Kinetic Theory of Gases - Grandinetti Theory of Gases Chapter 3 P. J. Grandinetti Chem. 4300 Aug. 28, 2017 P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 1

Kinetic Theory of GasesChapter 3

P. J. Grandinetti

Chem. 4300

Aug. 28, 2017

P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 1 / 45

History of ideal gas law

1662: Robert Boyle discovered with changing pressure at constanttemperature that the product of the pressure and volume of a gas atequilibrium is constant, pV = constant at constant T .

1780’s Jacques Charles found that the ratio of volume to temperaturewas also invariant when temperature was changed with the pressurekept constant, V /T = constant at constant p

1811 Amedeo Avogadro found the ratio of volume to amountremained constant with changing amount at fixed pressure andtemperature, V /n = constant at constant p and T .

1834 Emile Clapeyron combined the gas laws of Boyle, Charles, andAvogadro into the ideal gas equation of state,

pV = nRT

where R is the gas constant.


Bernoulli’s derivation of Boyle’s law, pV = constant.As early as 1738 Daniel Bernoulli proposed a microscopic kineticexplanation of Boyle’s law, but only after Clapeyron’s work did Bernoulli’skinetic theory gain widespread acceptance.

Bernoulli’s derivation

m

z

xy

Remembering that pressure is defined as force per unit area, we ask whatis the force of one gas molecule hitting a wall?


Force on wall is momentum change when a molecule hits it.

vy

-vy

The force along y is given by the ratio of the change in momentum to thetime between collisions.

Fy =∆p

∆t.

Assuming collisions with the walls are elastic we write

∆py = py ,final − py ,initial = (−mvy )− (mvy ) = −2mvy


Force of one molecule hitting one wall of the box.

We take ∆t = 2ℓ/vy as the time to travel the length of the box, hit wall,and travel back.

The average force of one molecule hitting one wall of the box is

Fy =∆py∆t

=−2mvy2ℓ/vy

= −mv2yℓ

.


Force of N molecules hitting all walls of the box.

The sum over N molecules hitting one wall is

FyN =

−mN∑j=1

v2yj

/ℓ.

If we add together the magnitude of all forces on the walls (i.e., ignoresigns) we get

Ftotal = 2m

ℓ

N∑j=1

(v2xj + v2yj + v2zj

)= 2

m

ℓ

N∑j=1

v2j ,

where v2j = v2xj + v2yj + v2zj .


Force of N molecules hitting all walls of the box.

Ftotal = 2m

ℓ

N∑j=1

v2j ,

We can define the mean square velocity as

v2 =1

N

N∑j=1

v2j ,

and get the total force on all walls of the box as

Ftotal = 2m

ℓNv2.


Pressure from N molecules inside the box.

The pressure is the force per unit area. The total area of the box walls is 6times the area of one wall: 6Awall.

Combining

p = Ftotal/(6Awall) and Ftotal = 2m

ℓNv2

we obtain

p = 2m

ℓNv2/(6Awall) =

Nmv2

3Awallℓ=

Nmv2

3V

where V is the volume of the box.

Rearranging we obtain we obtain Boyle’s law

pV =Nmv2

3


Molecular translational energy and temperature.With further rearrangement we find

pV =Nmv2

3=

2

3N

(1

2mv2

)=

2

3Nϵk ,

where ϵk = 12mv2 is the average kinetic energy per molecule.

Set N = NA for 1 mole of gas molecules

pV =2

3NAϵk =

2

3Ek ,

and define Ek as the kinetic energy of a mole of gas.Connecting this expression to the ideal gas law we find

pV =2

3Ek = RT

and discover

Ek =3

2RT .


Temperature is a quantity derived from energyThe kinetic energy of 1 mole of an ideal gas

Ek =3

2RT

This equation reveal the absolute nature of temperature. You can’t havenegative temperatures because you can’t have negative kinetic energy.


Molecular translational energy and temperature.

Ek =3

2RT .

If you raise the temperature of the gas then you increase the total energyof the gas.Dividing by NA we obtain the relationship on a per molecule basis

ϵk =3

2

R

NAT =

3

2kBT .

The Boltzmann constant is defined as kB = R/NA.


