Kinetic Theory of Gases Levine Chapter 14 - Chemistry Theory of Gases_____Levine Chapter 14 "So manyofthe properties of [gases]...can be deduced from the hypothesis that their minute

Kinetic Theory of Gases______________________________________Levine Chapter 14

"So many of the properties of [gases]...can be deduced from the hypothesis that their minute parts are in rapidmotion, the velocity increasing with the temperature...the relations between pressure, temperature and density in aperfect gas can be explained by supposing the particles move with uniform velocity in straight lines, striking againstthe sides of the containing vessel and thus producing pressure, James Clerk Maxwell, 1860

"In my opnion it would be a great tragedy for science if the theory of gases were temporarily thrown into oblivionbecause of a momentary hostile attitude toward it...I am conscious of being only an individual struggling weaklyagainst the stream of time. But it still remains in my power to contribute in such a way that, when the theory ofgases is again revived, not too much will have to be rediscovered. Ludwig Edward Boltzmann, 1898

Why Kinetic Theory

As stated in the goal below, collisions unite the three topics being discussed in this course. Furthermore

the kinetic-molecular theory of gases employs much of the mathematical theory and techniques which we

will use in quantum. And, perhaps most importantly, we will be introduced to a probability density func-

tion and how to use it. In quantum the probability density function for a system is directly related to the

system’s wave function ψ which is the solution to its Schrödinger equation.

GOAL: Predict macroscopic properties from microscopic motion (of molecules)

The most interesting event of a molecules’s existence occurs when it collides with

1. wall of container => pressure (Levine Ch 14)

2. other molecules => transport properties (EDR Chapter 24) or

3. photon =>

chemical reactions (EDR Chapters 25 and 26)

spectroscopy (quantum - the remaining 10 weeks of the course)

Background (14.1-14.3)

Vectors→v = vx

→e x + vy

→e y + vz

→e z

→v ⋅ →

v = |v2| cosθ = |v2| (θ = 0o => cosθ = 1)

= (vx→e x + vy

→e y + vz

→e z) ⋅ (vx

→e x + vy

→e y + vz

→e z)

= v2x + v2

y + v2z (14. 1) *14. 1) *

Note that the Cartesian unit vectors are orthogonal:→e i ⋅ →

e j = δ ij where the Kronecker delta function

δ ij = 1 if i = j and 0 otherwise.

Ideal Gas

The pressure that a gas exerts on the walls of its container is due to collisions of the gas molecules with

the walls. Since pressure is the force per unit area and using Newton’s second law

F = ma = m dv

dt=

d

dt(mv) =

dp

dt(14.5)

We only need to find the change in momentum ∆p over a time interval ∆t which can be found from the

change in momentum per collision, ∆pcoll times the number of collisions Ncoll with the wall in ∆t.

Ncoll/∆t is the collision frequency.

Consider a fixed volume V containing N gas molecules and an area A

on the wall of the container perpendicular to the x direction. The fig-

ure on the left shows a particular gas molecule of mass m and velocity→v colliding with A. Before the collision the particle has momentum

mvx along the x direction and momentum −mvx after collision yield-

ing a change in linear momentum in the x direction of −2mvx upon

EDR = Engel, Drobny, Reid, Physical Chemistry for the Life Sciences text

competition between immune cells and tumors, transport of water/molecules across biological membranes (osmosis), diffusion of neurotransmitters through cytoplasm, ion movement through biological membranes

-2-

collision. Note that the components of momentum along y and z

remain unchanged. Due to conservation of momentum this results in

a change in momentum of the container wall of 2mvx. To determine

the number of collisions with the area A in ∆t consider the volume element V ′ shown in the figure on the left defined by the area of the wall A times the length ∆x = vx∆t, V ′= Avx∆t. All molecules within a distance ∆x of the wall and traveling towards it with positive vx will collide with the wall. Since the gas molecules are randomly

distributed the fraction of molecules located within V ′ is just V ′/V. Then the total number of molecules that will collide with the wall in

∆t is ½ N(V ′/V) yielding a collision frequency

Ncoll

∆t=

1

2

NV ′V ∆t

=1

2

N

V

Avx

The factor of ½ accounts for the fact that the molecules can be traveling in the +x or −x direction with

equal probability. Note that the above collision frequency is a product of three terms

collision frequency = (number density) × (cross-sectional area) × (relative speed)

which we will encounter later this week. The pressure is then found from

P =F

A=

1

A

∆pcoll

Ncoll

∆t

=

1

A

2mvx

1

2

N

VAvx

=N

Vmv2

x (14.7)

