Kinetic Theory of Gases - PowerPoint Presentation

Slide1

Kinetic Theory of Gases

Dr. N. K. Shukla

Associate Professor

Department of Chemistry

Mahatma Gandhi P.G. College

Gorakhpur-273001Slide2

Part I

Connecting the microscopic with macroscopicSlide3

Macroscopic properties

Gas is made of atoms or molecules that occupy a much larger space than the size of the constituent particles.

The usual way to describe such a system is by measuring the pressure, volume and temperature. In addition, the mass, or the number of molecules is an indicator of the amount of matter in the container. Gas laws define the relationship between pressure, volume, temperature and the number of atoms/molecules.

These properties are called ‘macroscopic’, as they represent the overall state of the system, the word ‘macro’ meaning ‘large-scale.’Slide4

Atomistic Properties

Now ‘microscopic’, or more correctly, ‘atomistic’ properties of a gas have more to do with the atoms or molecules of the gas.

These include the position and distribution of the gas molecules in a given system (i.e. container) and the velocities with which these particles move around in the container.

Kinetic theory of gases attempts to establish a relationship between the macroscopic properties and atomistic properties. That way, we could predict macroscopic properties if we knew atomistic properties, or vice-versa.

To establish such a relation, however, we would need some basic assumptions.Slide5

Part II

Assumptions of kinetic theory of gasesSlide6

Here are some assumptions which help us derive the relation between the aforementioned properties:

A gas is made up of large number of particles which are constantly involved in random motion.

The particles do not interact with each other, except for collisions.

The particles undergo

perfectly elastic collisions

with each other and the walls of the container. This is to say that the particles do not lose any kinetic energy or momentum while colliding.

The time period of collision is negligible as compared to the time between collisions.

The size of the particles themselves is negligible as compared to the distance between such particles. That is to say that the atoms or molecules themselves occupy no volume.

There are no external forces on the gases (gravity, for example) so that density remains constant.Slide7

In addition, here are some more assumptions about the system under study:

We assume that the gas is in a cubical container with edge length

L

Each particle is defined by two variables; position and velocity

v

The position and velocity or a particle is expressed in accordance with the three dimensional co-ordinate system, as shown. The three axes are x-axis, y-axis and z-axis.

It is also assumed, for this derivation, that the gas in the container is of one type. That is to say that it contains atoms or molecules of the same type (same mass)Slide8

Part III

Relation between pressure and velocitySlide9

Let us assume a gas particle (atom or molecule) with velocity

v

moving in a certain direction. We consider the x-component in the velocity for now, denoted by

The particle moves until it collides with the wall of the container, and bounces back with the same velocity (elastic collision). Now, we wish to know the pressure exerted by the molecule on such a collision. Since pressure is nothing but force per unit area, we will try to formulate a relation for force exerted by a particle on collision.

 Slide10

Force is change in momentum by change in time. This equation is nothing but Newton’s second law. Since momentum is mv

, and the mass does not change, this equation becomes

Now, the change is velocity when the particle collides and bounces back will be 2 . Because it collides at a velocity and goes back with - . So the change is technically -2 . We ignore the negative sign here, since were are only concerned with the magnitude of the force exerted.

 Slide11

Now, the in our equation means the time between two collisions, since the time of the collision is negligible. The time taken would be a back and forth motion of the particle, when the particle collides on a wall, goes back and collides on the wall opposite to it, and then returns for another collision on the first wall. Distance travelled by the particle in this case would be

2L

, or twice the length of the container in x-direction. And we already know the velocity of the particle. Therefore,

 Slide12

Substituting this value in the equation of force, we get

This denotes the force exerted by one particle. For

N

number of particles, the force would be

And hence, the average force exerted by one particle would be

 Slide13

Solving further, we get

Note that is the average of squares, not the square of average.

To get pressure, we divide the above equation with the area of the wall, which is

This gives us

 Slide14

Relation between pressure and velocity

Now, the total velocity of particles is

In addition, the values of velocities in each direction would be equal. Therefore,

Our equation for pressure becomes

Dividing both sides by half, we get , where the portion of

RHS in brackets is the total kinetic energy of particles.

 Slide15

Total Internal Energy

This is an important result. Now we know that PV equals

NkT

(ideal gas law)

This makes the equation , where is

Botzmann’s

constant.

It shows that the kinetic energy of gas molecules depends directly upon the temperature of the gas.

This kinetic energy is also known as the Total Internal Energy for the gas, which is denoted by

U

.

or

 Slide16

REFERENCESPrinciples of Physical

Chemistry –

P.W. Atkins

Advance Physical Chemistry-

Gurtu

&

Snehi

Advance Physical Chemistry- D.N.

Bajpai

Modern Physical Chemistry- R.P.

Rastogi

, K. Singh, K.

Kishore

, V.K.

Srivastava

& M.L.

Srivastava

Essentials of Physical Chemistry-

Arun

Bahl, B.S. Bahl & G.D. Tuli