TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Telangana TSBIE TS Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory

Very Short Answer Type Questions

Question 1.
State the law of equipartition of energy. [TS Mar. ’18, ’16]
Answer:
Law of equipartition of energy :
The total energy of a gas is equally distributed in all possible energy modes, with each mode having an average energy equal to \(\frac{1}{2}\)K_BT. This is known as law of “equipartition of energy”.

Question 2.
Define mean free path. [AP Mar. ’19, ’18, ’17, ’15, May ’18; TS May ’18, Mar. ’17, ’15]
Answer:
Mean free path :
The average distance by can travel an atom or a molecule without colliding is called “mean free path”.
l = \(\frac{1}{\sqrt{2} \pi n d^2}\)
Where n is the number of molecules per unit volume and d is the diameter of the molecule.
(or)
Mean tee path of gas molecules is the aver-age distance covered by a molecule between two successive collisions.

Question 3.
How does kinetic theory justify Avogadro’s hypothesis and show that Avogadro Number in different gases is same?
Answer:
Avogadro’s law or Avogadro’s hypothesis:
It states that “equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules

Consider two gases distinguished by subscripts 1 and 2.

Let
m₁ and m₂ = masses of molecules of 1 and 2 respectively
N₁ and N₂ = no. of molecules of 1 and 2 respectively
P₁ and P₂ = pressures of 1 and 2 respectively
V₁ and V₂ = volumes of 1 and 2 respectively
T₁ and T₂ = absolute temperatures of 1 and 2 respectively
\(\overline{\mathrm{V}_1^2}\) and \(\overline{\mathrm{V}_2^2}\) = mean square velocities of molecules of 1 and 2 respectively.
According to kinetic theory of gases,
P₁V₁ = \(\frac{1}{3}\) m₁N₁ \(\overline{\mathrm{V}_1^2}\) and P₂V₂ = \(\frac{1}{3}\) m₂N₂ \(\overline{\mathrm{V}_2^2}\)
For equal volumes at the same pressure,
P₁V₁ = P₂ V₂
∴ \(\frac{1}{3}\) m₁N₁ \(\overline{\mathrm{V}_1^2}\) = \(\frac{1}{3}\) m₂N₂ \(\overline{\mathrm{V}_2^2}\) ……………. (1)

If the gases are at the same temperature, the average kinetic energies of their molecules are equal.
∴ equation (1) becomes : N₁ = N₂

Hence, equal volumes of all gases under similar conditions of temperature and pressure have the same number of molecules.

Question 4.
What are the units and dimensions of a specific gas constant? [AP Mar. ’14]
Answer:
For specific gas constant unit: Joule/Kelvin
Dimensional formula: r = \(\frac{PV}{T}\) =ML² T^-2 K^-1.

Question 5.
When does a real gas behave like an ideal gas? [AP Mar. ’19, ’14; TS Mar. ’19, ’16, May ’18, ’16; June ’15]
Answer:
No real gas is perfect or ideal. At extremely low pressures and high temperatures, some real gases (like H₂, O₂, N₂, He, etc.) obey the gas laws to a fair degree of accuracy and hence, behave as nearly ideal gas.

Question 6.
State Boyle’s Law and Charles Law. [AP Mar. 18, June 15; TS Mar. 15, May. 16]
Answer:
Boyle’s Law :
At constant temperature, the volume (V) of a given mass of a gas is inversely proportional to its pressure (P).
∴ V ∝ \(\frac{1}{P}\) or PV = constant = K.

Charles Law :
At constant pressure, the volume (V) of a given mass of a gas is directly proportional to its absolute temperature (T).
∴ V ∝ T or \(\frac{V}{T}\) = K (constant)

Question 7.
State Dalton’s law of partial pressures. [TS Mar. ’18, ’17; AP Mar. ’16, ’14]
Answer:
Dalton’s law of partial pressures :
For a mixture of non interacting ideal gases at same temperature and volume total pressure in the vessel is the sum of partial pressures of individual gases.
i.e., P = P₁ + P₂ + ………… where P is total pressure
P₁, P₂ ……….. etc, are individual pressures of each gas.

