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PROBABILITY DISTRIBUTIONS

PROBABILITY DISTRIBUTIONS. FINITE CONTINUOUS ∑ N g = N N v Δ v = N. PROBABILITY DISTRIBUTIONS. FINITE CONTINUOUS ∑ N g = N N v Δ v = N P g = N g /N ∫N v dv = N P v = N v /N. PROBABILITY DISTRIBUTIONS. FINITE CONTINUOUS

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PROBABILITY DISTRIBUTIONS

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  1. PROBABILITY DISTRIBUTIONS FINITECONTINUOUS ∑ Ng = N NvΔv = N

  2. PROBABILITY DISTRIBUTIONS FINITECONTINUOUS ∑ Ng = N NvΔv = N Pg = Ng /N ∫Nv dv = N Pv = Nv /N

  3. PROBABILITY DISTRIBUTIONS FINITECONTINUOUS ∑ Ng = N NvΔv = N Pg = Ng /N ∫Nv dv = N Normalized Pv = Nv /N ∑ Pg = 1 ∫Pv dv = 1

  4. PROBABILITY DISTRIBUTIONS FINITECONTINUOUS ∑ Ng = N NvΔv = N Pg = Ng /N ∫Nv dv = N Normalized Pv = Nv /N ∑ Pg = 1 ∫Pv dv = 1 < g> = ∑ g Pg < v > = ∫vPv dv

  5. PROBABILITY DISTRIBUTIONS FINITECONTINUOUS ∑ Ng = N NvΔv = N Pg = Ng /N ∫Nv dv = N Normalized Pv = Nv /N ∑ Pg = 1 ∫Pv dv = 1 < g> = ∑ g Pg < v > = ∫vPv dv <g2> = ∑ g2 Pg < v2> = ∫v2 Pv dv

  6. Velocity Distribution of Gases • Maxwell Velocity Distribution for gases is • N(v) dv = N4πv2 (m/2πkT)3/2 e –mv^2/2kTdv • where N is the number of molecules of mass m and temperature T.

  7. Velocity Distribution of Gases • Maxwell Velocity Distribution for gases is • N(v) dv = N4πv2 (m/2πkT)3/2 e –mv^2/2kTdv • where N is the number of molecules of mass m and temperature T. If one divides by N and changes the differential element dv to d3v = dvx dvy dvz ,

  8. Velocity Distribution of Gases • Maxwell Velocity Distribution for gases is • N(v) dv = N4πv2 (m/2πkT)3/2 e –mv^2/2kTdv • where N is the number of molecules of mass m and temperature T. If one divides by N and changes the differential element dv to d3v = dvx dvy dvz , then the normalized probability function F(v) is: • F(v) = (m/2πkT)3/2 e –mv^2/2kT

  9. Velocity Distribution of Gases • This velocity probability distribution has all the properties given before: ∫ F(v) d3v = 1

  10. Velocity Distribution of Gases • This velocity probability distribution has all the properties given before: ∫ F(v) d3v = 1 and the mean velocity and the mean of the square velocity are: <v> = ∫ v F(v) d3v <v2 > = ∫ v2F(v) d3v

  11. Velocity Distribution of Gases • This velocity probability distribution has all the properties given before: ∫ F(v) d3v = 1 and the mean velocity and the mean of the square velocity are: <v> = ∫ v F(v) d3v <v2 > = ∫ v2F(v) d3v (remember d3v means one must do a triple integration over dvx dvy dvz )

  12. Velocity Distribution of Gases • The results of this are: • <v> = √(8kT/(πm)) = 1.59 √kT/m

  13. Velocity Distribution of Gases • The results of this are: • <v> = √(8kT/(πm)) = 1.59 √kT/m • <v2> = √(3kT/m) = 1.73 √kT/m

  14. Velocity Distribution of Gases • The results of this are: • <v> = √(8kT/(πm)) = 1.59 √kT/m • <v2> = √(3kT/m) = 1.73 √kT/m • If one sets the derivative of the probability function to zero (as was done for the Planck Distribution) one obtains the most probable value of v

