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Kinetic Theory of Gases Chapter - Physics JEE Main

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Concepts of Kinetic Theory of Gases for JEE Main Physics

This chapter connects macroscopic aspects of gases like pressure and temperature to microscopic features like speed and kinetic energy of gas molecules. To describe the behaviour of gases, James Clark Maxwell, Rudolph, and Claussius created the Kinetic Theory of Gases.


So the question arises is what is the kinetic theory of gases. Basically, the kinetic theory of gases is a theoretical model that characterizes a gas's molecular composition in terms of a vast number of submicroscopic particles, such as atoms and molecules. It tells us about the behaviour of gases and also helps us in understanding the physical properties of gases at the molecular level. 


This chapter deals with the various assumptions and kinetic theory of gases postulates made while giving this theory. It provides us with concepts like the degree of freedom of a gas, various gas laws, expression for pressure exerted by an ideal gas and also the various types of relation between pressure, kinetic energy and temperature of a gas.


Now, let's move onto the important concepts and formulae related to JEE and JEE Main exams along with a few solved examples.


JEE Main Physics Chapters 2024


Important Topics of Kinetic Theory of Gases Chapter

  • Kinetic Theory of Gases Assumptions

  • Kinetic Theory of Gases Postulates

  • Pressure Exerted by an Ideal Gas

  • Most Probable Speed

  • Mean Speed or Average Speed

  • Root Mean square Speed

  • Degree of freedom

  • Law of Equipartition of Energy

  • Molar Heat Capacity

  • Expression of Mean Free Path


Kinetic Theory of Gases Important Concept for JEE Main

Name of the Concept

Key Points of the Concepts

1. Assumptions of the

Kinetic Theory of Gases

The Assumptions of the kinetic theory of gases are:
  • A gas consists of a large number of atoms or molecules which are a perfect elastic sphere.

  • Molecular density is uniform throughout the gas.

  • A molecule moves along a straight line between two successive collisions and the average straight distance covered between suc collisions is called mean free path.

  • The collisions are almost instantaneous.

2. Postulates of the Kinetic Theory of Gases

The postulate of the kinetic theory of gases are:
  • The size of the gas molecule is very small as compared to the distance between them.

  • The molecules of a gas are in a state of continuous random motion.

  • The collisions of molecules with themselves and with the walls of the vessel are perfectly elastic.

  • The molecules don’t exert any force of attraction or repulsion in each other except during collision.

  • The average kinetic energy of the gas changes with the temperature.

3. Pressure Exerted by an Ideal Gas

  • The pressure produced by the gas is caused due to constant bombardment of gas molecules against the container's walls and it is given as,

$P=\dfrac{1}{3}\dfrac{M}{V}\overline{v^2}=\dfrac{1}{3}\rho \overline{v^2}$

4. Most Probable Speed

  • Most probable speed of the molecule of a gas is that speed which is possessed by a maximum fraction of the total number of molecules of the gas and it is given as,

$v_{mp}=\sqrt{\dfrac{2K_BT}{m}}$


5. Mean Speed or Average Speed

  • Mean speed is the average speed with which a molecule of the gas moves and it is given as,

$v_{av}=\sqrt{\dfrac{8k_BT}{m\pi}}$

6. Root Mean square Speed

  • Root mean square speed of a gas molecule is defined as the square root of the mean of the squares of the random velocities of the individual molecule of a gas and it is given as,

$v_{rms}=\sqrt{\dfrac{3k_BT}{m}}$

7. Degree of freedom

  • The number of degrees of freedom of a system is defined as the total number of co-ordinates  or independent quantities required to describe completely the position and configuration of the system.

8. Law of Equipartition of Energy

  • According to this, for any system which is in thermal equilibrium, the total energy is distributed equally amongst all the degree of freedom, and energy associated with each molecule per degree of freedom is $\dfrac{1}{2}k_B T$, where $k_B$ is Boltzmann constant and $T$ is the temperature of the system.

