Grade 11 Physics Note

Kinetic theory of gases

The postulates of kinetic theory of gases;

The kinetic theory of gases (also known as kinetic-molecular theory) is a law that explains the behavior of a hypothetical ideal gas. According to this theory, gases are made up of tiny particles in random, straight line motion. They move rapidly and continuously and make collisions with each other and the walls. This was the first theory to describe gas pressure in terms of collisions with the walls of the container. The main postulates of kinetic theory of gases are as follows:

1. Every gas consists of a large number of very small particles called molecules.

2. The molecules of a gas move haphazardly colliding among themselves and with the walls of the container and change their direction.

3. The pressure exerted by a gas in a container is due to the collisions of its molecules with the walls of the container. The collisions among the molecules are perfectly elastic.

4. The molecules of a gas are widely separated from one another and there are sufficient empty spaces among them.

5. The molecules of a gas have no forces of attraction for each other.

6. The actual volume of molecules of a gas is negligible as compared to the volume occupied by the gas itself.

7. The kinetic energy of the gas molecules varies directly as the absolute temperature of a gas.


Molecular interpretation of temperature:

From the kinetic theory of gases, the average kinetic energy per molecule of the gases

                                     $\frac{1}{2}{\rm{m}}{{\rm{c}}^{\bar 2}} = \frac{3}{2}{\rm{k\: T\: }}$

This shows that average kinetic energy of the gas is directly proportional to the absolute temperature.

 

Ideal gas:

The Boyle’s law and Charles’s law can be combined to obtain a general relationship between pressure, volume and temperature of a given mass of gas. This relationship is called ideal gas.

Let the pressure, volume and temperature of a certain mass of a gas be denoted by P, V and T respectively,

Then by Boyle’s law, we have;

${\rm{P}} \propto \frac{1}{{\rm{v}}}$…… i) 

According to Charles’s law we have

${\rm{v}} \propto {\rm{T}}$…. ii)

Further the volume of a gas is directly proportional to the number of moles according to Avogadro’s law which states that “equal volumes of all gases contain equal number of molecules under same condition of temperature and pressure.”

So,

${\rm{V}} \propto {\rm{n\: }}$….. iii)

From i) ii) and iii) we get

V = nRT /P….. iv)

Where R is proportionality constant and is defined as universal gas constant i.e. the constant used when 1gram mole of each gas is taken under consideration.

From equation iv)

We have PV= nRT Hence proved.

 

One mole of a gas;

A convenient standard quantity is the mole, the mass of gas in grams equal to the molecular mass in amu. Avogadro's number is the number of molecules in a mole of any molecular substance. NA=Avogadro's number=6.0221 x 1023/mole.

 

Boyle’s law on the basis of kinetic theory of gases;

Boyle’s law states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional i.e

${\rm{p}} \propto \frac{1}{{\rm{v}}}$

or PV = K.

V and PV vs P respectively

The equation states that product of pressure and volume is a constant for a given mass of confined gas as long as the temperature is constant. For comparing the same substance under two different sets of condition, the law can be usefully expressed as

${{\rm{P}}_1}{{\rm{V}}_1} = {{\rm{P}}_2}{{\rm{V}}_2}{\rm{\: }}$

The equation shows that, as volume increases, the pressure of the gas decreases in proportion. Similarly, as volume decreases, the pressure of the gas increases.

 

The significance of Boltzmann constant:

For a gas, Boltzmann constant (K) is defined as the ratio of the universal gas constant to the Avorgado’s number. That is

                                                    ${\rm{k}} = \frac{{\rm{R}}}{{{{\rm{N}}_{\rm{A}}}}}$

We have R = 8.31 J${\rm{mo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}{\rm{\: and\: }}{{\rm{N}}_{\rm{A}}} = 6.02{\rm{*}}{10^{23}}{\rm{mo}}{{\rm{l}}^{ - 1}}$

Therefore,

${\rm{k}} = \frac{{\rm{R}}}{{{{\rm{N}}_{\rm{A}}}}} = \frac{{8.31}}{{6.02{\rm{*}}{{10}^{23}}}}$

$ = 1.38{\rm{*}}{10^{ - 23}}{\rm{J}}{{\rm{K}}^{ - 1}}$

 

The expression for the pressure exerted by gas:

A cubical vessel containing n number of molecule

 

Consider a gas of mass M having n number of molecules each of mass m contained in a cubical vessel of side l.

Let, the molecules numbering 123…,n are moving randomly with velocities c1, c2, c3, …cn

Now,

Each of these velocities has three components.

