Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT

n = PVRT

=1.0

5.8

Repulsive Forces

Attractive Forces

Maxwell distribution :In 1859, James Clerk Maxwell (1831 – 1879) worked out a formula for

the most probable distribution of speeds in a gas .

Molecules in a gas sample

move at a variety of speeds . Speed of

each molecule constantly changing

due to countless collisions(about 1

billion per second for each molecule).

At low temperature most molecules

move close to the average speed , at

higher temperature there is greater

distribution of speeds.

Maxwell distribution of speeds :

Just for your knowledge do not memorize

Boltzman Distribution. The behaviour of the gas molecules under the action of gravity.

(Harcourt school Publishers)

The Kinetic Theory

• The ideal gas law is an empirical law gives a macroscopic explanation of gas behavior.

• 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.

• Predicts experimental phenomena that haven't been observed. (Maxwell-Boltzmann Speed Distribution)

An ideal gas molecule in a cube

of sides L .

J. B. Callis,Washington University

Calculating force exerted on a container by Collision of a single particle

J. B. Callis,Washington University

axes..Cartesian three thealong velocity theof components

are and , and particle theof speed theis Where

2222

zyx

zyx

uuuu

uuuu

2222 2222

,

and for same theand 2

/1frequency collision The

)(2

uL

m

L

mu

L

mu

L

muF

Thus

FFL

umuF

tL

u

mumumumu

t

mu

t

ummaF

zyxtot

zyx

xx

x

xxxx

J. B. Callis,Washington University

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

V

umnN

V

umnNP

N

nnN

moleculeV

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 becan

sample gasgiven ain particles ofnumber theSince

)1(336

/2

2

J. B. Callis,Washington University

Kinetic Theory Relates the Kinetic Energy of the Particles to Temperature

nRTPV

Thus

RTKE

KEn

PV

V

nKEP

umNKE

avg

avgavg

Aavg

2

3

3

2or

3

2

2

1 2

J. B. Callis,Washington University

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,M is the molar mass expressed in kg/mol (to make the speed come out in units of m/s).

J. B. Callis,Washington University

Relationship of kinetic energy of one gas particle to temperature:

TN

RE

Ak

2

3

Conclusion:•Temperature is a measure of the molecular motion.• At the same temperature, all gases have the same average kinetic energy.

Gas diffusion and effusionGas diffusion and effusion

Graham’s law governs effusion Graham’s law governs effusion and diffusion of gas and diffusion of gas molecules. molecules. KE=1/2 mv2

Thomas Graham, 1805-Thomas Graham, 1805-1869. Professor in 1869. Professor in Glasgow and LondonGlasgow and London..

M of AM of B

Rate for B

Rate for A

M

1diffusionofRate

Rate of effusion is inversely Rate of effusion is inversely proportional to its molar massproportional to its molar mass..

Rate of effusion is inversely Rate of effusion is inverselyproportional to its molar massproportional to its molar mass..

Diffusion• Diffusion is 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).

• Effusion 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.

J. B. Callis,Washington University

GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION

• diffusiondiffusion is the gradual is the gradual mixing of molecules mixing of molecules of different gases.of different gases.

• effusioneffusion is the is the movement of molecules movement of molecules through a small hole through a small hole into an empty container.into an empty container.

The Process of Effusion J. B. Callis,Washington University

The inverse relation between

diffusion rate andmolar mass.

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

Due to it’s lightmass, ammonia

travels 1.46 timesas fast as

hydrogen chloride

J. B. Callis,Washington University

Relative Diffusion Rates of NH3 and HCl

J. B. Callis,Washington University

GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION

Molecules effuse thru holes in a rubber Molecules effuse thru holes in a rubber balloon, at a rate (= moles/time) that is balloon, at a rate (= moles/time) that is proportional to Tproportional to T

• inversely proportional to M.inversely proportional to M.

QuestionQuestion If you have 2 ballons flled with He & OIf you have 2 ballons flled with He & O2 2 Left

overnight at same T ,Which will effuse more?Which will effuse more?

He effuses more rapidly than OHe effuses more rapidly than O22 at same at same

T.T.

HeHe

OO22

Examples

Calculating Molecular Speeds :

Question :

Calculate the average speed of O2 in air at 20 oC .

1660 km/h! (1.6km=1mile).

Question :

What is the r.m.s. speed of SO2 atoms at 25°C?

urms = 340.78 ms-1

Calculation of Molecular Speeds and Kinetic Energies , T = 300 K

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

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.

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

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

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

Problem

Relative Diffusion Rate of NH3 compared to HCl:

RateNH3 =