9. Molar Heat Capacity

  • Molar heat capacity (C) is the heat required to raise the temperature of one mole of gas by one degree centigrade or kelvin and it is given as,

$C=\dfrac{\Delta Q}{n\Delta T}$

10. Expression of Mean Free Path

  • The path traversed by a molecule between two successive collisions with another molecule is called mean free path and it is given as,

$\lambda=\dfrac{k_BT}{\sqrt{2}\pi d^2n}$

where, $n$ is the number of molecules per unit volume, $d$ is diameter of each molecule of a gas, $T$ is the temperature and $k_B$ is the Boltzmann constant.

11. Avagadro’s Number

  • Avagadro’s number helps in establishing the amount of gas present in a specific space. This number is also known as a mole. 

1 mole = 6.0221415 x 1023

12. Kinetic Interpretation of Temperature

  • The Kinetic Interpretation of Temperature explains that temperature is the measure of how fast gas molecules move. In simple terms, when it's hot, gas molecules move faster, and when it's cold, they move slower.

\[\frac{1}{3}\frac{M}{NC^2}= \frac{R}{NT}=kT\]

Or,                \[\frac{1}{2}mC^2=\frac{3}{2} kT or \frac{1}{2}mC^2\alpha T\]

Where, 3/2 k is constant.


Kinetic Interpretation of Temperature Derivation

Before Deriving let's quickly discuss the Kinetic Interpretation of Temperature, The kinetic interpretation of temperature is a fundamental concept in physics that relates the temperature of a substance to the average kinetic energy of its constituent particles. According to the kinetic theory of gases, the average kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas. This means that hotter gases have faster-moving molecules, while cooler gases have slower-moving molecules. The kinetic interpretation of temperature can be extended to solids and liquids as well, where the average kinetic energy of the particles is associated with their vibrations or random movements, respectively.


To derive the kinetic interpretation of temperature, we start with the equation for the kinetic energy of a single particle:


$KE = \dfrac{1}{2}mv^2$


where KE is the kinetic energy of the particle


m is the mass of the particle


v is the speed of the particle


Next, we consider a large number of gas particles and assume that they are moving in random directions with different speeds. We can then calculate the average kinetic energy of these particles by taking an average over all of their speeds. This average kinetic energy is called the root mean square (rms) speed, and it is denoted by vrms:


$v_{rms} = \sqrt{\dfrac{(v_1^2 + v_2^2 + ... + v_n^2)}{n}}$


where $v_1, v_2, ..., v_n$ are the speeds of the n particles.

The rms speed is related to the average kinetic energy by the following equation:


$KE_{avg} = \dfrac{1}{2}m(v_{rms})^2$


where $KE_{avg}$ is the average kinetic energy of the particles.


Finally, we can relate the rms speed to the absolute temperature of the gas using the ideal gas law:


$PV = NkT$


where P is the pressure of the gas, V is the volume of the gas, N is the number of particles in the gas, k is Boltzmann's constant, and T is the absolute temperature of the gas.


Combining the above equations, we can derive the kinetic interpretation of temperature:


$KE_{avg} = \dfrac{3}{2}kT$


This equation states that the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. As discussed the Kinetic Interpretation of Temperature means that hotter gases have faster-moving molecules, while cooler gases have slower-moving molecules.


Kinetic Theory of Gases Formula:

The kinetic theory of gases is a model that describes the behavior of gases based on the motion of their constituent particles. It is based on the following assumptions:


  • Gases are composed of a large number of tiny, rapidly moving particles that are constantly colliding with each other and with the walls of the container.

  • The particles are assumed to be point-like and to have no attractive forces between them.

  • The collisions between the particles are assumed to be elastic, meaning that there is no loss of kinetic energy during the collisions.


These assumptions lead to a number of important formulas that can be used to describe the behavior of gases. Some of the KTG formulas are listed below.


List of Important Formulas for Kinetic Theory of Gases Chapter 

S.No.