The above arguments are based upon a single velocity component vx and v2x should be replaced by the

av erage ⟨v2x⟩ since the gas molecules have a distribution of speeds which we will shortly treat. Further-

more there is no preferred direction of motion for the molecules, space is isotropic, so

⟨v2⟩ = ⟨v2x⟩ + ⟨v2

y⟩ + ⟨v2z⟩ = 3 ⟨v2

x⟩. Finally from the ideal gas equation of state we can relate to temperature

PV = nRT = NkT =1

3Nm⟨v2⟩ (14.11)*

Recognizing that the average translational kinetic energy per molecule of mass m is given by ½ m⟨v2⟩(14.12) so that the total translational kinetic energy is ½ Nm⟨v2⟩ we find that PV has the units of energy

and can be expressed in terms of the total translational kinetic energy of the gas molecules

PV =2

3Etrans (14.13)*

Kinetic Theory of Gases (Boltzmann, Maxwell, Clausius, Joule) (14.4)

Assumptions

1. molecular volume is negligible compared to total volume (d < λ , mean free path)

2. large number of molecules in ceaseless random motion

3. rigid spheres - no forces of attraction or repulsion except during collision

4. all collisions are elastic - no energy lost

5. obey Newtons’s laws of motion (incorrect CV, need quantum)

Want distribution functions for speed and velocity

Consider fixed N , V , T

Newton's third law: re- sultant of the forces = 0 =>conservation of momentum

-3-

dNv

N= fraction of molecules with speeds

in the range v, v + dv

dNv

Nis 1) ∝ dv and 2) function of v

dNv

N= F(v)dv = probability (14. 23)

proportionality constant F(v)

distribution function molecular speeds

probability density (see in quantum)

Maxwell’s Derivation

1. velocity distribution is isotropic (independent of direction)

2. velocity components have independent probabilities (proof: quantum, particle in a box)

dNvx

N= fraction of molecules with x velocity

component in vx , vx + dvx

dNvx

Nis 1) ∝ dvx and 2) function of vx

dNvx

N= f (vx)dvx = probability (14. 26)

proportionality constant f(vx)

distribution function velocity component

probability density

velocity distribution is isotropic (Maxwell assumption) =>

dNvy

N= f (vy)dvy and

dNvz

N= f (vz)dvz (14.28)

What is the fraction of molecules that simultaneously have an x component of velocity in vx to vx + dvx ,

y component of velocity in vy to vy + dvy, and z component of velocity in vz to vz + dvz? Velocity com-

ponents are indepdendent (Maxwell) => probabilities indepdendent => product of probabilities is also

independent. So the joint probability is

dNvx ,vy,vz

N= f (vx) f (vy) f (vz)dvx dvy dvz (14.29)

where f (vx) f (vy) f (vz) is the distribution function for molecular velocities (probability density). This is

the probability that tip of vector→v lies in rectangular box at (vx , vy, vz) with sides of length dvx , dvy, dvz .

But dNvx ,vy,vz/N only depends upon |

→v→v | = speed vv.

Determining the Function of Speed, φφ (vv)

We will omit the mathematical details between equations 14.40 and 14.34. We are mostly interested in

understanding how to use the distribution function for averages and how to normalize it.

vydv

vx

vz

v

-4-

A separation of variables

1

vx

d ln f (vx)

dvx

= (same in vy)

= (same in vz)

= some constant, call it b

Normalizing Probability and Solving for b (normalization important for quantum)

d ln f (vx) = bvx dvx => f (vx) = A e bv2x /2 (14.34)

∞

−∞∫ f (vx) = 1 = A

∞

−∞∫e bv2

x /2 (14.35)

=> f (vx) = √ −b

2πe bv2

x /2 (14.36)

to solve for b

PV = nRT = NkT =1

3Nm⟨v2⟩ => ⟨v2⟩ =

3kT

mor vrms = ⟨v2⟩1/2 = √ 3kT

m

⟨v2x⟩ =

∞

−∞∫ v2

x f (vx)dvx

= √ −b

2π

∞

−∞∫ v2

xe bv2x /2dvx =

kT

m

=> b = −m

kT

Therefore

f (vx) = √ m

2π kTe−mv2

x /2kT = f (−vx) (14.42)*

Maxwell-Boltzmann Velocity Distributions, f(vx)

