# Kinetic Theory Of Gases

The behavior of the gases as stated by various gas laws can be well explained with the help of kinetic
theory of gases. It is explained basing on the molecular nature of matter and the kinetic (faster movement)
movement of gas molecules. Kinetic equation derived with the help of kinetic theory of gases gives a clear
explanation of various gas laws.

Assumptions made from kinetic theory of gases:

As per the kinetic theory of gases, an ideal gas exhibits the below listed properties.

- Gases are made up of a huge number of minute particles called molecules. Each of these molecules is separated from each other by a large distance. Thus, the actual volume occupied by each of these molecules is very less when compared to the total volume of the gas.
- All the molecules of a gas are in constant motion. They move in all directions in straight lines and collide with each other as well as the wall of the container.
- Due to the elastic nature of the gas molecules, there is no loss of energy when the molecules collide. However, energy is transferred in between the colliding molecules at the time of collision.
- Collision of the gas molecules with the walls of the container exerts pressure on the gas molecules. Greater the number of collisions of the gas molecules with the container greater will be the pressure.
- As the distance between the gas molecules is very high, attractive force between the gas molecules and between the gas molecules and the container is very less. Thus, the molecules move freely.
- The kinetic energy of the gas molecules is different and move at different speeds. However, the average kinetic energy of all the gas molecules is directly proportional to the absolute temperature. Increase in temperature increases the kinetic energy of the molecules and hence, they move at greater speed.

Explanation of the above assumptions:

**Assumption 1:**

When we compress a gas such as oxygen, nitrogen or hydrogen at normal temperature and pressure
(N.T.P), then the volume occupied by the molecules is just 0.014% of the total gas volume and the
rest of it is only an empty space.

**Assumption 2:**

We observe that in bright light passing through a narrow beam a large number of dust particles
moving in random motion. This is because of the collision between the gas molecules in the air
and the dust particles. During this collision, the gas molecules transfer some of their kinetic
energy to the dust particles which in turn move in zigzag motion termed the Brownian movement.

**Assumption 3:**

As we know that the gas molecules are in constant motion. This shows that the collisions
between the gas molecules are elastic in nature resulting in the transfer of energy. If
this is not the gas, every collision between the gas molecules results in loss of energy.
Finally, after certain number of collisions, all the gas molecules come to rest. But, this
is not happening. Hence, all the collisions between the gas molecules are elastic in nature
and there is no loss of energy.

**Assumption 4:**

When the gas molecules are held in a closed container, the rapidly moving gas molecules
collide with the walls of the container. This exerts some pressure on the gas molecules
resulting in the build of pressure within the container. This pressure is termed as
gaseous pressure.

**Assumption 5:**

When you open the lid of a perfume bottle, you can sense the pleasant smell within seconds
of opening of the lid. This is due to rapid spreading of the gas molecules. This shows that
the gas molecules are held freely without any attractive forces between them.

**Assumption 6:**

Whenever a gas is heated it evaporates very quickly. This is due to increase in the kinetic energy
of the gas molecules. Thus, the rate at which the gas molecules are moving is determined by the
temperature.

**Kinetic Gas Equation:**

Now, let us derive a kinetic gas equation basing on the above postulates.

Consider about N molecules of gas each having mass m are enclosed in a container with sides
measuring L cm. From the above postulates we know that gas molecules within a container move
at different speeds. However, the speed of a particular molecule at any moment of time can be
resolved into three components along the three axes of a vessel at right angles to each other.

Let us see how to do this for a single gas molecule having a velocity v. The three velocity
components of the molecule along the three axes X, Y and Z are given as v

_{x}, v_{y}, and v_{z}. The sum of these velocity components is made equivalent to the velocity v asV

^{2}= v

_{x}2 +, v

_{y}2, + v

_{z}2

Now, consider a molecule moving along with x-axis with a velocity of v

_{x}.

The momentum of the molecule before striking the face A is given as mv

_{x}.

