CHAPTER THIRTEEN

KINETIC THEORY

13.1 INTRODUCTION

Boyle discovered the law named after him in 1661. Boyle,

Newton and several others tried to explain the behaviour of

gases by considering that gases are made up of tiny atomic

particles. The actual atomic theory got established more than

150 years later. Kinetic theory explains the behaviour of gases

based on the idea that the gas consists of rapidly moving

atoms or molecules. This is possible as the inter-atomic forces,

which are short range forces that are important for solids

and liquids, can be neglected for gases. The kinetic theory

was developed in the nineteenth century by Maxwell,

Boltzmann and others. It has been remarkably successful. It

gives a molecular interpretation of pressure and temperature

of a gas, and is consistent with gas laws and Avogadro’s

hypothesis. It correctly explains specific heat capacities of

many gases. It also relates measurable properties of gases

such as viscosity, conduction and diffusion with molecular

parameters, yielding estimates of molecular sizes and masses.

This chapter gives an introduction to kinetic theory.

13.2 MOLECULAR NATURE OF MATTER

Richard Feynman, one of the great physicists of 20th century

considers the discovery that “Matter is made up of atoms” to

be a very significant one. Humanity may suffer annihilation

(due to nuclear catastrophe) or extinction (due to

environmental disasters) if we do not act wisely. If that

happens, and all of scientific knowledge were to be destroyed

then Feynman would like the ‘Atomic Hypothesis’ to be

communicated to the next generation of creatures in the

universe. Atomic Hypothesis: All things are made of atoms -

little particles that move around in perpetual motion,

attracting each other when they are a little distance apart,

but repelling upon being squeezed into one another.

Speculation that matter may not be continuous, existed in

many places and cultures. Kanada in India and Democritus

13.1 Introduction

13.2 Molecular nature of matter

13.3 Behaviour of gases

13.4 Kinetic theory of an ideal gas

13.5 Law of equipartition of energy

13.6 Specific heat capacity

13.7 Mean free path

Summary

Points to ponder

Exercises

Additional exercises

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324 PHYSICS

in Greece had suggested that matter may consist

of indivisible constituents. The scientific ‘Atomic

Theory’ is usually credited to John Dalton. He

proposed the atomic theory to explain the laws

of definite and multiple proportions obeyed by

elements when they combine into compounds.

The first law says that any given compound has,

a fixed proportion by mass of its constituents.

The second law says that when two elements

form more than one compound, for a fixed mass

of one element, the masses of the other elements

are in ratio of small integers.

To explain the laws Dalton suggested, about

200 years ago, that the smallest constituents

of an element are atoms. Atoms of one element

are identical but differ from those of other

elements. A small number of atoms of each

element combine to form a molecule of the

compound. Gay Lussac’s law, also given in early

century, states: When gases combine

chemically to yield another gas, their volumes

are in the ratios of small integers. Avogadro’s

law (or hypothesis) says: Equal volumes of all

gases at equal temperature and pressure have

the same number of molecules. Avogadro’s law,

when combined with Dalton’s theory explains

Gay Lussac’s law. Since the elements are often

in the form of molecules, Dalton’s atomic theory

can also be referred to as the molecular theory

of matter. The theory is now well accepted by

scientists. However even at the end of the

nineteenth century there were famous scientists

who did not believe in atomic theory !

From many observations, in recent times we

now know that molecules (made up of one or

more atoms) constitute matter. Electron

microscopes and scanning tunnelling

microscopes enable us to even see them. The

size of an atom is about an angstrom (10

-10

m).

In solids, which ar

e tightly packed, atoms are

spaced about a few angstroms (2 Å) apart. In

liquids the separation between atoms is also

about the same. In liquids the atoms are not

as rigidly fixed as in solids, and can move

around. This enables a liquid to flow. In gases

the interatomic distances are in tens of

angstroms. The average distance a molecule

can travel without colliding is called the mean

free path. The mean free path, in gases, is of

the order of thousands of angstroms. The atoms

are much freer in gases and can travel long

distances without colliding. If they are not

enclosed, gases disperse away. In solids and

liquids the closeness makes the interatomic force

important. The force has a long range attraction

and a short range repulsion. The atoms attract

when they are at a few angstroms but repel when

they come closer. The static appearance of a gas

Atomic Hypothesis in Ancient India and Greece

Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in

ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika

school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in

considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter.

It was argued that if matter could be subdivided without an end, there would be no difference between a

mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the

smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic

mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and

was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a

diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending

upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture

or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the

Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic

size, of the order of 10

–10

In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The

word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in

shape, size and other properties and this resulted in the different properties of the substances formed

by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each

other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held

together to form hard substances. The atoms of fire were thorny which is why it caused painful burns.

These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they

were intuitive conjectures and speculations not tested and modified by quantitative experiments - the

hallmark of modern science.

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KINETIC THEORY 325

is misleading. The gas is full of activity and the

equilibrium is a dynamic one. In dynamic

equilibrium, molecules collide and change their

speeds during the collision. Only the average

properties are constant.

Atomic theory is not the end of our quest, but

the beginning. We now know that atoms are not

indivisible or elementary. They consist of a

nucleus and electrons. The nucleus itself is made

up of protons and neutrons. The protons and

neutrons are again made up of quarks. Even

quarks may not be the end of the story. There

may be string like elementary entities. Nature

always has surprises for us, but the search for

truth is often enjoyable and the discoveries

beautiful. In this chapter, we shall limit ourselves

to understanding the behaviour of gases (and a

little bit of solids), as a collection of moving

molecules in incessant motion.

13.3 BEHAVIOUR OF GASES

Properties of gases are easier to understand than

those of solids and liquids. This is mainly

because in a gas, molecules are far from each

other and their mutual interactions are

negligible except when two molecules collide.

