# KINETIC THEORY OF GASES

## Kinetic Theory

Kinetic theory was developed by Maxwell, Boltzmann and Gibbs. It explains the behaviour of gases, based on the idea that, gases consists of a large number of atoms or molecules, which are in the state of continuous random motion and the interatomic forces binding the atoms are negligible.

## Success of Kinetic theory

Kinetic theory was successful due to,

• It gives a molecular interpretation of pressure and temperature of a gas.
• It is consistent with gas laws and Avogadro’s hypothesis.
• It is correctly explains specific heat capacities of many gases.
• It relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters such as molecular size and mass.

## Molecular nature of matter

Molecular nature of matter can be established by considering the following.

• Atomic Hypothesis: All the things are made of atoms – which are tiny particles that move around in perpetual motion, attracting each other when they are little distance apart but repelling upon being squeezed into one another. Kanada in India and Democritus’ in Greece has suggested the Atomic hypothesis.
• Atomic Theory: John Dalton proposed atomic theory and suggested that,

(i) The smallest constituents of an element are atoms.

(ii) Atoms of one element are identical but differ from those of other element.

(iii) A small number of atoms of each element combine to form a molecule of the compound.

• Avogadro’s Hypothesis and Gay Lussac’s Law were well explained based on the molecular nature of matter.

Note: Gay Lussac’s Law: When gases combine chemically to yield another gas, their volumes are in the ratio of small integers.

## Behaviour of Gas

Gases at low densities, low pressures and high temperatures obey the experimental result,

𝑃𝑉/𝑇 = 𝐾

Where 𝐾 is a constant and depends on mass of the gas, hence 𝐾 ∝ 𝑁 or 𝐾 = 𝑘𝐵𝑁 where 𝑁 → Number of molecules 𝑘𝐵 →Boltzman constant = 1.38 × 10−23𝐽𝐾−1

Implies that, 𝑷𝑽/𝑻 = 𝒌𝑩𝑵

The number of molecule per unit volume is same for all gases at a fixed temperature and pressure.

Explanation: We have, 𝑃1𝑉1/𝑇1𝑁1 = 𝑃2𝑉2/𝑇2𝑁2 = 𝑘B

If 𝑃, 𝑉 and 𝑇 are same, then 𝑁 is also same for all gases and this number is denoted as 𝑁𝐴, called Avogadro Number. In 22.4 litres of any gas, 𝑁𝐴 = 6.02 × 1023. Hence the equation becomes, 𝑷𝑽 = 𝒌𝑩𝑵𝑨𝑻

Note: The mass of 22.4 litres any gas is equal to its molecular weight in grams at 𝑆𝑇𝑃. This amount is called as 1 mole of substance.

## Gas Law

The relationship between any two physical quantities is used to specify the state of a gas keeping the third physical quantity constant is known as Gas law.

(i) Boyle’s Law: At constant temperature, pressure of a given mass of gas varies inversely with its volume. If 𝑇 is constant, 𝑃 ∝ 1/𝑉 or 𝑃𝑉 = constant.

(ii) Charles’ Law: At constant pressure, the volume of the gas is proportional to its absolute temperature. If 𝑃 is constant, 𝑉 ∝ 𝑇 or 𝑉/𝑇 = constant.

## Ideal Gas

A gas in which the molecules do not exert any attractive or repulsive force on each other is called an Ideal gas or Perfect gas. or A gas which obeys Boyle’s law and Charles’ law is called Ideal gas or Perfect gas.

## Ideal gas equation

For one mole of gas, we have

𝑃𝑉 = constant (Boyle’s law)

𝑉/𝑇 = constant (Charles’ law)

Combining these, we get 𝑃𝑉/𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

The constant is equal to 𝑅 and known as Universal gas constant, its value is 8.31𝐽𝑚𝑜𝑙−1𝐾−1. So, 𝑃𝑉 = 𝑅𝑇

For 𝜇 moles, 𝑷𝑽 = 𝝁𝑹𝑻 this is ideal gas equation.

