Kinetic Model of GasesSection 1.3 of Atkins, 6th Ed, 24.1 of Atkins, 7th Ed.21.1 of Atkins, 8th Ed., and 20.1 of Atkins, 9th Ed.

Basic Assumptions

Molecular SpeedsRMS SpeedMaxwell Distribution of SpeedsRelative Mean SpeedMost Probable Speed

Collision Frequency

Mean Free Path

Last updated: Sept. 14, 2018, Minor update on slide 14 for 11th edition users.

ρgas « ρliquid . ρsolid

βgas » βliquid , βsolid

- Low density:

- High compressibility:

- Exert an external pressure:

An external pressure must be applied to contain a gas

- Different gases diffuse into one another:

Different gases completely mix in a homogeneousfashion

Properties of Gases*

Kinetic Model: Basic Assumptions

In the kinetic model of gases, it is assumed that the onlycontribution to the energy of the gas comes from the kineticenergies of the atoms or molecules in the gas - interactionsbetween molecules (potential energy) make no contribution

1. Gas consists of atoms or molecules of mass mundergoing random, neverending motion

2. Molecular size is negligible: the distances over whichmolecules travel are much greater than molecular size

3. Molecules are treated as hard spheres: they makeperfectly elastic collisions with one another and thesides of the container - this means no energy istransferred to rotational, vibrational or electronic modes,nor to the walls - all energy is conserved for translation

pV '13nMc 2 , c ' ¢v 2¦1/2

Pressure, Volume and Kinetic Energy†

We will demonstrate that under the kinetic model of gases thatthe pressure and volume of the gas are related by:

where the molar mass of the molecules,M, is M = mNA, and c is the root meansquare speed of the molecules.

We will consider a molecule of mass m withan x-component of velocity vx. Uponcolliding with the wall, its linear momentumP = mvx changes to -mvx (changes by 2mvx)(assume no changes in y and z for now)

ΔP 'nNAAvxΔt

2V× 2mvx '

nMAv 2xΔt

V

Deriving the RMS SpeedDistance: vxΔtIn a small time interval, Δt, a molecule with velocity vx travels vxΔt along x(all molecules within this distance strike the wall during Δt)Volume: AvxΔtIf wall has area A, then all molecules in the volume V = AvxΔt hit the wall.Density: nNA/VThe number of particles n in the “container” of volume V, so in this smallvolume, there are nNA/V x AvxΔt particlesStatistics: ½At any given time, half of the particles move left, half move right, so theaverage number of collisions in Δt is ½nNAAvxΔt/VMomentum: 2mvxIn the interval Δt, the total momentum change is given by the product ofthe average number of collisions, and total change in momentum

F 'ΔPΔt

'nMAv 2

x

V

p 'FA

'nMv 2

x

V'

nM¢v 2x ¦

V

c 2 ' ¢v 2x ¦ % ¢v 2

y ¦ % ¢v 2z ¦ ' 3¢v 2

x ¦ , thus ¢v 2x ¦ '

13c 2

Deriving the RMS SpeedNewton’s Second Law of Motion: The rate of change of momentum isequal to force applied on an object: ΔP/Δt = F (kg m s-1/s = kg m s-2)

Pressure: Defined as force exerted over some area A:

On the RHS of the above equation, there are angular brackets, whichdenote the average of some value (since all molecules are not travellingat an identical speed). Since molecules move randomly as well, there issimilar behaviour along the y and z directions.

pV ' nM¢v 2x ¦ '

13nMc 2 , c ' ¢v 2¦1/2

c '3RTM

1/2

RMS Molecular Speed*The RMS speed, c2, depends only upon the temperature; thus at aconstant temperature, pV = constant (Boyle’s Law), so:

Under the ideal gas law, the RHS of the equation should be equal tonRT: algebraic substitution can be used to show that

The RMS speed is proportional to the square root of temperature, andinversely proportional to the square root of molar mass. On average:

- At higher temperatures, molecules travel faster- Heavier molecules travel slower than light molecules(Balloons)

*avogadro_3 (balloons)

f(v) ' 4π M2πRT

3/2v 2e &Mv

2 /2RT

Maxwell Distribution of SpeedsJames Clerk Maxwell (1831-1879), a Scottish physicist, didrevolutionary work in electromagnetism and the kinetic theory ofgases. By treating molecules of gases in rapid motion themstatistically he was able to formulate (1866), independently ofLudwig Boltzmann, the Maxwell-Boltzmann kinetic theory of gases.This theory showed that temperatures and heat involved onlymolecular movement. Philosophically, this theory meant a changefrom a concept of certainty--heat viewed as flowing from hot to cold--to one of statistics--molecules at high temperature have only ahigh probability of moving toward those at low temperature.This new approach did not reject the earlier studies ofthermodynamics; rather, it used a better theory of the basis ofthermodynamics to explain these observations and experiments.

In a real gas the speeds of individual molecules span wide ranges, withconstant collisions continually changing the molecular speeds. Maxwellshowed that the distribution of speeds of molecules in a gas can bewritten as an analytical equation:

Fraction between v1 and v2 ' mv2

v1

f(v) dv

Maxwell Distribution of Speeds**He demonstrated that the fraction ofmolecules that possess speeds in therange Δv = v to v + dv is proportional to thewidth of the range

Maxwell distributions show that:- Higher temperatures have broaderdistributions of molecular speeds- Lighter molecules have a broaderdistribution of speeds than heaviermolecules

To use Maxwell’s distribution equation, weneed to multiply it by Δv, or if Δv is notinfinitessimal, the definite integral belowmust be evaluated:

*maxwell_3, other Maxwell

Integral CalculusIntegrals: Give us the “area under the curve”

Two approaches:

1. Anti-derivative - if it is possible to write amathematical expression for a function thatdescribes it’s anti-derivative analytically(i.e., write a new function based on theinverse of differentiation)

Here we mathematically generate integrals!