Name of the Concept

Formula

1

Ideal Gas Law

PV = nRT

Where:

P is the pressure of gas

V is the volume of the gas

T is the temperature which is measured in kelvin

N is regarded as the moles of gas that are present

R is the constant, which is known as the universal gas constant

2

Pressure Exerted by an Ideal Gas

$P=\dfrac{1}{3}\dfrac{M}{V}\overline{v^2}=\dfrac{1}{3}\rho \overline{v^2}$

3

Most Probable Speed

$v_{mp}=\sqrt{\dfrac{2K_BT}{m}}$

4

Mean Speed or Average Speed

$v_{av}=\sqrt{\dfrac{8k_BT}{m\pi}}$

5

Root Mean square Speed

$v_{rms}=\sqrt{\dfrac{3k_BT}{m}}$

6

Degree of freedom

$f=3A-R$, where $A$ = number of particles in the system, $R$ = number of independent relations among the particles and $f$ = number of degrees of freedom of the system.

7

Law of Equipartition of Energy

$\dfrac{1}{2}k_B T$

8

Molar Heat Capacity

$C=\dfrac{\Delta Q}{n\Delta T}$

9

Expression of Mean Free Path

$\lambda=\dfrac{k_BT}{\sqrt{2}\pi d^2n}$

10

Heat Capacity at Constant Volume

$C_V=\dfrac{\text{d}U}{\text{d}T}=\dfrac{f}{2}R=\dfrac{R}{\gamma-1}$

11

Heat Capacity at Constant Pressure

$C_P=\dfrac{f}{2}R=\dfrac{\gamma R}{\gamma-1}$

12

Boltzmann’s Constant

$k_B = \dfrac{nR}{N}$

kB is the Boltzmann’s constant

R is the gas constant

n is the number of moles

N is the number of particles in one mole (the Avogadro number)

The value of Boltzmann’s constant is 1.38 x 10-23 J/K.

13

Total Translational K.E of Gas

K.E = $\dfrac{3}{2}nRT

n is the number of moles

R is the universal gas constant

T is the absolute temperature

14

Maxwell Distribution Law

Vrms > V> Vp

Vrms is the RMS speed.

V is the Average speed.

Vp is the most probable speed

15

RMS Speed ($V_{rms}$)

$V_{rms} = \sqrt{\dfrac{8kt}{m}} = \sqrt{\dfrac{3RT}{M}}$

R is the universal gas constant.

T is the absolute temperature.

M is the molar mass.

16

Internal Energy

$U = \left(\dfrac{f}{2}\right) nRT$

For n moles of an ideal gas.


JEE Main Kinetic Theory of Gases Solved Examples 

  1. An HCl molecule moves in three directions: rotationally, translationally, and vibrationally. If the rms velocity of HCl molecules in their gaseous phase is $v_{rms}$, $m$ be the mass, and the Boltzmann constant is $k_B$, then the temperature will be

a. $\dfrac{mv^2_{rms}}{6k_B}$

b. $\dfrac{mv^2_{rms}}{7k_B}$

c. $\dfrac{mv^2_{rms}}{3k_B}$

d. $\dfrac{mv^2_{rms}}{5k_B}$

Sol:

It is given that the HCl molecule moves in three directions therefore its degree of freedom is three and the rms velocity of HCl molecule is $v_{rms}$.

Now according to the equipartition energy theorem, the kinetic energy of the molecule must be equal to energy associated with each molecule in all degree of freedom, therefore we can write;

$\dfrac{1}{2}mv^2_{rms}=3\times \dfrac{1}{2} k_B T$

After rearranging and further solving it we can write;

$T=\dfrac{mv^2_{rms}}{3k_B}$

Therefore the temperature of the molecule will be $\dfrac{mv^2_{rms}}{3k_B}$ and hence option c is correct.


Key point: In order to solve this problem, the molecule's kinetic energy must equal the energy associated with each molecule in every degree of freedom.


  1. A tank used for filling helium balloons has a volume of $0.3\,m^3$ and contains $2.0$ mol of helium gas at $20^{\circ} C$. Assuming that helium behaves similarly to an ideal gas.

(a) How much is the gas molecule's total translational kinetic energy?

(b) What is the average kinetic energy per molecule?

Sol:

(a) It is given that,

Number of mol of a gas, $n=2.0$ mol

Temperature of the gas, $T=20^{\circ} C = (273+20)\,K = 293\,K$

Now, the expression of translational kinetic energy is,

$(K.E.)_{Trans}=\dfrac{3}{2}nRT$ …………(1)

Here in the above expression $R$ is the gas constant whose value is $8.31\,J.K^{-1}.mol^{-1}$.