Note:

(i) An ideal gas is a simple theoretical model of a gas, no real gas is truly Ideal.

(ii) We have 𝑃𝑉 = 𝑅𝑇 and 𝑃𝑉 = 𝑘𝐵𝑁𝐴𝑇

On comparison we get, 𝑅 = 𝑘𝐵𝑁𝐴

(iii) We have 𝜇𝑁𝐴 = 𝑁 and 𝜇𝑀0 = 𝑀

Where 𝑁 → Total number of molecules 𝑀 → Mass of the gas 𝑀0 →Molar mass 𝝁 = 𝑴/𝑴𝟎 = 𝑵/𝑵𝑨

## Assumptions of Kinetic Theory

1) Any gas consists of a very large number of identical particles called molecules, each having identical mass.

2) The molecules are considered to be rigid, perfectly elastic solid spheres of negligible size.

3) The molecules are in a state of random motion, moving in all directions with all possible velocities.

4) All the collisions between the molecules or between a molecule and the wall of the container are perfectly elastic.

5) The molecules exert no force of attraction or repulsion on each other and with the walls of the container, except during the collision.

6) The molecules during motion collide with one another and with the walls of the container. Between collisions, the molecules move in a straight line with uniform velocity. At each collision their velocity gets altered.

7) In steady state, the collisions do not affect the molecular density of the gas.

8) The molecules obey Newton’s laws of motion.

## Pressure of an Ideal gas

Consider a gas enclosed in a cube of side 𝑙. Let 𝑛 be the number of molecules per unit volume. Let a molecule of mass 𝑚 moving with velocity (𝑣𝑥, 𝑣𝑦, 𝑣𝑧) along 𝑥 −axis hits the face 𝐹1. Since the collision is elastic, the 𝑥 − component of velocity of the molecule is reversed whereas the 𝑦 and 𝑧 components remain unchanged.

Initial momentum of the molecule along 𝑥−axis = 𝑚𝑣𝑥

Final momentum of the molecule along 𝑥−axis = −𝑚𝑣𝑥

Change in momentum = −𝑚𝑣𝑥 − 𝑚𝑣𝑥 = −2𝑚𝑣𝑥

Momentum transferred to the wall = 2𝑚𝑣𝑥

To calculate pressure on the face 𝐹1, we have to calculate momentum transferred in time 𝑑𝑡. In time 𝑑𝑡, a molecule with 𝑥 −component of velocity 𝑣𝑥 will hit the face 𝐹1, if it is within a distance of 𝑣𝑥𝑑𝑡 from face 𝐹1. Then all the molecules within this distance will hit the face 𝐹1. Number of molecules within this volume = 𝑛𝐴𝑣𝑥𝑑𝑡

On the average half of these molecules are moving towards face 𝐹1 and half away from the face 𝐹1.

Number of molecules hitting the face 𝐹1 = (1/2)𝑛𝐴𝑣𝑥𝑑𝑡

Momentum transferred to the wall by these molecules 𝑑𝑝 = 2𝑚𝑣𝑥(1/2)𝑛𝐴𝑣𝑥𝑑𝑡𝑑𝑝 = 𝑛𝑚𝐴𝑣𝑥2𝑑𝑡

Force, 𝐹 = 𝑑𝑝/𝑑𝑡 = 𝑛𝑚𝐴𝑣𝑥2

Pressure, 𝑃 = 𝐹/𝐴 = 𝑛𝑚𝐴𝑣𝑥2/𝐴 = 𝑛𝑚𝑣𝑥2

Actually all the molecules in the gas do not have the same velocity. Therefore Pressure exerted by these molecules on face 𝐹1 is,

where

→ average of 𝑣𝑥2

But Inside the container the molecules are in random motion. By symmetry

Average of

Pressure,

where

→ mean of the square speed.