  15. Velocity Distribution of Gases • The results of this are: • <v> = √(8kT/(πm)) = 1.59 √kT/m • <v2> = √(3kT/m) = 1.73 √kT/m • If one sets the derivative of the probability function to zero (as was done for the Planck Distribution) one obtains the most probable value of v • vmost prob = √(2kT/m) = 1.41√kT/m

  16. Maxwell-Boltzmann Distribution • Molecules with more complex shape have internal molecular energy. Boltzmann realized this and changed Maxwell’s Distribution to include all the internal energy. FM (v)  FMB (v) FMB (v) = (1/Z) e –E/kT where Z = the normalization factor

  17. Maxwell-Boltzmann Distribution • Molecules with more complex shape have internal molecular energy.

  18. Maxwell-Boltzmann Distribution • Molecules with more complex shape have internal molecular energy. Boltzmann realized this and changed Maxwell’s Distribution to include all the internal energy. FM (v)  FMB (v)

  19. Maxwell-Boltzmann Distribution • Molecules with more complex shape have internal molecular energy. Boltzmann realized this and changed Maxwell’s Distribution to include all the internal energy. FM (v)  FMB (v) FMB (v) = (1/Z) e –E/kT where Z = the normalization factor

  20. MOLECULAR INTERNAL ENERGY • Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes.

  21. MOLECULAR INTERNAL ENERGY • Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. • EINT = < E > = ETRANS + EROT + EVIBR

  22. MOLECULAR INTERNAL ENERGY • Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. • EINT = < E > = ETRANS + EROT + EVIBR • ETRANS = < ETRANS > = ½ m <v2>

  23. MOLECULAR INTERNAL ENERGY • Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. • EINT = < E > = ETRANS + EROT + EVIBR • ETRANS = < ETRANS > = ½ m <v2> • EROT = ½ Ixωx2 + ½ Iyωy2 + ½ Izωz2

  24. MOLECULAR INTERNAL ENERGY • Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. • EINT = < E > = ETRANS + EROT + EVIBR • ETRANS = < ETRANS > = ½ m <v2> • EROT = ½ Ixωx2 + ½ Iyωy2 + ½ Izωz2 • Diatomic (2 axes) Triatomic (3 axes)

  25. MOLECULAR INTERNAL ENERGY • Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. • EINT = < E > = ETRANS + EROT + EVIBR • ETRANS = < ETRANS > = ½ m <v2> • EROT = ½ Ixωx2 + ½ Iyωy2 + ½ Izωz2 • Diatomic (2 axes) Triatomic (3 axes) • EVIBR = - ½ k x2VIBR (for each axis)

  26. INTERNAL MOLECULAR ENERGY • For a diatomic molecule then <E> = 5/2 kT

  27. INTERNAL MOLECULAR ENERGY • For a diatomic molecule then <E> = 5/2 kT • One of the basic principles used in the Kinetic Theory of Gases is that each degree of freedom has an average energy • of ½ kT.

  28. INTERNAL MOLECULAR ENERGY • For a diatomic molecule then <E> = 5/2 kT • One of the basic principles used in the Kinetic Theory of Gases is that each degree of freedom has an average energy • of ½ kT. Or <E> = (s/2) kT

  29. INTERNAL MOLECULAR ENERGY • For a diatomic molecule then <E> = 5/2 kT • One of the basic principles used in the Kinetic Theory of Gases is that each degree of freedom has an average energy • of ½ kT. Or <E> = (s/2) kT where s = the number of degrees of freedom

  30. INTERNAL MOLECULAR ENERGY • For a diatomic molecule then <E> = 5/2 kT • One of the basic principles used in the Kinetic Theory of Gases is that each degree of freedom has an average energy • of ½ kT. Or <E> = (s/2) kT where s = the number of degrees of freedom • This is called the • EQUIPARTION THEOREM

  31. INTERNAL MOLECULAR ENERGY • For dilute gases which still obey the ideal gas law, the internal energy is:

  32. INTERNAL MOLECULAR ENERGY • For dilute gases which still obey the ideal gas law, the internal energy is: • U = N<E> = (s/2) NkT

  33. INTERNAL MOLECULAR ENERGY • For dilute gases which still obey the ideal gas law, the internal energy is: • U = N<E> = (s/2) NkT • Real gases undergo collisions and hence can transport matter called diffusion.

