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PROBABILITY DISTRIBUTIONS
FINITE CONTINUOUS
∑ Ng = N Nv Δv = N
PROBABILITY DISTRIBUTIONS
FINITE CONTINUOUS
∑ Ng = N Nv Δv = N
Pg = Ng /N ∫Nv dv = N
Pv = Nv /N
PROBABILITY DISTRIBUTIONS
FINITE CONTINUOUS
∑ Ng = N Nv Δv = N
Pg = Ng /N ∫Nv dv = N
Normalized Pv = Nv /N
∑ Pg = 1 ∫Pv dv = 1
PROBABILITY DISTRIBUTIONS
FINITE CONTINUOUS
∑ Ng = N Nv Δv = N
Pg = Ng /N ∫Nv dv = N
Normalized Pv = Nv /N
∑ Pg = 1 ∫Pv dv = 1
< g> = ∑ g Pg < v > = ∫vPv dv
PROBABILITY DISTRIBUTIONS
FINITE CONTINUOUS
∑ Ng = N Nv Δv = 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
Velocity Distribution of Gases
• Maxwell Velocity Distribution for gases is
• N(v) dv = N 4πv2 (m/2πkT)3/2 e –mv^2/2kT dv
• where N is the number of molecules of mass m and temperature T.
Velocity Distribution of Gases
• Maxwell Velocity Distribution for gases is
• N(v) dv = N 4πv2 (m/2πkT)3/2 e –mv^2/2kT dv
• 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 ,
Velocity Distribution of Gases
• Maxwell Velocity Distribution for gases is
• N(v) dv = N 4πv2 (m/2πkT)3/2 e –mv^2/2kT dv
• 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
Velocity Distribution of Gases
• This velocity probability distribution has all
the properties given before:
∫ F(v) d3v = 1
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 > = ∫ v2 F(v) d3v
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 > = ∫ v2 F(v) d3v
(remember d3v means one must do a triple integration over dvx dvy dvz )
Velocity Distribution of Gases
• The results of this are:
• <v> = √(8kT/(πm)) = 1.59 √kT/m
Velocity Distribution of Gases
• The results of this are:
• <v> = √(8kT/(πm)) = 1.59 √kT/m
• <v2> = √(3kT/m) = 1.73 √kT/m
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
•
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
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
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. FM (v) FMB (v)
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
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.
• EINT = < E > = ETRANS + EROT + EVIBR
•
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>
•
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 Ix ωx2 + ½ Iy ωy
2 + ½ Iz ωz2
•
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 Ix ωx2 + ½ Iy ωy
2 + ½ Iz ωz2
• Diatomic (2 axes) Triatomic (3 axes)
•
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 Ix ωx2 + ½ Iy ωy
2 + ½ Iz ωz2
• Diatomic (2 axes) Triatomic (3 axes)
• EVIBR = - ½ k x2 VIBR (for each axis)
INTERNAL MOLECULAR ENERGY
• For a diatomic molecule then <E> = 5/2 kT
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.
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
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
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
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<E> = (s/2) NkT
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.
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>
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 D2 .
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.
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> = vRMS 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 D2 . If there is a collision then the molecule traveles a distance λ = vt. If one averages this
• <λ> = vRMS 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 = AvRMSτ =
MOLECULAR COLLISIONS
• The molecule sweeps out a volume which is V = AvRMSτ = σ vRMSτ
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
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)
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 .
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.
MOLECULAR COLLISIONS
• Thus τ = 1/ (√2 nσvRMS )
MOLECULAR COLLISIONS
• Thus τ = 1/ (√2 nσvRMS )
• and
<λ> = vRMS τ = 1/(√2 nσ)
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.
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:
• http://www.aip.org/history/einstein/brownian.htm
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
• 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 < R2 > = N L2
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 < R2 > = N L2
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 < R2 > = N L2
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 < R2 > = N L2
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
< R2 > = (t/τ) λ2
MOLECULAR COLLISIONS
• For diffusion to occur there must be a difference of concentration of a fluid
MOLECULAR COLLISIONS
• For diffusion to occur there must be a difference of concentration of a fluid
• Let J be the rate of diffusion
J = ΔNNEW /Δt
MOLECULAR COLLISIONS
• For diffusion to occur there must be a difference of concentration of a fluid
• Let J be the rate of diffusion
J = ΔNNEW /Δt
= D A ΔC/Δx
MOLECULAR COLLISIONS
• For diffusion to occur there must be a difference of concentration of a fluid
• Let J be the rate of diffusion
J = ΔNNEW /Δt
= D A ΔC/Δx
where D is the diffusion constant
MOLECULAR COLLISIONS
• For diffusion to occur there must be a difference of concentration of a fluid
• Let J be the rate of diffusion
J = ΔNNEW /Δt
= D A ΔC/Δx
where D is the diffusion constant and ΔC/Δx
is the concentration gradient in either moles/s or kg/s.
MOLECULAR COLLISIONS
• For diffusion to occur there must be a difference of concentration of a fluid
• Let J be the rate of diffusion
J = ΔNNEW /Δt
= D A ΔC/Δx
where D is the diffusion constant and ΔC/Δx
is the concentration gradient in either moles/s or kg/s. The above is Ficks Law
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