Gases at low pressures and high temperatures

much above that at which they liquefy (or

solidify) approximately satisfy a simple relation

between their pressure, temperature and volume

given by (see Chapter 11)

PV = KT (13.1)

for a given sample of the gas. Here T is the

temperature in kelvin or (absolute) scale. K is a

constant for the given sample but varies with

the volume of the gas. If we now bring in the

idea of atoms or molecules, then K is proportional

to the number of molecules, (say) N in the

sample. We can write K = N k . Observation tells

us that this k is same for all gases. It is called

Boltzmann constant and is denoted by k

1 1 2 2

P V P V

N T N T

= constant = k

(13.2)

if P, V and T are same, then N is also same for

all gases. This is Avogadro’s hypothesis, that the

number of molecules per unit volume is

the same for all gases at a fixed temperature and

pressure. The number in 22.4 litres of any gas

is 6.02 × 10

. This is known as Avogadro

number and is denoted by N

. The mass of 22.4

litres of any gas is equal to its molecular weight

in grams at S.T.P (standard temperature 273 K

and pressure 1 atm). This amount of substance

is called a mole (see Chapter 2 for a more precise

definition). Avogadro had guessed the equality of

numbers in equal volumes of gas at a fixed

temperature and pressure from chemical

reactions. Kinetic theory justifies this hypothesis.

The perfect gas equation can be written as

PV =

RT (13.3)

where

is the number of moles and R = N

is a universal constant. The temperature T is

absolute temperature. Choosing kelvin scale for

John Dalton (1766 – 1844)

He was an English chemist. When different types of atoms combine,

they obey certain simple laws. Dalton’s atomic theory explains these

laws in a simple way. He also gave a theory of colour

blindness.

Amedeo Avogadro (1776 – 1856)

He made a brilliant guess that equal volumes of gases

have equal number of molecules at the same

temperature and pressure. This helped in

understanding the combination of different gases in

a very simple way. It is now called Avogadro’s hypothesis (or law). He also

suggested that the smallest constituent of gases like hydrogen, oxygen and

nitrogen are not atoms but diatomic molecules.

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326 PHYSICS

absolute temperature, R = 8.314 J mol

–1

Here

M N

= =

(13.4)

where M is the mass of the gas containing N

molecules, M

is the molar mass and N

the

Avogadro’s number. Using Eqs. (13.4) and (13.3)

can also be written as

PV = k

NT or P = k

P (atm)

Fig.13.1 Real gases approach ideal gas behaviour

at low pressures and high temperatures.

where n is the number density, i.e. number of

molecules per unit volume. k

is the Boltzmann

constant introduced above. Its value in SI units

is 1.38 × 10

–23

J K

–1

Another useful form of Eq. (13.3) is

(13.5)

where

is the mass density of the gas.

A gas that satisfies Eq. (13.3) exactly at all

pressures and temperatures is defined to be an

ideal gas. An ideal gas is a simple theoretical

model of a gas. No real gas is truly ideal.

Fig. 13.1 shows departures from ideal gas

behaviour for a real gas at three different

temperatures. Notice that all curves approach

the ideal gas behaviour for low pressures and

high temperatures.

At low pressures or high temperatures the

molecules are far apart and molecular

interactions are negligible. Without interactions

the gas behaves like an ideal one.

If we fix

and T in Eq. (13.3), we get

PV = constant (13.6)

i.e., keeping temperature constant, pressure of

a given mass of gas varies inversely with volume.

This is the famous Boyle’s law. Fig. 13.2 shows

comparison between experimental P-V curves

and the theoretical curves predicted by Boyle’s

law. Once again you see that the agreement is

good at high temperatures and low pressures.

Next, if you fix P, Eq. (13.1) shows that V ∝ T

i.e., for a fixed pressure, the volume of a gas is

proportional to its absolute temperature T

(Charles’ law). See Fig. 13.3.

Fig.13.2 Experimental P-V curves (solid lines) for

steam at three temperatures compared

with Boyle’s law (dotted lines). P is in units

of 22 atm and V in units of 0.09 litres.

Finally, consider a mixture of non-interacting

ideal gases:

moles of gas 1,

moles of gas

2, etc. in a vessel of volume V at temperature T

and pressure P. It is then found that the

equation of state of the mixture is :

PV = (

+… ) RT (13.7)

i.e.

1 2

...

RT RT

V V

µ µ

= + +

(13.8)

= P

+ P

+ … (13.9)

Clearly P

R T/V is the pressure that

gas 1 would exert at the same conditions of

volume and temperature if no other gases were

present. This is called the partial pressure of the

gas. Thus, the total pressure of a mixture of ideal

gases is the sum of partial pressures. This is

Dalton’s law of partial pressures.

(

)

–1 –1

J mol K

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KINETIC THEORY 327

Fig. 13.3 Experimental T-V curves (solid lines) for

at three pressures compared with

Charles’ law (dotted lines). T is in units of

300 K and V in units of 0.13 litres.

We next consider some examples which give

us information about the volume occupied by

the molecules and the volume of a single

molecule.

Example 13.1 The density of water is 1000

kg m

–3

. The density of water vapour at 100 °C

and 1 atm pressure is 0.6 kg m

–3

. The

volume of a molecule multiplied by the total

number gives ,what is called, molecular

volume. Estimate the ratio (or fraction) of

the molecular volume to the total volume

occupied by the water vapour under the

above conditions of temperature and

pressure.

Answer For a given mass of water molecules,

the density is less if volume is large. So the

volume of the vapour is 1000/0.6 = 1/(6 ×10

-4

)

times larger. If densities of bulk water and water

molecules are same, then the fraction of

molecular volume to the total volume in liquid

state is 1. As volume in vapour state has

increased, the fractional volume is less by the

same amount, i.e. 6×10

-4

. t

Example 13.2 Estimate the volume of a

water molecule using the data in Example

13.1.

Answer In the liquid (or solid) phase, the

molecules of water are quite closely packed. The

density of water molecule may therefore, be

regarded as roughly equal to the density of bulk

water = 1000 kg m

–3

. To estimate the volume of

a water molecule, we need to know the mass of

a single water molecule. We know that 1 mole

of water has a mass approximately equal to

(2 + 16)g = 18 g = 0.018 kg.

Since 1 mole contains about 6 × 10

molecules (Avogadro’s number), the mass of

a molecule of water is (0.018)/(6 × 10

) kg =

3 × 10

–26

kg. Therefore, a rough estimate of the

volume of a water molecule is as follows :

Volume of a water molecule

= (3 × 10

–26

kg)/ (1000 kg m

–3

)

= 3 × 10

–29

= (4/3)

(Radius)

Hence, Radius ≈ 2 ×10

-10

m = 2 Å t

Example 13.3 What is the average

distance between atoms (interatomic

distance) in water? Use the data given in

Examples 13.1 and 13.2.

