PHYSICAL CHEMISTRY FOR ENGINEERSKinetic Theory of Gases

Dr. Nadia Adrus

Course outcome• Kinetic theory of gas motion and pressure• Distribution and effusion of gas particles• Molecular collisions

-2-Week 11-12

It is expected that the students will be able to• Discuss the translation motion of gas

particles. • Explain the gas molecular distribution and

effusion.• Justify molecular collision including

frequency and distance particle travel between collisions.

Outline of this Lecture-3-

Kinetic theory of gas motion and pressure

Distribution and effusion of gas particles

Molecular collisions

Previous: Macroscopic (i.e., large quantity) behaviour of gases – P, V & T.

This chapter: Kinetic molecular theory of gases attempts to explain the behaviour of gases on a molecular level.

Consider gases at low P (ideal gas) – starting point for the discussion of its transport properties.

Introduction-4-

nRTPV

No. of moles in the ideal gas

Gas constant, R = 8.314 J/mol.K

Temp. (K)

Ideal Gas Law

Note: N, total number of molecules; NA, Avogadro’s number: the number of atoms, molecules, etc, in a mole of a substance: NA=6.02 x 1023/mol.

In the kinetic model of gases – assume that the only contribution to the energy of the gas is from kinetic energies of the molecules.

Ideal Gas Law-5-

The kinetic model is based on 3 assumptions:

1. The gas consists of molecules of mass m in ceaseless random motion.

2. The size of the molecules is negligible, in the sense that their diameters are much smaller than the average distance travelled between collisions.

3. The molecules interact only through brief, infrequent, and elastic collisions

Note: An elastic collision is a collision in which the total translational kinetic energy of the molecules is conserved.

Assumptions of Kinetic Theory-6-

Kinetic Theory of Gas Motion &

Pressure

2nMc31PV

From the assumptions of the kinetic model, P & V of the gas are related by,

Molar mass of the molecules, M = mNA

rms speed of the molecules

Note: rms, root mean square

212vc

Pressure & Molecular Speed-8-

Consider particle of mass m that is travelling with a component of velocity vx (parallel to the x-axis)

Collides with the wall

Reflected

Its linear momentum changes from mvx –mvx

Þ x-comp. momentum Þ 2mvx (y- & z-compt. unchanged)

Many molecules collide with the wall in an interval t

Total change of momentum: (Product of the change in momentum of each molecule) (No. of molecules that reach the wall during the interval)

Derivation: Molecular Speed-9-

Consider, wall has area A Þ All the particles in a volume,

A vxt will reach the wall

Number of density of particles,

nNA/VTotal amount of molecules in the container

Avogadro’s constant

Number of molecules in the volume,

(nNA/V) Avxt

Only those close enough to the surface hit it in time t, those within the distance vx t

Momentum Change-10-

Assume: surface is normal to the x-axis, half the molecules of speed vx move toward the surface.

Average number of collisions with the wall

V

tAvnN21 xA

Total momentum change,

VtnMAv

VtAvnmN2mv2V

tAvnNchange Momentum2

x2

xAx

xA

Particle mass

Total Momentum Change-11-

Force,

VnMAvF

tchange momentumchange momentum of Rate

2xÞ

Pressure

VnMv

AFP

2x

Molecules travels with the same velocity

VvnM

AFP

2x

Force & Pressure-12-

Speed of a single molecule, v

2z

2y

2x

2 vvvv 212vc

2z

2y

2x

2 vvvvc

Molecules are moving randomly,

22x

2x

2 c31vv3c Þ

2nMc31PV

VvnM

P2

x

substitute

Pressure in Terms of C-13-

Video

Website: http://www.youtube.com/watch?v=YSTRa27a3BQ ; http://www.youtube.com/watch?v=iMFwuHDu8dA

Video-14-

2nMc31PV

AmNM2A vmV

nN31P

2vc

2

2

vm21

VN

32

vmVN

31P

NnNA

22 mc21vm2

1KE

KEVN

32P

nRTPV Tk23

NnRT

23KE B

Boltzmann constant, kB=R/NA= 1.38 10–23

Note: Molar mass, M = mNA; N = nNA

Average Translational Kinetic Energy-15-

If c of the molecules depends only on temp.,

constantPV

Boyle’s law (const. T)

nRTPV 2nMc31nRTPV

Ideal gas

21

M3RTc

Conclusion: the c of the molecules of a gas is proportional to the square root of the T and inversely proportional to the square root of the M.

