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 = 𝑘_𝐵𝑁 where 𝑁 → Number of molecules 𝑘_𝐵 →Boltzman constant = 1.38 × 10⁻²³𝐽𝐾⁻¹

Implies that, 𝑷𝑽/𝑻 = 𝒌_𝑩𝑵

Avogadro Hypothesis

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 = 𝑘_B

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

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.

A graph of (𝑃𝑉/𝜇𝑇) versus 𝑃 is as shown. From graph, it can be concluded that, a real gas behaves as an ideal gas at high temperatures.

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 = 𝑘_𝐵𝑁_𝐴𝑇

On comparison we get, 𝑅 = 𝑘_𝐵𝑁_𝐴

(iii) We have 𝜇𝑁_𝐴 = 𝑁 and 𝜇𝑀₀ = 𝑀

Where 𝑁 → Total number of molecules 𝑀 → Mass of the gas 𝑀₀ →Molar mass 𝝁 = 𝑴/𝑴_𝟎 = 𝑵/𝑵_𝑨

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 (𝑣_𝑥, 𝑣_𝑦, 𝑣_𝑧) 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 = 𝑚𝑣_𝑥

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

Change in momentum = −𝑚𝑣_𝑥 − 𝑚𝑣_𝑥 = −2𝑚𝑣_𝑥

Momentum transferred to the wall = 2𝑚𝑣_𝑥

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 𝑣_𝑥 will hit the face 𝐹₁, if it is within a distance of 𝑣_𝑥𝑑𝑡 from face 𝐹₁. Then all the molecules within this distance will hit the face 𝐹₁. Number of molecules within this volume = 𝑛𝐴𝑣_𝑥𝑑𝑡

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)𝑛𝐴𝑣_𝑥𝑑𝑡

Momentum transferred to the wall by these molecules 𝑑𝑝 = 2𝑚𝑣_𝑥(1/2)𝑛𝐴𝑣_𝑥𝑑𝑡𝑑𝑝 = 𝑛𝑚𝐴𝑣_𝑥²𝑑𝑡

Force, 𝐹 = 𝑑𝑝/𝑑𝑡 = 𝑛𝑚𝐴𝑣_𝑥²

Pressure, 𝑃 = 𝐹/𝐴 = 𝑛𝑚𝐴𝑣_𝑥²/𝐴 = 𝑛𝑚𝑣_𝑥²

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 𝑣_𝑥²

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 = 𝑘_𝐵𝑁𝑇

Therefor𝑒 𝐸 = (3/2)𝑘_𝐵𝑁𝑇

𝑬/𝑵 = (𝟑/𝟐)𝒌_𝑩𝑻

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

Expression for RMS speed

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).

∴ 𝐸_𝑡 = (1/2)𝑚𝑣_𝑥² + (1/2)𝑚𝑣_𝑦² + (1/2)𝑚𝑣_𝑧²

(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

∴ 𝐸_𝑡 = (1/2)𝑚𝑣_𝑥² + (1/2)𝑚𝑣_𝑦² + (1/2)𝑚𝑣_𝑧² + (1/2)𝐼_𝑦𝜔_𝑦 + (1/2)𝐼_𝑧𝜔_𝑧

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 = 𝐸_𝑡 + 𝐸_𝑟 + 𝐸_𝑣 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)𝑘_𝐵𝑇.

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

𝐸_𝑡 = (1/2)𝑚𝑣_𝑥² + (1/2)𝑚𝑣_𝑦² + (1/2)𝑚𝑣_𝑧²

The average energy of the gas is given by,

〈𝐸_𝑡〉 = 〈(1/2)𝑚𝑣_𝑥²〉 + 〈(1/2)𝑚𝑣_y²〉 + 〈(1/2)𝑚𝑣_z²〉

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

〈(1/2)𝑚𝑣_𝑥²〉 + 〈(1/2)𝑚𝑣_y²〉 + 〈(1/2)𝑚𝑣_z²〉 = (3/2)𝑘_𝐵𝑇

〈(1/2)𝑚𝑣_𝑥²〉 = (1/2)𝑘_𝐵𝑇, 〈(1/2)𝑚𝑣_y²〉 = (1/2)𝑘_𝐵𝑇, 〈(1/2)𝑚𝑣_z²〉 = 𝑘_𝐵𝑇

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)𝑘_𝐵𝑇𝑁_𝐴

𝑈 = (3/2)𝑅𝑇

𝐶_𝑉 = 𝑑𝑈/𝑑𝑇 = (3/2)𝑅

𝐶_𝑃 − 𝐶_𝑉 = 𝑅

𝐶_𝑃 = 𝑅 + 𝐶_𝑉 = 𝑅 + (3/2)𝑅 = (5/2)𝑅

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

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

𝑈 = (5/2)𝑘_𝐵𝑇𝑁_𝐴

𝑈 = (5/2)𝑅𝑇

𝐶_𝑉 = 𝑑𝑈/𝑑𝑇 = (5/2)𝑅

𝐶_𝑃 − 𝐶_𝑉 = 𝑅

𝐶_𝑃 = 𝑅 + 𝐶_𝑉 = 𝑅 + (5/2)𝑅 = (7/2)𝑅

The ratio of specific heats, 𝜸 = 𝑪_𝑷/𝑪_𝑽 = (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)𝑘_𝐵𝑇 + (3/2)𝑘_𝐵𝑇 + 𝑓𝑘_𝐵𝑇)𝑁_𝐴

𝑈 = ((3/2) + (3/2) + 𝑓 )𝑁_𝐴𝑘_𝐵𝑇

𝑈 = (3 + 𝑓)𝑅𝑇

𝐶_𝑉 = 𝑑𝑈/𝑑𝑇 = (3 + 𝑓)𝑅

𝐶_𝑃 = 𝑅 + 𝐶_𝑉

𝐶_𝑃 = 𝑅 + (3 + 𝑓)𝑅 = (4 + 𝑓)𝑅

𝜸 = 𝑪_𝑷/𝑪_𝑽 = (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)𝑘_𝐵𝑇 = 𝑘_𝐵𝑇

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

If there are 𝑁_𝐴 atoms, total energy per mole, 𝑈 = 3𝑘_𝐵𝑇𝑁_𝐴

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 = 3𝑘_𝐵𝑇. Water has three atoms, so 𝑈 = 3 × 3𝑘_𝐵𝑇𝑁_𝐴 = 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 = 〈𝑣_𝑟〉𝜏

But 〈𝑣_𝑟〉 = 〈𝑣〉/√2

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

𝒍 = 𝟏/√𝟐𝒏𝝅𝒅^𝟐

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).