Transport Properties of a Perfect Gas

PERGERAKAN MOLEKUL

Transport properties of a perfect gas/Sifat-sifat perpindahan dari gas ideal

Berlian Sitorus, Ph.D

Universitas Tanjungpura

Istilah-istilah dalam pembahasan molekul yang bergerak

q Sifat transport (transport property), kemampuan suatu zat untuk mentransfer materi, energi, atau sifat-sifat lain; dari satu tempat ke tempat lain

q Difusi (diffusion), migrasi dari materi karena adanya gradien konsentrasi.

q Konduksi termal (thermal conduction), migrasi energi karena adanya gradien suhu.

q Konduksi elektrik (electric conduction), migrasi dari muatan elektrik sepanjang sebuah gradien potensial listrik.

q Viskositas (viscosity), migrasi dari momentum linier ke bawah gradien kecepatan.

q Efusi (effusion), munculnya gas dari wadah melalui lubang kecil

Molecules in

Motion/Pergerakan Molekul

Transport properties of a perfect gas

Motion in liquids

Diffusion

Transport properties of a perfect gas

The transport of matter and physical properties can be described by a set of closely related empirical equations. For a gas, it is possible to understand the form of these equations by building a model based on the kinetic theory of gases discussed in previous chapter.

With this approach, the rate of diffusion the rate of thermal conduction, viscosity, and effusion can all be related to quantities arising from the kinetic theory.

Transport property

A transport property is a process by which matter or an attribute of matter, such as momentum, is carried through a medium from one location to another.

The rate of transport is commonly expressed in terms of an equation that is an empirical summary of

experimental observations.

These equations apply to all kinds of properties and

media and can be adapted to the discussion of transport

properties of gases.

Transport property-cont’d

qIn such cases, the kinetic theory of gases provides simple expressions that show how the rates of transport of these properties depend on the pressure and the temperature.

qThe most important concept from the kinetic theory is the mean free path, 𝝀, average distance a molecule travels between collisions.

qAccording to equation at a temperature T and a Pressure, P

𝜆 = ^!"_#$ (1) Mean free path [kinetic theory]

qThe parameter, 𝜎, is the collision cross-section of a molecule, a measure of the target area it presents in a collision.

qAnother important result from kinetic theory is the mean speed of molecules of molar mass M at a temperature, T, which is given by eqn

𝑣_%&'( = ^)*"_+, ^-// (2) Mean speed[kinetic theory]

•

The ‘phenomenological’

equation

If matter is flowing (as in diffusion), the matter flux is reported as so many molecules per square metre per second (number or amount m^-2.s^-1).

The rate of migration of a property is measured by its flux, J, the quantity of that property passing through a given area in a given time interval divided by the area and the duration of the interval.

A ‘phenomenological equation’ is an equation that summarizes empirical observations on phenomena without, initially at least, being based on an understanding of the molecular processes responsible for the property. Such equations are encountered commonly in the study of fluids

The ‘phenomenological’

equation

Experimental observations on transport properties show that the flux of a property is usually proportional to the first derivative of a related quantity.

The flux, J, may be positive or negative: the significance of its sign is discussed.

The total quantity of a property transferred through a given area A in a given time interval Δt is |J|A.Δt.

If the property migrating is energy (as in thermal conduction), then the energy flux is expressed in joules per square metre per second (J m^-2.s^-1), and so on.

The ‘phenomenological’ equation

qFor example, the flux of matter diffusing parallel to the z-axis of a container is found to be proportional to the gradient of the concentration along the same direction:

𝐽 𝑚𝑎𝑡𝑡𝑒𝑟 ∝ ^0𝒩₀₂ (3) Fick’s first law of diffusion

where 𝒩 is the number density of particles, with units number per metre cubed (m³).

qThe proportionality of the flux of matter to the concentration gradient is sometimes called Fick’s first law of diffusion: the law implies that diffusion is faster when the concentration varies steeply with position than when the concentration is nearly uniform.

qThere is no net flux if the concentration is uniform (^0𝒩⁄₀₂ = 0).

qSimilarly, the rate of thermal conduction (the flux of the energy associated with thermal motion) is found to be proportional to the temperature gradient:

