Samacheer Class 11 Physics - Kinetic Theory of Gases

(a) \(\frac { PV }{ kT }\)
Hint:
The area under the graph will give total number of gas molecules in the system.
n = \(\frac { PV }{ RT }\) R = k
n = \(\frac { PV }{ kT }\)

The microscopic origin of pressure was proposed by considering a thermodynamic system as a collection of molecules. By the kinetic theory of gases, the pressure is linked to the velocity of molecules (v) and number density (\(\frac {N}{ V }\))
p = \(\frac { 1 }{ 3 }\)\(\frac { N }{ V }\)mv²
Where v – velocity of molecular
\(\frac { N }{ V }\) – number density

The average K.E per molecule \(\overline{\mathrm{KE}}=\epsilon=\frac{3}{2} k \mathrm{T}\)
The equation implies that the temperature of a gas is a measure of the average translational K.E. per molecule of the gas.

The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Hence all molecules of the gases escape from the surface of the Moon easily.

(i) The root mean square speed (rms):
v rms = \(\sqrt{\frac{3 R T}{M}}\)
(ii) Average speed:
\(\overline{v}\) = 1.60\(\sqrt{\frac{k \mathrm{~T}}{m}}\)
(iii) Most probable speed:
v mp = 1.41\(\sqrt{\frac{k \mathrm{~T}}{m}}\)

The minimum number of independent coordinates needed to specify the position and configuration of a thermo-dynamical system in space is called the degree of freedom of the system.

According to kinetic theory, the average kinetic energy of a system of molecules in thermal equilibrium at temperature T is uniformly distributed to all degrees of freedom (x or y or z directions of motion) so that each degree of freedom will get \(\frac {1}{ 2 }\) kT of energy. This is called law of equipartition of energy.

PV = \(\frac {2}{ 3 }\)U
For a fixed pressure, the volume of the gas is proportional to internal energy of the gas.
(or)
Average kinetic energy is directly proportional to absolute temperature. It is implied that,
V ∝ T
(or) \(\frac {V}{ T }\) = constant.

We know that,
PV = \(\frac {2}{ 3 }\)U
But the internal energy of an ideal gas is equal to N times the average kinetic energy (∈) of each molecule.
U = N∈
For a fixed temperature, the average translational kinetic energy ∈ will remain constant. It implies that,
PV = \(\frac {2}{ 3 }\)N∈
Thus, PV = constant

Avogadro’s law states that, at constant temperature and pressure, equal volumes of all gases contain the same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases,
From equation,
where \(\overline{v_{1}^{2}}\) and \(\overline{v_{2}^{2}}\) are the mean square speed for two gases and N 1 and N 2 are the number of gas molecules in two different gases.
At the same temperature, the average kinetic energy per molecule is the same for two gases.
\(\frac {1}{ 2 }\)m 1 \(\overline{v_{1}^{2}}\) = \(\frac {1}{ 2 }\)m 2 \(\overline{v_{2}^{2}}\) … (2)
Dividing the equation (1) by (2) we get, N 1 = N 2.
This is Avogadro’s law. It is sometimes referred to as Avogadro’s hypothesis or Avogadro’s Principle.

• Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase.
• It is the reason why the smell of hot sizzling food reaches several metre away than smell of cold food.
• The mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

According to kinetic theory, any particle suspended in a liquid or gas is continuously bombarded from all directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner.
III. Long Answer Questions:

• All the molecules of a gas are identical, elastic spheres.
• The molecules of different gases are different.
• The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
• The molecules of a gas are in a state of continuous random motion.
• The molecules collide with one another and also with the walls of the container.
• These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
• Between two successive collisions, a molecule moves with uniform velocity.
• The molecules do not exert any force of attraction or repulsion on each other except during collision.
• The molecules do not possess any potential energy and the energy is wholly kinetic.
• The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
• These molecules obey Newton’s laws of motion even though they move randomly.