Answer : A given mass of water in vapour state

has 1.67×10

times the volume of the same mass

of water in liquid state (Ex. 13.1). This is also

the increase in the amount of volume available

for each molecule of water. When volume

increases by 10

times the radius increases by

1/3

or 10 times, i.e., 10 × 2 Å = 20 Å. So the

average distance is 2 × 20 = 40 Å. t

Example 13.4 A vessel contains two non-

reactive gases : neon (monatomic) and

oxygen (diatomic). The ratio of their partial

pressures is 3:2. Estimate the ratio of (i)

number of molecules and (ii) mass density

of neon and oxygen in the vessel. Atomic

mass of Ne = 20.2 u, molecular mass of O

= 32.0 u.

Answer Partial pressure of a gas in a mixture is

the pressure it would have for the same volume

and temperature if it alone occupied the vessel.

(The total pressure of a mixture of non-reactive

gases is the sum of partial pressures due to its

constituent gases.) Each gas (assumed ideal)

obeys the gas law. Since V and T are common to

the two gases, we have P

V =

RT and P

V =

RT, i.e. (P

) = (

). Here 1 and 2 refer

to neon and oxygen respectively. Since (P

) =

(3/2) (given), (

) = 3/2.

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328 PHYSICS

(i) By definition

= (N

) and

= (N

)

where N

and N

are the number of molecules

of 1 and 2, and N

is the Avogadro’s number.

Therefore, (N

) = (

) = 3/2.

(ii) We can also write

= (m

) and

) where m

and m

are the masses of

1 and 2; and M

and M

are their molecular

masses. (Both m

and M

; as well as m

and

should be expressed in the same units).

and

are the mass densities of 1 and

2 respectively, we have

= = = ×













m V

3 20.2

0.947

2 32.0

= × =

13.4 KINETIC THEORY OF AN IDEAL GAS

Kinetic theory of gases is based on the molecular

picture of matter. A given amount of gas is a

collection of a large number of molecules

(typically of the order of Avogadro’s number) that

are in incessant random motion. At ordinary

pressure and temperature, the average distance

between molecules is a factor of 10 or more than

the typical size of a molecule (2 Å). Thus,

interaction between molecules is negligible and

we can assume that they move freely in straight

lines according to Newton’s first law. However,

occasionally, they come close to each other,

experience intermolecular forces and their

velocities change. These interactions are called

collisions. The molecules collide incessantly

against each other or with the walls and change

their velocities. The collisions are considered to

be elastic. We can derive an expression for the

pressure of a gas based on the kinetic theory.

We begin with the idea that molecules of a

gas are in incessant random motion, colliding

against one another and with the walls of the

container. All collisions between molecules

among themselves or between molecules and the

walls are elastic. This implies that total kinetic

energy is conserved. The total momentum is

conserved as usual.

13.4.1 Pressure of an Ideal Gas

Consider a gas enclosed in a cube of side l. Take

the axes to be parallel to the sides of the cube,

as shown in Fig. 13.4. A molecule with velocity

, v

) hits the planar wall parallel to yz-

plane of area A (= l

). Since the collision is elastic,

the molecule rebounds with the same velocity;

its y and z components of velocity do not change

in the collision but the x-component reverses

sign. That is, the velocity after collision is

(-v

, v

) . The change in momentum of the

molecule is: –mv

– (mv

) = – 2mv

. By the

principle of conservation of momentum, the

momentum imparted to the wall in the collision

= 2mv

To calculate the force (and pressure) on the

wall, we need to calculate momentum imparted

to the wall per unit time. In a small time interval

∆t, a molecule with x-component of velocity v

will hit the wall if it is within the distance v

∆t

from the wall. That is, all molecules within the

volume Av

∆t only can hit the wall in time ∆t.

But, on the average, half of these are moving

towards the wall and the other half away from

the wall. Thus, the number of molecules with

velocity (v

, v

) hitting the wall in time ∆t is

½A v

∆t n, where n is the number of molecules

per unit volume. The total momentum

transferred to the wall by these molecules in

time ∆t is:

Q = (2mv

) (½ n A v

∆t ) (13.10)

The force on the wall is the rate of momentum

transfer Q/∆t and pressure is force per unit

area :

P = Q /(A ∆t) = n m v

(3.11)

Actually, all molecules in a gas do not have

the same velocity; there is a distribution in

velocities. The above equation, therefore, stands

for pressure due to the group of molecules with

speed v

in the x-direction and n stands for the

number density of that group of molecules. The

Fig. 13.4 Elastic collision of a gas molecule with

the wall of the container.

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KINETIC THEORY 329

total pressure is obtained by summing over the

contribution due to all groups:

P = n m

(13.12)

where

is the average of v

. Now the gas

is isotropic, i.e. there is no preferred direction

of velocity of the molecules in the vessel.

Therefore, by symmetry,

= (1/3) [

] = (1/3)

(13.13)

where v is the speed and

denotes the mean

of the squared speed. Thus

P = (1/3) n m

(13.14)

Some remarks on this derivation. First,

though we choose the container to be a cube,

the shape of the vessel really is immaterial. For

a vessel of arbitrary shape, we can always choose

a small infinitesimal (planar) area and carry

through the steps above. Notice that both A and

∆t do not appear in the final result. By Pascal’s

law, given in Ch. 10, pressure in one portion of

the gas in equilibrium is the same as anywhere

else. Second, we have ignored any collisions in

the derivation. Though this assumption is

difficult to justify rigorously, we can qualitatively

see that it will not lead to erroneous results.

The number of molecules hitting the wall in time

∆t was found to be ½ n Av

∆t. Now the collisions

are random and the gas is in a steady state.

Thus, if a molecule with velocity (v

, v

)

acquires a different velocity due to collision with

some molecule, there will always be some other

molecule with a different initial velocity which

after a collision acquires the velocity (v

, v

If this were not so, the distribution of velocities

would not remain steady. In any case we are

finding

. Thus, on the whole, molecular

collisions (if they are not too frequent and the

time spent in a collision is negligible compared

to time between collisions) will not affect the

calculation above.