After putting the values of the quantities in the equation (1) we get;

$(K.E.)_{Trans}=\dfrac{3}{2}(2.0)(8.31)(293)$

On further simplification we get;

$(K.E.)_{Trans}=7.3 \times 10^{3}\,J$

Hence, the total translational energy of the molecule of the gas is $7.3 \times 10^{3}\,J$.


(b) The average kinetic energy per molecule of a gas is $\dfrac{3}{2}k_B T$. Therefore, we can write;

$\dfrac{1}{2}m\overline{v^2}=\dfrac{1}{2}m\overline{v^2_{rms}}=\dfrac{3}{2}k_B T$

After putting the values of known quantities we get;

$\dfrac{1}{2}m\overline{v^2}=\dfrac{3}{2}(1.38 \times 10^{-23})(293)$

After simplification,

$\dfrac{1}{2}m\overline{v^2}=6.07 \times 10^{-21}\,J$

Hence, the average kinetic energy per molecule of a gas is $6.07 \times 10^{-21}\,J$.


Key point: The formula of total translational kinetic energy and average kinetic energy is an important key to solve this problem.


Previous Year Questions from JEE Paper

  1. Two gases - argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140), have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to  (JEE Main 2018)

a. 1.83

b. 4.67

c. 2.3

d. 1.09

Sol:

According to question the given quantities are;

Atomic diameter of argon, $d_1= 2r_1 = 2 \times 0.07 =0.14\,nm$

Atomic weight of argon, $M_1 = 40$

Atomic diameter of xenon, $d_2 = 2r_2 = 2 \times 0.1 =0.2\,nm$

Atomic weight of xenon, $M_1 = 140$

To find, the ratio of mean free time of argon to xenon gas i.e, $\dfrac{\tau_1}{\tau_2}.

The formula for the mean free path is as follows:

$\lambda=\dfrac{k_BT}{\sqrt{2}\pi d^2n}$ …………(i)

And the expression of mean free time is given as,

$\tau=\dfrac{\lambda}{v}$...........(ii)

After observing equations (i) and (ii) we can conclude that,

$\tau \varpropto \dfrac{\sqrt{M}}{d^2}$

Therefore, we can write,

$\dfrac{\tau_1}{\tau_2}=\dfrac{\sqrt{M_1}}{d^2_1}\dfrac{d^2_2}{\sqrt{M_2}}$

After putting the values of know quantities we get;

$\dfrac{\tau_1}{\tau_2}=\dfrac{\tau_1}{\tau_2}=\dfrac{\sqrt{40}}{(0.14)^2}\dfrac{(0.2)^2}{\sqrt{140}}$

On further simplification we get;

$\dfrac{\tau_1}{\tau_2} = 1.09$

Hence, the option d is correct.


Key point: The formula of mean free path and mean free time are essential for solving this problem.

  

  1. For a given gas at $1$ atm pressure, the rms speed of the molecule is $200\,m/s$ at $127^{\circ}$. At $2$ atm pressure and at $227^{\circ}$, the rms speed of the molecules will be: (JEE Main 2018)

a. $80\,m/s$

b. $100 \sqrt{5}\,m/s$

c. $80 \sqrt{5}\,m/s$

d. $100\,m/s$

Sol:

Given that,

At $1$ atm pressure, the rms speed of the molecule i.e, $v’_{rms}=200\,m/s$ and temperature, $T_1=127^{\circ}= (273+127)K=400K$.

Similarly, $2$ atm pressure, the temperature, $T_2=227^{\circ}= (273+227)K=500K$ and we have to find the rms speed of the molecule. Let us consider it as $v’’_{rms}$.

