The Kinetic Molecular Theory of Gasesand

Effusion and Diffusion

Chemistry 142 B

Autumn Quarter, 2004

J. B. Callis, Instructor

Lecture #15

The Ideal Gas Law

• A purely empirical law - solely the consequence of experimental observations

• Explains the behavior of gases over a limited range of conditions.

• A macroscopic explanation. Says nothing about the microscopic behavior of the atoms or molecules that make up the gas.

The Kinetic Theory

• Starts with a set of assumptions about the microscopic behavior of matter at the atomic level.

• Supposes that the constituent particles (atoms) of the gas obey the laws of classical physics.

• Accounts for the random behavior of the particles with statistics, thereby establishing a new branch of physics - statistical mechanics.

• Offers an explanation of the macroscopic behavior of gases.• Predicts experimental phenomena that haven't been

observed. (Maxwell-Boltzmann Speed Distribution)

The Four Postulates of the Kinetic Theory

• A pure gas consists of a large number of identical molecules separated by distances that are great compared with their size.

• The gas molecules are constantly moving in random directions with a distribution of speeds.

• The molecules exert no forces on one another between collisions, so between collisions they move in straight lines with constant velocities.

• The collisions of molecules with the walls of the container are elastic; no energy is lost during a collision. Collisions with the wall of the container are the cause of pressure.

An ideal gas molecule in a cube of sides L.

Cartesian Coordinate System for the Gas Particle

Cartesian Components of a Particle’s Velocity

n.informatio ldirectiona

no conveys andquantity scalar a is that Note

axes.Cartesian

principal three thealong velocity theof components

are and , and particle theof speed theis Where

2222

u

uuuu

uuuu

zyx

zyx

Component of velocity along the x axis.

Calculation of the Force Exerted on a Container by Collision of a Single Particle

2222 2222

,

and for simalarly and 2

frequency collision that theRecalling

2

uL

m

L

mu

L

mu

L

muF

Thus

FFL

umuF

L

u

mumu

t

mu

t

ummaF

zyxtot

zyx

xx

x

xx

Calculation of the Pressure in Terms of Microscopic Properties of the Gas Particles

V

umnN

V

umnNP

N

nnN

V

um

L

um

L

Lum

Area

FP

uL

mF

A

A

A

A

Tot

Tot

tot

22

2

3

2

2

2

2

21

3

2

3

number, sAvogadro' is and moles

ofnumber theis where, as expressed be can

sample gas givena in particles ofnumber theSince

336

/2

2

The Kinetic Theory Relates the Kinetic Energy of the Particles to Temperature

RTKE

TKEn

PV

V

nKEP

umNKE

avg

avgavg

Aavg

2

3

3

2or

3

2

2

1 2

Mean Square Speed and Temperature

By use of the facts that (a) PV and nRT have units of energy, and (b) the square of the component of velocity of a gas particle striking the wall is on average one third of the mean square speed, the following expression may be derived:

M

RTurms

3

where R is the gas constant (8.31 J/mol K), T is the temperature in kelvin, and M is the molar mass expressed in kg/mol (to make the speed come out in units of m/s).

Finally, we obtain the relationship of kinetic energy of one gas phase particle to temperature:

TN

RE

Ak

2

3

Note that: (a) Temperature is a measure of the molecular motion, and (b) the kinetic energy is only a function of temperature. At the same temperature, all gases have the same average kinetic energy.

Path of one particle in the gas.

Thus far we have discussed the random nature of molecular motion in terms of the average (root mean square) speed. But how is this speed distributed? The kinetic theory predicts the distribution function for the molecular speeds. Below we show the distribution of molecular speeds for N2 gas at three temperatures.

Molecular Speed Distribution

The Mathematical Description of the Maxwell-Boltzmann Speed Distribution

Kin re temperatu

and J/K, 101.38066 constant sBoltzman'

kgin particle of mass ,/in peed :where

24)(

23-

22 2

T

k

msmsu

euTk

muf

B

Tkmu

B

B

f(u)du gives the fraction of molecules that have speeds between u and u + u.

