## 1. Kinetic theory

Introduction

As all molecules of a gas are in a state of rapid and continuous motion, various properties of a gas like pressure, temperature, energy are explained and the kinetic theory was developed by Scottish physicist James Maxwell and Austrian Physicist Ludwig Boltzmann.

Kinetic theory of gases is based on the molecular picture of matter. It correlates the macroscopic properties (like pressure and temperature) of gases to microscopic properties like speed, kinetic energy of gas molecules.

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 interatomic forces, which are short range forces that are important for solids and liquids, can be neglected for gases.

Dalton’s Atomic Theory

• Atomic hypothesis was given by many scientists. According to which everything in this universe is made up of atoms.
• Atoms are little particles that move around in a perpetual order attracting each other when they are little distance apart.
• But if they are forced very close to each other then they rebel.

For example: - Consider a block of gold. It consists of molecules that are constantly moving.

• Dalton’s atomic theory is also referred to as the molecular theory of matter. This theory proves that matter is made up of molecules which in turn are made up of atoms.

According to Gay Lussac’s law - when gases combine chemically to yield another gas, their volumes are in ratios of small integers.

Avogadro’s law states that the equal volumes of all gases at equal temperature and pressure have the same number of molecules.

Conclusion: - All these laws proved the molecular nature of gases.

Dalton’s molecular theory forms the basis of Kinetic theory.

Behaviour of Gases

Gases at low pressures and high temperatures much above that at which they liquefy (or solidify) approximately satisfy a relation between their pressure, temperature and volume:

P V= K T . This is the universal relation which is satisfied by all gases. Where P, V, T are pressure, volume and temperature resp. and K is the constant for a given volume of gas. It varies with the volume of gas. K=NKB

Where N=number of molecules and kB = Boltzmann Constant and its value never change.

From above two equations we have

P V=NKBT⇒  PV/NT=KB Which is the same for all gases.

Consider there are 2 gases :- (P1, V,1, T1) and (P2, V2, T2) where P, V and T are pressure, volume and temperature respectively.

This is Avogadro’s hypothesis, that the number of molecules per unit volume is same for all gases at a fixed temperature and pressure

Conclusion: - This relation is satisfied by all gases at low pressure and high temperature.

According to Avogadro’s hypothesis, the number of molecules per unit volume is the same for all gases at a fixed P and T.

NA= 6.022 ×1023. It is a universal value.

Experimentally it has been found that the mass of 22.4 litres of any gas is equal to molecular weight in grams at standard temperature and pressure.

Perfect Gas Equation

Perfect gas equation is given by PV=μRT,

Where P, V are pressure, volume, T =absolute temperature, μ = number of moles and R =universal gas constant.

R= kBNA where kB = Boltzmann constant and NA = Avogadro’s number

This equation tells about the behaviour of gas in a particular situation. If a gas satisfies this equation then the gas is known as Perfect gas or an ideal gas.

Ideal gas: A gas that satisfies the perfect gas equation exactly at all pressures and temperatures.  Ideal gas is a theoretical concept.

• No real gas is truly ideal. A gas which is ideal is known as real gas.
• Real gases approach the ideal gas behaviour for low pressures and high temperatures

Dalton’s Law of partial pressures

Dalton’s law of partial pressure states that the total pressure of a mixture of ideal gases is the sum of partial pressures.

Consider if there are several ideal gases mixed together in a vessel, then the total pressure of that vessel is equal to the sum of partial pressure.

Partial pressure is the pressure exerted by a particular gas if only that gas is present in the vessel.

For example: - Consider if in a vessel there is a mixture of 3 gases, A,B and C.So the partial pressure of A is equal to pressure exerted only by A and considering B and C are not present.

Similarly partial pressure of B is equal to the pressure exerted only by B and considering A and C are not there.  And Similarly for C.

According to Dalton’s law the total pressure of mixture is sum of partial pressure of A, partial pressure of B and partial pressure of C

Therefore P= P1+P2+---total pressure due to the mixture of gases is equal to the sum of the partial pressure of the gas.