Let, u1, v1 and w1 be the components of velocity c1 along the x, y, z-axis, 

Then,

                                       ${\rm{c}}{1^2} = {\rm{u}}{1^2} + {\rm{v}}{1^2} + {\rm{w}}{2^2}$

……………………………………

……………………………………

${\rm{c}}_{\rm{n}}^2 = {\rm{u}}_{\rm{n}}^2 + {\rm{v}}_{\rm{n}}^2 + {\rm{w}}_{\rm{n}}^{2{\rm{\: }}}$

The momentum of the molecule along the x-axis before striking the wall is mu1. Since the collision is elastic, the momentum of the molecules along the x-axis is (-mu1) = 2 mu1

The force exerted by the molecule numbering 1 on the face given by

                                                                    $\frac{{{\rm{mu}}{1^2}}}{{\rm{l}}}$

Similarly, the forces exerted by the molecules numbering 2, 3, n etc given by

                            $\frac{{{\rm{mu}}{2^2}}}{{\rm{l}}},\frac{{{\rm{mu}}{3^2}}}{{\rm{l}}}, \ldots ..\frac{{{\rm{m}}{{\rm{u}}^2}{\rm{n}}}}{{\rm{l}}},{\rm{etc}}$

The net force exerted by the molecules on the face is

${{\rm{F}}_{\rm{x}}} = \frac{{{\rm{mu}}{1^2}}}{{\rm{l}}} + \frac{{{\rm{mu}}{2^2}}}{{\rm{l}}} +  \ldots .. + \frac{{{\rm{mu}}_{\rm{n}}^2{\rm{\: }}}}{{\rm{l}}}$

${{\rm{F}}_{\rm{x}}} = \frac{{\rm{m}}}{{\rm{l}}}\left( {{\rm{u}}{1^2} + {\rm{\: u}}{2^2} +  \ldots . + {\rm{u}}{{\rm{n}}^2}} \right)$….1

The pressure exerted by the molecules on the face is given by

\${{\rm{p}}_{\rm{x}}} = \frac{{{{\rm{F}}_{\rm{x}}}}}{{{{\rm{l}}^2}}} = \frac{{\rm{m}}}{{{{\rm{l}}^3}}}\left( {{\rm{u}}{1^2} + {\rm{\: u}}{2^2} +  \ldots  + {\rm{u}}_{\rm{n}}^2} \right) \ldots  \ldots .2$

The pressures on the faces due to all the molecules are

$\frac{{\rm{m}}}{{{{\rm{l}}^3}}}\left( {{\rm{v}}{1^2} + {\rm{\: v}}{2^2} +  \ldots  + {\rm{v}}_{\rm{n}}^2} \right) \ldots  \ldots .3$

${{\rm{p}}_{\rm{y}}} = \frac{{\rm{m}}}{{{{\rm{l}}^3}}}({\rm{w}}{1^2} + {\rm{\: w}}{2^2} +  \ldots .. + {\rm{w}}_{\rm{n}}^2 \ldots  \ldots 4$

The pressures in all directions on the cube are equal.

Then,

The mean pressure (P) on the cube is given by

${\rm{P}} = \frac{{{{\rm{P}}_{\rm{X}}} + {\rm{\: }}{{\rm{P}}_{\rm{Y}}} + {\rm{\: }}{{\rm{P}}_{\rm{Z}}}}}{3}{\rm{\: }}$

$\frac{{\rm{m}}}{{3{{\rm{l}}^3}}}\left[ {\left( {{\rm{u}}{1^2} + {\rm{v}}{1^2} + {\rm{w}}{1^2}} \right) + \left( {{\rm{u}}_2^2 + {\rm{v}}_2^2 + {\rm{w}}_2^2} \right) +  \ldots  + \left( {{\rm{u}}_{\rm{n}}^2 = {\rm{v}}_{\rm{n}}^2 + {\rm{w}}_{\rm{m}}^2} \right)} \right]$

$\frac{{\rm{m}}}{{3{{\rm{l}}^3}}}\left( {{\rm{c}}{1^2} + {\rm{c}}{2^2} + {\rm{c}}{3^2} +  \ldots  + {\rm{c}}_{\rm{n}}^2} \right)$

Let,

${{\rm{c}}^2} = \frac{{\left( {{\rm{c}}{1^2} + {\rm{c}}{2^2} + {\rm{c}}{3^2} +  \ldots  + {\rm{c}}_{\rm{n}}^2} \right)}}{{\rm{n}}} \ldots .5$

 

Then, c is called root mean square speed of the molecules.

                                   ${\rm{P}} = \frac{{{\rm{mn}}}}{{3{{\rm{l}}^3}}}{\rm{*}}{{\rm{c}}^2}$

Now,

Volume of the cube V = l3 and mass of gas M= mn

                                      ${\rm{P}} = \frac{{{\rm{mn}}}}{{3{\rm{V}}}}{\rm{*}}{{\rm{c}}^2}$

Or, ${\rm{P}} = \frac{{{\rm{m}}{{\rm{c}}^2}}}{{3{\rm{V}}}} \ldots  \ldots  \ldots  \ldots  \ldots  \ldots ..6$

Now,

$\frac{{\rm{m}}}{{\rm{V}}} = \frac{{{\rm{mass}}}}{{{\rm{volume}}}}$ = p (density of the gas). Thus we have

${\rm{P}} = \frac{1}{3}{\rm{\: }}p{{\rm{c}}^2} \ldots ..7$

Equation 6 and 7 are the expression for the pressure exerted by gas.