The distribution ofmolecular speeds for N2

at three temperatures

Features of the Speed DistributionThe most probable speed is at the peak of the

curve.

The most probable speed increases as the temperature increases.

The distribution broadens as the temperature increases.

Relationship between molar mass andmolecular speed

Features of the Speed Distribution

The most probable speed increases as the molecular mass decreases.

The distribution broadens as the molecular mass decreases.

Calculation of Molecular Speeds and Kinetic Energies

Molecule H2 CH4 CO2

MolecularMass (g/mol)

2.016 16.04 44.01

Kinetic Energy (J/molecule)

6.213 x 10 - 21 6.213 x 10 - 21 6.213 x 10 - 21

Velocity (m/s)

1,926 683.8 412.4

T = 300 K

The Three Measures of the Speed of a Typical Particle

M

RTuu

M

RTu

M

RTu

avg

mp

rms

8

2

3

Various Ways to Summarize the ‘Mean’ Speed.

Problem 15-1: Calculate the Kinetic Energy of (a) a Hydrogen Molecule

traveling at 1.57 x 103 m/sec, at 300 K.

Mass =

KE =

KE =

KE =

Problem 15-1 Kinetic Energies for (b) CH4 and (c) CO2 at 200 K

(b) For Methane, CH4 , u = 5.57 x 102 m/s

KE =

(c) For Carbon Dioxide, CO2 , u = 3.37 x 102 m/s

KE =

Note

• At a given temperature, all gases have the same molecular kinetic energy distributions,

and

• the same average molecular kinetic energy.

Visualizing the Kinetic-Molecular Theory

http://comp.uark.edu/~jgeabana/mol_dyn/

EffusionEffusion is the process whereby a gas escapes from its container through a tiny hole into an evacuated space. According to the kinetic theory a lighter gas effuses faster because the most probable speed of its molecules is higher. Therefore more molecules escape through the tiny hole in unit time.

This is made quantitative in Graham's Law of effusion: The rate of effusion of a gas is inversely proportional to the square root of its molar mass.

M

1effusionofRate

The Process of Effusion

Effusion CalculationProblem 15-2: Calculate the ratio of the effusion rates of ammonia and hydrochloric acid.Approach: The effusion rate is inversely proportional to square root of molecular mass, so we find the molar ratio of each substance and take its square root. The inverse of the ratio of the square roots is the effusion rate ratio. Numerical Solution:

HCl = NH3 =

RateNH3 =

Diffusion

• The movement of one gas through another by thermal random motion.

• Diffusion is a very slow process in air because the mean free path is very short (for N2 at STP it is 6.6x10-8 m). Given the nitrogen molecule’s high velocity, the collision frequency is very high also (7.7x109 collisions/s).

• Diffusion also follows Graham's law:

M

1diffusionofRate

Diffusion of agas particlethrough aspace filledwith otherparticles

NH3(g) + HCl(g) = NH4Cl(s)

HCl = 36.46 g/mol NH3 = 17.03 g/mol

Problem 15-3: Relative Diffusion Rate of NH3 compared to HCl:

RateNH3 =

The inverse relation betweendiffusion rate andmolar mass.

NH3(g) + HCl(g) NH4Cl(s)

Due to it’s lightmass, ammonia travels 1.46 timesas fast as hydrogen chloride

Relative Diffusion Rates of NH3 and HCl

Problem 15-4: Gaseous Diffusion Separation of Uranium 235 / 238

235UF6 vs 238UF6

Separation Factor =

after Two runs

after approximately 2000 runs235UF6 is > 99% Purity !

Y - 12 Plant at Oak Ridge National Lab

Answers to Problems in Lecture #15

1. (a), (b) and (c) are all the same value: 4.13 x 10- 21 J / molecule

2. RateNH3 = RateHCl x 1.463

3. RateNH3 = RateHCl x 1.463

4. S = 1.0086