Whenever a molecule strikes the walls of a container, it will bounce with same speed but in opposite direction i.e. - v

_{x}Thus, the momentum after striking the surface is given as - mv

_{x}

The change in momentum before and after striking is given as mv

_{x}- (-mv

_{x}) = 2mv

_{x}

For the molecule to strike the same face again it has to go to the opposite face and come back. Thus,
the molecule has to travel a distance of 2Lcm. (as the length of each side is Lcm).

The time required for colliding on the same face again is given as

Distance 2L

---------- = ----- sec.

Velocity v

---------- = ----- sec.

Velocity v

_{x}Number of collisions made by the molecule on the face A per sec is given as

V

------ per sec.

2L

_{x}------ per sec.

2L

As seen above the change in momentum for one collision is given as 2mv

_{x}

v

For ----- collisions the change in momentum is given as --------------- = ----------

2L 2L L

_{x}2mv_{x}.x v_{x}m v_{x}^{2}For ----- collisions the change in momentum is given as --------------- = ----------

2L 2L L

Likewise, the change in momentum per second due to collision on other surface is given as

m v

= ------

L

_{x}^{2}= ------

L

Now, the total change in momentum per second due to collision of one molecule on two faces along the x-axis is given as

m v

---------- = -----

L L

_{x}^{2}m v_{x}^{2}---------- = -----

L L

2m v

= ------

L

_{x}^{2}= ------

L

Similarly, the change in momentum due to collisions of the molecule on the surfaces y and z is given as

2 m v

= ---------- and ----- respectively.

L L

_{y}^{2}2 m v_{z}^{2}= ---------- and ----- respectively.

L L

Thus, the total change in momentum per second on all the six faces along the three axes is given as

2m v

------- + ------- + -------

L L L

_{x}^{2}m v_{y}^{2}2 m v_{z}^{2}------- + ------- + -------

L L L

2m

= --------(v

L

= --------(v

_{X}^{2}+ v_{Y}^{2}+ v_{z}^{2})L

2mv

= --------as ((v

L

^{2}= --------as ((v

_{X}^{2}+ v_{ Y}^{2}+ v_{ z}^{2}= v^{2})L

As there are N molecules of a gas each moving with a different velocity, the velocity of each molecule contributes to change in momentum per second on all the six faces along the three axes.

Thus, the total change in momentum due to all molecules on all the six faces along the three axes is given as

2m

= --------(v

L

= --------(v

_{1}^{2}+ v_{ 2}^{2}+ v_{ 3}^{2}+-------------+ v_{ n}^{2})L

By multiplying the numerator and denominator with N, the total number of molecules gives

2mN

= --------((v

L

= --------((v

_{1}^{2}+ v_{ 2}^{2}+ v_{ 3}^{2}+-------------+ v_{ n}^{2}) / N )L

But, ((v

_{1}

^{2}+ v

_{ 2}

^{2}+ v

_{ 3}

^{2}+-------------+ v

_{ n}

^{2}) / N is equal to u

^{2}which is root mean square velocity.

Thus, the total change of momentum per second due to N molecules is given as

2m Nu

------

L

^{2}------

L

The change momentum per second is termed as force F.

Hence,

2m Nu

F = ------

L

^{2}F = ------

L

Total force

Now, pressure P is given as force per unit area. = --------------

Total area

Now, pressure P is given as force per unit area. = --------------

Total area

2m Nu

= ----------

L x area

^{2}= ----------

L x area

The area of six faces is given as 6L

^{2}

2m Nu

Thus P is given as --------------- = ----------

L x 6L

^{2}2m Nu^{2}Thus P is given as --------------- = ----------

L x 6L

^{2}L^{3}
We know that L

^{3}= volume, V.
mNu

Thus P = ------- or --- mNu

3L 3

^{2}1Thus P = ------- or --- mNu

^{2}which is the kinetic gas equation.3L 3

^{}
Though the above equation Is derived for a cubic vessel it is applicable for a vessel of any shape as the
total volume is considered to be made up of a number of small cubes.

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