 

An expression for average kinetic energy per molecule of an ideal gas:

Consider one mole of an ideal gas at pressure P, volume V and temperature T.

From equation of state of the gas, we have

P V = R T…..1

Again,

The kinetic theory of gas

p V =$\frac{1}{3}{\rm{M\: }}\overline {{{\rm{c}}^2}}  \ldots  \ldots .2$

From equation 1 and 2 we get

                                           $\frac{1}{3}{\rm{M\: }}\overline {{{\rm{c}}^2}}  = {\rm{RT}}$

                       $\frac{1}{2}{\rm{M\: }}\overline {{{\rm{c}}^2}}  = \frac{3}{2}{\rm{RT}} \ldots ..3$

Here,

    $\frac{1}{2}{\rm{M\: }}{{\rm{c}}^2}{\rm{\: is\: the\: average\: kinetic\: energy\: of\: one\: mole}}.$

So average kinetic energy per mole of gas is $\frac{2}{3}{\rm{\: R\: T}}.$

If the mass of a molecule is m and N is the number of molecules in one mole of gas,

Equation 3 can be written as

                               $\frac{1}{2}{\rm{mN\: }}\overline {{{\rm{c}}^2}}  = \frac{3}{2}{\rm{RT}}$

Or, ${\rm{\: }}\frac{1}{2}{\rm{m\: }}\overline {{{\rm{c}}^2}}  = \frac{3}{2}\frac{{\rm{R}}}{{\rm{N}}}{\rm{T}}$

                                                                                $\frac{3}{2}{\rm{kT\: }}$

Where, $\frac{{\rm{R}}}{{\rm{N}}} = {\rm{k\: is\: the\: boltzmann\: constant}}.$

Average kinetic energy = $\frac{3}{2}{\rm{kT\: }}$

 

 

Dalton’s law of partial pressure:

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
Pressure Total = Pressure Gas 1 + Pressure Gas 2 + Pressure Gas 3 + ... Pressure Gas n

If the total pressure is known and the number of moles of each component gas are known, the partial pressure can be computed as;

Px = P Total ( nx / n Total )
where
Px = partial pressure of gas x P Total = total pressure of all gases nx = number of moles of gas x nTotal = number of moles of all gases This relationship applies to ideal gases, but also can be used in real gases at some cases.

Let us consider three gases A,B and C at pressure P1, P2, and P3 and volumes V1,V2,V3 respectively be mixed together to form a gaseous mixture at constant temperature. If the partial pressure of the gases A,B and C are Pa, Pb and Pc respectively, Boyles law can be applied for their calculation. For partial pressure of gas A: Pa(V1+V2+V3)

=P1V1 i.e. Pa = P1V1(V1+V2+V3)

Similarly for that of B Pb = $\frac{{{{\rm{P}}_2}{{\rm{V}}_2}}}{{{{\rm{V}}_1} + {{\rm{V}}_2} + {{\rm{V}}_3}}}{\rm{\: }}$and for that of C : Pc = $\frac{{{{\rm{P}}_3}{{\rm{V}}_3}}}{{{{\rm{V}}_1} + {{\rm{V}}_2} + {{\rm{V}}_3}}}$

According to Dalton’s law of partial pressure, the total pressure of the mixture of the gases P will be the sum of individual pressure as P = $\frac{{{{\rm{P}}_1}{{\rm{V}}_1} + {{\rm{P}}_2}{{\rm{V}}_2} + {{\rm{P}}_3}{{\rm{V}}_3}}}{{{{\rm{V}}_1} + {{\rm{V}}_2} + {{\rm{V}}_3}}}$

This above expression is the expression of Dalton’s law of partial pressure.

 

 

A gas is heated in a vessel at constant volume. Pressure increases:

A gas is heated in a vessel at constant volume. Pressure increases, because Pressure of gas is directly proportional to temperature at constant volume.

 

Absolute zero temperature:

The temperature, at which volume or pressure of any gas reduces to zero, is referred to as absolute zero temperature. Its value is equal to at 0 degrees Kelvin or -273 degrees Celsius or at -460 degrees Fahrenheit. 

 

Bicycles tubes burst in summer:

This is due to the kinetic molecular theory, during the summer where temperatures higher, the air particles have more energy (high temp. = higher average kinetic energy). 
With higher energy, they are moving about inside the tire with higher speed and collisions are more forceful Thus, these particles tend to occupy more space, and so the tire is forced to expand, bicycles tubes burst in summer. 


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