  34. INTERNAL MOLECULAR ENERGY • For dilute gases which still obey the ideal gas law, the internal energy is: • U = N<E> = (s/2) NkT • Real gases undergo collisions and hence can transport matter called diffusion. The average distance a molecule moves between collisions is <λ>

  35. COLLISIONS OF MOLECULES • Let D be the diameter of a molecule.

  36. COLLISIONS OF MOLECULES • Let D be the diameter of a molecule. The collision cross section is merely the cross-sectional area σ = π D2 .

  37. COLLISIONS OF MOLECULES • Let D be the diameter of a molecule. The collision cross section is merely the cross-sectional area σ = π D2 . If there is a collision then the molecule traveles a distance λ = vt.

  38. COLLISIONS OF MOLECULES • Let D be the diameter of a molecule. The collision cross section is merely the cross-sectional area σ = π D2 . If there is a collision then the molecule traveles a distance λ = vt. If one averages this • <λ> = vRMSτ where τ = mean collision time.

  39. COLLISIONS OF MOLECULES • Let D be the diameter of a molecule. The collision cross section is merely the cross-sectional area σ = π D2 . If there is a collision then the molecule traveles a distance λ = vt. If one averages this • <λ> = vRMSτ where τ = mean collision time. During this time there are N collisions in a volume V.

  40. MOLECULAR COLLISIONS • The molecule sweeps out a volume which is V = AvRMSτ =

  41. MOLECULAR COLLISIONS • The molecule sweeps out a volume which is V = AvRMSτ = σ vRMSτ

  42. MOLECULAR COLLISIONS • The molecule sweeps out a volume which is V = AvRMSτ = σ vRMSτ Since there are number / volume (ndens ) molecules undergoing a collision then the average number of collisions per unit time is τ = 1/N

  43. MOLECULAR COLLISIONS • The molecule sweeps out a volume which is V = AvRMSτ = σ vRMSτ Since there are number / volume (ndens ) molecules undergoing a collision then the average number of collisions per unit time is τ = 1/N = 1/(nV/τ)

  44. MOLECULAR COLLISIONS • The molecule sweeps out a volume which is V = AvRMSτ = σ vRMSτ Since there are number / volume (ndens ) molecules undergoing a collision then the average number of collisions per unit time is τ = 1/N = 1/(nV/τ) = 1/nσ vRMS .

  45. MOLECULAR COLLISIONS • The molecule sweeps out a volume which is V = AvRMSτ = σ vRMSτ Since there are number / volume (ndens ) molecules undergoing a collision then the average number of collisions per unit time is τ = 1/N = 1/(nV/τ) = 1/nσ vRMS . However, both molecules are moving and this increases the velocity by √ 2.

  46. MOLECULAR COLLISIONS • Thus τ = 1/ (√2 nσvRMS )

  47. MOLECULAR COLLISIONS • Thus τ = 1/ (√2 nσvRMS ) • and <λ> = vRMS τ = 1/(√2 nσ)

  48. MOLECULAR COLLISIONS • Thus τ = 1/ (√2 nσvRMS ) • and <λ> = vRMS τ = 1/(√2 nσ) In 1827 Robert Brown observed small particles moving in a suspended atmosphere. This was later hypothesised to be due to collisions by gas molecules.

  49. MOLECULAR COLLISIONS • The movement of these particles was observed to be random and was similar to the mathematical RANDOM WALK problem.

  50. MOLECULAR COLLISIONS • The movement of these particles was observed to be random and was similar to the mathematical RANDOM WALK problem. See the link below: • http://www.aip.org/history/einstein/brownian.htm

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