Now the formula of rms speed of the molecule is,

$c=\sqrt{\dfrac{3k_BT}{m}}$

From above expression we can conclude that,

$v_{rms}\propto\sqrt{T}$

Hence we can write;

$\dfrac{v’_{rms}}{v’’_{rms}}=\dfrac{\sqrt{T_1}}{\sqrt{T_2}}$

After putting the values of known quantities we get;

$\dfrac{200}{v’’_{rms}}=\dfrac{\sqrt{400}}{\sqrt{500}}$

On further simplification we can get;

$v’’_{rms} = \dfrac{\sqrt{500}}{\sqrt{400}} \times 200$

$v’’_{rms} = \dfrac{\sqrt{5}}{2} \times 200$

$v’’_{rms} = 100 \sqrt{5}\,m/s$

Hence option b is correct.


Key point: Understanding the root mean square velocity formula is essential for resolving such issues.


Practice Questions

  1. An ideal gas occupies a volume of $4\,m^3$ at a pressure of $4 \times 10^6\,Pa$. Calculate the energy of the gas. (Ans: $24 \times 10^6\,J$)

  2. Calculate the number of molecules in 2 c.c. of the perfect gas at $27^\circ$ at a pressure of 10 cm of mercury.Mean kinetic energy of each molecule at $27^\circ=7\times 10^{-4}\,erg$ and acceleration due to gravity is $980\,cm/s^2$ . (Ans:  $5.71 \times 10^8$)


JEE Main Physics Kinetic Theory of Gases Study Materials

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JEE Main Physics Study and Practice Materials

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Conclusion

The "Kinetic Theory of Gases" chapter is a fundamental cornerstone in the realm of JEE Main preparation. It unveils the microscopic world of gas particles, providing essential insights into their behavior and properties. Understanding concepts like the ideal gas law, kinetic energy, and the distribution of molecular speeds equips students to solve complex problems in physics and chemistry. Beyond the exam, this knowledge plays a pivotal role in comprehending various real-world phenomena. The chapter fosters an appreciation for the interplay between macroscopic and microscopic behavior, laying a strong foundation for success in JEE Main and a deep understanding of the physical world.

FAQs on Kinetic Theory of Gases Chapter - Physics JEE Main

1. What is the weightage of the kinetic theory of gases in JEE?

From this chapter, approximately 2-3 questions are asked every year and thus leading to the approximate weightage of 3-4% in the exam.

2. Is it necessary to study the kinetic theory of gases for JEE?

Fundamentally, the kinetic theory of gases is a crucial topic since it provides the foundation for understanding the gaseous state of matter. It's also useful for learning the fundamentals of thermodynamics.

3. Is it truly beneficial to practice prior year questions for the JEE exam?

Practicing prior year questions from any chapter helps us to understand which topic is important from an exam point of view. It also gives us an idea about the difficulty level of the topics that have been asked in the chapter. Therefore, it is important to practice the previous year's questions for a better understanding of important topics of the chapter and also try to make your own kinetic theory of gases notes for the last revision before the exam.

4. What is Kinetic Interpretation of Temperature?

The kinetic interpretation of temperature is a fundamental concept in physics that relates the temperature of a substance to the average kinetic energy of its constituent particles. According to the kinetic theory of gases, the average kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas. This means that hotter gases have faster-moving molecules, while cooler gases have slower-moving molecules. The kinetic interpretation of temperature can be extended to solids and liquids as well, where the average kinetic energy of the particles is associated with their vibrations or random movements, respectively.

5. What are the assumptions of kinetic theory of gases?

The kinetic theory of gases is a model that describes the behavior of gases at the molecular level. It is based on the following assumptions:

Assumption

Explanation

Molecules are point-like

This means that the molecules are so small that their size can be neglected.

Molecules are in constant random motion

This means that the molecules are always moving in all directions at different speeds.

Collisions are perfectly elastic

This means that the molecules do not lose any energy during collisions.

Interactions between molecules are negligible except during collisions

This means that the molecules only interact with each other when they collide.

Average kinetic energy is proportional to temperature

This means that the molecules move faster when the temperature is higher.

6. What are the postulate of kinetic theory of gases?

The postulates of kinetic theory of gases are:


  1. Gas molecules are in a constant state of random motion.

  2. Gas molecules are very small and are far apart from each other.

  3. All collisions between gas molecules are elastic.

  4. The average kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas.

  5. The pressure of a gas is caused by the collisions of gas molecules with the walls of the container.

  6. The more collisions there are, the higher the pressure.