kinetic theory of gases and...

Kinetic theory of gases and applications

Massimo Mella, [email protected] 2.65F

•Energetics of particle movement•Collisions and reactions•Statistics and probabilities•Statistical Thermodynamics

Tuesday, November 15, 2011

mailto:[email protected]

mailto:[email protected]

Kinetic theory of gases and applications: I

Ludwig Boltzmann died suicidal drowning in the gulf of Trieste in 1906 because only a few scientists believed in atoms as key agent in physical transformations.

However, his ideas have survived him and nowadays represent a cornerstone of our understanding of both physical and chemical transformations.

In this endeavor, he was not alone, though. James. C. Maxwell, Joshua W. Gibbs and others played a role.


Kinetic theory of gases and applications: II

pV = nkBT Equation of states: represents the relationship between p, V and T for a physical system

p =F · n

A

A

F

n

If we accept that a gas is composed by atoms moving around chaotically, we could interpret the pressure as due to atoms colliding on the container surfaces.


Kinetic theory of gases and applications: III

F =dpdt

= mdvdt

= ma

vx

vy

v

vx

vyv

m∆vx = 2mvx



Assumptions of GKT:

F =dpdt

= mdvdt

= ma

vx

vy

v

vx

vyv

m∆vx = 2mvx



Assumptions of GKT: The molecular volume is negligible with respect to the

container volume.

F =dpdt

= mdvdt

= ma

vx

vy

v

vx

vyv

m∆vx = 2mvx




container volume. Molecules move quickly and in straight lines.

F =dpdt

= mdvdt

= ma

vx

vy

v

vx

vyv

m∆vx = 2mvx




container volume. Molecules move quickly and in straight lines. Molecules do not interact.

F =dpdt

= mdvdt

= ma

vx

vy

v

vx

vyv

m∆vx = 2mvx




container volume. Molecules move quickly and in straight lines. Molecules do not interact. Molecules are constantly moving chaotically and elastically.

collide with walls and other molecules.

F =dpdt

= mdvdt

= ma

vx

vy

v

vx

vyv

m∆vx = 2mvx




container volume. Molecules move quickly and in straight lines. Molecules do not interact. Molecules are constantly moving chaotically and elastically.

collide with walls and other molecules. Pressure is due to molecular collision on the walls.

F =dpdt

= mdvdt

= ma

vx

vy

v

vx

vyv

m∆vx = 2mvx


Kinetic theory of gases and applications: IV

A molecule collides with the wall in Δt only if its distance is less than vxΔt.

If the gas is composed of N molecules in a total volume V, the number of molecule hitting the wall on the right during Δt is

So the force due to those collision is

and the pressure is

However, not all molecules would have the same speed, so we need the average value

ncoll =Avx∆tN

2V

Fx = ncoll2mvx/∆t = mAv2x∆tN/(V ∆t) = mAv2

xN/V

p = Fx/A = mv2xN/V

p̄ = mv̄2xN/V = mN�v2

x�/V Boyle’s law!


Kinetic theory of gases and applications: V

To compute the average value, we need to know the fraction of molecules that have their speed in the interval [vx,vx +dvx] (AKA their speed distribution)

Number of people with height hi

hi

nih̄ =

�i nihi�i ni

=�

i

pihi

Fraction (or probability) of people with height hi

h̄ =�

i

pihi =�

i

pi

δhhiδh =

�

i

ρi(δh)hiδh

Interval widthProbability density

h̄ = limδh→0

�

i

ρi(δh)hiδh ≡�

ρ(h)hdh�

ρ(h)dh = 1with


Kinetic theory of gases and applications: VI

➡Directions x, y and z are identical since space is isotropic.•Distribution along x, y and z must be identical (p is the same over all walls)•Distribution along x, y and z must be symmetric (depend on |vx| or vx2)•Chaotic movement implies that vx is independent of vy, etc..•Total speed probability should not depend on the direction ( F function of v2= vx2+vy2+vz2))

F (vx, vy, vz) = f(vx)f(vy)f(vz) ∝ e−βv2xe−βv2

ye−βv2z

f(vx) = (β/π)1/2e−βv2x Normalized



Consider a gas composed of N particles.



ye−βv2z





We want the probability distribution of molecular speeds F(vx,vy,vz)



ye−βv2z






The function F(vx,vy,vz) dvxdvydvz gives us the probability for a particle to have

speeds vx , vy and vz



ye−βv2z






The function F(vx,vy,vz) dvxdvydvz gives us the probability for a particle to have

speeds vx , vy and vz

James Clerk Maxwell, during 1859, derives F(vx,vy,vz) with a clever reasoning.



ye−βv2z



Kinetic theory of gases and applications: VII

�v2x� =

� ∞

−∞v2

xf(vx)dvx =� ∞

−∞v2

x(β/π)1/2e−βv2xdvx = 1/(2β)

p̄ = mv̄2xN/V = mN�v2

x�/V = mN/(2βV )

β = mNA/(2RT ) = m/(2kBT )

�Ekin,x� =m

2�v2

x� = m/(4β) =kBT

2

Then, identifying

one gets

but

Conclusion:using classical mechanics we have derived the mechanical pressure and using informations on the shape of f(vx) we have obtained a link between temperature and movement for a ideal gas.Tuesday, November 15, 2011

Kinetic theory of gases and applications: VIII

Increas ing temperature , distribution maximum moves t o w a r d h i g h e r s p e e d s (warmer gases move faster)

Decreasing mass, distribution maximum moves toward higher speeds (lighter gases move faster).