Expression for the mean square speed of molecules

Note: Molar mass, M = mNA; N = nNA

Root Mean Square Speed-16-

Actual gas:• Speeds of individual molecules span a wide range• Collisions in the gas continually redistribute the speeds among

the molecules.

Before a collision: a molecule may be travelling rapidly.

After a collision: it may be accelerated to a very high speed, only to be slowed again by the next collision.

The Maxwell Velocity Distribution-17-

FRACTION OF MOLECULES that have speeds in the range v to (v + dv) is proportional to the width of the range, Þ Written f(v)dv

2RTMv223

2evRT2M4vf

Distribution of speeds

Distribution of Speed-18-

Narrow range of speed

Wide range Þ Integrate

vvf

Evaluate at the speed on interest

Width of the range of speeds of interest

dvvfv to v range the in Fraction 2

1

v

v21

Calculate the Fraction of Molecules-19-

2z

2y

2x mv2

1mv21mv2

1E zyxzyx aaaa

2kTmv2kTmv2kTmv

kTmv21mv21mv21kTE

2z

2y

2x

2z

2y

2x

eeKeKeKef

Const. of proportionality (at const. T)

zyx vfvfvff

Factorize

Fraction of molecules in the velocity range vx to dvx, vy to dvy,& vz to dvz

2kTmv1

z

2kTmv1y

2kTmv1x

2z

2y

2x

eKvfeKvfeKvf

3

3

3

Derivation: Maxwell Distribution Speed-20-

1dvvf xx

Note that molecule must have a velocity somewhere in the range –< vx

,

2kTmv1x

2xeKvf 3 21

31x

2kTmv31mkT2KdveK1 2

x

adxe 2ax

Use standard integral

2323

RT2M

kT2mK

x2RTMv

21

x dveRT2Mvf 2

x

2kTmv1x

2xeKvf 3

Maxwell-Boltzmann velocity distribution

Determination of K-21-

Surface area, 4v2

Probability that a molecule has a velocity in the range vx to vx + dvx, vy to vy + dv , vz to vz + dvz is

zyx

2kTvvvM23

zyx dvdvdveRT2Mvvvf 2

z2

y2

x

2RTMv223

2evRT2M4vf

2z

2y

2x

2 vvvv

dvv4 2

Maxwell speed distribution

2kTMv223

2evkT2m4vf

or

Distribution of Speeds-22-

2121

m2kT

M2RTc

2121

m8kT

M8RTc

2121212

m3kT

M3RTvc

BA

BA21

rel mmmmμμ

8kTc

c*: speed at which df(v)/dv = 0

c: square root of the 2nd moment of distribution

c̅rel: two dissimilar molecules of masses mA & mB

c̅: Average value

Note: k=R/NA, Boltzmann’s constant; , reduced mass

*Previously discussed *Will be discussed later

c*, c̅, c & c̅rel-23-

Some important speeds in the Maxwell speed distribution

Most probable speed, c*

Average speed, c̅

Root-mean square

speed, c

Relative mean

speed, c̅rel

To compare only certain aspects of the distributionÞ Such as the speed at which the distribution is maximized, or

at the average speed

Comparative Values for Speed Distributions-24-

Most probable speed

Average speed

Root-mean square speed

Comparison of c, c̅, c -25-

Compute νmp, νave, and νrms for O2 at 300 and 500 K. How would

your answers change for H2?