𝐽 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑚𝑜𝑡𝑖𝑜𝑛 ∝ ^0𝒯₀₂ (4) Flux of energy qA positive value of J signifies a flux towards positive z; a negative value of J

signifies a flux towards negative z.

qBecause matter flows down a concentration gradient, from high concentration to low concentration, J is positive if ( ⁄^0𝒩 ₀₂) is negative) -- Fig 1.

qTherefore, the coefficient of proportionality in eqn (3) must be negative, and it is written as −𝐷 :

𝐽 𝑚𝑎𝑡𝑡𝑒𝑟 = −𝐷 ^0𝒩₀₂ (5) Fick’s first law in terms of the diffusion coefficient.

qThe constant 𝐷 is the called the diffusion coefficient; its SI units are metre squared per second (m² s^-1).

Fig. 1 The flux of particles down a concentration gradient. Fick’s first law states that the flux of matter is proportional to the concentration gradient at that point.

Transport properties

Transport properties

Table 1. Transport properties of gases at 1 atm * q Energy migrates down a temperature gradient, and the

same reasoning leads to

𝐽 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑚𝑜𝑡𝑖𝑜𝑛 = −𝜅^0𝒯₀₂ (6) Flux of energy in terms of the coefficient of thermal

conductivity

where 𝜿 (kappa) is the coefficient of thermal conductivity.

q The units of 𝜅 are joules per kelvin per metre per second (J K^-1 m^-1 s^-1) or, because 1 J s^-1 =1 W, watts per kelvin per metre (W K^-1 m^-1).

q Some experimental values are given in Table 1.

q Viscosity arises from the flux of linear momentum.

q To see the connection, consider a fluid in a state of Newtonian flow, in which a series of layers move past one another and, in this case, in the x-direction (Fig.2).

q The layer next to the wall of the vessel is stationary, and the velocity of successive layers varies linearly with distance, z, from the wall.

q Molecules ceaselessly move between the layers and bring with them the x-component of linear momentum they possessed in their original layer. A layer is retarded by molecules arriving from a more slowly moving layer because such molecules have a lower momentum in the x-direction.

Fig.2 The viscosity of a fluid arises from the transport of linear momentum. In this illustration the fluid is undergoing Newtonian (laminar) flow in the x-direction, and particles bring their initial momentum when they enter a new layer.

Transport properties

Transport properties

q A layer is accelerated by molecules arriving from a more rapidly moving layer. The net retarding effect is interpreted as the viscosity of the fluid

q Because the retarding effect depends on the transfer of the x-component of linear momentum into the layer of

interest, the viscosity depends on the flux of this x- component in the z-direction.

q The flux of the x-component of momentum is proportional to d𝑣₄/d𝑧, where 𝑣₄ is the velocity in the x-direction; the flux can therefore be written

𝐽 𝑥 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = −𝜂⁰⁵₀₂^! (7)

momentum flux in terms of the coefficient of viscosity

Coefficient of viscosity

q

The constant of proportionality,

𝜂

, is the coefficient of viscosity (or simply ‘the

viscosity’). Its units are kilograms per metre per second (kg m

^-1

s

^-1

).

q

Viscosities are often reported in the non-SI unit poise (P), with 1 P = 10

^-1

kg m

^-1

s

^-1

.

q

Some experimental values are given in Table 1.

q

Although it is not strictly a transport property, closely related to diffusion is effusion, the

escape of matter through a small hole.

Table 1. Transport properties of gases at 1 atm *

The essential empirical observations on effusion are summarized by Graham’s law of effusion, which states that the rate of effusion is inversely proportional to the square root of the molar mass, M.

Brief illustration 1

The Transport Parameters

q

The kinetic theory of gases can be used to derive expressions for the diffusion characteristics of a perfect gas.

q

All the expressions depend on knowing the

collision flux, 𝛧_!

,which is the rate at which molecules strike a region in the gas (this region may be an imaginary window, apart of a wall, or a hole in a wall).

q

Specifically, the collision flux is the number of collisions divided by the area of the region and the duration of the time interval.

q

Its dependence on pressure and temperature can be

derived from the kinetic theory.

The Transport Parameters

DIFFUSION

COEFFICIENT THERMAL

CONDUCTIVITY VISCOSITY EFFUSION

How is that done?

