Chapter 21

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 Tcan 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 it’s 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 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.

7. 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.

8. 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

9. 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

10. 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:

11. 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

14. 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-molsample of oxygen gas is confined to a 5.00-Lvessel at a pressure of 8.00 atm. Find the average translational kinetic energy of an oxygen molecule under these conditions.

16. A 2.00-molsample of oxygen gas is confined to a 5.00-Lvessel 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

19. 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

20. Theorem of Equipartition of Energy • Each degree of freedom contributes ½kBTto 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 ΔTis the same for each process, ΔEintis 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 = nCVΔTfor constant-volume processes Q = nCPΔTfor 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 • QconstantP > QconstantVfor given values of n and ΔT

25. 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

26. Ideal Monatomic Gas • Therefore, • ΔEintis a function of T only • At constant volume, Q = ΔEint = nCVΔT This applies to all ideal gases, not just monatomic ones

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

28. Monatomic Gases • In a constant-pressure process, ΔEint = Q + Wand • 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

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

30. Ratio of Molar Specific Heats • We can also define • Theoretical values of CV, CP, and gare 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) at300 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 kJof energy is transferred to the air by heat.

33. A 1.00-mol sample of air (a diatomic ideal gas) at300 K, confined in a cylinder under a heavy piston, occupies a volume of5.00 L. Determine the final volume of the gas after4.40 kJof 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) at300 K, confined in a cylinder under a heavy piston, occupies a volume of5.00 L. Determine the final volume of the gas after4.40 kJof 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 CPand CVare approximately equal

36. Adiabatic Processes for an Ideal Gas • Assume an ideal gas is in an equilibrium state and so PV = nRTis valid • The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV g= constant • g = CP / CVis assumed to be constant during the process • All three variables in the ideal gas law (P, V, T) can change during an adiabatic process

37. 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

38. Equipartition of Energy • Rotational motion about the various axes also contributes We can neglect the rotation around the yaxis since it is negligible compared to thexand z axes

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

40. 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 nRTand 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

42. 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

43. 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

45. 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

46. 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

47. 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

48. 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 3nRTandCv = 3 R