Example 2

Determine vmp, vave, and vrms for the following species at 298 K:

a) Ne b) Kr c) CH4

Note: J kg–1 ≡ m2 s–2

Example 1-26-

For N2 at 298 K, what fraction of molecules has a speed between

200 and 300 m s–1? What is this fraction if the gas temperature is

500 K?

Example 4

Compare the average speed and average kinetic energy of O2

with that of CCl4 at 298 K.

Example 3-27-

Molecular Collisions

Quantitative picture of the events that take place in a collection of gaseous molecules. • Frequency of collisions?• Distance between successive collisions?• Rate of collisions per unit volume?

A pair of molecules will collide whenever the centers of the 2 molecules come within a distance d (collision diameter) of 1 another

No collision Collision occurs

d

Intermolecular Collisions-29-

Consider the particle interest is moving & that all other molecule are stationary

Particle of interest sweeps out a collisional cylinderÞ Determines the no. of collisions the particles undergoes per unit of

time

Collision occur between the particle of interest & other particles that are positioned within the cylinder.

The Collision Cylinder-30-

v2

v1

v2

v1

v2

v190

c 2v2

vvvc1/2

1/222

21relrel

1/2

M8RTvc

2112

1/2

12relrel M

1M1

μ1withμ

8RTvc

The reduced mass, μ of the two identical particles is m/2 & therefore

To determine the collision frequency, Z, we have to consider the relative speed of the colliding particles.

Relative Mean Speed, c̅rel-31-

Gas comprised of one type of particle, = m1/2

Number of collisions an individual molecule undergoes with other collisional partners

1/2

1

A11/2

1

111 M

8RT2σRTNP

m8kT2σV

Nz

1/22ave2cyl2

12 μ8kTσV

Ndt

dtσvVN

dtV

VNz

Collisional cross section (Table 16.1)

Individual Particle Collision Frequency -32-

2 type of gas molecule

For gas consisting of only one type of particle is

1/2

1

2A1

1/2

1

21

11 M8RTσRT

NP21

m8kTσV

N21Z

1/2A2A1

12 μ8kTσRT

NPRTNPZ

Total number of collisions that occur for all gases

Total Collision Frequency-33-

Gas molecules encounter collisions with other gas molecules & with the walls of the container.

Define as the average distance between successive molecular collisions

1211

ave

1211

avezz

vdtzz

dtvλ

ave1

ave

11

ave

σv2VN

vzvλ

σP2

kTNP

RTσ2

1λA1

1 type particle, N2 = 0 Þz12 = 0

The Mean Free Path-34-

a) A standard rotary pump is capable of producing a vacuum on the

order of 10–3 Torr. What is the single-particle collisional

frequency and mean free path for N2 at this pressure and 298

K?

b) A cryogenic pump can produce a vacuum on the order of 10–10

Torr. What is the collisional frequency and mean free path

for N2 at this pressure and 298 K?

Example 5-35-

A gas under pressure goes (escapes) from one compartment of a container to another by passing through a small opening.

Rate, r at which molecules pass through a small hole of area A,

2121A

c kT2PA

RTM2APNAZr

m

Collisional rate = dNc/dt

Note: dNc, no. of particles that hit the wall of the container

Collisional flux

Effusion-36-

Effusion of Gas Particles

Graham’s Law: estimate the ratio of the effusion rates for 2 different gases.

Effusion rate of gas 1 Þr1. Effusion rate of gas 2 Þ r2.

211

Ac,11 RTM2

APNAZr

21

2

Ac,22 RTM2

APNAZr

1

A

2

A

1

2

RTM2APN

RTM2APN

rr

2

1

1

2MM

rr

Effusion Ratio-38-

a) How many molecules strike a 1-cm2 surface during 1 min if the

surface is exposed to O2 at 1 atm and 298 K?

b) Ultrahigh vacuum studies typically employ pressures on the order

of 10–10 Torr. How many collisions will occur at this pressure at

298 K?

Example 6-39-