13.4.2 Kinetic Interpretation of Temperature

Equation (13.14) can be written as

PV = (1/3) nV m

(13.15a)

Founders of Kinetic Theory of Gases

James Clerk Maxwell (1831 – 1879), born in Edinburgh,

Scotland, was among the greatest physicists of the nineteenth

century. He derived the thermal velocity distribution of molecules

in a gas and was among the first to obtain reliable estimates of

molecular parameters from measurable quantities like viscosity,

etc. Maxwell’s greatest achievement was the unification of the laws

of electricity and magnetism (discovered by Coulomb, Oersted,

Ampere and Faraday) into a consistent set of equations now called

Maxwell’s equations. From these he arrived at the most important

conclusion that light is an

electromagnetic wave.

Interestingly, Maxwell did not

agree with the idea (strongly

suggested by the Faraday’s

laws of electrolysis) that

electricity was particulate in

nature.

Ludwig Boltzmann

(1844 – 1906) born in

Vienna, Austria, worked on the kinetic theory of gases

independently of Maxwell. A firm advocate of atomism, that is

basic to kinetic theory, Boltzmann provided a statistical

interpretation of the Second Law of thermodynamics and the

concept of entropy. He is regarded as one of the founders of classical

statistical mechanics. The proportionality constant connecting

energy and temperature in kinetic theory is known as Boltzmann’s

constant in his honour.

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330 PHYSICS

PV = (2/3) N x ½ m

(13.15b)

where N (= nV) is the number of molecules in

the sample.

The quantity in the bracket is the average

translational kinetic energy of the molecules in

the gas. Since the internal energy E of an ideal

gas is purely kinetic*,

E = N × (1/2) m

(13.16)

Equation (13.15) then gives :

PV = (2/3) E (13.17)

We are now ready for a kinetic interpretation

of temperature. Combining Eq. (13.17) with the

ideal gas Eq. (13.3), we get

E = (3/2) k

NT (13.18)

or E/ N = ½ m

= (3/2) k

T (13.19)

i.e., the average kinetic energy of a molecule is

proportional to the absolute temperature of the

gas; it is independent of pressure, volume or

the nature of the ideal gas. This is a fundamental

result relating temperature, a macroscopic

measurable parameter of a gas

(a thermodynamic variable as it is called) to a

molecular quantity, namely the average kinetic

energy of a molecule. The two domains are

connected by the Boltzmann constant. We note

in passing that Eq. (13.18) tells us that internal

energy of an ideal gas depends only on

temperature, not on pressure or volume. With

this interpretation of temperature, kinetic theory

of an ideal gas is completely consistent with the

ideal gas equation and the various gas laws

based on it.

For a mixture of non-reactive ideal gases, the

total pressure gets contribution from each gas

in the mixture. Equation (13.14) becomes

P = (1/3) [n

+ n

+… ] (13.20)

In equilibrium, the average kinetic energy of

the molecules of different gases will be equal.

That is,

½ m

= ½ m

= (3/2) k

so that

P = (n

+ n

+… ) k

T (13.21)

which is Dalton’s law of partial pressures.

From Eq. (13.19), we can get an idea of the

typical speed of molecules in a gas. At a

temperature T = 300 K, the mean square speed

of a molecule in nitrogen gas is :

–26

4.65 10

6.02 10

= = = ×

kg.

= 3 k

T / m = (516)

-2

The square root of

is known as root mean

square (rms) speed and is denoted by v

rms

( We can also write

as < v

>.)

rms

= 516 m s

-1

The speed is of the order of the speed of sound

in air. It follows from Eq. (13.19) that at the same

temperature, lighter molecules have greater rms

speed.

Example 13.5 A flask contains argon and

chlorine in the ratio of 2:1 by mass. The

temperature of the mixture is 27 °C. Obtain

the ratio of (i) average kinetic energy per

molecule, and (ii) root mean square speed

rms

of the molecules of the two gases.

Atomic mass of argon = 39.9 u; Molecular

mass of chlorine = 70.9 u.

Answer The important point to remember is that

the average kinetic energy (per molecule) of any

(ideal) gas (be it monatomic like argon, diatomic

like chlorine or polyatomic) is always equal to

(3/2) k

T. It depends only on temperature, and

is independent of the nature of the gas.

(i) Since argon and chlorine both have the same

temperature in the flask, the ratio of average

kinetic energy (per molecule) of the two gases

is 1:1.

(ii) Now ½ m v

rms

= average kinetic energy per

molecule = (3/2) ) k

T where m is the mass

of a molecule of the gas. Therefore,

(

)

( )

(

)

( )

(

)

Cl Cl

Ar Ar

rms

m M

= =

70.9

39.9

=1.77

where M denotes the molecular mass of the gas.

(For argon, a molecule is just an atom of argon.)

Taking square root of both sides,

(

)

( )

rms

= 1.33

You should note that the composition of the

mixture by mass is quite irrelevant to the above

* E denotes the translational part of the internal energy U that may include energies due to other degrees of

freedom also. See section 13.5.

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KINETIC THEORY 331

calculation. Any other proportion by mass of

argon and chlorine would give the same answers

to (i) and (ii), provided the temperature remains

unaltered. t

Example 13.6 Uranium has two isotopes

of masses 235 and 238 units. If both are

present in Uranium hexafluoride gas which

would have the larger average speed ? If

atomic mass of fluorine is 19 units,

estimate the percentage difference in

speeds at any temperature.

Answer At a fixed temperature the average

energy = ½ m <v

> is constant. So smaller the

mass of the molecule, faster will be the speed.

The ratio of speeds is inversely proportional to

the square root of the ratio of the masses. The

masses are 349 and 352 units. So

349

/ v

352

= ( 352/ 349)

1/2

= 1.0044 .

Hence difference

∆

= 0.44 %.

[

235

is the isotope needed for nuclear fission.

To separate it from the more abundant isotope

238

U, the mixture is surrounded by a porous

cylinder. The porous cylinder must be thick and

narrow, so that the molecule wanders through

individually, colliding with the walls of the long

pore. The faster molecule will leak out more than

Maxwell Distribution Function

In a given mass of gas, the velocities of all molecules are not the same, even when bulk

parameters like pressure, volume and temperature are fixed. Collisions change the direction

and the speed of molecules. However in a state of equilibrium, the distribution of speeds is

constant or fixed.