Kinetic interpretation of Temperature (Relation between Kinetic energy and temperature)

We have

Here

is the average translational kinetic energy of the molecules of the gas.

The internal energy of the gas is purely kinetic and is given by,

Equation (i) becomes, 𝑃𝑉 = (2/3)𝐸

But we have, 𝑃𝑉 = 𝑘𝐵𝑁𝑇

Therefor𝑒 𝐸 = (3/2)𝑘𝐵𝑁𝑇

𝑬/𝑵 = (𝟑/𝟐)𝒌𝑩𝑻

Where 𝐸/𝑁 → average kinetic energy of a molecule. Implies that, 𝑬/𝑵 ∝ 𝑻 or 〈𝑬𝒕〉 ∝ 𝑻

The average kinetic energy is directly proportional to the absolute temperature.

## Definition for Temperature

Temperature is defined as the average kinetic energy of a molecule.

## RMS speed of a gas molecule

It is the square root of the mean of the square of the velocities of individual molecules of the gas.

We have

## Degrees of freedom

The number of co-ordinates required to specify the configuration and position of a gas molecule is called Degrees of freedom. or It is the total number of independent ways in which the gas molecule can absorb the energy.

## Degrees of freedom of a gas molecule

A gas molecule can possess translational kinetic energy, rotational kinetic energy and vibrational energy.

(i) A mono atomic gas: A monoatomic gas molecule consists of a single atom. It can have translational motion in any direction in 3-dimensional space. Therefore it has 3 translational degrees of freedom (n=3).

∴ 𝐸𝑡 = (1/2)𝑚𝑣𝑥2 + (1/2)𝑚𝑣𝑦2 + (1/2)𝑚𝑣𝑧2

(ii) Diatomic Gas: The molecule consists of two atoms bound to each other. Assuming that the diatomic molecule is rigid, it has,

a) 3 translational degrees of freedom. (Each along 𝑥, 𝑦 and 𝑧 −axis)

b) 2 rotational degrees of freedom. (One along 𝑦 and other 𝑧 −axis)

c) No vibrational energy. (because molecule is rigid)

Therefore 𝑛 = 3 + 2 = 5

∴ 𝐸𝑡 = (1/2)𝑚𝑣𝑥2 + (1/2)𝑚𝑣𝑦2 + (1/2)𝑚𝑣𝑧2 + (1/2)𝐼𝑦𝜔𝑦 + (1/2)𝐼𝑧𝜔𝑧

Generally, number of degrees of freedom of a molecule is equal to total number of co-ordinates required to specify the positions of the molecule minus the number of independent relations between the atoms or molecules.

𝒏 = 𝟑𝑵 − 𝒓

(iii) Triatomic molecule (non-linear structure),

𝑛 = 3𝑁 − 𝑟

𝑁 = 3 and 𝑟 = 2 implies that, 𝑛 = 3(3) − 3 = 6

A non-linear rigid triatomic molecule has 3 translational degrees of freedom, 3 rotational degrees of freedom and no vibrational degrees of freedom.

## Internal energy of an Ideal gas

Internal energy of an ideal gas is sum of potential and kinetic energies of all the gas molecules. Denoted by 𝑈.

𝑈 = 𝐾 + 𝑉

For an ideal gas, potential energy is zero, since no intermolecular forces between the molecules and kinetic energy may be translational, rotational and vibrational.

𝑉 = 0 𝑎𝑛𝑑 𝐾 = 𝐸𝑡 + 𝐸𝑟 + 𝐸𝑣 then 𝑈 = 𝐾

## Law of Equipartition of energy

In thermal equilibrium, the total energy is equally distributed in all possible degrees of freedom and the average energy in each degree of freedom is equal to (1/2)𝑘𝐵𝑇.