Deriving an expression for the collision flux qConsider a wall of area A perpendicular

to the x-axis (Fig. 3). In the following calculation, note that for a perfect gas the equation of state 𝑝𝑉 = 𝑛𝑅𝑇 can be used to relate the number density, 𝒩, to the pressure by :

𝒩 = ( 𝑁

𝑉 = n𝑁

_!

𝑉 = 𝑛𝑁

_!

𝑃

𝑛𝑅𝑇 = 𝑝/𝑘𝑇 qIn the final equality 𝑅 = 𝑁

_!

𝑘 was used,

where 𝑘 is Boltzmann’s constant.

Fig.3 A molecule will reach the wall on the right within an interval Δ𝑡, if it is within a distance 𝑣_!Δ𝑡 of the wall and travelling to the right.

Step 1 : Identify the number of molecules that will strike an area

q

If a molecule has

𝑣_" > 0

(that is, it is travelling in the direction of positive

𝓍

), then it will strike the wall

within an interval

∆𝑡

if it lies within a distance

𝑣_"∆𝑡

of the wall.

q

Therefore, all molecules in the volume

𝐴𝑣_"∆𝑡,

and with a positive

𝓍

-component of velocity, will strike the wall in the interval

∆𝑡

.

q

The total number of collisions in this interval is

therefore

𝒩𝐴𝑣_"∆𝑡

, where

𝒩

is the number density of

molecules.

Step 2 Take into account the range of velocities

qThe velocity, 𝑣₄, has a range of values described by the probability distribution 𝑓(𝑣₄) given in eqn 8 :

𝑓 𝑣₄ = _/+!"^{% -//} 𝑒^6%5!"//!" (8)

with 𝑓 𝑣₄ d𝑣₄ 𝑖𝑠 the probability of finding a molecule with a component of velocity between 𝑣₄ and 𝑣₄ + d𝑣₄.

qThe total number of collision is found by summing 𝒩𝐴𝑣₄∆𝑡 over all positive values of 𝑣₄ (because only molecules with a positive component of velocity are moving towards the area of interest) with each value of 𝑣₄ being weighted by the probability of it

occurring :

Number of collisions = 𝒩𝐴𝑣₄∆𝑡 ∫₇⁸𝑣₄𝑓 𝑣₄ d𝑣₄

qThe collision flux is the number of collisions divided by 𝐴 and ∆𝑡, so 𝑍₉ = 𝒩 ∫₇⁸𝑣₄𝑓 𝑣₄ d𝑣₄

Step 3 :

Evaluate the Integral

Because

∫_#^$𝑣_"𝑓 𝑣_" d𝑣_" = ^%

&'()

*/&

∫_#^$𝑣_"𝑒^,%-"#/&() d𝑣_" = ⁽⁾

&'%

*/&

It follows that

𝑍_! = 𝒩 𝑘𝑇 2𝜋𝑚

*/&

= 𝑝 𝑘𝑇

𝑘𝑇 2𝜋𝑚

*/&

and therefore

𝑍

_L

=

^M

NOPQR ^!/# (9)

collision flux in terms of pressure [perfect gas]

Integral G.2

𝒩 = 𝑝/𝑘𝑇

Step 4 Develop an alternative

expression in terms of the mean speed

qThe mean speed is given by eqn 2 as

𝑣_$%&' = ^()*_+, ^-/. = ^(/*_+$ ^-/.

where 𝑅 = 𝑁₀𝑘 and 𝑀 = 𝑁₀𝑚

qIt follows that _.+$^{/* -/.} = ^-

1𝑣_$%&' and therefore 𝑍₂ = 𝒩 ^/*

.+$

-/.can be expressed as

𝑍₂ = ^-₁𝒩𝑣_$%&' (10)

Collision flux [perfect gas]

Step 4 (continued)

Reminder:

𝑍_! = ^.