Average molecular speed

Since1

2mv2 =

3

2kBT ,

then the root mean square speed, crms =√

v2, is related to temperatureand molecular mass, m, or molar mass, M, according to

crms =

√3kBT

m=

√3RT

M.

As temperature increases the average kinetic energy of the moleculesincreases and thus the average molecular speed increases.We also see that as the molecular mass increases for a fixed averagekinetic energy the average molecular speed must decrease.


Average molecular speed

Example

Calculate the root mean square speed for a mole of vanillin molecules atroom temperature.

Solution

Since Vanillin has the chemical formula C8H8O3 with a molecular weightof 152.1 g/mol we obtain

crms =

√3R(300 K)

152.1 g/mol≈ 221 m/s.

If vanillin has such a high speed why does it take so long for the scent totravel across a room?


Maxwell Distribution Laws

In 1859 James Clerk Maxwell worked out the probability distribution ofmolecular velocities, f (v⃗), for gas molecules as perfectly elastic spheres.

Maxwell assumed that the distribution of velocities in each direction wereuncorrelated, that is, f (v⃗) can be written as the product of threeindependent distributions

f (v⃗) = f (vx)f (vy )f (vz).

He also reasoned that the distribution of velocities is independent ofdirection, implying that f (v⃗) should only depend on the magnitude of thevelocities,

f (vx)f (vy )f (vz) = ϕ(v2x + v2y + v2z ).

This is an example of a functional equation: an equation in which theunknowns are functions.

P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 14 / 45

How do we solve this functional equation?

f (vx)f (vy )f (vz) = ϕ(v2x + v2y + v2z ).

Product of the functions on left gives the sum of their variables as thefunction argument on the right.

A function, f (vi ), that satisfies this functional equation is

f (vi ) = ae−bv2i .

Putting this function into our functional equation gives

ϕ(v2x + v2y + v2z ) = a3e−b(v2x+v2

y+v2z )


Normalizing Maxwell’s distribution for molecular velocities

As the f (vi ) are probability distributions we require them to be normalized,∫ ∞

−∞f (vi )dvi = 1.

This leads to a =√

b/π and

f (vi ) =

√b

πe−bv2

i .

Taken together Maxwell’s probability distribution becomes

f (v⃗) =

(b

π

)3/2

e−b(v2x+v2

y+v2z ).


Maxwell’s distribution law for molecular velocitiesFrom Bernoulli’s kinetic theory we learned that v2 = 3kBT/m.This mean square speed should also be obtained from our probabilitydistribution according to

v2 =

∫ ∞

−∞(v2x + v2y + v2z )f (vx)f (vy )f (vz)dvxdvydvz .

Substituting our normalized solutions for f (vi ) we obtain

v2 =

∫ ∞

−∞(v2x + v2y + v2z )

(b

π

)3/2

e−b(v2x+v2

y+v2z )dvxdvydvz .

Evaluating this integral and setting it equal to 3kBT/m yieldsb = m/(2kBT ). Thus we find

f (v⃗) =1√(2π)3

(m

kBT

)3/2

e−12(v2

x+v2y+v2

z )/(kBT/m).

This is Maxwell’s distribution law for molecular velocities.P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 17 / 45

Maxwell’s distribution law for molecular velocities

Maxwell’s distribution law is a three dimensional Gaussian distributioncentered on v⃗ = 0.

The distribution for one component of the velocity vector for N2 gas atthree different temperatures.

-1500 -1000 -500 0 500 1000 1500

0.0005

0.0000

0.0010

0.0015

0.0020

0.0025

100 K

300 K

1000 K

velocity/ m/s

N2 gas

/ s/m


Maxwell’s distribution law for molecular speeds


Maxwell’s distribution law for molecular speedsSpeed is the magnitude of a velocity vector. To get the speed distributionwe transform Maxwell’s velocity distribution into spherical coordinates,

c =√v2x + v2y + v2z , cos θ =

vzc, tanϕ =

vyvx

.

With this change of variables we find

f (v⃗) = f (c , θ, ϕ) =1√(2π)3

(m

kBT

)3/2

e−12c2/(kBT/m).

This is independent of θ and ϕ we we put it into the normalization∫ ∞

0

∫ π

0

∫ 2π

0f (c , θ, ϕ)c2dc sin θ dθ dϕ = 1

and integrate away the θ and ϕ to obtain

f (c) =

√2

π

(m

kBT

)3/2

c2e−12c2/(kBT/m).