Increas ing temperature , distribution maximum moves toward higher speeds

Notice:1.Ussain Bolt runs at ~10 m/s2.A car at 60 mph runs at ~27 m/s3.Speed of sound in air is ~300 m/s


Kinetic theory of gases and applications: IX

The simple ideas of a bimolecular reaction in the gas phase:consider the gas phase bimolecular reaction of the form

The transformation of A and B (R's) into C and D (P's) it is likely to happen when A and B are in close contact; only in this situation they could easily exchange atoms between themselves.If this idea is correct, then, in order to predict the rate of the reaction (i.e. the rate at which A and B disappear while C and D appear), one must provide an estimate for the collision frequency between A and B. To do so, let us now suppose that both A and B are spherical objects with radius rA and rB,

respectively, and that A molecules are stationary while B molecules are moving with a velocity distribution described by the Boltzmann distribution at temperature T.

A + B → C + D


Kinetic theory of gases and applications: IX

From the picture, we see that B, while traveling across the gas for a time Δt, may collide with all the molecule whose centre of mass is inside a cylinder of radius d=rA+rB, and of

height <v>Δt. The number of A molecules inside this volume (hence the number of collisions) is given by their concentration (CA=NA/V)

times the volume of the cylinder:

�v�∆tπ(rA + rB)2Ca = �v�∆tσCa

Molecules standing still

�v� = (mB

2πkBT)3/2

�dv |v| e−

mBv2

2kBT = 4π(mB

2πkBT)3/2

�dv |v|3 e−

mBv2

2kBT = (8kBT

πmB)1/2


Kinetic theory of gases and applications: X

Substituting <v> in the previous formula and dividing by the time span to get the number of collision for unit time, one gets

There are, however, two adjustments that we should make to this equation: first, A molecules are not standing still; second, the above formula gives the frequency of collision for a single B molecule, but the system contains many more B molecule.To correct for the first point, it is enough to use the reduced mass of A and B (1/μAB=1/

μA+1/μB) that comes from considering the “relative velocity” of the two species; for the

second point, it is enough to multiply the above formula by the concentration of B to obtain the collision frequency density (per unit volume):

This is similar to the familiar expression for the rate of a bimolecular reaction as a function of the concentrations of A and B; in other words, we can write kAB as

(assume σ=12.5*10-20 m2, T=300 K, μAB=100*10-3 kg/mol -> 3.15*10-17 m3 s-1

z = σCA(8kBT

πmB)1/2

ZAB = σCACB(8kBT

πmB)1/2

kAB = σ(8kBT

πmB)1/2


Kinetic theory of gases and applications: XIMoving away from simple collision theory- activated and oriented collisions:the equation just derived indicates that, assuming a reactive event every collision, the rate of

the reaction should increase with T following a T1/2 law. This is far from being correct, as shown in the following experimental results for the reactions

Indeed, the semi-logarithmic plot in the figure indicates that the rate constants increase following an exponential behavior against 1/kBT, which is not present in our “naive”

prediction using simple collision theory.

CH3 + O2 → CH3O + O

CH3 + O2 → CH2O + OH

Arrhenius proposed that the difference in behaviour could be explained by assuming that reactive encounters take place only between molecules that posses at least a minimum amount of energy relative kinetic energy EA; the fraction of molecules that posses such

amount of energy is proportional to exp[-EA/kBT] (prove it!). The

slow change proportional to T1/2 is hidden by the fast variation due to the exponential term over small temperature ranges.


Kinetic theory of gases and applications: XIIWith this assumption, one can measure the rate constant for a reaction at various T and extract the activation energy from a semi-logarithmic plot as the one shown before (mmhhh, what we really want to do is to compute the activation energy and predict the rate, but let us explore this possibility first).

There is still an ingredient missing, which is the molecular cross section. A simple way to estimate this quantity is to use information coming from transport measurements that give an indication of how large is a molecule.

Let us assume for the time being that we posses all these pieces of information and let us compute the rate for a given reaction to compare with experimental results. In doing do we would discover that our predicted rates are always too large with respect to the experimental ones.

This fact is usually explained using the concept of oriented collision:a collisional event can be reactive if and only if the two molecules are oriented in a suitable way (e.g. an exchanging atom should found itself facing the incoming acceptor molecule).

Very often, this spatial requirement is quite demanding (think about a relatively large molecule that has many possible conformations), so that the ratio between predicted and measured constants can be quite large. In other words, it seems quite difficult to make rate constant predictions unless we explore the fundamentals of the dynamics involved in the reactive encounter.


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