Kinetic Theory of an Ideal Gas

Basis and assumptions of Kinetic Theory: -

1. Molecules of gas are in incessant random motion, colliding against one another and with the walls of the container.
1. All collisions are elastic. And total Kinetic energy and momentum are conserved.  In case of an elastic collision total Kinetic energy and momentum before collision is equal to the total Kinetic energy and momentum after collision.
2. The density and the distribution of the molecules is uniform throughout the gas.
3. Between two collisions a molecule moves in a straight path with uniform velocity. But when they come closer they experience the intermolecular forces and as a result their velocities change.
4. There are no intermolecular forces between the molecules of gas except during collisions.
5. There will be no force ,between the molecules. As a result molecules are moving freely as per Newton's first law of motion.
6. At ordinary temperature and pressure the molecular size is very small as compared to intermolecular distance between them.

In the above pictures we can see that molecules moving randomly first and then molecules colliding with each other and change their direct.

### The Pressure of an Ideal Gas Based on Kinetic Theory

• Consider a container in the shape of a cube that is filled with an ideal gas. Only one molecule will be considered; the molecule collides with the container's walls and bounces back.
• Let the molecule's velocity while moving be (vx, vy, vz).
• The velocity of the molecule as it bounces back will be (-vx, vy, vz).
• The change in momentum = Pf – Pi where Pf = final momentum and Pi = initial momentum)
•   Pf-Pi= =mvx-mvx= -2 mvx
• The wall receives this change in momentum as a result of the contact.
• One molecule's momentum delivered to the wall in a collision= 2mvx

However, because there are so many molecules, we must calculate the overall momentum transferred to the wall by them all.

To figure out how many molecules hit the wall, do the following:

• The area of the wall will be ‘A’. Therefore, in time Δt within a distance of  AvxΔt  all the molecules can hit the wall.
• If n be the number of molecules per unit volume and on average, half of the molecules will hit the wall and half of them will move away from the wall. Therefore,  will hit the wall.
• The total momentum will be:.
• The force exerted on the wall is equal to the rate of change of momentum which will be equal to
• The Pressure on the wall is equal to  This is true for molecules having velocity vx

#### Note:

• The velocity of all the molecules in the gas will not be the same. The velocities of each will be different.
• As a result, the following equation is valid for pressure due to a group of molecules moving at vx in the x-direction, where n is the number density of that group of molecules.

As a result, the total pressure owing to all such groups may be calculated by adding the contributions due to each molecule.

• Because the gas is isotropic, the molecules travel at random, meaning that their velocity can be in any direction.
• Therefore, the pressure is equal to   where v2 is the average square speed.

### Kinetic Interpretation of Temperature

A molecule's average kinetic energy is proportional to the absolute temperature of the gas. It is unaffected by the ideal gas's pressure, volume, or nature.

For the equation:  , by multiplying both sides by V we will get

Also  nV= N ( total number of molecules)

After simplifying the above equation

Here

‘N’ is the number of molecules in a sample.

EN=12mv2=32kBT ......equation (3)

Therefore, the above equation depicts the average kinetic energy.

So, kinetic energy is directly proportional to the temperature. So, temperature can be identified as a molecular quantity.

### Kinetic Theory: Consistent With Ideal Gas Equation and Gas Laws

1. It is consistent with the ideal gas equation:

For the kinetic gas equation:

For an ideal gas, its internal energy is directly proportional to the temperature. This depicts that internal energy of an ideal gas is only dependent on its temperature, not on pressure or volume.

2. When Kinetic theory is consistent with Dalton’s Law of partial pressure:

The equation for Kinetic theory

If the mixture of gases is present in the vessel then

The average Kinetic energy of the molecules of different gases at equilibrium will be equal

Then the total pressure P is given by

So we have Total pressure P of mixture of gases as equal to the sum of partial pressure of individual gasesP= P1 + P2+ P3............ This is known as Dalton’s law of partial pressure.

## 2. Law of equipartition of energy

Law of Equipartition of energy: Degrees of Freedom

Degrees of Freedom can be defined as independent displacements or rotations that specify the orientation of a body or system. 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 to. If constrained to move along a line, it needs just one coordinate to locate it.

For example:-Consider a room and if we tie a thick rope from one wall to another.

Take a ball which is moving straight on the rope. The ball has only 1 degree of freedom. It can move only in one particular dimension. Consider if the ball is on the floor which is two-dimensional, then the ball can move along 2 directions. The ball has 2 degrees of freedom.