Distributions are very important and useful when dealing with systems containing large

number of objects. As an example consider the ages of different persons in a city. It is not

feasible to deal with the age of each individual. We can divide the people into groups: children

up to age 20 years, adults between ages of 20 and 60, old people above 60. If we want more

detailed information we can choose smaller intervals, 0-1, 1-2,..., 99-100 of age groups. When

the size of the interval becomes smaller, say half year, the number of persons in the interval

will also reduce, roughly half the original number in the one year interval. The number of

persons dN(x) in the age interval x and x+dx is proportional to dx or dN(x) = n

dx. We have

used n

to denote the number of persons at the value of x.

Maxwell distribution of molecular speeds

In a similar way the molecular speed distribution gives the number of molecules between

the speeds v and v+ dv. dN(v)

= 4p N a

–bv

dv = n

dv. This is called Maxwell distribution.

The plot of n

against v is shown in the figure. The fraction of the molecules with speeds v and

v+dv is equal to the area of the strip shown. The average of any quantity like v

is defined by

the integral <v

> = (1/N ) ∫ v

dN(v)

ªª

ª(3k

T/m)

which agrees with the result derived from

more elementary considerations.

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332 PHYSICS

the slower one and so there is more of the lighter

molecule (enrichment) outside the porous

cylinder (Fig. 13.5). The method is not very

efficient and has to be repeated several times

for sufficient enrichment.]. t

When gases diffuse, their rate of diffusion is

inversely proportional to square root of the

masses (see Exercise 13.12 ). Can you guess the

explanation from the above answer?

Fig. 13.5 Molecules going through a porous wall.

Example 13.7 (a) When a molecule (or

an elastic ball) hits a ( massive) wall, it

rebounds with the same speed. When a ball

hits a massive bat held firmly, the same

thing happens. However, when the bat is

moving towards the ball, the ball rebounds

with a different speed. Does the ball move

faster or slower? (Ch.6 will refresh your

memory on elastic collisions.)

(b) When gas in a cylinder is compressed

by pushing in a piston, its temperature

rises. Guess at an explanation of this in

terms of kinetic theory using (a) above.

pushes a piston out and expands. What

would you observe ?

(d) Sachin Tendulkar used a heavy cricket

bat while playing. Did it help him in

anyway ?

Answer (a) Let the speed of the ball be u relative

to the wicket behind the bat. If the bat is moving

towards the ball with a speed V relative to the

wicket, then the relative speed of the ball to bat

is V + u towards the bat. When the ball rebounds

(after hitting the massive bat) its speed, relative

to bat, is V + u moving away from the bat. So

relative to the wicket the speed of the rebounding

ball is V + (V + u) = 2V + u, moving away from

the wicket. So the ball speeds up after the

collision with the bat. The rebound speed will

be less than u if the bat is not massive. For a

molecule this would imply an increase in

temperature.

You should be able to answer (b) (c) and (d)

based on the answer to (a).

(Hint: Note the correspondence, pistonà bat,

cylinder à wicket, molecule à ball.) t

13.5 LAW OF EQUIPARTITION OF ENERGY

The kinetic energy of a single molecule is

2 2 2

1 1 1

2 2 2

t x y z

mv mv mv

= + +

(13.22)

For a gas in thermal equilibrium at

temperature T the average value of energy

denoted by <

> is

2 2 2

1 1 1 3

2 2 2 2

t x y z B

mv mv mv k T

= + + =

(13.23)

Since there is no preferred direction, Eq. (13.23)

implies

1 1

2 2

x B

mv k T

1 1

2 2

y B

mv k T

1 1

2 2

z B

mv k T

(13.24)

A molecule free to move in space needs three

coordinates to specify its location. If it is

constrained to move in a plane it needs two; and

if constrained to move along a line, it needs just

one coordinate to locate it. This can also be

expressed in another way. We say that it has

one degree of freedom for motion in a line, two

for motion in a plane and three for motion in

space. Motion of a body as a whole from one

point to another is called translation. Thus, a

molecule free to move in space has three

translational degrees of freedom. Each

translational degree of freedom contributes a

term that contains square of some variable of

motion, e.g., ½ mv

and similar terms in

and v

. In, Eq. (13.24) we see that in thermal

equilibrium, the average of each such term is

½ k

T .

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KINETIC THEORY 333

Molecules of a monatomic gas like argon have

only translational degrees of freedom. But what

about a diatomic gas such as O

or N

? A

molecule of O

has three translational degrees

of freedom. But in addition it can also rotate

about its centre of mass. Figure 13.6 shows the

two independent axes of rotation 1 and 2, normal

to the axis joining the two oxygen atoms about

which the molecule can rotate*. The molecule

thus has two rotational degrees of freedom, each

of which contributes a term to the total energy

consisting of translational energy

and

rotational energy

2 2 2 2 2

1 1 2 2

1 1 1 1 1

2 2 2 2 2

t r x y z

mv mv mv I I

ε ε ω ω

+ = + + + +

(13.25)

Fig. 13.6 The two independent axes of rotation of a

diatomic molecule

where

and

are the angular speeds about

the axes 1 and 2 and I

, I

are the corresponding

moments of inertia. Note that each rotational

degree of freedom contributes a term to the

energy that contains square of a rotational

variable of motion.

We have assumed above that the O

molecule

is a ‘rigid rotator’, i.e., the molecule does not

vibrate. This assumption, though found to be

true (at moderate temperatures) for O

, is not

always valid. Molecules, like CO, even at

moderate temperatures have a mode of vibration,

i.e., its atoms oscillate along the interatomic axis

like a one-dimensional oscillator, and contribute

a vibrational energy term

to the total energy:

ky=













t r v

ε ε ε

= + + ε

(13.26)

where k is the force constant of the oscillator

and y the vibrational co-ordinate.

Once again the vibrational energy terms in

Eq. (13.26) contain squared terms of vibrational

variables of motion y and dy/dt .

At this point, notice an important feature in

Eq.(13.26). While each translational and

rotational degree of freedom has contributed only

one ‘squared term’ in Eq.(13.26), one vibrational

mode contributes two ‘squared terms’ : kinetic

and potential energies.

