Chapter 21: Molecules in motion

• Diffusion: the migration of matter down a concentration gradient.

• Thermal conduction: the migration of energy down a temperature gradient.

• Electric conduction: the migration of electric charge along an electrical potential gradient.

• Viscosity: the migration of linear momentum down a velocity gradient.

• Effusion: the emergence of a gas from a container through a small hole.

21.1 The kinetic model of gases

• Three 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 traveled between collisions.

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

(Elastic collision: a collision in which the total translational kinetic energy of the molecules is conserved.)

Pressure and Molecular speeds

PV = 1/3 nMc2 (21.1)

where M = mNA, the molar mass of the molecules,

c is the root mean square speed of the molecules:

c2 = vx2 + vy

2 + vz2 c = < vx

2 + vy2 + vz

2 >1/2 (21.2)

Relationship between temperature and the root mean square speed

• Provided that the root mean square speed of the molecules depends only on the temperature:

pV = constant at constant temperature

• In comparison with Boyle’s law, one gets

c = (21.3)

• The root mean square speed of the gas molecules is proportional to the square root of temperature and inversely proportional to the square root of the molar mass.

2/13

M

RT

Maxwell distribution of speeds (21.4) Fraction in the range v1 to v2 equals

2

1

v

vdvvf )(

RTMvevRT

Mvf 2/2

2/32

24)(

Expression of molecular speeds

• Mean speed

• The most probable speed

• Relative mean speed:

(reduced mass)

2/18

M

RTc

212

/*

M

RTc

2121 8

2/_

/_

kT

ccrel

BA

BA

mm

mm

Measuring molecular speed

The collision frequency

• Collision diameter: the actual diameter of the molecule.

• Collision frequency (z): the number of collisions made by one molecule divided by the time interval during which the collisions are counted.

• Collision cross-section: σ = πd2

• N is number density= kT

Pc

t

NtcZ relrel

)(

_

Mean free path

• Mean free path, λ, the average distance a molecule travels between collisions.

• How does temperature or pressure affect the mean free path ?

• The distance between collisions is determined by the number of molecules present in the given volume, not by the speed at which they travel.

P

kT

z

c

212 /

_

21.2 Collisions with walls and surfaces

• The collision flux, Zw, the number of collisions with the area in a given time interval divided by the area and the duration of the interval.

• Collision frequency can be obtained by multiplication of the collision flux by the area of interest.

NcmkT

pZw

4

1

)2( 2/1

21.3 The rate of effusion

• Graham’s law of effusion: the rate of effusion is inversely proportional to the square root of the molar mass.

• Vapor pressures of liquids and solids can be measured based on the above equation. ( Knudsen method)

210

210

022 // )()( MRT

NpA

mkT

pAAZeffusionofrate A

w

Example 21.2 Caesium (m.p. 29oC, b.p. 686 oC) was introduced into a container and heated to 500 oC. When a hole of diameter 0.500mm was opened in the container for 100s, a mass loss of 385 mg was measured. Calculate the vapor pressure of liquid cesium at 500K.

Solution: Despite the effusion, the vapor pressure is constant inside the container because the hot liquid metal replenishes the vapor. Consequently, the effusion rate is constant!

The mass loss Δm in an interval Δt is related to the collision flux by:Δm = ZwA0mΔt

Where A0 is the area of the hole and m is the mass of one

atom.

plug in numbers, one gets p = 11k Pa

tmA

mZW

0

tmA

m

mkT

p

0

212 /)(

• Self-test 21.2: How long would it take 1.0 g of Cs atoms to effuse out of the over under the same conditions as listed in example 21.2? (260 s)

• Self-test: There is 1.0 g of Cs solid in the effusion oven, how long does it take to effuse out of the oven?

• Step 1: Derive an expression that shows how the pressure of a gas inside an effusion oven varies with time if the oven is not replenished as the gas escapes.

• step 2: calculate the derivative of P:

step 3: the rate of change of the number of molecules is equal to the collision frequency with the hole multiplied by the area of the hole:

so

step 4: integrate the above equation P = P0e-t/τ

(Remark: how does the temperature (or the size of the hole) affect the decrease of pressure?)

dt

dN

V

kT

dt

dP

210

20 /)( mKT

pAwAZ

dt

dN

V

PA

m

kT

dt

dP 0

2/1

2

0

212

A

V

kT

m/

21.4 Transport properties of a perfect gas

• Experimental observations on transport properties shows that the flux of a property is proportional to the first derivative of other related properties.

• The flux of matter is proportional to the first derivative of the concentration (Fick’s first law of diffusion): J(matter)

• The rate of thermal conduction is proportional to the temperature gradient: J(energy)

• J(matter) = D is called the diffusion coefficient (m2s-1);

dz

dN

dz

dT

dz

dND

• J(energy) = - k dT/dz k is called the coefficient of thermal conductivity (J K-1 m-1 s-1)

• J(x-component of momentum) = , η is the coefficient of viscosity.

•

dz

dvx

Table 21.3

Diffusion

_cD

3

1

As represented by the above Figure, on average the molecules passing through the area A at z = 0 have traveled about one free path.

The average number of molecules travels through the imaginary window A from Left to Right during an interval Δt is

ZwA Δt (L→R) Because Zw = So A Δt (L→R) The average number of molecules travels through the imaginary window A from Right to

Left during an interval Δt is A Δt (R →L)

The net number of molecules passing through the window A along the z direction is:

A Δt - A Δt

By definition the flux of molecules along z direction can be calculate as J(z) = ( A Δt - A Δt )/(A Δt )

J(z) = The number density N(-λ) and N(λ) can be represented by number density N(0) at z =0

N(-λ) = N(0) - λ N(λ) = N(0) + λ

Therefore: J(z) =

then we get D = (different from what we expected)

cN )( 4

1

cN )( 4

1

cN )(4

1

cN )( 4

1

cN )(4

1

cN )( 4

1 cN )(4

1

cNcN )()( 4

1

4

1

0)(dz

dN0)(

dz

dN

02

1)(

dz

dNc

c2

1

A factor of 2/3 needs to be introduced.

So we get

D =

c3

1

Thermal conduction

k =

where CV,m is the molar heat capacity at constant volume.

Because λ is inversely proportional to the molar concentration of the gas, the thermal conductivity is independent of the concentration of gas, and hence independent of the gas pressure.

One exception: at very low pressure, where the mean free path is larger than the size of the container.

][,_

ACc mV3

1

dz

dTkenergyJ )(

• J(x-component of momentum) = , η is the coefficient of viscosity.

dz

dvx

Viscosity

• The viscosity is independent of

the pressure.• Proportional to T1/2

][_AcM

3

1

Measuring the viscosity

• Poiseuille’s formula: 0

422

21

16 pl

rpp

dt

dV

)(

Calculations with Poiseuille’s formula

• Example: In a poiseuille flow experiment to measure the viscosity of air at 298K, the sample was allowed to flow through a tube of length 100cm and internal diameter 1.00mm. The high-pressure end was at 765 Torr and the low-pressure end was at 760Torr. The volume was measured at the latter pressure. In 100s, 90.2cm3 of air passed through the tube.

• Solution: Reorganize Poseuille’s equation:

)/(

)(

dtdVlp

rpp

0

422

21

16

)(.

).

().().(

).(}).().{(

21114

351

4422

1110821

10010029

31337601000116

1000531337603133765

skgmPaskgm

sm

Pam

mPaPa