Kinetic theory of gases derivation

What are the principles of the kinetic molecular theory?

The common behavior of different gases suggested that the internal structure in all gases must be similar. Study in the seventeenth-century explained Gassendhi and Hooke explained the physical properties of gases on the assumption of the existence of rapidly moving independent minute particles. Bernoulli in 1738 the first time explained the properties of gases on a mechanical basis.

In solid the molecules or particles are held very closely together and are entirely devoid of any translatory motion. If heat is supplied to solid, it takes the form of vibrational motion with the rise of temperature. With the further increases the thermal energy, the vibrational motion rises to such extent the molecules break down to transform into a liquid.

Further absorption of heat causes the particles to brack away from the restraining forces holding together and the particles move away from the liquid to the gaseous state. The gases then are essentially composed freely moving particles.

These basic ideas were at the root of the theory to explain the behavior of gases called the kinetic theory of gases. In the nineteenth century, the effort of Joule, Kronig, Clausius, Boltzmann, and Maxwell, the theory succeeded in attaining a rigid mathematical form.

Fortunately, such theory has been developed for the formulation of the kinetic gas equation based upon certain postulates which are supposed to be applicable to an ideal gas.

Postulates of the kinetic theory of gases in chemistry

  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. Gas molecules within the container are moving in all directions with a verity of speeds. Some are very fast while others are slow.
    Due to random motion, the gas molecules are executing collision with the walls of the container called wall collision and with themselves called an intermolecular collision.
  3. These collisions are perfectly elastic and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.
  4. Gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
  5. There exists no intermolecular attraction especially at low pressure, that is one molecule that can exert pressure independent of the influence of other molecules.
  6. The pressure exerted by a gas is due to the uniform wall collisions. 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, the average kinetic energy of the gas molecules remains fixed at a given temperature.

How does the kinetic theory 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. That is why we find an increase of pressure with a rise in temperature at constant volume. This is Charles’s law.

What is the RMS speed of gas molecules?

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}

The kinetic gas equation for an ideal gas

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.

The molecules are moving constantly with different velocities in different directions bombarding on the walls of the cube.

Gas molecules, N1 have velocity C1, N2 have velocity C2, N3 have velocity C3, and so on.

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

The component velocity of gas molecules in a cubic container

The component velocity of gas molecules

Kinetic molecular theory of gases
Component velocity

∴ C12 = Cx2 + Cy2 + Cz2

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

Change of momentum along X-direction for a single collision,

= m Cx – (- m Cx) = 2 m Cx.

Rate of change of momentum for this collision,

=2mC_{X}\times \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 momentum imparted on all six walls of the container by the collision of the gas molecules,

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

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

{\color{DarkBlue} =\frac{2mC_{1}^{2}}{l}}

For N1 molecules, change of momentum

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

Consider all the molecules of the gas present in the cubic container, 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 \frac{N_{1}C_{1}^{2}+N_{2}C_{2}^{2}+N_{3}C_{3}^{2}...}{N}

{\color{DarkBlue} \therefore \frac{2mNC_{RMS}^{2}}{l}}

where CRMS = root means square velocity of the gases.

According to Newton’s second law of motion, the rate of change of momentum due to wall collision is equal to force developed within the gas molecules.

P\times 6l^{2}=\frac{2mNC_{RMS}^{2}}{l}

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

{\color{DarkBlue} \therefore PV=\frac{1}{3}mNC_{RMS}^{2}}

where l3 = volume of the cubic container.

How do you determine the density of a gas?

P = \frac{1}{3}\times \frac{mN}{V}\times C_{RMS}^{2}

{\color{DarkBlue} \therefore P=\frac{1}{3}dC_{RMS}^{2}}

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

This equation is valid for any shape of the gas container.

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 molecule 105 cm sec-1.

Solution
Numer of gas molecules (N) = 1023

Mass of the gas molecule (m) = 10-22 gm

= 10-25 Kg

Volume (V) = 1 dm3 = 10-3 m3

CRMS = 105 cm sec-1 = 103 m sec⁻¹.

Kinetic gas equation,

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

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

\therefore P=\frac{10^{-25}\times 10^{23}\times 10^{-6}}{3\times 10^{3}}

∴ P = 0.333 × 107 Pascal

RMS velocity kinetic theory of gases

The kinetic relations and ideal gas law may be used to formulate the velocity of the gas molecules

Kinetic relation for a 1-mole ideal gas

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

where mN = mN₀ = M.

Ideal gas law for one-mole gas

PV = RT

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

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

{\color{DarkBlue} or,C_{RMS}^{2}=3RT}

{\color{DarkBlue} \therefore C_{RMS}=\sqrt{\frac{3RT}{M}}}

Thus root means square velocity depends on the molecular weight and temperature of the gas molecules.

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}}

M = 32 gm mol-1

T = 27° C = (273+27)K = 300 K

C_{RMS}=\sqrt{\frac{3\times 8.314\times 10^{7}\times 300}{32}}

∴ CRMS = 48356 cm sec-1

Why do hydrogen molecules move faster than oxygen?

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

Hence at a given temperature,

\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.

How do you find the kinetic energy of a gas?

The average kinetic energy defined as,

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

Kinetic gas equation

PV=\frac{1}{3}mNC_{RMS}^{2}=\frac{2}{3}N\times \frac{1}{2}mC_{RMS}^{2}

{\color{DarkBlue} \therefore PV=\frac{2}{3}N\bar{E}}

Ideal gas law for 1-mole gas

PV = RT and N = N₀.

\therefore RT=\frac{2}{3}N_{0}\bar{E}

{\color{DarkBlue} or, \bar{E}=\frac{3}{2}\times \frac{R}{N_{0}}\times T=\frac{3}{2}KT}k = R/N0 = Boltzmann constant of gas

= 1.38 × 10-23 J K-1

Total kinetic energy for 1 mole of the gas molecules

E_{Total}=N_{0}\times \bar{E}=N_{0}\times \frac{3}{2}\times \frac{RT}{N_{0}}{\color{DarkBlue} \therefore E_{Total}=\frac{3}{2}RT}

Average kinetic energy dependent on temperature only and average kinetic energy independent of the nature of the gas.

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

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

{\color{DarkBlue} \therefore E_{Total}=\frac{3}{2}RT=\frac{3}{2}\times 2\times 300}

= 900 cal mol-1

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

= w/M mol

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 molecules 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

{\color{DarkBlue} C_{RMS}=\sqrt{\frac{3RT}{M}}}

=\sqrt{\frac{3\times 101325\times 22.4\times 10^{-3}}{17\times 10^{-3}}}

= 632 m sec-1