2. Sum - the area under a function may alsobe calculated as the sum of many small(infinitessimal) elements (these are oftenevaluated numerically by some iterativeprocess on a computer)

Here we give physical meaning to integrals!

f(vx,vy,vz)dvxdvydvz

f(vx,vy,vz) ' f(vx) f(vy) f (vz)

f (vx) ' f (v 2x ), f (vx,vy,vz) ' f (v 2

x ) f (v 2y ) f (v 2

z )

f (vx,vy,vz) ' f(v 2x %v

2y %v

2z ) ,

thus f (v2x %v

2y %v

2z ) ' f(v 2

x ) f(v2y ) f(v

2z )

Deriving the Maxwell Distribution: 1The probability that a molecule has velocity in the range:vx to vx+dvx, vy to vy+dvy, vz to vz+dvz is proportional to widths anddepends on velocity components:

Cartesian (x,y,z) velocity components are orthogonal (independent)

Equal probability of molecule having velocity +*vx* to +*vx*dvx or -*vx* to -*vx*dvx, so f(vx) must be related to square of vx

Molecular speed is independent of direction of individual components

f (vx) ' Ke ±ζv 2x

m4

&4

f(vx) dvx ' 1

1 ' Km4

&4

e &ζv2x dvx ' K π

ζ

1/2

Deriving the Maxwell Distribution: 2An exponential function is ideal here, since: eaeb = e(a+b); thus,

where K and ζ (zeta) are constants, and f(vx) = f(vx2) (same for y and z)

Probability that a molecule has an extremely high velocity is low,therefore we can discard the solution with +ζvx

2 in the exponentTo determine K, molecule must have v in range -4 to 4:

Substitute above expression for f(vx) and integrate wrt vx:

Therefore, K = (ζ/π)1/2 (solution of above definite integral comes froma table of integrals, found in CRC Handbook or basic calculustextbooks)

¢v 2x ¦ ' m

4

&4

v 2x f(vx)dvx '

ζπ

1/2

m4

&4

v 2x e

&ζv 2x dvx

'ζπ

1/2× 1

2πζ3

1/2

'12ζ

c ' (¢v 2x ¦ % ¢v 2

x ¦ % ¢v 2x ¦)1/2 '

32ζ

1/2

f(vx) 'M

2πRT

1/2e &Mv

2x /2RT

Deriving the Maxwell Distribution: 3In order to find ζ, the RMS Speed must be calculated. In the next step,we calculate the expectation value for vx

2:

Since we know c = (3RT/M)1/2, ζ = M/2RT. Therefore

We are very close to the final solution now!

f(vx,vy,vz)dvxdvydvz ' f (vx) f(vy) f(vz)dvxdvydvz

'M

2πRT

3/2e &Mv 2/2RTdvxdvydvz

f(v) ' 4π M2πRT

3/2v 2e &Mv

2 /2RT

Deriving the Maxwell Distribution: 4Probability that a molecule has speed in the range vi to vi+dvi (i = x,y,z)

Since speed is not dependent upon direction, the probabilities that thevelocities are in the volume element dvxdvydvz on the RHS can bewritten as the volume of a spherical shell of radius v, 4πv2dv. Thus,

The mean speed, c, and the most probable speed, c*, can be evaluated by similar methods (please try Example and Self Test 1.6 inAtkins, 6th Ed. or Example and Self Test 24.1 in Atkins, 7th Ed., 21.1 in 8th Ed., 20.1 in the 9th Ed., 1.6 in the 10th Ed., Ex. 1B.1 in 11th Ed.)

c '3RTM

1/2

f (v) ' 4π M2πRT

3/2v 2e &Mv

2 /2RT

c '8RTπM

1/2

c ( '2RTM

1/2

Summary of Speeds

Root mean square speed:

Maxwell distribution of speeds:

Mean speed:

Most probable speed:

crel ' 21/2c '8kTπµ

1/2, µ '

mAmBmA%mB

Relative Mean SpeedThe relative mean speed is the speed with which one moleculeapproaches another, and is given by

where k is the Boltzmann constant, k = R/NA, and μ is called thereduced mass (expression is used for molecules of different masses)

What does crel tell us?

If molecules move in same direction, therelative mean speed is 0.If molecules are on a direct collisioncourse, relative mean speed is 2v.If molecules have orthogonal pathways,the relative mean speed is 21/2v (this is themost typical speed)

z 'σcrelNV

'σcrelpkT

, σ ' πd 2

Collision frequencyKinetic model allows us to:- calculate the frequency with which molecules collide- calculate the distance traveled by molecules between collisions- calculate molecular transport properties and reactivities in gasesWhen molecules are within distance d, the collision diameter, they aresaid to collide with one another. The kinetic model gives the collisionfrequency, z, which is the number of collisions in some time interval

σ is the collision cross-section of the molecules

ccrelΔt

In interval )t, moleculesweeps a tube ofdiameter 2d and lengthcrelΔt. Each hit changesthe direction of the path,but the “volume” of thepath is the same

λ 'cz

'kT

21/2σp

Mean Free PathIf the collision frequency is known, then one can calculate the meanfree path, λ, which is the average distance between collisions

If a molecule collides with frequency z, it is in traveling at a mean speedof c for a time 1/z (substituting in for z):

c

The kinetic model works if d << λ (molecules are far from one another)

Example: For N2 gas at 1 atm and 25oC (F = 0.43 nm2)

z = 7 x 109 s-1 (collisions about every ns or so)

λ = 70 nm (103 molecular diameters)

c = 350 m s-1