV R
• This also applies to any ideal gas
• CP 5/2 R 20.8 J/mol . K

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Ratio of Molar Specific Heats

• We can also define
• Theoretical values of CV , CP , and g are in
  excellent agreement for monatomic gases
• But they are in serious disagreement with the
  values for more complex molecules
• Not surprising since the analysis was for
  monatomic gases

31
Sample Values of Molar Specific Heats
32

• A 1.00-mol sample of air (a diatomic ideal gas)
  at 300 K, confined in a cylinder under a heavy
  piston, occupies a volume of 5.00 L. Determine
  the final volume of the gas after 4.40 kJ of
  energy is transferred to the air by heat.

33

• A 1.00-mol sample of air (a diatomic ideal gas)
  at 300 K, confined in a cylinder under a heavy
  piston, occupies a volume of 5.00 L. Determine
  the final volume of the gas after 4.40 kJ of
  energy is transferred to the air by heat.

The piston moves to keep pressure constant

34

• A 1.00-mol sample of air (a diatomic ideal gas)
  at 300 K, confined in a cylinder under a heavy
  piston, occupies a volume of 5.00 L. Determine
  the final volume of the gas after 4.40 kJ of
  energy is transferred to the air by heat.

35
Molar Specific Heats of Other Materials

• The internal energy of more complex gases must
  include contributions from the rotational and
  vibrational motions of the molecules
• In the cases of solids and liquids heated at
  constant pressure, very little work is done since
  the thermal expansion is small and CP and CV are
  approximately equal

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Adiabatic Processes for an Ideal Gas

• Assume an ideal gas is in an equilibrium state
  and so PV nRT is valid
• The pressure and volume of an ideal gas at any
  time during an adiabatic process are related by
  PV g constant
• g CP / CV is assumed to be constant during the
  process
• All three variables in the ideal gas law (P, V, T
  ) can change during an adiabatic process

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Equipartition of Energy

• With complex molecules, other contributions to
  internal energy must be taken into account
• One possible way to energy change is the
  translational motion of the center of mass

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Equipartition of Energy

• Rotational motion about the various axes also
  contributes
• We can neglect the rotation around the y axis
  since it is negligible compared to the x and z
  axes

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Equipartition of Energy

• The molecule can also vibrate
• There is kinetic energy and potential energy
  associated with the vibrations

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Equipartition of Energy

• The translational motion adds three degrees of
  freedom
• The rotational motion adds two degrees of freedom
• The vibrational motion adds two more degrees of
  freedom
• Therefore, Eint 7/2 nRT and CV 7/2 R

41
Agreement with Experiment

• Molar specific heat is a function of temperature
• At low temperatures, a diatomic gas acts like a
  monatomic gas CV 3/2 R

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Agreement with Experiment

• At about room temperature, the value increases to
  CV 5/2 R
• This is consistent with adding rotational energy
  but not vibrational energy
• At high temperatures, the value increases
  to CV 7/2 R
• This includes vibrational energy as well as
  rotational and translational

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Complex Molecules

• For molecules with more than two atoms, the
  vibrations are more complex
• The number of degrees of freedom is larger
• The more degrees of freedom available to a
  molecule, the more ways there are to store
  energy
• This results in a higher molar specific heat

44
Quantization of Energy

• To explain the results of the various molar
  specific heats, we must use some quantum
  mechanics
• Classical mechanics is not sufficient
• In quantum mechanics, the energy is proportional
  to the frequency of the wave representing the
  particle
• The energies of atoms and molecules are
  quantized

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Quantization of Energy

• This energy level diagram shows the rotational
  and vibrational states of a diatomic molecule
• The lowest allowed state is the ground state

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Quantization of Energy

• The vibrational states are separated by larger
  energy gaps than are rotational states
• At low temperatures, the energy gained during
  collisions is generally not enough to raise it to
  the first excited state of either rotation or
  vibration

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Quantization of Energy

• Even though rotation and vibration are
  classically allowed, they do not occur
• As the temperature increases, the energy of the
  molecules increases
• In some collisions, the molecules have enough
  energy to excite to the first excited state
• As the temperature continues to increase, more
  molecules are in excited states

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Quantization of Energy

• At about room temperature, rotational energy is
  contributing fully
• At about 1000 K, vibrational energy levels are
  reached
• At about 10 000 K, vibration is contributing
  fully to the internal energy

49
Molar Specific Heat of Solids

• Molar specific heats in solids also demonstrate a
  marked temperature dependence
• Solids have molar specific heats that generally
  decrease in a nonlinear manner with decreasing
  temperature
• It approaches zero as the temperature approaches
  absolute zero

50
DuLong-Petit Law

• At high temperatures, the molar specific heats
  approach the value of 3R
• This occurs above 300 K
• The molar specific heat of a solid at high
  temperature can be explained by the equipartition
  theorem
• Each atom of the solid has six degrees of
  freedom
• The internal energy is 3nRT and Cv 3 R