Question 8.
Define absorptive power of a body. What is the absorptive power of a perfect black body? [AP May ’14]
Answer:
Absorptive power of a body is defined as the ratio of energy absorbed by the body within the wave length range of λ and λ + dλ to the total energy flux following on the body.
Absorptive power,

Question 9.
Pressure of an ideal gas in container is independent of shape of the container – explain. [AP June ’15]
Answer:
Pressure exerted by a gas is due to continuous bombardment of gaseous molecules with the walls of the container. During each collision, certain momentum is transferred to the walls of the container and this transfer is independent of its shapes because area A and time interval ∆t do not appear in the final formula. Hence, pressure of an ideal gas in a container is independent of the shape of the container.

Question 10.
Explain the concept of degrees of freedom for molecules of a gas.
Answer:
The number of degrees of freedom of a dynamical system is defined as the total number of co-ordinates or independent quantities required to describe completely the position and configuration of the system.

Question 11.
What is the expression between the pressure and kinetic energy of a gas molecule?
Answer:
The pressure exerted by an ideal gas is numerically equal to \(\frac{2}{3}\) rd of the mean kinetic energy of translation per unit volume of gas.
P = \(\frac{2}{3}\)E

Question 12.
The absolute temperature of a gas is increased 3 times. What will be the increase in rms velocity of the gas molecule? [TS Mar. ’19, June ’15]
Answer:
The relation between r.m.s. velocity and absolute temperature of a gas is C ∝ √T.
Therefore, the r.m.s. velocity becomes √3 C.
Hence, increase in r.m.s. velocity
√3 C – C = 0.732 C = 73.2 %

Short Answer Questions

Question 1.
Explain the kinetic interpretation of Temperature.
Answer:
According to kinetic theory of gases, the pressure P exerted by one mole of an ideal gas is given by

Thus, the square root of the absolute temperature of an ideal gas is directly proportional to root mean square velocity of its molecules.
Also, from eqn. (1)

But, \(\frac{1}{2}\)mc² is average translational K.E per molecule of a gas.

Therefore, average kinetic energy of translation per molecule of a gas is directly proportional to the absolute temperature of the gas.

Question 2.
How can specific heat capacity of monoatomic, diatomic, and polyatomic gases be explained on the basis of Law of equipartition of Energy? [AP May ’17. ’16; Mar. ’13]
Answer:
From law of equipartition of energy, energy per each degree of freedom is \(\frac{1}{2}\) K_BT Per atom or molecule.

1) Monoatomic gas has three degrees of freedom.

But molar specific heat at constant volume C_v = \(\frac{dU}{dT}\)

2) A diatomic gas has 3 translational and two rotational degrees of freedom.
∴ Kinetic energy per molecule U₁ = 5.\(\frac{1}{2}\)K_B.T
For one gram mole total energy U = \(\frac{5}{2}\)K_B.T.N_A
Molar-specific heat at constant volume

3) A polyatomic gas has three translational, three rotational, and at least one vibrational degrees of freedom.
∴ Kinetic energy per molecule
U₁ = 3.\(\frac{1}{2}\)K_B.T + 3.\(\frac{1}{2}\)K_B.T + f = (3 + f)K_BT
f = Number of vibrational degrees of freedom

Kinetic energy
= per gram mole of molecules
= U₁N_A = U = (3 + f)K_B.N_A.T = (3 + f)RT
Molar specific heat C_v = \(\frac{dU}{dT}\) = (3 + f) R
∴ C_P = (u + f)R
∴ For polyatomic gases C_v = (3 + f) R & C_P = (4 + f)R

Question 3.
Explain the concept of absolute zero of temperature on the basis of kinetic theory.
Answer:
According to kinetic theory of gases, the pressure ‘P* exerted by one mole of an ideal gas is,

Absolute zero:
When T = 0, from equation (1), C = 0.

Hence, absolute zero of temperature may be defined as that temperature at which the root mean square velocity (C) of the gas molecules reduces to zero.

It means, molecular motion ceases at absolute zero.

This definition holds in cases of an ideal gas or perfect gas.