Consider if we throw the ball in space which is 3 dimensional. Then the ball can move in 3 dimensions. Therefore degree of freedom tells us in how many ways a body can move or rotate or vibrate.

Categories of Degrees of Freedom

1. Translational degrees of freedom.

2. Rotational degrees of freedom.

3. Vibrational degrees of freedom.

Translational degree of freedom:-

Translation means motion of the body as a whole from one point to another.

For example:

• Consider the oxygen molecule; it has 2 oxygen atoms which are bonded together. The 2 oxygen atoms along with the bond are considered as the whole body. When the body as a whole is moving from one point to another is known as translational.
• Consider a molecule which is free to move in space and so it will need 3 coordinates(x, y, and z) to specify its location. Therefore it has 3 degrees of freedom.
• Similarly a molecule which is free to move in a plane which is 2 dimensional and so it needs 2 coordinates to specify its location. Therefore it has 2 degrees of freedom.
• Similarly a molecule which is free to move in line needs 1 coordinate to specify its location. Therefore it has 1 degree of freedom.
• Molecules of monatomic gas have only translational degrees of freedom. This means gases which have only one atom. For example:-Helium atom it consists of only one He atom. It will have translational degrees of freedom.
• Each translational degree of freedom contributes a term that contains a square of some variable of motion.

The variable of motion means the velocity (vx, vy,vz )

The term   will contribute to energy. This is Kinetic energy which is involved with the motion of the molecule from one point to another.

In thermal equilibrium, the average of each such term is

Rotational Degree of freedom

Independent rotations that specify the orientation of a body or system. There is rotation of one part of the body with respect to the other part.

Rotational degree of freedom happens only in diatomic gas. Diatomic molecules have rotational degrees of freedom in addition to translational degrees of freedom.

It is possible in diatomic molecules as 2 atoms are connected together by a bond. So the rotation of one atom w.r.t. to another atom.

For example: - Two oxygen atoms joined together by a bond. There are two perpendicular axes. There are 2 rotations possible along the two axes. They have 3 translational degrees of freedom and also 2 rotational degrees of rotation.

Therefore Rotational degree of freedom contributes a term to the energy that contains the square of a rotational variable of motion.

Rotational variable of motion comes from angular momentum ω.

Linear velocity is (vx,vy,vz) . Whereas angular velocity is (ωx, ωy, ωz).

These are 3 rotational degrees of freedom along the 2 perpendicular axes.

The total energy contribution due to the degrees of freedom for oxygen molecules.  There will be 3 translational degree of freedom and two rotational degree of freedom

Kinetic energy contribution from translational motion is given by

Kinetic energy contribution from rotational motion is given by

Vibrational degree of freedom

Some molecules have a mode of vibration, i.e. atoms oscillate along the inter-atomic axis like a one-dimensional oscillator. This vibration is observed in some molecules.  For example - CO atoms oscillate along the interatomic axis like a one-dimensional oscillator.

Consider two 2 atoms vibrating along the inter-atomic axis. The vibrational energy terms contain squares of vibrational variables of motion.

Total vibrational energy term

Where, = Kinetic energy and   = Potential energy and k = force constant one-dimensional oscillator.

The vibrational degree of freedom contributes 2 terms.

(1) In the first figure rotational motion along two axes perpendicular to line joining two particles (here y and z directions) is shown.

(2) In the second figure vibrational motion along line joining the two atoms is shown

Comparison between 3 energy modes

Law of Equipartition of energy

According to this law, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having average energy equal to  .

1. Each translational degree of freedom contributes
2. Each rotational degree of freedom contributes
3. Each vibrational degree of freedom contributes

Specific Heat Capacity for monoatomic gases Monoatomic gases will only have a translational degree of freedom. Maximum they can have is three translational degrees of freedom. Each degree of freedom will contribute  . Therefore 3 degrees of freedom will contribute

• By using law of equipartition of energy, the total internal energy of 1 mole of gas  .
• Specific heat capacity at constant volume  .
• For an ideal gas CP - CV=R , By using above equation
• Ratio of specific heats γ=CP/CV=(5/3).

Specific Heat of Diatomic gases (rigid)

A rigid diatomic gas means they will have translational as well as rotational degrees of freedom but not vibrational. They are rigid oscillators.