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Molar Specific Heat of Solids

• As T approaches 0, the molar specific heat
  approaches 0
• At high temperatures, CV becomes a constant at 3R

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Specific Heat and Molar Specific Heat of Some
Solids and Liquids
Substance c, kJ/kg?K c, J/molK
Aluminum 0.900 24.3
Bismuth 0.123 25.7
Copper 0.386 24.5
Gold 0.126 25.6
Ice (-100C) 2.05 36.9
Lead 0.128 26.4
Silver 0.233 24.9
Tungsten 0.134 24.8
Zink 0.387 25.2
Mercury 0.140 28.3
Water 4.18 75.2

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• The molar mass of copper is 63.5 g/mol. Use
  the Dulong-Petit law to calculate the specific
  heat of copper.

54

• The molar mass of copper is 63.5 g/mol. Use
  the Dulong-Petit law to calculate the specific
  heat of copper.
• The molar specific heat is the heat capacity
  per mole
• The Dulong-Petit law gives c in terms of R
• c 3R

This differs from the measured value of 0.386
kJ/kg?K given in Table by less than 2
55

• The specific heat of a certain metal is measured
  to be 1.02 kJ/kg?K. (a) Calculate the molar mass
  of this metal, assuming that the metal obeys the
  Dulong-Petit law. (b) What is the metal?

56

• The specific heat of a certain metal is measured
  to be 1.02 kJ/kg?K. (a) Calculate the molar mass
  of this metal, assuming that the metal obeys the
  Dulong-Petit law. (b) What is the metal?
• (a)

(b) The metal must be magnesium, which has a
molar mass of 24.31 g/mol
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Boltzmann Distribution Law

• The motion of molecules is extremely chaotic
• Any individual molecule is colliding with others
  at an enormous rate
• Typically at a rate of a billion times per
  second
• We add the number density nV (E )
• This is called a distribution function
• It is defined so that nV (E ) dE is the number
  of molecules per unit volume with energy between
  E and E dE

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Number Density and Boltzmann Distribution Law

• From statistical mechanics, the number density is
  
• This equation is known as the Boltzmann
  Distribution Law
• It states that the probability of finding the
  molecule in a particular energy state varies
  exponentially as the energy (½mv2) divided by kBT

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Distribution of Molecular Speeds

• The observed speed distribution of gas molecules
  in thermal equilibrium is shown in figure. The
  number of molecules having speeds in the range v
  to vdv is equal to the area of the shaded
  rectangle, Nvdv.
• The function Nv approaches zero as v approaches
  infinity.

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Distribution of Molecular Speeds

• As indicated in figure, the average speed is
  somewhat lower than the rms speed. The most
  probable speed vmp is the speed at which the
  distribution curve reaches a peak.

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Distribution Function

• The fundamental expression that describes the
  distribution of speeds in N gas molecules is
• m is the mass of a gas molecule, kB is
  Boltzmanns constant and T is the absolute
  temperature

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Speed Distribution

• The peak shifts to the right as T increases
• This shows that the average speed increases with
  increasing temperature
• The asymmetric shape occurs because the lowest
  possible speed is 0 and the highest is infinity

63
Speed Distribution

• The distribution of molecular speeds depends both
  on the mass and on temperature
• The speed distribution for liquids is similar to
  that of gases

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Evaporation

• Some molecules in the liquid are more energetic
  than others
• Some of the faster moving molecules penetrate the
  surface and leave the liquid
• This occurs even before the boiling point is
  reached
• The molecules that escape are those that have
  enough energy to overcome the attractive forces
  of the molecules in the liquid phase
• The molecules left behind have lower kinetic
  energies
• Therefore, evaporation is a cooling process

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Mean Free Path

• The molecules move with constant speed along
  straight lines between collisions
• The average distance between collisions is called
  the mean free path
• The path of an individual molecule is random
• The motion is not confined to the plane of the
  paper

66
Mean Free Path

• A molecule moving through a gas collides with
  other molecules in a random fashion
• This behavior is sometimes referred to as a
  random-walk process
• The mean free path increases as the number of
  molecules per unit volume decreases

67
Mean Free Path

• The mean free path is related to the diameter of
  the molecules and the density of the gas
• We assume that the molecules are spheres of
  diameter d
• No two molecules will collide unless their paths
  are less than a distance d apart as the molecules
  approach each other

68
Mean Free Path

• The mean free path, l, equals the average
  distance vavg?t traveled in a time interval ?t
  divided by the number of collisions that occur in
  that time interval

69
Collision Frequency

• The number of collisions per unit time is the
  collision frequency
• The inverse of the collision frequency is the
  collision mean free time