Question 4.
Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Answer:
According to kinetic theory of gases, the pressure ‘P’ exerted by one mole of an ideal gas is
P = \(\frac{1}{3}\) ρC² where p is density of the gas.
∴ P = \(\frac{1}{3}\frac{M}{V}\)C² ⇒ PV = \(\frac{1}{3}\)MC²
But PV = RT for one mole of ideal gas

But, \(\frac{1}{2}\) mC² is averaSe translational kinetic energy per molecule of a gas.

Therefore, average kinetic energy of molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Question 5.
Two thermally insulated vessels 1 and 2 of volumes V₁ and V₂ are joined with a valve and filled with air at temperatures (T₁, T₂) and pressures (P₁, P₂) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessels at equilibrium.
Answer:
According to Standard gas equation

As the vessels are thermally insulated and no work is done, total energy remains conserved.
Therefore, \(\frac{3}{2}\) (P₁V₁ + P₂V₂) = \(\frac{3}{2}\) P(V₁ + V₂)
where P is resultant pressure

For mixture of two gases,
(µ₁ + µ₂)RT = P(V₁ + V₂)
where T is resultant temperature.

Substitute equation (1) and equation (2) in the above equation, we have

This is the final equilibrium temperature of the air in the vessels.

Question 6.
What is the ratio of r.m.s speed of Oxygen and Hydrogen molecules at the same temperature?
Answer:
The r.m.s. speed of oxygen molecules at temperature T is

The r.m.s. speed of Hydrogen molecules at same temperature (T) is
C_H = \(\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}}}}\) But molecular weight of hydrogen = 2
Ratio of r.m.s. speed of Oxygen and Hy-drogen molecules is
∴ C_H = \(\sqrt{\frac{3 \mathrm{RT}}{2}}\)
Ratio of r.m.s. speed of Oxygen and Hydrogen molecules is

∴ C_ghy : C_H = 1 : 4

Question 7.
Four molecules of a gas have speeds 1,2,3 and 4 km/s. Find the rms speed of the gas molecule.
Answer:
Given, C₁ = 1 km/s ; C₂ = 2 km/s
C₃ = 3 km/s ; C₄ = 4 km/s

The root mean square speed of molecules is

C_rms = 2.74 km/s

Question 8.
If a gas has ‘f degrees of freedom, find the ratio of C_p and C_v.
Answer:
Suppose a polyatomic gas molecule has ‘f degrees degrees of freedom (0 and ratio of Cp and of freedom.
∴ Internal energy of one gram mole of the gas is
U = f × \(\frac{1}{2}\)k_BT × N_A = \(\frac{f}{2}\)RT

This is the relation between number of degrees of freedom (f) and ratio of C_p and C_v & (γ)

Question 9.
Calculate the molecular K.E of 1 gram of Helium (Molecular weight 4) at 127°C. Given R = 8.31 J mol^-1C^-1.
Answer:
n = number of moles of Helium = \(\frac{1gm}{4gm mol^{-1}}\) = 0.25 mole
R = 8.314 J mol^-1 KT^-1,
T = 127°C = 127 + 273 = 400 K
∴ Kinetic energy = \(\frac{3}{2}\) nRT = \(\frac{3}{2}\) × 0.25 mol × 8.314 J mol^-1 K^-1 × 400K = 1247.1 J

Question 10.
When pressure increases by 2%, what is the percentage decrease in the volume of a gas, assuming Boyle’s law is obeyed?
Answer:
Let ‘P’ be the initial pressure and ‘V’ be the volume.
When pressure is increased by 2%, new pressure, P¹ = P + \(\frac{2}{100}\)P ⇒ P¹ = \(\frac{102}{100}\)P
According to Boyle’s law, PV = constant ⇒ PV = P’V’

% decrease in volume = \(\frac{2}{100}\) × 100 = 1.96%

∴ When pressure increases by 2%, the volume decreases by 1.96%.

Long Answer Questions

Question 1.
Derive an expression for the pressure of an ideal gas in a container from Kinetic Theory and hence give Kinetic Interpretation of Temperature.
Answer:
Consider a cube of side ‘l’ and an ideal gas is filled in it. Let the co-ordinates X, Y and Z will coincide with the sides of the cube. Let velocities of gas molecules along these directions are V_x, V_y and V_z. Consider a gas molecule moving along X – axis with a velocity V_x. Its motion is perpendicular to Y – Z plane. Let the gas molecule suffered elastic collision with Y – Z plane and bounced back. In this case the velocities V_y and V_z are not considered because collision is along X – direction only.