&'%() ^$/# (9)

q

According to eqn 9, the collision flux increases with pressure simply because increasing the pressure increases the number density and hence the number of collisions.

q

The flux decreases with increasing mass of the molecules because heavy molecules move more slowly than light molecules.

q

Caution, however, is needed with the interpretation of the role of temperature: it is wrong to conclude that because

𝑇^*/&

appears in the denominator that the collision flux decreases with increasing temperature.

q

If the system has constant volume, the pressure increases with temperature (

𝑝 ∝ 𝑇

), so the collision flux is in fact proportional to

𝑇/𝑇^*/& = 𝑇^*/&

and increases with temperature

(because the molecules are moving faster).

Brief illustration 2

The Transport Parameters

DIFFUSION

COEFFICIENT THERMAL

CONDUCTIVITY VISCOSITY EFFUSION

A. Diffusion coefficient

q The first application of the result in eqn (10) is to use it to find an expression for the net flux of molecules arising from a

concentration gradient.

𝑍₉ = ^-_:𝒩𝑣_%&'( (10)

qDeriving an equation for the netflux of matter Consider the arrangement depicted in Fig. 4.

The molecules passing through the area A at z= 0 have travelled an average of about one mean free path, 𝜆, since their last collision.

Fig. 4 The calculation of the rate of diffusion of a gas considers the net flux of molecules through a plane of area, A, as a result of molecules arriving from an average distance away in each direction.

Step 1 : Set up expressions for the flux

in each direction

Step 2 Evaluate the net flux

(11)

q This equation shows that the flux is proportional to the gradient of the concentration, in agreement with the empirical observation expressed by Fick’s law, eqn (3)

q At this stage it looks as though a value of the diffusion coefficient can be picked out by comparing eqns (11) and (5), so obtaining 𝐷 =

-⁄

/𝜆𝑣_%&'(

q It must be remembered, however, that the calculation is quite crude, and is little more than an assessment of the order of magnitude of D.

Step 2 Evaluate the net flux (cont’d)

q

One aspect that has not been taken into account is illustrated in Fig. 5, which shows that although a molecule may have begun its journey very close to the window, it could have a long flight before it gets there.

q

Because the path is long, the molecule is likely to collide before reaching the window, so it ought not to be counted as passing through the window.

q

Taking this effect into account results in the appearance of a factor of 2/3 representing the lower flux. The modification results in

𝐷 = ^*_/𝜆𝑣_%012 (12)

Diffusion coefficient

Fig. 5 One issue ignored in the simple treatment is that some molecules might make a long flight to the plane even though they are only a short perpendicular distance away from it. A molecule taking the longer flight has a higher chance of colliding during its journey

Physical

interpretation

There are three points to note about eqn (12)

q

The mean free path, 𝜆, decreases as the pressure is increased (eqn 1), so D decreases with

increasing pressure and, as a result, the gas molecules diffuse more slowly.

q

The mean speed,

𝑣_%012

increases with the temperature (eqn 2), so D also increases with temperature. As a result, molecules in a hot gas diffuse more quickly than those when the gas is cool (for a given concentration gradient).

q

Because the mean free path increases when the collision cross-section, 𝜎, of the molecules

decreases, the diffusion coefficient is greater for

small mol ecules than for large molecules.

B. Thermal conductivity

q

According to the equipartition theorem, each molecule carries an average energy

ℰ = 𝑣𝑘𝑇

where

𝑣

depends on the number of quadratic contributions to the energy of the molecule and is a number of the order of 1.

q

For atoms, which have three translational degrees of freedom,

𝑣 = ^/_&

.

q

When one molecule passes through the imaginary window, it transports that average energy.

q

An argument similar to that used for diffusion can be used to discuss the transport of energy through this window.

Deriving an expression for the thermal conductivity

q

Assume that the number density is uniform but that the temperature, and hence the average energy of the molecules, is not.

q

Molecules arrive from the left after travelling a mean free path from their last collision in a hotter region, and therefore arrive with a higher energy.

q

Molecules also arrive from the right after

travelling a mean free path from a cooler

region, and hence arrive with lower energy.

Step 1 Write expressions for

the forward,

reverse, and net

energy flux

Step 2 Express the energy gradient as a temperature gradient

q

This equation shows that the energy flux is

proportional to the temperature gradient, which is the desired result.

q

As before, the constant is multiplied by

^&_/

to take long flight paths into account, and comparison of this equation with eqn (6) shows that

−𝜅 = ^*_/𝑣𝑘𝑣_%012𝜆𝒩 (13) Thermal conductivity

With these substitutions, eqn 13 becomes

Yet another form is found by starting with eqn 13 recognizing that 𝒩 = 𝑝/𝑘𝑇 and then using the expression for D in eqn 12.