A rigid diatomic molecule will have 3 translational degrees of freedom and 2 rotational degrees of freedom. Total 5 degrees of freedom.

• By law of equipartition of energy, each degree of freedom will contribute .
• Therefore 5 degree of freedom will contribute Therefore the total internal energy of 1 mole of gas
• Specific heat capacity at constant volume .
• Specific heat capacity at constant pressure of a rigid diatomic is given as .
• Ratio of specific heats γ=Cp/Cv= (7/5).

Specific Heat of Diatomic gases (non-rigid)

A non-rigid diatomic gas has translational, rotational as well as vibrational degrees of freedom.

There will be 3 translational degrees of freedom and 2 rotational degrees of freedom and 1 vibrational degree of freedom.

• Total contribution by translational =  , rotational  and vibrational =kBT.
• Total contribution from 6 degree of freedom =
• Total Internal energy for 1 mole Specific heat at constant volume Cv=dU/dT = (7/2) R.

Specific heat at constant pressure Cp=Cv+R= (9/2) R.

Ratio of specific heat γ= Cp/Cv = (9/7)

There are two independent axes of rotation (1) and (2) normal to the axis joining the two oxygen molecule. It has 3 translational and 2 rotational degrees of freedom

Specific Heat Capacity for polyatomic gases

Polyatomic gases will have 3 translational degree of freedom, 3 rotational degrees of freedom and ‘f’ number of vibrational modes.

• Total internal energy of 1 mole of gas = ((3/2) + (3/2) +f) RT = (3 + f) RT.
• Specific heat at constant volume Cv=dU/dT = (3 + f) R
• Specific heat at constant pressure Cp=CV+R=(4 + f) R
• Ratio of specific heat γ= Cp/CV = (4 + f)/(3 + f)

Specific Heat Capacity for solids

Consider there are N atoms in a solid. Each atom can oscillate about its mean position. Therefore vibrational degree of freedom = kBT

• In one-dimensional average energy=kBT, in three-dimensional average energy =3kBT
• Therefore total internal energy (U) of 1 mole of solid = 3KBTxN= 3RT
• At constant pressure, ΔQ = ΔU + PΔV change in volume is very less in solids .Therefore ΔV = 0. So we have finally ΔQ = ΔU for solids.
• Specific heat at constant volume
• Specific heat at constant pressure  as ΔQ = ΔU, Therefore CV=dU/dT=3R
• Therefore CP = CV = 3R

Specific Heat Capacity of water

Consider water as solid, so it will have ‘N’ number of atoms. Therefore for each atom average energy =3kBT

• Number of molecules in H2O= 3 atoms.
• Total internal energy  U=3kBT×3×NA=9RT
• CV = CP = 9R.

Conclusion on Specific heat

• According to classical mechanics, the specific heat which is calculated based on the degree of freedom should be independent of temperature.
• However  T→0,degree of freedom becomes inefficient.
• This shows classical mechanics is not enough; as a result quantum mechanics came into play.
• According to quantum mechanics minimum non-zero energy is required for a degree of freedom to come into play.
• Specific heats of all substances approach zero as T→0.

Mean free path

Mean free path is the average distance between the two successive collisions.

Inside the gas there are several molecules which are randomly moving and colliding with each other. The distance which a particular gas molecule travels without colliding is known as the mean free path.

Expression for mean free path

Consider each molecule of gas is a sphere of diameter (d).The average speed of each molecule is <v>.

Suppose the molecule suffers collision with any other molecule within the distance (d). Any molecule which comes within the distance range of its diameter will have a collision with that molecule.

The volume within which a molecule suffers collision =<v>Δtπd2.

Let number of molecules per unit volume =n

Therefore the total number of collisions in time Δt =<v>Δtπd2(n)

Rate of collision  = <v>πd2n.

Suppose time between collision  T= 1/(<v>d2)

Average distance between collision = T<v>= 1/(πd2n).

1/(πd2n), this value was modified and a factor was introduced.

Mean free path (l) = .

Conclusion: - Mean free path depends inversely on:

1. Number density (number of molecules per unit volume)
2. Size of the molecule.

The volume swept by a molecule in time Δtin which any molecule will collide with it.