# Molecular kinetic energy theory of gases

## Kinetic molecular theory of gases-Average kinetic energy

The kinetic energy of gases first time developed by Bernoulli in 1738 to explain the molecular properties of gases on the basis of mechanical theory. These ideas were at the root of the theory to explain the kinetic behavior of gases.

Thus in the nineteenth century, the effort of Joule, Kronig, Clausius, Boltzmann, and Maxwell, gives the formula of the kinetic theory of gases.

In solid the molecules or particles are held very closely together and are entirely devoid of any translatory motion.

If a specific heat supplied to solid, it takes the form of vibrational motion with the rise of temperature.

But with the further increases the thermal energy, the vibrational motion rises to such extent the molecules break down to transform into a liquid.

### Kinetic molecular theory of gases postulates

Fortunately, such theory has been developed for the formula of the kinetic equation for gases based upon certain postulates.

1. Gas molecules are composed of very small discrete particles. In any one gas, all the molecules are of the same size and mass, but these differ gas to gas.
2. Within the container, the gas molecules are moving in all directions with a verity of speeds. Some are very fast while others are slow.
Thus due to random motion, the gas molecules are executing two types of collisions. When it collided with the walls of the container called wall collision but with themselves called an intermolecular collision.
3. These collisions are perfectly elastic. Thus there occurs no loss of kinetic energy or momentum of the molecules by this collision.
4. Gas molecules are assumed to be point masses. Hence the size of the gas molecules is very small in comparison to the distance where they travel.
5. Especially at low pressure, the gas molecules have no intermolecular attraction. Thus one molecule can exert pressure independent of the influence of other molecules.
6. The pressure exerted by a gas is due to the uniform wall collisions. Hence higher the frequency of the wall collision greater will be the pressure of the gas.
7. This explains Boyle’s law since when the volume is reduced at a constant temperature, wall collision becomes more frequent and pressure is increased.
8. Through the molecular velocity constantly changing due to the intermolecular collision but the average kinetic energy of the gas molecules remains fixed at a given temperature.

### Kinetic theory of gases formula derivation

Let us take a cubic gas container with edge length l containing N molecules of gas of mass m and RMS speed CRMS at temperature T and pressure P. Thus the molecules are moving constantly with different velocities in different directions bombarding on the walls of the cube.

Among these molecules, N1 has velocity C1, N2 has velocity C2, N3 has velocity C3, and so on.

If we concentrate our discussion on a single molecule among N1 that has resultant velocity C1 and the component velocities are Cx, Cy, Cz.

∴ C12 = Cx2 + Cy2 + Cz2

The above picture shows that the molecule will collide walls A and B of the container with the component velocity Cx and other opposite faces by Cy and Cz.

Thus the change of momentum along X-direction for a single collision,

= m Cx – (- m Cx) = 2 m Cx

Hence rate change of momentum

$=2mC_{X}\times&space;\frac{C_{X}}{l}$

$=\frac{2mC_{X}^{2}}{l}$

Similarly, along Y and Z directions, the rate of change of momentum for the molecule 2 mCy2/l and 2 mCz2/l respectively.

Hence the net change of momentum imparted on all six walls of the container by this collision of gases

$=\frac{2mC_{x}^{2}}{l}+\frac{2mC_{y}^{2}}{l}+\frac{2mC_{z}^{2}}{l}$

$=\frac{2m}{l}\left&space;(&space;C_{x}^{2}+C_{y}^{2}+C_{z}^{2}&space;\right&space;)$

${\color{DarkBlue}&space;=\frac{2mC_{1}^{2}}{l}}$

Thus for N1 molecules, change of momentum

$=\frac{2mN_{1}C_{1}^{2}}{l}$

If we consider all the molecules of the gas present in this cubic container than the total rate of change of momentum,

$=\frac{2mN_{1}C_{1}^{2}}{l}+\frac{2mN_{2}C_{2}^{2}}{l}+\frac{2mN_{3}C_{3}^{2}}{l}+...$

$=\frac{2mN}{l}\times&space;\frac{N_{1}C_{1}^{2}+N_{2}C_{2}^{2}+N_{3}C_{3}^{2}...}{N}$

$=&space;\frac{2mNC_{RMS}^{2}}{l}$

where CRMS = root means square velocity

But according to Newton’s second law of motion, the rate of change of momentum due to wall collision equal to force developed within the gases.

$P\times&space;6l^{2}=\frac{2mNC_{RMS}^{2}}{l}$

$or,&space;P\times&space;l^{3}=\frac{1}{3}mNC_{RMS}^{2}$

${\color{DarkBlue}&space;\therefore&space;PV=\frac{1}{3}mNC_{RMS}^{2}}$

where l3 = V = volume of the container.

This gas formula derived from the kinetic theory of gases and this equation valid for any shape of the container.

#### The density of the gases from the kinetic equation

$P&space;=&space;\frac{1}{3}\times&space;\frac{mN}{V}\times&space;C_{RMS}^{2}$

${\color{DarkBlue}&space;\therefore&space;P=\frac{1}{3}dC_{RMS}^{2}}$

where (mN/V) is the density(d) of the gases.