This is Maxwell’s distribution law for molecular speeds.P. J. Grandinetti (Chem. 4300) Maxwell Distribution Laws Aug. 28, 2017 20 / 45


f (c) =

√2

π

(m

kBT

)3/2

c2e−12c2/(kBT/m).

5000 1000 1500 2000

0.0005

0.0000

0.0010

0.0015

0.0020

0.0025

0.0030

0.0035

speed/ m/s

100 K

300 K

1000 K

N2 gas

/ s/m



f (c) =

√2

π

(m

kBT

)3/2

c2e−12c2/(kBT/m).

5000 1000 1500 2000 2500

0.0005

0.0000

0.0010

0.0015

0.0020

0.0025

speed/ m/s

Ar

Ne

He

300 K

/ s/m


Kinetic Theory of Gases Simulation


http://www.grandinetti.org/web-app-kinetic-theory-gases

Mean speed

With Maxwell’s distribution law for molecular speeds we can calculate themean speed

c =

∫ ∞

0cf (c)dc =

√2

π

(m

kBT

)3/2

· 12

(2kBT

m

)2

,

and obtain

c =

√8kBT

πm=

√8RT

πM.

Note that the mean speed, c, is smaller than the root mean square speed,crms.


Most probable speedTo calculate the most probable speed we need to find the value of c wheref (c) is a maximum. So we set df (c)/dc = 0 and solve for c to obtain

cmode =

√2kBT

m=

√2RT

M.

10 2 3 4 5

0.1

0.0

0.2

0.3

0.4

0.5

0.6

speed /


Fraction of molecules with speeds between c1 and c2.The fraction of molecules with speeds between c1 and c2 is obtained byintegrating Maxwell speed distributions between these two limits,

δN

N=

∫ c2

c1

f (c)dc .

10 2 3 4 5

0.1

0.0

0.2

0.3

0.4

0.5

0.6

speed /


Miller and Kusch experiments

In the 1955 Miller and Kusch published the first convincing measurementsof the speed distribution for K and Tl atoms in the gas phase. For eachfixed rotation speed only molecules with a small range of speeds can travelfrom the furnace to the detector. With the dimensions given in theinstrument diagram the selected speed is v0 = ωl/ϕ.


Miller and Kusch experiments

0.20

5

10

15

20

0.6 0.80.4 1.0 1.2 1.4 1.6 1.8 2.0speed / speed /

0.2 0.6 0.80.4 1.0 1.2 1.4 1.6 1.8 2.0

Inte

nsity

0

5

10

15

20

Inte

nsity

Run 31Run 60Run 57

Run 99Run 97


Distribution of kinetic energies

Homework

Derive the distribution of kinetic energies,

f (ϵk) =2√π

(1

kBT

)3/2

ϵk1/2e−ϵk/(kBT ),

from Maxwell’s speed distribution.


Collisions and Mean Free Path


Collisions and Mean Free Path

It is through molecular collisions that molecules react.

Let’s use the kinetic theory of gases to get the collision rate of molecules.

How close the molecules have to come to be considered a collision?

Let’s assume a spherical molecule of diameter d .

2dd

d

effective collision area = collision cross section = π d2

The effective collision area is called the collision cross section, σcs, givenby σcs = πd2.

P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 31 / 45

Collision Rate

How many collisions occur in a given time t?

A molecule moving at c will sweep out a collision volume given be

Collision Volume = collision area× length = (σcs) (ct)

How many target molecules inside this volume?We multiply the number density, N/V , by the volume to find the numberof molecules inside the collision volume.

# collision in time t inside collision volume = (N/V )(σcs)(ct)


Collision Rate

How many collisions occur in a given time t?

WAIT! We incorrectly assumes the target molecules are not moving, thatis, v⃗all others = 0.

A proper treatment uses the relative speed of the colliding molecules, c relnot c . One can show (see notes) that c rel =

√2 c .

Thus we add a factor of√2 to correct our previous expression,

# collisions in time t inside collision volume = (N/V )︸︷︷︸number density

× (σcs√2 ct)︸︷︷︸

collision volume

.