Let us consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.
During each collision, the molecules impart certain momentum, to the wall. Because momentum, is transferred by molecules, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas.
Let us consider a molecule of mass m moving with a velocity v having components (v x, v y, v z ) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. The components of velocity of the molecule after collision are (-V x, V y, V z ).
The x-component of momentum of the molecule before collision = mv x.
The x-component of momentum of the molecule after collision = – mv x.
According to law of conservation of linear momentum, the change in momentum of the wall = 2 mv x
The molecules within the distance of v x ∆t from the right side wall and moving towards the right will hit the wall in the time interval ∆t.
The number of molecules that will hit the right side wall in a time interval At is equal to the product of volume (Av x ∆t) and number density of the molecules (n). Here A is area of the wall and n is number of molecules per unit volume (\(\frac { N }{ V }\)). It is assumed that the number density is the same throughout the cube.
Not all the n molecules will move to the right, hence on an average only half of then n molecules move to the right and the other half moves towards left side.
The number of molecules that hit the right side wall in a time interval.
∆t = \(\frac { n }{ 2 }\)Av x ∆t … (2)
In the same interval of time ∆t, the total momentum transferred by the molecules,
∆p = \(\frac { n }{ 2 }\)Av x ∆t x 2mv x
= Av² x mn∆t … (3)
From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall is given by
F = \(\frac { ∆p }{ ∆t }\) = nmAv² x … (4)
Pressure, P = force the area of the wall,
P = \(\frac { F }{ A }\) = nmv² x … (5)
Since all the molecules are moving completely in random manner, they do not have same speed, so the term v² x can be replaced by the average \(\overline{v_{x}^{2}}\) in equation (5)
P = nm\(\overline{v_{x}^{2}}\) … (6)
Since the gas is assumed to move in a random direction, it has no preferred direction of motion. It is implied that the molecule has the same average speed in all three directions. So, \(\overline{v_{x}^{2}}\) = \(\overline{v_{y}^{2}}\) = \(\overline{v_{z}^{2}}\) The mean square speed is obtained from,

We know that,
P = \(\frac { 1 }{ 3 }\) \(\frac { N }{ V }\) m\(\overline{v^{2}}\)
PV = \(\frac { 1 }{ 3 }\) Nm\(\overline{v^{2}}\) … (1)
Comparing the equation (1) with ideal gas equation PV = NkT,
NkT = \(\frac { 1 }{ 3 }\) Nm\(\overline{v^{2}}\)
KT = \(\frac { 1 }{ 3 }\) m\(\overline{v^{2}}\) … (2)
Multiply the above equation by 3/2 on both sides,
\(\frac { 3 }{ 2 }\)KT = \(\frac { 1 }{ 2 }\) m\(\overline{v^{2}}\) … (3)
R.H.S. of the equation (3) is called average kinetic energy of a single molecule (\(\overline{KE}\)).
The average kinetic energy per molecule
\(\overline{KE}\) = ∈ = \(\frac { 3 }{ 2 }\)KT … (4)
It is implied from equation (3) that the temperature of a gas is a measure of the average translational kinetic energy per molecule of the gas.

Monoatomic molecule:
A monoatomic molecule has only three translational degrees of freedom by virtue of its nature.
∴ f = 3
Example: Helium, Neon, Argon.
Diatomic molecule: There are two cases.
(i) At Normal temperature: A molecule of a diatomic gas consists of two atoms bound to each other by a force of attraction, the center of mass lies in the center of the diatomic molecule. So, the motion of the center of mass requires three translational degrees of freedom.
In addition, the diatomic molecule can be rotated about three mutually perpendicular axes.
In addition, the diatomic molecule can be rotated about three mutually perpendicular axes.
But the moment of inertia about its own axis of rotation is negligible. Hence, it has only two rotational degrees of freedom. So totally there are five degrees of freedom.
f = 5
(ii) At high temperature: At a very high temperature such as 5000 K, the diatomic molecules possess additional two degrees of freedom due to vibrational motion [one due to kinetic energy of vibration and the other is due to potential energy].
So totally there are seven degree of freedom.
f = 7
Example: Hydrogen, Nitrogen, Oxygen.
(i) Linear triatomic molecule:
The linear triatomic molecule has three translational degrees of freedom. It has two rotational degrees of freedom because it is similar to a diatomic molecule except there is an additional atom at the center.
At normal temperature, linear triatomic molecules will have five degrees of freedom. It has two additional vibrational degrees of freedom at high temperatures.
So a linear triatomic molecule has seven degrees of freedom.
Example: Carbon dioxide.
(ii) Non-linear triatomic molecule: It has three translational degrees of freedom and three rotational degrees of freedom about three mutually orthogonal axes. So, the total degrees of freedom.
f = 6
Example: Water, Sulphur dioxide.

