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KTG and Thermodynamics

Kinetic Theory of Gases and Thermodynamics - JEE Advanced Notes

Kinetic Theory of Gases and Thermodynamics - Detailed Notes

Introduction

Kinetic Theory of Gases (KTG) and Thermodynamics are fundamental topics for JEE aspirants. Understanding the microscopic behavior of gas molecules and the laws governing energy, heat, and work is essential. In this post, we will cover every concept in detail with examples to help you master these topics up to the JEE Advanced level.

Kinetic Theory of Gases (KTG)

Basic Assumptions of Kinetic Theory

  • Gases consist of a large number of molecules in random motion.
  • Molecules of a gas are point masses with no internal structure.
  • Collisions between molecules are perfectly elastic.
  • No intermolecular forces act between the molecules except during collisions.

Key Formulas in KTG

Average Kinetic Energy: \( \frac{3}{2} k_B T \)

Where \( k_B \) is the Boltzmann constant and \( T \) is the temperature.

Root Mean Square Velocity: \( v_{rms} = \sqrt{\frac{3k_B T}{m}} \)

Thermodynamics

First Law of Thermodynamics (FLOT)

The First Law of Thermodynamics states that the heat added to a system is used for increasing its internal energy and doing work.

\( \Delta Q = \Delta U + \Delta W \)

Where:

  • \( \Delta Q \) = Heat added to the system
  • \( \Delta U \) = Change in internal energy
  • \( \Delta W \) = Work done by the system

Work Done by Each Thermodynamic Process

The work done by a gas depends on the type of thermodynamic process it undergoes. Here’s a summary for each process:

1. Isothermal Process (\( T = \text{constant} \))

For an ideal gas, the work done in an isothermal process is given by:

\( W = nRT \ln\left(\frac{V_f}{V_i}\right) \)

Where \( V_f \) and \( V_i \) are the final and initial volumes.

2. Adiabatic Process (\( Q = 0 \))

In an adiabatic process, no heat is exchanged with the surroundings. The work done is given by:

\( W = \frac{P_i V_i - P_f V_f}{\gamma - 1} \)

Where \( \gamma \) is the adiabatic index, \( P_i, V_i \) and \( P_f, V_f \) are the initial and final pressures and volumes.

3. Isochoric Process (\( V = \text{constant} \))

In an isochoric process, the volume remains constant, and hence no work is done:

\( W = 0 \)

4. Isobaric Process (\( P = \text{constant} \))

In an isobaric process, the work done is given by:

\( W = P(V_f - V_i) \)

5. Cyclic Process

In a cyclic process, the net change in internal energy over one complete cycle is zero, and the work done is equal to the net heat added:

\( W = \Delta Q \)

Example Problem on Work Done

Example:

An ideal gas expands isothermally from a volume of 2 L to 6 L at a temperature of 300 K. Calculate the work done by the gas. Assume 1 mole of gas.

Solution:

For an isothermal process, the work done is:

\( W = nRT \ln\left(\frac{V_f}{V_i}\right) \)

Substituting the values:

\( W = 1 \times 8.314 \times 300 \times \ln\left(\frac{6}{2}\right) \)
\( W = 2494.2 \times \ln(3) = 2739.6 \, \text{J} \)
Solve JEE Mains PYQs on Thermodynamics

Conclusion

Mastering Kinetic Theory of Gases and Thermodynamics requires a solid understanding of both concepts and applications. Keep practicing with more problems, especially from previous year JEE papers, to build confidence.

NEET - physics 2024

 Q. A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π × 10–⁷ SI units)                                                         (NEET-2024) 1. 44 mT (2)  (3)  (4)  2. 4.4 T 3. 4.4 mT 4.None of these  Show Answer.   
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