Chapter 03 - Kinetic Theory of Gases · collisions. Fy = Δpy Δt vy-vy Linear momentum is conserved in collision with wall ... Average molecular speed Given 1 2 mv2 = 3 2 kBT deﬁne

Chapter 03Kinetic Theory of Gases

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 03: Kinetic Theory of Gases

History of ideal gas law1662: Robert Boyle discovered with changing pressure at constant temperature thatproduct of pressure and volume of a gas at equilibrium is constant,

pV = constant at constant T

1780s: Jacques Charles found that ratio of volume to temperature was also invariantwhen temperature was changed with pressure kept constant,

V∕T = constant at constant p

1811: Amedeo Avogadro found ratio of volume to amount remained constant withchanging amount at fixed pressure and temperature,

V∕n = constant at constant p and T

1834: Emile Clapeyron combined gas laws of Boyle, Charles, and Avogadro into idealgas equation of state,

pV = nRTwhere R is gas constant.


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

m

z

xy

Remember pressure is defined as force per unit area.What is the force of one gas molecule hitting a wall?


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

Force along y is given by ratio of change in momentum to time betweencollisions.

Fy =Δpy

Δt

vy

-vy

Linear momentum is conserved in collision with wall

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

Time to travel length of box, hit wall, and travel back is Δt = 2𝓁∕vy

Average force of 1 molecule hitting 1 wall of box is Fy =Δpy

Δt=

−2mvy

2𝓁∕vy= −

mv2y

𝓁P. J. Grandinetti Chapter 03: Kinetic Theory of Gases

Force of N molecules hitting all walls of the box.Sum over N molecules hitting one wall is

FyN= −m

𝓁

N∑𝛼=1

v2y𝛼

Add magnitude (i.e., ignore signs) of all forces on all 6 walls (top, bottom, left, right, front, back)

Ftotal = 2m𝓁

N∑𝛼=1

v2x𝛼+ 2m

𝓁

N∑𝛼=1

v2y𝛼+ 2m

𝓁

N∑𝛼=1

v2z𝛼

= 2m𝓁

N∑𝛼=1

(v2

x𝛼+ v2

y𝛼+ v2

z𝛼

)⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

v2𝛼

= 2m𝓁

N∑𝛼=1

v2𝛼

Define mean square velocity as

v2 = 1N

N∑𝛼=1

v2𝛼 or Nv2 =

N∑𝛼=1

v2𝛼

and write total force on all 6 walls of box asFtotal = 2m

𝓁Nv2


Pressure from N molecules inside the box.Pressure is force per unit area. Total area of box walls is 6 times area of 1 wall: Atotal = 6Awall.

p = Ftotal∕Atotal = Ftotal∕(6Awall) Substituting previous result: Ftotal = 2m𝓁

Nv2

gives

p = 2m𝓁

Nv2∕(6Awall) =Nmv2

3Awall𝓁= Nmv2

3Vwhere V = Awall𝓁 is volume of box

Rearranging gives Boyle’s law (pV = constant)

pV = Nmv2

3= 2

3N(1

2mv2

)= 2

3N𝜖k, where 𝜖k is mean kinetic energy of molecule

Rewriting in terms of moles, i.e., N = nNA where NA is Avogradro constant,

pV = 23

nNA𝜖k = 23

nEk where Ek is kinetic energy of 1 mole of gas


Temperature is a quantity derived from energyFinally connect to Ideal gas law:

pV = nRT = 23

nEk

and we discoverEk =

32

RT kinetic energy of 1 mole of ideal gas

Equation reveals true nature of temperature—reflects kinetic energy of atoms andmolecules.Can’t have negative temperatures because can’t have negative kinetic energy.Raising gas temperature increases kinetic energy of gas molecules and vice versa.Dividing by NA we obtain relationship on per molecule basis

𝜖k =EkNA

= 32

RNA

T = 32

kBT

kB = R∕NA = 1.38064852 × 10−23 J/K is defined as Boltzmann constant.P. J. Grandinetti Chapter 03: Kinetic Theory of Gases

Average molecular speed

Given12

mv2 = 32

kBT

define root mean square speed, crms =√

v2, and obtain

crms =√

3kBTm

=√

3RTM

crms is related to temperature and molecular mass, m, or molar mass, M

Molecular speeds increase with increasing temperature.Molecular speeds decrease with increasing molecular mass.


Average molecular speed

ExampleCalculate the rms speed for a mole of vanillin molecules at room temperature.

SolutionSince Vanillin has chemical formula C8H8O3 with a molecular weight of 152.1 g/mol we obtain

crms =√

3RTM

=

√3R(300 K)

152.1 g/mol≈ 221 m/s ≈ 500 mph

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


Maxwell Distribution Laws

James Clerk Maxwell1831-1879

What is parent probability distribution function formolecular velocities, p(v⃗)molecular speeds, p(c)molecular energies, p(E)?


Maxwell Distribution LawsIn 1859 James Clerk Maxwell worked out the probability distribution of molecular velocities,f (v⃗), for gas molecules as perfectly elastic spheres.

Maxwell assumed that distribution of velocities in each direction were uncorrelated, that is,f (v⃗) can be written as product of 3 independent distributions

f (v⃗) = f (vx) f (vy) f (vz)

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

f (vx) f (vy) f (vz) = 𝜙(v2x + v2

y + v2z )

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


How do we solve this functional equation?

f (vx) f (vy) f (vz) = 𝜙(v2x + v2

y + v2z )

Product of functions on left must give sum of their variables as function argument on right.