Explanation: The translational kinetic energy of a single molecule is,

𝐸𝑡 = (1/2)𝑚𝑣𝑥2 + (1/2)𝑚𝑣𝑦2 + (1/2)𝑚𝑣𝑧2

The average energy of the gas is given by,

〈𝐸𝑡〉 = 〈(1/2)𝑚𝑣𝑥2〉 + 〈(1/2)𝑚𝑣y2〉 + 〈(1/2)𝑚𝑣z2

From Kinetic theory, 𝐸/𝑁 = 〈𝐸〉 = (3/2)𝑘𝐵𝑇

〈(1/2)𝑚𝑣𝑥2〉 + 〈(1/2)𝑚𝑣y2〉 + 〈(1/2)𝑚𝑣z2〉 = (3/2)𝑘𝐵𝑇

〈(1/2)𝑚𝑣𝑥2〉 = (1/2)𝑘𝐵𝑇, 〈(1/2)𝑚𝑣y2〉 = (1/2)𝑘𝐵𝑇, 〈(1/2)𝑚𝑣z2〉 = 𝑘𝐵𝑇

## Specific heat capacity of gas

(i) Monoatomic gases: Molecule of monoatomic gas has 3 degrees of freedom. Total internal energy of the monoatomic gas,

𝑈 = (3/2)𝑘𝐵𝑇𝑁𝐴

𝑈 = (3/2)𝑅𝑇

𝐶𝑉 = 𝑑𝑈/𝑑𝑇 = (3/2)𝑅

𝐶𝑃 − 𝐶𝑉 = 𝑅

𝐶𝑃 = 𝑅 + 𝐶𝑉 = 𝑅 + (3/2)𝑅 = (5/2)𝑅

The ratio of specific heats, 𝜸 = 𝑪𝑷/𝑪𝑽 = (𝟓𝑹/𝟐) × (𝟐/𝟑𝑹) = 𝟓/𝟑

(ii) Diatomic gases: Molecules of diatomic gas has 5 degrees of freedom.

𝑈 = (5/2)𝑘𝐵𝑇𝑁𝐴

𝑈 = (5/2)𝑅𝑇

𝐶𝑉 = 𝑑𝑈/𝑑𝑇 = (5/2)𝑅

𝐶𝑃 − 𝐶𝑉 = 𝑅

𝐶𝑃 = 𝑅 + 𝐶𝑉 = 𝑅 + (5/2)𝑅 = (7/2)𝑅

The ratio of specific heats, 𝜸 = 𝑪𝑷/𝑪𝑽 = (𝟕𝑹/𝟐) × (𝟐/𝟓𝑹) = 𝟕/𝟓

(iii) Polyatomic gases: A poly molecule has 3 translational, 3 rotational and 𝑓 number of vibrational energy.

𝑈 = ((3/2)𝑘𝐵𝑇 + (3/2)𝑘𝐵𝑇 + 𝑓𝑘𝐵𝑇)𝑁𝐴

𝑈 = ((3/2) + (3/2) + 𝑓 )𝑁𝐴𝑘𝐵𝑇

𝑈 = (3 + 𝑓)𝑅𝑇

𝐶𝑉 = 𝑑𝑈/𝑑𝑇 = (3 + 𝑓)𝑅

𝐶𝑃 = 𝑅 + 𝐶𝑉

𝐶𝑃 = 𝑅 + (3 + 𝑓)𝑅 = (4 + 𝑓)𝑅

𝜸 = 𝑪𝑷/𝑪𝑽 = (𝟒 + 𝒇)/(𝟑 + 𝒇)

All the above values are in good agreement with experimental values.

## Specific heat capacities of solids

Every atom in a solid vibrates about its equilibrium position. A one dimensional oscillator has two degrees of freedom, one translational and one vibrational.