Each quadratic term occurring in the

expression for energy is a mode of absorption of

energy by the molecule. We have seen that in

thermal equilibrium at absolute temperature T,

for each translational mode of motion, the

average energy is ½ k

T. The most elegant

principle of classical statistical mechanics (first

proved by Maxwell) states that this is so for each

mode of energy: translational, rotational and

vibrational. That is, in equilibrium, the total

energy is equally distributed in all possible

energy modes, with each mode having an average

energy equal to ½ k

T. This is known as the law

of equipartition of energy. Accordingly, each

translational and rotational degree of freedom

of a molecule contributes ½ k

T to the energy,

while each vibrational frequency contributes

2 × ½ k

T = k

T , since a vibrational mode has

both kinetic and potential energy modes.

The proof of the law of equipartition of energy

is beyond the scope of this book. Here, we shall

apply the law to predict the specific heats of

gases theoretically. Later, we shall also discuss

briefly, the application to specific heat of solids.

13.6 SPECIFIC HEAT CAPACITY

13.6.1 Monatomic Gases

The molecule of a monatomic gas has only three

translational degrees of freedom. Thus, the

average energy of a molecule at temperature

T is (3/2)k

T . The total internal energy of a mole

of such a gas is

* Rotation along the line joining the atoms has very small moment of inertia and does not come into play for

quantum mechanical reasons. See end of section 13.6.

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334 PHYSICS

3 3

2 2

B A

U k T N RT

= × =

(13.27)

The molar specific heat at constant volume,

, is

(monatomic gas) =

RT (13.28)

For an ideal gas,

– C

= R (13.29)

where C

is the molar specific heat at constant

pressure. Thus,

R (13.30)

The ratio of specific heats

= =

(13.31)

13.6.2 Diatomic Gases

As explained earlier, a diatomic molecule treated

as a rigid rotator, like a dumbbell, has 5 degrees

of freedom: 3 translational and 2 rotational.

Using the law of equipartition of energy, the total

internal energy of a mole of such a gas is

5 5

2 2

B A

U k T N RT

= × =

(13.32)

The molar specific heats are then given by

(rigid diatomic) =

R, C

R (13.33)

(rigid diatomic) =

(13.34)

If the diatomic molecule is not rigid but has

in addition a vibrational mode

U k T k T N RT

B B A

= +













7 9 9

, ,

2 2 7

v p

C R C R

= = =

R (13.35)

13.6.3 Polyatomic Gases

In general a polyatomic molecule has 3

translational, 3 rotational degrees of freedom

and a certain number (f) of vibrational modes.

According to the law of equipartition of energy,

it is easily seen that one mole of such a gas has

U =







T +

T + f k

T N

i.e.,C

= (3 + f ) R, C

= (4 + f ) R,

(

)

(

)

4 +

3 +

(13.36)

Note that C

– C

= R is true for any ideal

gas, whether mono, di or polyatomic.

Table 13.1 summarises the theoretical

predictions for specific heats of gases ignoring

any vibrational modes of motion. The values are

in good agreement with experimental values of

specific heats of several gases given in Table 13.2.

Of course, there are discrepancies between

predicted and actual values of specific heats of

several other gases (not shown in the table), such

as Cl

, C

and many other polyatomic gases.

Usually, the experimental values for specific

heats of these gases are greater than the

predicted values as given in Table13.1 suggesting

that the agreement can be improved by including

vibrational modes of motion in the calculation.

The law of equipartition of energy is, thus, well

Nature of

Gas

(J mol

)

(J mol

)

- C

(J mol

)

Monatomic

12.5 20.8 8.31 1.67

Diatomic 20.8 29.1 8.31 1.40

Triatomic 24.93 33.24 8.31 1.33

Table 13.1 Predicted values of specific heat

capacities of gases (ignoring

vibrational modes)

Table13.2 Measured values of specific heat

capacities of some gases

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KINETIC THEORY 335

verified experimentally at ordinary

temperatures.

Example 13.8 A cylinder of fixed capacity

44.8 litres contains helium gas at standard

temperature and pressure. What is the

amount of heat needed to raise the

temperature of the gas in the cylinder by

15.0 °C ? (R = 8.31 J mo1

–1

Answer Using the gas law PV =

RT, you can

easily show that 1 mol of any (ideal) gas at

standard temperature (273 K) and pressure

(1 atm = 1.01 × 10

Pa) occupies a volume of

22.4 litres. This universal volume is called molar

volume. Thus the cylinder in this example

contains 2 mol of helium. Further, since helium

is monatomic, its predicted (and observed) molar

specific heat at constant volume, C

= (3/2) R,

and molar specific heat at constant pressure,

= (3/2) R + R = (5/2) R. Since the volume of

the cylinder is fixed, the heat required is

determined by C

. Therefore,

Heat required = no. of moles × molar specific

heat × rise in temperature

= 2 × 1.5 R × 15.0 = 45 R

= 45 × 8.31 = 374 J. t

13.6.4 Specific Heat Capacity of Solids

We can use the law of equipartition of energy to

determine specific heats of solids. Consider a

solid of N atoms, each vibrating about its mean

position. An oscillation in one dimension has

average energy of 2 × ½ k

T = k

T . In three

dimensions, the average energy is 3 k

T. For a

mole of solid, N = N

, and the total

energy is

U = 3 k

T × N

= 3 RT

Now at constant pressure ∆Q = ∆U + P∆V

= ∆U, since for a solid ∆V is negligible. Hence,

Q U

C R

T T

∆ ∆

= = =

∆ ∆

(13.37)

Table 13.3 Specific Heat Capacity of some

solids at room temperature and

atmospheric pressure

As Table 13.3 shows the prediction generally

agrees with experimental values at ordinary

temperature (Carbon is an exception).

13.6.5 Specific Heat Capacity of Water

We treat water like a solid. For each atom average

energy is 3k

T. Water molecule has three atoms,

two hydrogen and one oxygen. So it has

U = 3 × 3 k

T × N

= 9 RT

and C = ∆Q/ ∆T =∆ U / ∆T = 9R .