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

f (vi) = ae−bv2i

Putting this function into functional equation gives

𝜙(v2x + v2

y + v2z ) = a3e−b(v2

x+v2y+v2

z )

Need to determine a and b.


Normalizing Maxwell’s distribution for molecular velocitiesAs the f (vi) are probability distributions we require them to be normalized,

∫∞

−∞f (vi)dvi = 1 so ∫

∞

−∞ae−bv2

i dvi = 1

requires that a =√

b∕𝜋 and

f (vi) =√

b𝜋

e−bv2i

Taken together Maxwell’s probability distribution then becomes

f (v⃗) = f (vx) f (vy) f (vz) =( b𝜋

)3∕2e−b(v2

x+v2y+v2

z )

Still need to determine b.


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 probability distribution

v2 = ∫ (v2x + v2

y + v2z ) f (v⃗) dv⃗ = 3kBT∕m

Substituting our normalized solution for f (v⃗) we obtain

v2 = ∫∞

−∞ ∫∞

−∞ ∫∞

−∞(v2

x + v2y + v2

z )( b𝜋

)3∕2e−b(v2

x+v2y+v2

z )dvx dvy dvz = 3kBT∕m

Solving this 3D integral equation requires b = m∕(2kBT), and finally obtain

f (v⃗) = 1√(2𝜋)3

(m

kBT

)3∕2e−

12 (v

2x+v2

y+v2z )∕(kBT∕m)

This is Maxwell’s distribution law for molecular velocities: A 3D Gaussian probabilitydistribution with a standard deviation of

√kBT∕m.


Maxwell’s distribution law for molecular velocitiesMaxwell’s distribution law is a 3D Gaussian distribution centered on v⃗ = 0.

Distribution for one component of velocity vector for N2 gas at 3 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 magnitude of velocity vector. To get speed distribution transform Maxwell’s velocitydistribution into spherical coordinates,

c =√

v2x + v2

y + v2z , cos 𝜃 =

vz

c, tan𝜙 =

vy

vx.

With this change of variables we find

f (v⃗) = f (c, 𝜃, 𝜙) = 1√(2𝜋)3

(m

kBT

)3∕2e−

12 c2∕(kBT∕m)

This is independent of 𝜃 and 𝜙 so if we put it into the normalization

∫∞

0 ∫𝜋

0 ∫2𝜋

0f (c, 𝜃, 𝜙)c2dc sin 𝜃 d𝜃 d𝜙 = 1

we can integrate away 𝜃 and 𝜙 and obtain

f (c) =√

2𝜋

(m

kBT

)3∕2c2e−

12 c2∕(kBT∕m) Maxwell’s distribution law

for molecular speeds.



f (c) =√

2𝜋

(m

kBT

)3∕2c2e−

12 c2∕(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) =√

2𝜋

(m

kBT

)3∕2c2e−

12 c2∕(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


Mean speedWith Maxwell’s distribution law for molecular speeds

f (c) =√

2𝜋

(m

kBT

)3∕2c2e−

12 c2∕(kBT∕m)

we can calculate the mean speed

c = ∫∞

0cf (c)dc =

√2𝜋

(m

kBT

)3∕2⋅

12

(2kBT

m

)2

which simplifies to

c =√

8kBT𝜋m

=√

8RT𝜋M

Note, mean speed, c, is smaller than root mean square speed, crms =√

3kBTm

.


Most probable speed

f (c) =√

2𝜋

(m

kBT

)3∕2c2e−

12 c2∕(kBT∕m)

To calculate most probablespeed need to find c value wheref (c) is maximum.Set df (c)∕dc = 0, solve for c toobtain

cmode =√

2kBTm

=√

2RTM

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.Fraction of molecules with speeds between c1 and c2 is obtained by integrating Maxwell speeddistributions between these two limits,

𝛿NN

= ∫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 /P. J. Grandinetti Chapter 03: Kinetic Theory of Gases

Web Apps by Paul Falstad

Kinetic Theory of Gases Simulation


http://www.falstad.com/gas/fullscreen.html

1955 - Miller and Kusch experimentally verify Maxwell’s molecular speed distribution.


Miller and Kusch experiments

In 1955 Miller and Kusch published firstconvincing measurements of speeddistribution for K and Tl atoms in gasphase.For each fixed rotation speed onlymolecules with a small range of speeds cantravel from the furnace to the detector.With dimensions given in instrumentdiagram the selected speed is v0 = 𝜔l∕𝜙.Measuring intensity as a function ofrotation speed gives the atomic speeddistribution.



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

Potassium vapor Thallium vapor




Distribution of kinetic energies

Homework

Given

f (c) =√

2𝜋

(m

kBT

)3∕2c2e−

12 c2∕(kBT∕m)

derive distribution of kinetic energies,

f (𝜖k) =2√𝜋

(1

kBT

)3∕2𝜖k

1∕2e−𝜖k∕(kBT)

50 10 15 20 25 30

0.05

0.00

0.10

0.15

0.20

0.25

0.30

0.35

1000 K

300 K

100 K

Energy/ 10-21 J

10-2

1 J


Documents

Chapter 03 - Kinetic Theory of Gases · collisions. Fy = Δpy Δt vy-vy Linear momentum is conserved in collision with wall ... Average molecular speed Given 1 2 mv2 = 3 2 kBT deﬁne