JEE Main 2026 Thermodynamics and Kinetic Theory Practice Problems with Solutions

Aindrila

Updated On: November 20, 2025 12:02 PM

Boost your JEE Main 2026 Physics score! Practice 15+ Thermodynamics and Kinetic Theory problems with full solutions in this article.

JEE Main 2026 Thermodynamics and Kinetic Theory Practice Problems with Solutions

Thermodynamics and the Kinetic Theory of Gases are among the most conceptual and scoring chapters in JEE Main Physics 2026. To improve your exam preparation and get an idea of the type of questions asked in the exam, you need to solve practice problems regularly. This will help you overcome your fear and strengthen your basics. In this article, we have provided Thermodynamics & Kinetic Theory questions with their solutions that are important for JEE Main 2026 . Download the PDF link given below and start solving the problems on your own. If you are stuck with a question, check the solutions in the downloaded file.

Also Check - How to Boost Your JEE Main 2026 Revision with Online Test Series & Digital Tools

JEE Main 2026 Thermodynamics & Kinetic Theory Practice Problems with Solutions

Check here some of the Thermodynamics & Kinetic Theory JEE Main 2026 with solutions.

Q1. A gas expands from volume V1​ = 1L to V2 = 3 L following a polytropic process PV1.5 = constant. The initial pressure is 2 × 105 Pa. Find the work done by the gas.

Solution

For a polytropic process PVn = constant,

W = P1V1 - P2V2 / n - 1

Using P1V11.5 = P2V21.5,

P2​ = P1 ​(​V1/V2​​)1.5 = 2 × 105 (1/3​)1.5 = 3.85 × 104 Pa

W = (2 X 105) (1 X 10-3) - (3.85 X 104) (3 X 10-3) / 1.5 -1

W = (200−115.5) / 0.5 = 169J

Answer: W = 169J

Q2. For an ideal gas, show that CP - CV = R

Solution

From the first law of thermodynamics:

dQ = dU + PdV

At constant volume: dV = 0 ⇒ dQv = dU = nCv​dT

At constant pressure: dQp = dU + PdV = nCp​dT

For an ideal gas, PV = nRT ⇒ PdV + VdP = nRdT

At constant pressure, dP = 0 ⇒ PdV = nRdT

Substitute into the first law: nCp​dT = nCv​dT + nRdT

⇒ Cp​ − Cv​ = R

Q3. An ideal gas absorbs 500 J of heat and does 200 J of work. Find the change in internal energy.

Solution

From the first law, ΔU = Q−W = 500 − 200 = 300 J

ΔU = 300 J

Answer: ΔU = 300 J

Also Check - Strategies to Boost Both Speed and Accuracy in JEE Main 2026

JEE Main 2026 Thermodynamics & Kinetic Theory Sample Questions PDF

Students who want to get a good score in JEE Main 2026 need to solve Thermodynamics & Kinetic Theory. We have shared the practice problems with solutions in this PDF.

Quick Links:

JEE Main 2026 Syllabus JEE Main Practice Test

JEE Main 2026 Thermodynamics & Kinetic Theory: Most Important Sub Topics

The weightage of JEE Main Thermodynamics is 10%. You can check the JEE Main 2026 Thermodynamics & Kinetic Theory syllabus in the table below:

Chapter

Topics

Thermodynamics

  • Thermal equilibrium and the concept of temperature

  • Zeroth law of thermodynamics, heat, work and internal energy.

  • The first law of thermodynamics, isothermal and adiabatic processes.

  • The second law of thermodynamics: reversible and irreversible processes.

Kinetic Theory

  • Equation of state of a perfect gas, work done on compressing a gas

  • Kinetic theory of gases: assumptions, the concept of pressure, kinetic interpretation of temperature

  • RMS speed of gas molecules, degrees of freedom, law of equipartition of energy and applications to specific heat capacities of gases, mean free path

  • Avogadro's number

Students need to solve the Thermodynamics & Kinetic problems that are provided in this article. You can also look for JEE Main previous year question papers with solutions PDF , sample papers or mock test papers for practising. Good luck!

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