Change in momentum along X-direction is final momentum (-mV_x) – Initial momentum (mV_x) = -2mV_x ………… (1)

Let area of one side of the cube is A. Then during the time ∆t only the molecules at a distance of V_x ∆t will collide with the walls. Let number of gas molecules in the volume AV∆t are say ‘n’. In these molecules half of the molecules will move towards the wall and remaining will move away from the wall.
∴ Amount of momentum transferred to the wall = Q
= 2mV_x × number of molecules collided with wall
∴ Q = 2mV_x(\(\frac{1}{2}\)nAV_x∆t)

Pressure on Y – Z plane =

If behaviour of gas is isotropic that is equal in all directions then V_x = V_y = V_z
∴ Root mean square velocity of gas

∴ Pressure of gas P = \(\frac{1}{3}\)nm\(\overline{\mathrm{V^2}}\)
From Pascal’s law pressure is same throughout the container so pressure in any direction (say x, y or z) is same.

Kinetic interpretation of temperature :
From gas equation

For Ideal gas internal energy is purely kinetic energy.

But \(\frac{E}{N}=\frac{1}{2}\)mV² and \(\frac{E}{N}=\frac{3}{2}\) K_BT

∴ \(\frac{E}{N}\) ∝ T is the average kinetic energy of gas molecule is proportional to absolute temperature of gas. But it does not depend on volume of container.

Additional Problems

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP, Take the diameter of an oxy-gen molecule to be 3 Å.
Solution:
Here, diameter, d – 3Å,
r = \(\frac{d}{2}=\frac{3}{2}\)Å = \(\frac{3}{2}\) × 10^-8
Molecular volume, V = \(\frac{4}{3}\) πr³ . N, where N is
Avogadro’s number = \(\frac{4}{3}\times\frac{22}{7}\) (1.5 × 10^-8)³ × (6.023 × 10²³) = 8.52 cc.
Actual volume occupied by 1 mole of oxygen at STP, V’ = 22400 cc
∴ \(\frac{V}{V’}=\frac{8.52}{22400}\) = 3.8 × 10^-4 ≈ 4 × 10^-4

Question 2.
Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°Q. Show that it is 22.4 litres.
Solution:
For one mole of an ideal gas, PV = RT
∴ V = \(\frac{RT}{P}\)
Put R = 8.31 .1 mole^-1 K^-1, T = 273 K,
P = 1 atmosphere = 1.013 × 10⁵ Nm^-2

= 0.0224 × 10⁶ cc = 22400 cc = 22.4 litre.

Question 3.
An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?
Solution:
V₁ = 1.0 cm³ = 1.0 × 10^-6m³;
T₁ = 12°C = 12 + 273 = 285 K;
P₁ = 1 atm. + h₁ p g = 1.01 × 10⁵ + 40 × 10³ × 9.8 = 493000 Pa.
When the air bubble reaches at the surface of lake, then
V₂ = ? ; T₂ = 35°C = 35 + 273 = 308 K ;
P₂ = 1 atm. = 1.01 × 10⁵ Pa

Question 4.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a he-lium gas atom at – 20°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Solution:
Let C and C’ be the r.m.s. velocities of argon and a helium gas atoms at temperature TK and T’ K respectively.
Here, M = 39.9 ; M’ = 4.0 ; T = ?
T’ = -20 + 273 = 253 K

Question 5.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R₁/R₂ = (M₂/ M₁)^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]
Solution:
According to Graham’s law of diffusion, = \(\frac{r_1}{r_2}=\sqrt{\frac{\mathrm{M}_2}{\mathrm{M}_1}}\)
Where, r₁ = diffusion rate of hydrogen = 28.7 cm³ s^-1
r₂ = diffusion rate of unknown gas = 7.2 cm³ s^-1
M₁ = molecular mass of hydrogen = 2 u
M₂ = ?