1D:dNvx

N= f (vx)dvx

2D:dNvx vy

N= f (vx) f (vy)dvx dvy

3D:dNvx vyvz

N= f (vx) f (vy) f (vz)dvx dvy dvz

Distribution function, f(vx), for vx in N2 at 300 Kf (vx), s/m

(14.22)

To evaluate this use integrals 1 and 2 in Table 14.1; n = 0, a = −b /2.

∫e bv2x /2dvx = 2 ∫e bv2

x /2dv∞∞

−∞ 0

2πb

=

dvx dvx

x

1/2

∫ e bv2x /2dvx = 2 ∫ v2

x ebv2

x /2dv∞∞

−∞ 0

xv2x

To evaluate this use integrals 1 and 3 in Table 14.1; n = 1, a = −b /2.

−

2πb

=

1/2

− 3

vx (m/s)

f(vx) is a normalized Gaussian function

_

_

EDR (24.5)

-5-

Want Distribution Function for Speed, F(v)

f (vx) = √ m

2π kTe−mv2

x /2kT => f (vx) f (vy) f (vz)dvx dvy dvz =

m

2π kT

3/2

e−m(v2x+v2

y+v2z)/2kT dvx dvy dvz

=

m

2π kT

3/2

e−mv2/2kT dvx dvy dvz (14.43)

Determining the Integral Volume Element, dvx dvy dvz

1. (Proper Mathematical Treatment) Jacobian of the transformation dvx dvy dvz → dvdθ dφ with volume

element

dvx dvy dvz =∂(vx , vy, vz)

∂(v,θ , φ )dvdθ dφ

where the Jacobian of the transformation is the determinant

∂(vx , vy, vz)

∂(v,θ , φ )=

∂vx

∂v∂vx

∂θ∂vx

∂φ

∂vy

∂v∂vy

∂θ∂vy

∂φ

∂vz

∂v∂vz

∂θ∂vz

∂φ

defined in spherical polar coordinates

vx = v sinθ cos φ v: 0 → ∞vy = v sinθ sin φ θ : 0 → πvz = v cosθ φ : 0 → 2π

2. Levine shell volume calculation, dV , of 14.44

dV = V (v + dv) − V (v)

= 4/3 π (v + dv)3 − 4/3π v3

= 4/3 π [v3 + 3v2dv + 3v(dv)2 + (dv)3] − 4/3 π v3

= 4π v2dv + 4π v(dv)2 + 4/3 π (dv)3

where, since dv is small, (dv)2 and (dv)3 are even smaller and the last two terms can be ignored.

3. (Our Approach) Boundary area of volumes in space - if a volume V in space is defined by one vari-

able, say r, then dV /dr is the boundary area (surface area) of the volume.

V = 4/3 π v3 =>dV

dv= 4π v2 = area => dV = 4π v2dv

Maxwell-Boltzmann Speed Distributions, F(v)

dNv

N= F3(v)dv = f (vx) f (vy) f (vz) 4π v2 dv

(14.45)*

dNv

N= F2(v)dv = f (vx) f (vy) 2π v dv

dNv

N= F1(v)dv = 2 f (vx) dv

Distribution function, F(v) , for v in N2 at 300 K

v (m/s)

F (v), s/m

vydv

vx

vz

v

-6-

Applications of the Maxwell-Boltzmann Distribution (14.5)

Overview of the Distribution

From the mathematical form of the distribution of a velocity component or the speed distribution we can

assess how temperature and molecular mass affect the distribution, both the position of the maximum and

the width of the distribution. Recognizing (perhaps from Analytical) that f (vx) is a Gaussian which in

standard form is given as

1

σ √ 2πe−x2/2σ 2

where the full width at half the maximum of the peak is 2√ 2 ln 2σ . Basically the width is given by σ , the

standard deviation. For f (vx) σ = kT /m.

