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Last updated date: 28th Jan 2023

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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.

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

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.

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.

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.

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.

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: $24 \times 10^6\,J$)

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:

**$5.71 \times 10^8$**)

In this article we have discussed the various assumptions and state the postulates of the kinetic theory of gases. We have also discussed the answer to the question, **“what is the kinetic theory of gases?”**. We've also discussed a gas molecule's most likely velocity, average velocity, and root mean square velocity. We also examined mean free path, the law of equipartition of energy, and specific heat capacity.

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JEE Main 2023 January and April Session exam dates and revised schedule have been announced by the NTA. JEE Main 2023 January and April Session will now be conducted on 24-Jan-2023 to 31-Jan-2023 and 6-Apr-2023 to 12-Apr-2023, and the exam registration closes on 12-Jan-2023 and Apr-2023. You can check the complete schedule on our site. Furthermore, you can check JEE Main 2023 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.

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Last updated date: 28th Jan 2023

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NTA has announced the JEE Main 2023 January session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2023 January and April session Application Form is available on the official website for online registration. Besides JEE Main 2023 January and April session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. April Session's details will be updated soon by NTA.

Last updated date: 28th Jan 2023

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It is crucial for the the engineering aspirants to know and download the JEE Main 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Main 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Main 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.

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JEE Main 2023 Study Materials: Strengthen your fundamentals with exhaustive JEE Main Study Materials. It covers the entire JEE Main syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Main study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Main exam 2023 with Vedantu right on time.

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Download JEE Main Question Papers & Answer Keys of 2022, 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Main Question Paper are provided language-wise along with their answer keys. We also offer JEE Main Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Main Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Main examinations.

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In order to prepare for JEE Main 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2023 exam so that they can grab the top rank in the all India entrance exam.

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JEE Main 2023 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2023 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.

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NTA is responsible for the release of the JEE Main 2023 January and April Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2023 January and April Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.

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NTA will release the JEE Main 2023 January and April sessions exam dates on the official website, i.e. {official-website}. Candidates can directly check the date sheet on the official website or https://jeemain.nta.nic.in/. JEE Main 2023 January and April sessions is expected to be held in February and May. Visit our website to keep updates of the respective important events of the national entrance exam.

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JEE Main 2023 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in IIT colleges. JEE Main 2023 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2023 state rank list on the official website or on our site.

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Want to know which Engineering colleges in India accept the JEE Main 2023 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.

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FAQ

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.

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