1. The Kinetic Theory of Gases
Chapter 21
The Kinetic Theory of Gases

2. 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

3. 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 don’t 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.

4. 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 molecule’s energy and momentum.

5. 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

6. Pressure and Kinetic Energy
Since the collision is elastic, the y - component of molecule’s 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 molecule’s velocity will not change until it hits the left-hand wall and its x - velocity will again reverses.
Δt = 2d / vx

7. 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:

8. Pressure and Kinetic Energy
is just the average of the squares of x – components of velocities:
Since,

9. 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:

10. 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

11. 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

12. A 2. 00-mol sample of oxygen gas is confined to a 5
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.

13. A 2. 00-mol sample of oxygen gas is confined to a 5
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.

14. 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

15. 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

16. 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

17. 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

18. 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

19. Root Mean Square Speed
The root mean square (rms) speed is the square root of the average of the squares of the speeds
Square, average, take the square root
Solving for vrms we find
M is the molar mass and M = mNA

20. Some Example vrms Values
At a given temperature, lighter molecules move faster, on the average, than heavier molecules

21. Molar Specific Heat
Several processes can change the temperature of an ideal gas
Since ΔT is the same for each process, ΔEint is also the same
The heat is different for the different paths
The heat associated with a particular change in temperature is not unique

22. 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

23. Molar Specific Heat Molar specific heats:
Q = nCV ΔT for constant-volume processes
Q = nCP ΔT for constant-pressure processes
Q (constant pressure) must account for both the increase in internal energy and the transfer of energy out of the system by work
Qconstant P > Qconstant V for given values of n and ΔT

24. 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

25. Ideal Monatomic Gas ΔEint is a function of T only At constant volume,
Therefore,
ΔEint is a function of T only
At constant volume,
Q = ΔEint = nCV ΔT
This applies to all ideal gases, not just monatomic ones

26. 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

27. Monatomic Gases CP – CV = R
In a constant-pressure process, ΔEint = 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

28. Monatomic Gases CP = 5/2 R = 20.8 J/mol . K CP – CV = R
This also applies to any ideal gas
CP = 5/2 R = 20.8 J/mol . K

29. 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

30. Sample Values of Molar Specific Heats

31. 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.

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.
The piston moves to keep pressure constant:

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.

34. 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

35. 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

36. 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

37. 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

38. Equipartition of Energy
The molecule can also vibrate
There is kinetic energy and potential energy associated with the vibrations

39. 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

40. 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

41. 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

42. 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

43. 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 frequency
The energies of atoms and molecules are quantized

44. 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

45. 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

46. 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

47. Quantization of Energy
At about room temperature, rotational energy is contributing fully
At about 1000 K, vibrational energy levels are reached
At about K, vibration is contributing fully to the internal energy

48. 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

49. 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

50. Molar Specific Heat of Solids
As T approaches 0, the molar specific heat approaches 0
At high temperatures, CV becomes a constant at ~3R

51. 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
Alcohol (ethyl)
2.4
111
Mercury
0.140
28.3
Water
4.18
75.2

52. The molar mass of copper is 63. 5 g/mol
The molar mass of copper is 63.5 g/mol. Use the Dulong-Petit law to calculate the specific heat of copper.

53. The molar mass of copper is 63. 5 g/mol
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 lawgives c* in terms of R:
c*= 3R
This differs from the measured value of kJ/kg·K given in Table by less than 2%

54. The specific heat of a certain metal is measured to be 1. 02 kJ/kg·K
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?

55. The specific heat of a certain metal is measured to be 1. 02 kJ/kg·K
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 g/mol

56. 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

57. Number Density and Boltzmann Distribution Law
From statistical mechanics, the number density is
nV (E ) = noe –E /kBT
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 divided by kBT

58. Distribution of Molecular Speeds
The observed speed distribution of gas molecules in thermal equilibrium is shown at right
NV is called the Maxwell-Boltzmann speed distribution function

59. 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 Boltzmann’s constant and T is the absolute temperature

60. Most Probable Speed
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

61. 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

62. 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

63. 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

64. 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

65. 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
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

67. Mean Free Path
The mean free path, ℓ, equals the average distance vΔt traveled in a time interval Δt divided by the number of collisions that occur in that time interval:

68. 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