Using the Distribution

One of the major applications of a distribution function for some variable x is to calculate the average

value of any function of x. To be specific for some function of speed, h(v)

⟨h(v)⟩ =∞

0

∫ h(v)F(v)dv

There are three ways to quantify typical speeds of particles in thermal equilibrium: av erage speed ⟨v⟩,root-mean-square speed, √ ⟨v2⟩, and most probable speed, vmp.

EX 1. What is the average speed? (Hint: you will need integral 6 in Table 14.1 of Levine)

⟨v⟩ =∞

0

∫ vF(v)dv =∞

0

∫ v

m

2π kT

3/2

e−mv2/2kT 4π v2dv

as T ↑, v↑ and the distrubution broadens as m↑, v↓ and the distrubution narrows

-7-

EX 2. What is the root-mean-square speed? (Hint: integral 3 in Table 14.1)

⟨v2⟩ =0

∫ v2F(v)dv =0

∫ v2

m

2π kT

∞ ∞ 3/2

e−mv2/2kT 4π v2dv

EX 3. What is the most probable speed?

Maxwell-Boltzmann Translational Kinetic Energy Distributions

Especially as we go into chemical kinetics, the kinetic energy that molecules possess will be more impor-

tant than their speed. The 3D speed distribution given on p. 5 (14.45) can be transformed into a distribu-

tion of kinetic energy by substituting all occurrences of speed v by ε tr using the relationship ε tr = ½ mv2.

ε tr =1

2mv2 => v = tr√ 2ε

m=> dv =

1

m √ 2

m

ε tr

dε tr

dNv

N=

m

2π kT

3/2

e−mv2/2kT 4π v2dv =>dNε tr

N=

m

2π kT

3/2

e−ε tr/kT 4π2ε tr

m

1

m

2ε tr

m

−1/2

dε tr

= 2π

1

π kT

3/2

e−ε tr/kT ε tr1/2dε tr (14.52)

vmp⟨v⟩

sqrt(⟨v 2⟩)

-8-

Collisions of Molecules with a Wall and Effusion (14.6)On p. 2 of these notes, in our nonrigorous derivation of the pressure for an ideal gas, we determined that

the rate which molecules with x-component of velocity vx collided with an area A of the wall was

Ncoll

∆t=

1

2

N

V

Avx

The argument in this derivation was based upon all of the molecules having a single x-component of

velocity vx . To correct this, note that the number of molecules with x-componet of velocity between vx

and vx + dvx is Nf (vx)dvx (14.26). Then the collision frequency with the area A of the wall becomes

dNvx

dt=

N

V

A f (vx)vx dvx

and the total collision frequency with the wall W is found by integrating over vx

dNW

dt=

N

V

A

∞

0

∫ f (vx)vx dvx (14.54)

=

N

V

A√ m

2π kT

∞

0

∫ e−mv2x /2kT vx dvx

=

N

V

A√ kT

2π m

=

N

V

A

1

2⟨v⟩1D

=

N

V

A

1

4⟨v⟩3D

where in the last line the relationship ⟨v⟩1D = ½ ⟨v⟩3D was used. The frequency of molecular collisions

with the wall, ZW (rate of wall collisions - note that the units are those of a frequency, s−1), is

ZW =dNW

dt

=1

4

N

V

A⟨v⟩ (14.56)

Introduction to Collisions Between Molecules, 14.7

By rewriting the above wall collision frequency we will uncover a physical relationship common to all

collisions, in particular, the rate of intermolecular collisions.

zbb =

Nb

V

σ ⟨v(µbb)⟩ (14.62)

Zbb, the total number of collisions between b molecules per unit volume per unit time is

To evaluate this use integral 5 in Table 14.1 with a = m/2kT.

∫e − mv2x /2kT vxdvx =

∞

0

kT

m

CHEM 343: Eq 3 of Low Pressure Effusion of Gases experimental writeup.