This is the value observed and the agreement

is very good. In the calorie, gram, degree units,

water is defined to have unit specific heat. As 1

calorie = 4.179 joules and one mole of water

is 18 grams, the heat capacity per mole is

~ 75 J mol

-1

~ 9R. However with more

complex molecules like alcohol or acetone the

arguments, based on degrees of freedom, become

more complicated.

Lastly, we should note an important aspect

of the predictions of specific heats, based on the

classical law of equipartition of energy. The

predicted specific heats are independent of

temperature. As we go to low temperatures,

however, there is a marked departure from this

prediction. Specific heats of all substances

approach zero as T à0. This is related to the

fact that degrees of freedom get frozen and

ineffective at low temperatures. According to

classical physics, degrees of freedom must

remain unchanged at all times. The behaviour

of specific heats at low temperatures shows the

inadequacy of classical physics and can be

explained only by invoking quantum

considerations, as was first shown by Einstein.

Quantum mechanics requires a minimum,

non-zero amount of energy before a degree of

freedom comes into play. This is also the reason

why vibrational degrees of freedom come into play

only in some cases.

13.7 MEAN FREE PATH

Molecules in a gas have rather large speeds of

the order of the speed of sound. Yet a gas leaking

from a cylinder in a kitchen takes considerable

time to diffuse to the other corners of the room.

The top of a cloud of smoke holds together for

hours. This happens because molecules in a gas

have a finite though small size, so they are bound

to undergo collisions. As a result, they cannot

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336 PHYSICS

move straight unhindered; their paths keep

getting incessantly deflected.

Fig. 13.7 The volume swept by a molecule in time ∆t

in which any molecule will collide with it.

Suppose the molecules of a gas are spheres

of diameter d. Focus on a single molecule with

the average speed <v>. It will suffer collision with

any molecule that comes within a distance d

between the centres. In time ∆t, it sweeps a

volume

<v> ∆t wherein any other molecule

will collide with it (see Fig. 13.7). If n is the

number of molecules per unit volume, the

molecule suffers n

<v> ∆t collisions in time

∆t. Thus the rate of collisions is n

<v> or the

time between two successive collisions is on the

average,

= 1/(n

<v> d

) (13.38)

The average distance between two successive

collisions, called the mean free path l, is :

l = <v>

= 1/(n

) (13.39)

In this derivation, we imagined the other

molecules to be at rest. But actually all molecules

are moving and the collision rate is determined

by the average relative velocity of the molecules.

Thus we need to replace <v> by <v

> in Eq.

(13.38). A more exact treatment gives

(

)

1/ 2

l n d

(13.40)

Let us estimate l and

for air molecules with

average speeds <v> = ( 485m/s). At STP

n =

(

)

( )

–3

0.02 10

22.4 10

= 2.7 × 10

-3.

Taking, d = 2 × 10

–10

= 6.1 × 10

–10

and l = 2.9 × 10

–7

m ≈ 1500d (13.41)

Seeing is Believing

Can one see atoms rushing about. Almost but not quite. One can see pollen grains of a flower being

pushed around by molecules of water. The size of the grain is ~ 10

-5

m. In 1827, a Scottish botanist

Robert Brown, while examining, under a microscope, pollen grains of a flower suspended in water

noticed that they continuously moved about in a zigzag, random fashion.

Kinetic theory provides a simple explanation of the phenomenon. Any object suspended in water is

continuously bombarded from all sides by the water molecules. Since the motion of molecules is random,

the number of molecules hitting the object in any direction is about the same as the number hitting in

the opposite direction. The small difference between these molecular hits is negligible compared to the

total number of hits for an object of ordinary size, and we do not notice any movement of the object.

When the object is sufficiently small but still visible under a microscope, the difference in molecular

hits from different directions is not altogether negligible, i.e. the impulses and the torques given to the

suspended object through continuous bombardment by the molecules of the medium (water or some

other fluid) do not exactly sum to zero. There is a net impulse and torque in this or that direction. The

suspended object thus, moves about in a zigzag manner and tumbles about randomly. This motion

called now ‘Brownian motion’ is a visible proof of molecular activity. In the last 50 years or so molecules

have been seen by scanning tunneling and other special microscopes.

In 1987 Ahmed Zewail, an Egyptian scientist working in USA was able to observe not only the

molecules but also their detailed interactions. He did this by illuminating them with flashes of laser

light for very short durations, of the order of tens of femtoseconds and photographing them. ( 1 femto-

second = 10

-15

s ). One could study even the formation and breaking of chemical bonds. That is really

seeing !

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KINETIC THEORY 337

As expected, the mean free path given by

Eq. (13.40) depends inversely on the number

density and the size of the molecules. In a highly

evacuated tube n is rather small and the mean

free path can be as large as the length of the

tube.

Example 13.9 Estimate the mean free path

for a water molecule in water vapour at 373 K.

Use information from Exercises 13.1 and

Eq. (13.41) above.

Answer The d for water vapour is same as that

of air. The number density is inversely

proportional to absolute temperature.

25 25 –3

273

2.7 10 2 10 m

373

n = × × = ×

Hence, mean free path

–7

4 10 m

l = ×

Note that the mean free path is 100 times the

interatomic distance ~ 40 Å = 4 ×10

-9

m calculated

earlier. It is this large value of mean free path that

leads to the typical gaseous behaviour. Gases can

not be confined without a container.

Using, the kinetic theory of gases, the bulk

measurable properties like viscosity, heat

conductivity and diffusion can be related to the

microscopic parameters like molecular size. It

is through such relations that the molecular

sizes were first estimated.

SUMMARY

1. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature

(T ) is

PV =

RT = k

where

is the number of moles and N is the number of molecules. R and k

are universal

constants.

R = 8.314 J mol

–1

, k

= 1.38 × 10

–23

J K

–1

Real gases satisfy the ideal gas equation only approximately, more so at low pressures

and high temperatures.

2. Kinetic theory of an ideal gas gives the relation

P n m v

where n is number density of molecules, m the mass of the molecule and

is the

mean of squared speed. Combined with the ideal gas equation it yields a kinetic

interpretation of temperature.

1 3

2 2

m v k T

(

)

1/2

rms

v v=

k T

This tells us that the temperature of a gas is a measure of the average kinetic energy

of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at

a fixed temperature the heavier molecule has the lower average speed.

3. The translational kinetic energy

E =

NT.