Monoatomic molecule: Average kinetic energy of a molecule
= [\(\frac { 3 }{ 2 }\)kT]
(i) Total energy of a mole of gas
= \(\frac { 3 }{ 2 }\)
(ii) For one mole, the molar specific heat at constant volume
C V = \(\frac { dU }{ dT }\) = \(\frac { d }{ dT }\)[\(\frac { 3 }{ 2 }\)RT]
C V = [latex]\frac { 3 }{ 2 }[/latex]R
C p = C V + R = \(\frac { 3 }{ 2 }\)R + R
= \(\frac { 5 }{ 2 }\) R
(iii) The ratio of specific heats,
Diatomic molecule: Average kinetic energy of a diatomic molecule at low temperature \(\frac { 5 }{ 2 }\)
(i) Total energy of one mole of gas
= \(\frac { 5 }{ 2 }\)kT x N A = \(\frac { 5 }{ 2 }\)RT
(Here, the total energy is purely kinetic)
(ii) For one mole specific heat at constant volume
(iii) Energy of a diatomic molecule at high temperature is equal to \(\frac { 7 }{ 2 }\) RT.
C V = \(\frac { dU }{ dT }\) = [ \(\frac { 7 }{ 2 }\)RT] = [\(\frac { 7 }{ 2 }\) R
∴C P = C V + R = \(\frac { 7 }{ 2 }\) R + R
C P = \(\frac { 9 }{ 2 }\)R
It is noted that the C V and C P are higher for diatomic molecules than the monoatomic molecules. It is implied that to increase the temperature of diatomic gas molecules by 1 °C it require more heat energy than monoatomic molecules.
Triatomic molecule:
(i) Linear molecule: Energy of one mole
= \(\frac { 7 }{ 2 }\)KT x N A = \(\frac { 7 }{ 2 }\)RT
(ii) Non-Linear molecule: Energy of one mole
= \(\frac { 6 }{ 2 }\)KT x N A = \(\frac { 6 }{ 2 }\)RT
C V = \(\frac { dU }{ dT }\) = 3R
C P = C V + R
= 3 R + R = 4R
∴ γ = \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) = \(\frac { 4R }{ 3R }\)
= \(\frac { 4 }{ 3 }\)
= 1.33

(i) All molecules in any gas move with different velocities in random directions.
(ii) Each molecule collides with every other molecule and their speed is exchanged.
(iii) Let us calculate the rms speed of each molecule and not the speed of each molecule which is rather difficult.
(iv) In general our interest is to find how many gas molecules have the range of speed from v to v + dv.
(v) This is given by Maxwell’s speed distribution function.
N v = 4πN(\(\frac { m }{ 2πkT }\)) 3/2 v² \(e^{-m v^{2} / 2 k T}\)
The above expression is graphically shown as follows
(vi) For a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed, and most probable speed are indicated in the Figure.
(vii) It is found that the rms speed is greatest among the three.
(viii) The area under the graph will give the total number of gas molecules in the system.
(ix) From the speed distribution graph for two different temperatures, it is found that, as temperature increases, the peak of the curve is shifted to the right.
(x) It is implied that the average speed of each molecule will increase. But the area under each graph is the same. Since it represents the total number of gas molecules.

