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PROBABILITY DISTRIBUTIONS

PROBABILITY DISTRIBUTIONS

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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 Speed Distribution for N molecules with speeds within v and v+dv is

  7. Velocity Distribution of Gases • Maxwell Speed Distribution for N molecules with speeds within v and v+dv is • dN =N f(v) dv

  8. Velocity Distribution of Gases • Maxwell Speed Distribution for N molecules with speeds within v and v+dv is • dN =N f(v) dv • f(v) = dN/N = 4/(√π)(m/2kT)3/2 v2 e –mv^2/2kT

  9. Velocity Distribution of Gases • Maxwell Speed Distribution for N molecules with speeds within v and v+dv is • dN =N f(v) dv • f(v) = dN/N = 4/(√π)(m/2kT)3/2 v2 e –mv^2/2kT • where N is the number of molecules of mass m and temperature T.

  10. Velocity Distribution of Gases • This velocity probability distribution has all the properties given before: ∫ f(v) dv = 1

  11. Velocity Distribution of Gases • This velocity probability distribution has all the properties given before: ∫ f(v) dv = 1 and the mean velocity and the mean of the square velocity are: <v> = ∫ v f(v) dv <v2 > = ∫ v2f(v) dv

  12. Velocity Distribution of Gases • This velocity probability distribution has all the properties given before: ∫ f(v) dv = 1 and the mean velocity and the mean of the square velocity are: <v> = ∫ v f(v) dv <v2 > = ∫ v2f(v) dv (remember dv means one must do a triple integration over dvx dvy dvz )

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

  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

  16. 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

  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 (E)

  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 (E) FMB (E) = C(2E/m)1/2e –E/kT where C = Maxwell distribution constant C = 4/(√π)(m/2kT)3/2

  20. Maxwell-Boltzmann Distribution • Now N f(v) dv = N F(E)dE • So this is: • f(v) dv = C v2 e –mv^2/2kT dv • F(E)dE = C (2E/m)1/2 (1/m) e –E/kT dE • This may be simplified to: • F(E) = 2/(√π) (1/kT)3/2 (E)1/2 e -E/kT

  21. Maxwell-Boltzmann Distribution • This distibution function may be used to find <E>, <E2> and Emost prob . • Also the M-B Energy distribution function can be thought of as the product of two factors.( In the language of statistical mechanics) This is the product of the density of states ~ √E and the probability of a state being occupied (The Boltzmann factor) e –E/kt .

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

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

  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>

  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

  26. 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)

  27. 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)

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

  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

  31. 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

  32. 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

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

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

  35. 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.

  36. 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> The Mean Free Path.

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

  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 .

  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.

  40. 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.

  41. 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 travels 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.

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

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

  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

  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)

  46. 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 .

  47. 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.

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

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

  50. 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.