Z1: Collision rate for single molecule inside the collisionvolume

Finally, we can calculate Z1, the collision rate for a single molecule insidethe collision volume, by dividing by time, t, to get

Z1 = σcs√2c

(N

V

).

Bu what is the total collision rate for all N molecules in a gas?


ZN : Total collisions per unit time per unit volumeFor all N molecules inside a total volume, V , the total collision rate is

ZN =

[σcs

√2c

(N

V

)]︸︷︷︸

1

×[1

2N

]︸︷︷︸

2

×[1

V

]︸︷︷︸

3

.

1 is the single molecule collision rate.

2 is a scaling up by N molecules. Use 1/2 to avoid double counting.

3 is to get collisions occurring inside the total volume (not collisionvolume).

Altogether we obtain

ZN =1

2σcs

√2c

(N

V

)2

,

as the collisions per unit time per unit volume.P. J. Grandinetti (Chem. 4300) Collisions and Mean Free Path Aug. 28, 2017 35 / 45

Mean Free Path


Mean free path: λ

Definition

Mean free path, λ, is average distance a particle travels between collisionsin the gas phase.

We calculate λ according to

λ = c∆t

where c ≡ average speed, and∆t ≡ time between collisions.

∆t is the inverse of the collision rate, that is, 1/Z1. Thus we obtain

λ =c

Z1=

1

σcs√2(N/V )

.


Mean free path in terms of p and T

We can eliminate (N/V ) from

λ =1

σcs√2(N/V )

.

by rearranging the ideal gas law, pV = nRT ,

N

V=

pNA

RT=

p

kBT,

and expressing the mean free path in terms of pressure and temperature as

λ =kBT√2σcsp

.


Example

Calculate the mean free path for a vanillin molecules at room temperatureand pressure. Take σcs = 0.5 nm2 for vanillin.

λ =kBT√2σcsp

=kB(300 K)√

2(0.5 nm2)(1 atm)= 57.81 nm.

For vanillin we find a mean free path of λ ≈ 60 nm, which explains why ittakes so long for the scent to travel across a room.


Effusion

P. J. Grandinetti (Chem. 4300) Effusion Aug. 28, 2017 40 / 45

Effusion

Effusion

The rate at which gas escapes through a pin hole into vacuum or lowpressure region. Pin hole diameter must be smaller than mean free path.

High pressure side Vacuum side


Graham’s law of effusionDetermined empirically by Scottish physical chemist Thomas Graham in 1848.

Definition

Graham’s law of effusion: Rate ofeffusion varies inversely with squareroot of the molecular weight,

rate1

rate2=

√M2

M1

where M1 and M2 are the molarmasses of the two gases.


Effusion Rate

How many molecules pass through pin hole in a given time interval,∆t?

There is a range of velocities in the gas and those with fast (high) vxvalues can be further away and still pass through the pin hole while thosewith a slower (low) vx value will need to be closer to pass through the pinhole during the same ∆t.

A = Area of pin hole

x

y

z

Only molecules with

vx > 0 can move towards thepin hole, and

if it is within vx∆t from the pinhole during ∆t, and

if it is inside a volume vx∆tAduring ∆t.


Effusion RateAverage number of molecules passing through a pin hole of area A in ∆t is(

N

V

)·[∫ ∞

0vx f (vx)dvx

]·∆t · A,

where f (vx) is Maxwell’s velocity distribution for vx .The rate of effusion, ZA, is the number of molecules escaping per unit areaper unit time (i.e, divide above Eq by ∆t · A),

ZA =

(N

V

)∫ ∞

0vx f (vx)dvx .

Using the Maxwell-Boltzmann distribution we obtain

ZA =

(N

V

)(m

2πkBT

)1/2 ∫ ∞

0vxe

−mv2x /(2kBTdvx

which becomes

ZA =

(N

V

)(kBT

2πm

)1/2


Effusion Rate

Using pV = nRT = NkBT gives the effusion rate

ZA =p

(2πmkBT )1/2.

Taking the ratio at a constant temperature,

Z1

Z2=

√m2

m1

we find agreement with Graham’s law of effusion.


Kinetic Theory of Gases - Grandinetti Theory of Gases Chapter 3 P. J. Grandinetti Chem. 4300 Aug. 28, 2017 P. J. Grandinetti (Chem. 4300) Kinetic Theory of Gases Aug. 28, 2017 1

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