This leads to a relation

PV =

4. The law of equipartition of energy states that if a system is in equilibrium at absolute

temperature T, the total energy is distributed equally in different energy modes of

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338 PHYSICS

absorption, the energy in each mode being equal to ½ k

T. Each translational and

rotational degree of freedom corresponds to one energy mode of absorption and has

energy ½ k

T. Each vibrational frequency has two modes of energy (kinetic and potential)

with corresponding energy equal to

2 × ½ k

T = k

5. Using the law of equipartition of energy, the molar specific heats of gases can be

determined and the values are in agreement with the experimental values of specific

heats of several gases. The agreement can be improved by including vibrational modes

of motion.

6. The mean free path l is the average distance covered by a molecule between two successive

collisions :

= l

n d

where n is the number density and d the diameter of the molecule.

POINTS TO PONDER

1. Pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid.

Any layer of gas inside the volume of a container is in equilibrium because the pressure

is the same on both sides of the layer.

2. We should not have an exaggerated idea of the intermolecular distance in a gas. At

ordinary pressures and temperatures, this is only 10 times or so the interatomic distance

in solids and liquids. What is different is the mean free path which in a gas is 100

times the interatomic distance and 1000 times the size of the molecule.

3. The law of equipartition of energy is stated thus: the energy for each degree of freedom

in thermal equilibrium is ½ k

T. Each quadratic term in the total energy expression of

a molecule is to be counted as a degree of freedom. Thus, each vibrational mode gives

2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the

energy 2 × ½ k

T = k

4. Molecules of air in a room do not all fall and settle on the ground (due to gravity)

because of their high speeds and incessant collisions. In equilibrium, there is a very

slight increase in density at lower heights (like in the atmosphere). The effect is small

since the potential energy (mgh) for ordinary heights is much less than the average

kinetic energy ½ mv

of the molecules.

5. < v

> is not always equal to ( < v >)

. The average of a squared quantity is not necessarily

the square of the average. Can you find examples for this statement.

EXERCISESEXERCISES

EXERCISES

13.113.1

13.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen

gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

13.213.2

13.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard

temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4

litres.

13.313.3

13.3 Figure 13.8 shows plot of PV/T versus P

for 1.00×10

–3

kg of oxygen gas at two

different temperatures.

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KINETIC THEORY 339

Fig. 13.8Fig. 13.8

Fig. 13.8

(a) What does the dotted plot signify?

(b) Which is true: T

> T

or T

< T

axis?

(d) If we obtained similar plots for 1.00×10

–3

kg of hydrogen, would we get the same

value of PV/T

at the point where the curves meet on the y-axis? If not, what mass

of hydrogen yields the same value of PV/T

(for low pressure high temperature

region of the plot) ? (Molecular mass of H

= 2.02 u, of O

= 32.0 u,

R = 8.31 J mo1

–1

13.413.4

13.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and

a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge

pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of

oxygen taken out of the cylinder (R = 8.31 J mol

–1

, molecular mass of O

= 32 u).

13.513.5

13.5 An air bubble of volume 1.0 cm

rises from the bottom of a lake 40 m deep at a

temperature of 12 °C. To what volume does it grow when it reaches the surface,

which is at a temperature of 35 °C ?

13.613.6

13.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water

vapour and other constituents) in a room of capacity 25.0 m

at a temperature of

27 °C and 1 atm pressure.

13.713.7

13.7 Estimate the average thermal energy of a helium atom at (i) room temperature

(27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature

of 10 million kelvin (the typical core temperature in the case of a star).

13.813.8

13.8 Three vessels of equal capacity have gases at the same temperature and pressure.

The first vessel contains neon (monatomic), the second contains chlorine (diatomic),

and the third contains uranium hexafluoride (polyatomic). Do the vessels contain

equal number of respective molecules ? Is the root mean square speed of molecules

the same in the three cases? If not, in which case is v

rms

the largest ?

13.913.9

13.9 At what temperature is the root mean square speed of an atom in an argon gas

cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar

= 39.9 u, of He = 4.0 u).

13.10

13.1013.10

13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a

cylinder containing nitrogen at 2.0 atm and temperature 17

C. Take the radius of a

nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the

molecule moves freely between two successive collisions (Molecular mass of N

28.0 u).

(JK)

–1

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340 PHYSICS

Additional ExercisesAdditional Exercises

Additional Exercises

13.1113.11

13.11 A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm

long mercury thread, which traps a 15 cm column of air. What happens if the tube

is held vertically with the open end at the bottom ?

13.1213.12

13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of

28.7 cm

–1

. The diffusion of another gas under the same conditions is measured to

have an average rate of 7.2 cm

–1

. Identify the gas.

[Hint : Use Graham’s law of diffusion: R

= ( M

)

1/2

, where R

, R

are diffusion

rates of gases 1 and 2, and M

and M

their respective molecular masses. The law is

a simple consequence of kinetic theory.]

13.1313.13

13.13 A gas in equilibrium has uniform density and pressure throughout its volume. This

is strictly true only if there are no external influences. A gas column under gravity,

for example, does not have uniform density (and pressure). As you might expect, its

density decreases with height. The precise dependence is given by the so-called law

of atmospheres

= n

exp [ -mg (h

– h

)/ k

where n

, n

refer to number density at heights h

and h

respectively. Use this

relation to derive the equation for sedimentation equilibrium of a suspension in a

liquid column:

= n

exp [ -mg N

(

ρ′

) (h

–h

)/ (ρ RT)]

where

is the density of the suspended particle, and

ρ′

, that of surrounding medium.

is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes

principle to find the apparent weight of the suspended particle.]

13.1413.14

13.14 Given below are densities of some solids and liquids. Give rough estimates of the

size of their atoms :

[Hint : Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the

known value of Avogadro’s number. You should, however, not take the actual numbers

you obtain for various atomic sizes too literally. Because of the crudeness of the

tight packing approximation, the results only indicate that atomic sizes are in the

range of a few Å].

Substance Atomic Mass (u) Density (10

Kg m

-3

)

Carbon (diamond) 12.01 2.22

Gold 197.00 19.32

Nitrogen (liquid) 14.01 1.00

Lithium 6.94 0.53

Fluorine (liquid) 19.00 1.14

2020-21