The average energy = 2×(1/2)𝑘𝐵𝑇 = 𝑘𝐵𝑇

Each atom in a solid can be treated as three dimensional harmonic oscillators. Hence average energy of an atom in a solid = 3𝑘𝐵𝑇

If there are 𝑁𝐴 atoms, total energy per mole, 𝑈 = 3𝑘𝐵𝑇𝑁𝐴

𝑈 = 3𝑅𝑇

At constant pressure we have 𝑑𝑄 = 𝑑𝑈 + 𝑃𝑑𝑉

But 𝑑𝑉 ≈ 0 for solids, then 𝑑𝑄 = 𝑑𝑈

𝑪 = 𝒅𝑼/𝒅𝑻 = 𝒅𝑸/𝒅𝑻 = 𝟑𝑹

Specific heat capacity of water

Water can be treated as solid. Then for each atom average energy = 3𝑘𝐵𝑇. Water has three atoms, so 𝑈 = 3 × 3𝑘𝐵𝑇𝑁𝐴 = 9𝑅𝑇

𝑪 = 𝒅𝑼/𝒅𝑻 = 𝒅𝑸/𝒅𝑻 = 𝟗𝑹

𝐶 ≈ 75𝐽𝑚𝑜𝑙−1𝐾−1

## Free path

The distance travelled by a gas molecule between two successive collisions is known as free path. Mean free path: The average distance travelled by a molecule between two successive collisions is called mean free path.

## Expression for Mean free path

Consider a gas containing 𝑛 molecules per unit volume. Let 𝑑 be the diameter of the each molecule. Consider one such molecule is in motion and other are at rest. The moving molecule will collide with all those molecules whose centres are at a distance 𝑑 from the centre of the molecule. Let 〈𝑣〉 be the average speed and distance travelled by the molecule in time ∆𝑡 is 〈𝑣〉∆𝑡. This molecule collide with all the molecules whose centres lie on the cylinder of radius 𝑑. Number of molecules in this volume,

𝑁 = 𝑛𝜋𝑑2〈𝑣〉∆𝑡

Number of collisions in this volume = Number of molecules = 𝑛𝜋𝑑2〈𝑣〉∆𝑡

Rate of collision = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠/∆𝑡 = 𝑛𝜋𝑑2〈𝑣〉∆𝑡/∆𝑡 = 𝑛𝜋𝑑2〈𝑣〉

Average time between two successive collisions, 𝜏 = 1/𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛

𝜏 = 1/𝑛𝜋𝑑2〈𝑣〉

Average distance between two successive collisions – mean free path, 𝑙 = 〈𝑣〉𝜏

But actually all molecules are moving. So collision rate is determined by average relative velocity of the molecule. Therefore, mean free path, 𝑙 = 〈𝑣𝑟〉𝜏

But 〈𝑣𝑟〉 = 〈𝑣〉/√2

Mean free path, 𝑙 = (〈𝑣〉/√2)𝜏 = (〈𝑣〉/√2)(1/𝑛𝜋𝑑2〈𝑣〉)

𝒍 = 𝟏/√𝟐𝒏𝝅𝒅𝟐

Note: From above equation it is observed that the mean free path is inversely proportional to number of molecules per unit volume, (𝑛) and size of the molecule (𝑑).

## Suggested Questions

One Mark.

1. Define degrees of freedom of a molecule.

2. What is mean free path of a gas molecule?

3. How does an average kinetic energy of a gas molecule depend on the absolute temperature?

4. State Charles’ law.

5. What is ideal gas?

6. Mention the degrees of freedom for a triatomic gas molecule.

7. Name a factor on which internal energy of the gas depends.

Two Marks.

1. Mention an expression for pressure of an ideal gas and explain the symbol used.

Three Marks.

1. Mention any three postulates of kinetic theory of gases.

2. Mention the expression for average kinetic energy of a molecule in terms of absolute temperature.

3. Derive

with usual notation.

4. Define degrees of freedom. Mention the degrees of freedom for (a) monoatomic gas. (b) diatomic molecule.

5. State and explain the law of equipartition of energy.

6. State the law of equipartition of energy. Write an expression for the energy associated with diatomic molecule.

7. Derive the expression for the specific heat capacity of solids.

Five Marks.

1. Derive the relation between kinetic energy of a gas molecule and its absolute temperature.

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