Physical interpretation

To interpret eqn 12 and 13 note that:

q The mean free path is 𝜆, s inversely proportional to the pressure, but the number density, 𝒩, is proportional to the pressure 𝒩 = _!"^; . It follows that the product 𝜆𝒩, which appears eqn 12, is independent of pressure, and so therefore is the thermal conductivity.

q The thermal conductivity is greater for gases with a high heat capacity (eqn 13) because a given temperature gradient then corresponds to a steeper energy gradient.

q The physical reason for the pressure independence of the thermal conductivity is that the conductivity can be expected to be large when many molecules are available to transport the energy, but the presence of so many molecules limits their mean free path and they cannot carry the energy over a great distance.

q These two effects cancel. The thermal conductivity is indeed found experimentally to be independent of the pressure, except when the pressure is very low, and then 𝜅 ∝ 𝑝.

q At very low pressures it is possible for 𝜆 to exceed the dimensions of the apparatus, and the distance over which the energy is transported, is then determined by the size of the container and not by collisions with the other molecules present.

q The flux is still proportional to the number of carriers, but the length of the journey no longer depends on 𝜆, so 𝜅 ∝[ 𝐽 which implies that 𝜅 ∝ 𝑝

C. Viscosity

If the momentum in the x-direction depends on z as

𝑚𝑣_"(𝑧)

then molecules moving from the right in Fig. 6 (from a fast layer to a slower one) transport a momentum

𝑚𝑣_"(𝜆)

to their new layer at z=0; those travelling from the left transport

𝑚𝑣_"(−𝜆)

to it. This picture can be used to

build an expression for the coefficient of viscosity.

Fig. 6 The calculation of the viscosity of a gas examines the net x-component of momentum brought to a plane from faster and slower layers a mean free path away in each direction.

Deriving an expression for the viscosity

q

The strategy is the same as in the previous derivations, but now the property being

transported is the linear

momentum of the layers.

Step 2 Identify the coefficient of viscosity

q

The flux is proportional to the velocity gradient, in line with the phenomenological equation. Comparison of this expression with eqn

7, and multiplication by ^&_/

in the normal way, leads to

(14)

q Two alternative forms of this expression (after using

𝑚𝑁_< = 𝑀 and eqn 12 , 𝐷 = ^-₌𝜆𝑣_%&'( are

𝜂 = 𝑀𝐷 𝐽 (15)

𝜂 = ^;,>_*" (16)

where [J] is the molar concentration of the gas molecules J and M is their molar mass.

Reminder :

𝐽 𝑥 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = −𝜂 ⁰⁵₀₂^! (7)

The physical interpretation of eqn for viscocity is as follows:

q

The increase of viscosity with temperature is explained when it is recalled that at high temperatures the molecules travel more quickly, so the flux of momentum is greater.

q

In contrast, `as discussed in Topic 16B, the viscosity of a liquid decreases with increase in temperature because

intermolecular interac tions must be

overcome.

D. Effusion

q

Because the mean speed of molecules is inversely proportional to

𝑀^$⁄#

, the rate at which they strike the area of the hole through which they are effusing is also inversely

proportional to

𝑀^$⁄#

, in accord with Graham’s law.

q

However, by using the expression for the collision flux, a more detailed expression for the rate of effusion can be obtained and used to interpret effusion data in a more

sophisticated way.

(17)

qWhen a gas at a pressure p and temperature T is separated from a vacuum by a small hole, the rate of escape of its molecules is equal to the rate at which they strike the area of the hole, which is the product of the collision flux and the area of the hole, 𝐴₇

q Equation (17) is the basis of the Knudsen method for the determination of the vapour pressures of liquids and solids,

• particularly of substances with very low vapour pressures and

• which cannot be measured directly..

q Thus, if the vapour pressure of a sample is p, and it is enclosed in a cavity with a small hole, then the rate of loss of mass from the

container is proportional to p.

Calculating the vapour pressure from a mass loss

(17)

THE END

Terimakasih