PROBABILITY DISTRIBUTIONS FINITE CONTINUOUS ∑ N g = N N v Δv = N

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

PROBABILITY DISTRIBUTIONS FINITE CONTINUOUS ∑ N g = N N v Δv = N P g = N g /N ∫N v dv = N Normalized P v = N v /N ∑ P g = 1 ∫P v dv = 1

PROBABILITY DISTRIBUTIONS FINITE CONTINUOUS ∑ N g = N N v Δv = N P g = N g /N ∫N v dv = N Normalized P v = N v /N ∑ P g = 1 ∫P v dv = 1 = ∑ g P g = ∫vP v dv

PROBABILITY DISTRIBUTIONS FINITE CONTINUOUS ∑ N g = N N v Δv = N P g = N g /N ∫N v dv = N Normalized P v = N v /N ∑ P g = 1 ∫P v dv = 1 = ∑ g P g = ∫vP v dv = ∑ g 2 P g = ∫v 2 P v dv

Velocity Distribution of Gases Maxwell Speed Distribution for N molecules with speeds within v and v+dv is

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

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 v 2 e –mv^2/2kT

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 v 2 e –mv^2/2kT where N is the number of molecules of mass m and temperature T.

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

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 f(v) dv = ∫ v 2 f(v) dv

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 f(v) dv = ∫ v 2 f(v) dv (remember dv means one must do a triple integration over dv x dv y dv z )

Velocity Distribution of Gases The results of this are: = √(8kT/(πm)) = 1.59 √kT/m

Velocity Distribution of Gases The results of this are: = √(8kT/(πm)) = 1.59 √kT/m = √(3kT/m) = 1.73 √kT/m

Velocity Distribution of Gases The results of this are: = √(8kT/(πm)) = 1.59 √kT/m = √(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

Velocity Distribution of Gases The results of this are: = √(8kT/(πm)) = 1.59 √kT/m = √(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 v most prob = √(2kT/m) = 1.41√kT/m

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

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. f M (v)  F MB (E)

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. f M (v)  F MB (E) F MB (E) = C(2E/m) 1/2 e –E/kT where C = Maxwell distribution constant C = 4/(√π)(m/2kT) 3/2

Maxwell-Boltzmann Distribution Now N f(v) dv = N F(E)dE So this is: f(v) dv = C v 2 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

Maxwell-Boltzmann Distribution This distibution function may be used to find, and E most 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.

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

MOLECULAR INTERNAL ENERGY Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. E INT = = E TRANS + E ROT + E VIBR

MOLECULAR INTERNAL ENERGY Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. E INT = = E TRANS + E ROT + E VIBR E TRANS = = ½ m

MOLECULAR INTERNAL ENERGY Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. E INT = = E TRANS + E ROT + E VIBR E TRANS = = ½ m E ROT = ½ I x ω x 2 + ½ I y ω y 2 + ½ I z ω z 2

MOLECULAR INTERNAL ENERGY Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. E INT = = E TRANS + E ROT + E VIBR E TRANS = = ½ m E ROT = ½ I x ω x 2 + ½ I y ω y 2 + ½ I z ω z 2 Diatomic (2 axes) Triatomic (3 axes)

MOLECULAR INTERNAL ENERGY Diatomic, Triatomic and more complex molecules can rotate and vibrate about their symmetrical axes. E INT = = E TRANS + E ROT + E VIBR E TRANS = = ½ m E ROT = ½ I x ω x 2 + ½ I y ω y 2 + ½ I z ω z 2 Diatomic (2 axes) Triatomic (3 axes) E VIBR = - ½ k x 2 VIBR (for each axis)

INTERNAL MOLECULAR ENERGY For a diatomic molecule then = 5/2 kT

INTERNAL MOLECULAR ENERGY For a diatomic molecule then = 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.

INTERNAL MOLECULAR ENERGY For a diatomic molecule then = 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 = (s/2) kT

INTERNAL MOLECULAR ENERGY For a diatomic molecule then = 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 = (s/2) kT where s = the number of degrees of freedom

INTERNAL MOLECULAR ENERGY For a diatomic molecule then = 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 = (s/2) kT where s = the number of degrees of freedom This is called the EQUIPARTION THEOREM

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

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

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

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

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

COLLISIONS OF MOLECULES Let D be the diameter of a molecule. The collision cross section is merely the cross- sectional area σ = π D 2.

COLLISIONS OF MOLECULES Let D be the diameter of a molecule. The collision cross section is merely the cross- sectional area σ = π D 2. If there is a collision then the molecule traveles a distance λ = vt.

COLLISIONS OF MOLECULES Let D be the diameter of a molecule. The collision cross section is merely the cross- sectional area σ = π D 2. If there is a collision then the molecule traveles a distance λ = vt. If one averages this = v RMS τ where τ = mean collision time.

COLLISIONS OF MOLECULES Let D be the diameter of a molecule. The collision cross section is merely the cross- sectional area σ = π D 2. If there is a collision then the molecule travels a distance λ = vt. If one averages this = v RMS τ where τ = mean collision time. During this time there are N collisions in a volume V.

MOLECULAR COLLISIONS The molecule sweeps out a volume which is V = Av RMS τ =

MOLECULAR COLLISIONS The molecule sweeps out a volume which is V = Av RMS τ = σ v RMS τ

MOLECULAR COLLISIONS The molecule sweeps out a volume which is V = Av RMS τ = σ v RMS τ Since there are number / volume (n dens ) molecules undergoing a collision then the average number of collisions per unit time is τ = 1/N

MOLECULAR COLLISIONS The molecule sweeps out a volume which is V = Av RMS τ = σ v RMS τ Since there are number / volume (n dens ) molecules undergoing a collision then the average number of collisions per unit time is τ = 1/N = 1/(nV/τ)

MOLECULAR COLLISIONS The molecule sweeps out a volume which is V = Av RMS τ = σ v RMS τ Since there are number / volume (n dens ) molecules undergoing a collision then the average number of collisions per unit time is τ = 1/N = 1/(nV/τ) = 1/nσ v RMS.

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

MOLECULAR COLLISIONS Thus τ = 1/ (√2 nσv RMS )

MOLECULAR COLLISIONS Thus τ = 1/ (√2 nσv RMS ) and = v RMS τ = 1/(√2 nσ)

MOLECULAR COLLISIONS Thus τ = 1/ (√2 nσv RMS ) and = v RMS 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.

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

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: tmhttp:// tm

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: tmhttp:// tm Also click on the Essay on Einstein Brownian Motion.

MOLECULAR COLLISIONS Each of the distances moved by the molecules is L, because of the possibility of positive and negative directions it is best to calculate the = N L 2

MOLECULAR COLLISIONS Each of the distances moved by the molecules is L, because of the possibility of positive and negative directions it is best to calculate the = N L 2 where N is the number of steps or collisions.

MOLECULAR COLLISIONS Each of the distances moved by the molecules is L, because of the possibility of positive and negative directions it is best to calculate the = N L 2 where N is the number of steps or collisions. Since there are N collisions in a time t t = Nτ

MOLECULAR COLLISIONS Each of the distances moved by the molecules is L, because of the possibility of positive and negative directions it is best to calculate the = N L 2 where N is the number of steps or collisions. Since there are N collisions in a time t t = Nτ so in the above equation = (t/τ) λ 2