Let us consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume. It is assumed that only one molecule is in motion and all others are at rest as shown in the figure.
If a molecule moves with average speed v in a time t, the distance travelled is vt. In this time t, let us consider the molecule to move in an imaginary cylinder of volume itd2vt. It collides with any molecule whose center is within this cylinder. Hence, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equal to πd²vt. The total path length divided by the number of collisions in time t is the mean free path.
It is assumed that only one molecule is moving at a time and other molecules are at rest. But in actual practice all the molecules are in random motion. Hence the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct expression for mean free path,
λ = \(\frac{1}{\sqrt{2} n \pi d^{2}}\)
It is implied from the equation that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases. Hence it decreases the distance travelled by the molecule before collisions.
Rearranging the equation (2) using ‘m’ (mass of the molecule)
∴ λ = \(\frac{m}{\sqrt{2} \pi d^{2} m n}\)
But mn – mass per unit volume = ρ (density of the gas)

The random (Zig – Zag path) motion of pollen suspended in a liquid is called Brownian motion. Brownian motion is due to the bombardment of suspended particles by molecules of the surrounding fluid. Einstein gave the systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.
According to kinetic theory, any particle suspended in a liquid or gas is continuously bombarded from all directions in such a way that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner.
Factors affecting Brownian Motion:
* Brownian motion increases with increasing temperature.
* It decreases with bigger particle size, high viscosity, and density of the liquid (or) gas.
IV. Numerical Problems:

Let, T = T 1 = 273K
RMS velocity,
C = \(\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}\) … (1)
When the RMS velocity is tripled,
3C = \(\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}}}\) … (2)
Dividing equation (2) by (1) we get
\(\frac { 3C }{ C }\) = \(\sqrt{\frac{3 \mathrm{RT}_{2} / \mathrm{M}}{3 \mathrm{R} \times 273 / \mathrm{M}}}\)
3 = \(\sqrt{\frac{\mathrm{T}_{2}}{273}}\)
Squaring on both side
9 = \(\frac{\mathrm{T}_{2}}{273}\)
∴ T 2 = 273 x 9 = 2457K
Temperature T 2 = 2457K

Temperature T = 80 + 273 = 353K
Pressure P = 5 x 10 -10 N/m²
K B = \(\frac { R }{ N }\) ∴ R = K B N
P = nRT = nK B NT
Number of molecules,
Number of molecules = 1.02 x 10 11

We know that \(\frac { 1 }{ 2 }\) mV²rms = \(\frac { 3 }{ 2 }\) kT
V rms = 1.842 x 10³ ms -1
The value of ve is less than the velocity of gas present on the moon. Hence moon cannot have an atmosphere.

Velocity = 2 x 10³ms -1
P = 26.72 x 10 -7 x 8 x 2 x 10³
= 427.5 x 10 -4 kg ms -1
Component of momentum normal to wall,
P C = P cosθ
Angle θ = 30°
P C = 427.5 x 10 -4 cos30°
= 427.5 x 10 -4 x \(\frac{\sqrt{3}}{2}\)
= 370.2 x 10 -4 kg ms -1

Meyer’s relation is,
C P – C V = R
∴ C P = C V + R
γ = \(\frac{C_{P}}{C_{V}}\)
= \(\frac{C_{V}+R}{C_{V}}\)
γ = 1 + \(\frac{R}{C_{V}}\)
∴ C V = \(\frac { R }{ r-1 }\)
For monoatomic gas, R

One moles of an ideal gas at S.T.P occupies a volume of 22.4 x 10 -3 m³.
In terms of number of molecules and radius the mean free path at S.T.P can be written as,

Since oxygen is a diatomic molecule with 5 degrees of freedom,
Degree of freedom of molecules in 2 moles of oxygen = f 1 = 2N x 5 = 10 N.
Since argon is a monoatomic molecule, degrees of freedom of molecules in 4 moles of argon f 2 = 4N x 3 = 12 N.
∴ Total degrees of freedom of the mixture = f = f 1 + f 2
= 10 N+ 12N
= 22N.
According to the principle of law of equation partition energy, associated with each degree of freedom of a molecule = \(\frac { 1 }{ 2 }\)kT.

Boltzmann’s constant K B = 1.38 x 10 -23 Jk -1
K B = \(\frac { R }{ N }\)
∴ R = K B N
Now P= nRT = nK B NT
∴The number of molecules in the room PV
= nN = \(\frac{\mathrm{PV}}{\mathrm{TK}_{\mathrm{B}}}\)
Temperature = 27 + 273 = 300K
= \(\frac{1.013 \times 10^{5} \times 25}{300 \times 1.38 \times 10^{-23}}\)
= 6.117 x 10 26 molecules
= 6.1 x 10 26 molecules