Problem
Calculate the pressure exerted by 1023 gas molecules each of the mass 10-22 gm in a container of volume 1 dm3. Given RMS speed of the gas 105 cm sec-1.

Solution

Numer of gas molecules (N) = 1023,
mass (m) = 10-22 gm = 10-25 Kg,
volume (V) = 1 dm3 = 10-3 m3
and CRMS = 105 cm sec-1 = 103 m sec⁻¹.

The formula for the kinetic theory of  gases,

$PV=\frac{1}{3}mNC_{RMS}^{2}$

$or,P=\frac{mNC_{RMS}^{2}}{3V}$

$\therefore&space;P=\frac{10^{-25}\times&space;10^{23}\times&space;10^{-6}}{3\times&space;10^{3}}$

∴ P = 0.333 × 107 Pascal

#### Kinetic theory to explain Charles law

When temperature raised, the molecules would move more vigorously resulting in a larger number of impacts on the wall of the container at constant volume.

Thus it increases the pressure with the rising temperature at constant volume. This is Charles’s law for gases.

### Definition of RMS velocity

RMS or root mean square speed is defined as the square root of the average of the squares of speeds.

$C_{RMS}=\sqrt\frac{{N_{1}C_{1}^{2}+N_{2}C_{2}^{2}+N_{3}C_{3}^{2}+...}}{N}$

#### Kinetic theory of gases RMS velocity

The kinetic theory and ideal gas law may be used to formulate the RMS velocity of the gases.

The kinetic formula for a 1-mole ideal gases

$PV=\frac{1}{3}mNC_{RMS}^{2}$

where mN = mN₀ = M.

From the ideal gas equation, PV = RT

$RT=\frac{1}{3}MC_{RMS}^{2}$

$or,C_{RMS}^{2}=3RT$

${\color{DarkBlue}&space;\therefore&space;C_{RMS}=\sqrt{\frac{3RT}{M}}}$

From the above equation, RMS velocity depends on the molar mass and temperature of the gases.

Problem
Calculate the root mean square velocity of oxygen molecules at 27⁰C?

Solution

RMS velocity of a gas molecule

$C_{RMS}=\sqrt{\frac{3RT}{M}}$

For oxygen, M = 32 gm mol-1

T = 27° C = (273+27)K = 300 K

$\therefore&space;C_{RMS}=\sqrt{\frac{3\times&space;8.314\times&space;10^{7}\times&space;300}{32}}$

or, CRMS = 48356 cm sec-1

#### The ratio of RMS velocity of hydrogen to oxygen

At a given temperature ratio of the RMS velocity of hydrogen to oxygen

$\frac{C_{RMS_{H_{2}}}}{C_{RMS_{O_{2}}}}=\sqrt{\frac{M_{O_{2}}}{M_{H_{2}}}}=\sqrt{\frac{32}{2}}=4$

∴ RMS velocity of hydrogen = 4 × RMS velocity of oxygen.

### The average kinetic energy of gases

The average kinetic energy defined as,

$\bar{E}=\frac{1}{2}mC_{RMS}^{2}$

Agin from the kinetic gas equation

$PV=\frac{1}{3}mNC_{RMS}^{2}=\frac{2}{3}N\times&space;\frac{1}{2}mC_{RMS}^{2}$

$\therefore&space;PV=\frac{2}{3}N\bar{E}$

For an ideal gas, PV = RT and N = N0

$\therefore&space;RT=\frac{2}{3}N_{0}\bar{E}$

${\color{DarkBlue}&space;or,&space;\bar{E}=\frac{3}{2}\times&space;\frac{R}{N_{0}}\times&space;T=\frac{3}{2}KT}$where k = R/N0 = Boltzmann constant
= 1.38 × 10-23 J K-1

Thus the average kinetic energy dependent on temperature only but independent of the nature of the gas.

#### The kinetic energy of gases formula

The total kinetic energy = Avogadro number × average kinetic energy.

$\therefore&space;E_{Total}=N_{0}\times&space;\bar{E}=N_{0}\times&space;\frac{3}{2}\times&space;\frac{RT}{N_{0}}$${\color{DarkBlue}&space;or,&space;E_{Total}=\frac{3}{2}RT}$

Problem
What is the average kinetic energy per molecule of methane gas at a temperature of 27⁰C?

Solution
Kinetic energy for 1 mole of the ideal gas molecule,

${\color{DarkBlue}&space;\therefore&space;E_{Total}=\frac{3}{2}RT=\frac{3}{2}\times&space;2\times&space;300}$

= 900 cal mol-1

w gm of ideal gas = (w/M) mol
= w/M mol

Thus the kinetic energy of w gm of an ideal gas molecule at 27⁰C

= (w/M) × 900 calories
= 900w/M calories.

Problem
How to calculate the RMS speed of the ammonia at N.T.P?

Solution

At N.T.P,
V = 22.4 dm3 mol-1 = 22.4 × 10-3 m3 mol-1
P = 1 atm = 101325 Pa
M = 17 × 10-3 Kg mol-1

$C_{RMS}=\sqrt{\frac{3RT}{M}}$

$=\sqrt{\frac{3\times&space;101325\times&space;22.4\times&space;10^{-3}}{17\times&space;10^{-3}}}$

= 632 m sec-1