_

Z = (number density) × (cross sectional area) × (relative speed)

Then, zbb, the number of collisions of one b molecule with other b molecules per unit time is

The flux J of a physical property (mass, energy, momen- tum, etc) is the rate at which that property crosses a unit area. Then, in effusion, the number of molecules which strike the hole in the wall per unit time and per unit area is

J = EDR (24.4) dNW

dt

1

A__ ____

-9-

Zbb =1

2

Nb

V

2

σ ⟨v(µbb)⟩ (14.64)

(where the factor of one-half is necessary so as not to count every collision twice), zbc, the number of col-

lisions of one b molecule with c molecules per unit time is

zbc =

Nc

V

σ bc⟨v(µbc)⟩ (14.61)

and Zbc, the total number of collisions between b and c molecules per unit volume per unit time is

Zbc =

Nb

V

Nc

V

σ bc⟨v(µbc)⟩ (14.63)

where σ is the collisional cross sectional area and µ (which we will encounter again in quantum) is the reduced mass

σ bc = π (rb + rc)2 = π [(db + dc)/2]2 = π d2bc

1

µbc

=1

mb

+1

mc

Collisions Between Molecules

The most interesting events in the life of a molecule occur when it makes a collision with another mole-

cule. Chemical reactions between molecules depend upon such collisions. Important transport proper-

ties, through which kinetic energy (by thermal conductivity), mass (by diffusion), momentum (by viscos-

ity), and charge (by ionic conductivity) are transferred from one point to another involve collisions

between molecules (EDR Chapter 24). A brief simplified account of molecular collisions will be given to rationalize the above formula for Zbc, neglecting the distribution of molecular velocities.

Consider two kinds of molecules, b and c, which interact as rigid spheres with radii rb and rc and diame-

ters db and dc. The contact between two such molecules is shown below. A collision occurs whenever

the distance between centers becomes as small as dbc = (db + dc)/2. Imagine a sphere of radius dbc cir-

cumscribed about the center of b. Whenever the center of a c molecule comes within this sphere, c can be

said to make a rigid-sphere collision with b.

Suppose that all the c molecules are at rest and the b molecule is moving through the volume of stationary

c molecules with an average speed ⟨vb⟩. We can imagine that the moving b molecule in unit time sweeps

out a cylindrical volume π d2bc⟨vb⟩. If Nc/V is the number of c molecules per unit volume, there will be

π d2bc⟨vb⟩Nc/V centers of c molecules encountered per unit time in the sweep by the b molecule. We can

thus estimate the collision frequency for one b molecule as

zbc(est ) =

Nc

V

σ bc⟨vb⟩

where σ bc is the cross sectional area, π d2bc. If there are (Nb/V ) molecules of b per unit volume, the total

collision frequency between b and c molecules would be

Hybc I

Hybc I dt

cc

c

b

bc

rb � rc db � dc

1

µ=

1

m1+

1

m2+

1

mN...+

For N particles 1/(reduced mass) is

___ ___ ______

hard sphere model

-10-

Zbc(est ) =

Nb

V

Nc

V

σ bc⟨vb⟩

This estimate neglects the effect of the excluded volumes of the molecules (similar to the van der Waals

correction to the equation of state of an ideal gas), but at low pressures this correction would be small.

A much more important error in this estimate comes from the assumption that the c molecules are station-

ary while the b molecules sweep through the volume. Actually, it is the speed of b relative to c that deter-

mines the frequency of collisions. This relative speed vbc is the magnitude of the vector difference

between the velocities of b and c. As shown below, the magnitude vbc depends on the angle between vb

and vc (as given by the law of cosines from trigonometry).

vbc = (v2b + v2

c − 2vbvc cosθ )1/2

Thus, the expressions for the collision frequencies should be corrected to employ the average relative

speed, ⟨vbc⟩

zbc =

Nc

V

σ bc⟨vbc⟩

Zbc =

Nb

V

Nc

V

σ bc⟨vbc⟩ (14.63)

In the rigorous derivation of the molecular collision frequency (collision frequency handout) one finds that

⟨vbc⟩ = ⟨v(µbc)⟩ = √ 8kT

π µbc

where µbc is the reduced mass. Note that if we were to consider a volume of a single kind of gas where

mb = mc = m in the reduced mass formula, the average relative speed becomes

⟨vbb⟩ = ⟨v(µbb)⟩ = √2√ 8kT

π m= √2⟨v⟩

An important quantity in kinetic theory is the average distance a molecule travels between two successive

intermolecular collisions. This is called the mean free path (we will encounter this in EDR Chapter 24.2). The average number of collisions experienced by one molecule in unit time is zbb. The distance the

molecule has traveled in this unit time is ⟨vb⟩. Therefore, recognizing that distance = velocity × time and

time is just the reciprocal of the collision f requency, the mean free path λ is

λ =⟨vb⟩zbb

=1

√2σ (N /V )(14.67)