Download Free PDF of Kinetic Theory for Class 11 Physics Chapter 12
FAQs on CBSE Class 11 Physics Chapter 12 Kinetic Theory – NCERT Solutions 2025-26
1. How are the NCERT solutions for deriving the pressure of an ideal gas (P = 1/3 ρv²) structured?
The NCERT solutions for this derivation provide a clear, step-by-step method as per the CBSE guidelines. The typical structure is as follows:
- Step 1: State the key assumptions of the kinetic theory for an ideal gas in a cubical container.
- Step 2: Calculate the change in momentum of a single molecule colliding elastically with one wall.
- Step 3: Determine the force exerted by that molecule by calculating the rate of change of momentum.
- Step 4: Calculate the total pressure by summing the forces from all N molecules and dividing by the area.
- Step 5: Establish the final relation P = (1/3)ρv², linking macroscopic pressure to microscopic molecular speed.
2. What is the correct method for calculating the root mean square (RMS) speed of gas molecules in an NCERT problem?
To solve NCERT problems for RMS speed correctly, follow these steps:
- Identify the formula: The standard formula is v_rms = √(3RT/M).
- Check the units: Ensure the temperature (T) is in Kelvin (K), the molar mass (M) is in kilograms per mole (kg/mol), and the universal gas constant (R) is 8.314 J/(mol·K).
- Substitute values: Carefully place the given values into the formula. A common error is forgetting to convert temperature from Celsius to Kelvin or mass from g/mol to kg/mol.
- Calculate the result: Solve the equation to find the RMS speed in meters per second (m/s).
3. How should a student apply the law of equipartition of energy to find the specific heat of gases in NCERT questions?
The law of equipartition of energy simplifies finding specific heats. The method is:
- Determine degrees of freedom (f): For a monoatomic gas (like Ar, He), f = 3. For a diatomic gas (like O₂, N₂) at normal temperatures, f = 5.
- Calculate Internal Energy (U): The total internal energy for one mole of gas is U = (f/2)RT.
- Find Molar Specific Heat at Constant Volume (Cv): Cv is the rate of change of U with respect to T, so Cv = dU/dT = (f/2)R.
- Find Molar Specific Heat at Constant Pressure (Cp): Use the relation Cp = Cv + R.
This step-by-step approach is crucial for solving related numericals in the NCERT exercises.
4. Do the NCERT Solutions for Chapter 12 explain how to solve numericals on mean free path?
Yes, the solutions provide a complete method for solving mean free path problems. The typical solution demonstrates these steps:
- State the formula: Clearly write the formula for mean free path, λ = 1 / (√2 * π * d² * n).
- Define variables: Explain what each term represents (λ is mean free path, d is molecular diameter, and n is the number density).
- Substitute and solve: Show the correct substitution of given values into the formula to arrive at the final answer, paying close attention to units.
5. Why is it essential to use the Kelvin scale for temperature when solving problems from NCERT Chapter 12?
Using the Kelvin scale is non-negotiable in kinetic theory for a fundamental reason. The kinetic energy of gas molecules is directly proportional to the absolute temperature. The Kelvin scale is an absolute scale where 0 K represents zero kinetic energy. Using Celsius or Fahrenheit, which are relative scales, would break this direct proportionality and lead to incorrect results in formulas like the ideal gas law (PV=nRT) and the RMS speed equation.
6. How do the basic assumptions of kinetic theory simplify the solutions for ideal gas problems in the NCERT textbook?
The assumptions are a key problem-solving tool. By assuming that gas molecules have negligible volume and no intermolecular forces, we can directly apply the simple ideal gas equation (PV = nRT). This avoids using complex equations of state (like the van der Waals equation) that account for real gas behaviour, thus making the calculations straightforward and aligned with the Class 11 syllabus.
7. What is a common mistake students make when solving for the total kinetic energy of a diatomic vs. a monoatomic gas?
A common mistake is using the wrong number for the degrees of freedom (f). Students often forget that:
- A monoatomic gas (like Helium) only has 3 translational degrees of freedom (f=3).
- A diatomic gas (like Nitrogen) has 3 translational and 2 rotational degrees of freedom at moderate temperatures, making f=5.
Using f=3 for a diatomic gas is a frequent error that leads to an incorrect calculation of its total internal energy (U = (f/2)nRT).
8. How does understanding 'degrees of freedom' help solve specific heat capacity questions in the NCERT exercises more efficiently?
Understanding degrees of freedom (f) offers a direct shortcut. Instead of memorising separate formulas, you can derive specific heats from one principle. Knowing 'f' allows you to instantly find the molar specific heat at constant volume with Cv = (f/2)R and at constant pressure with Cp = ((f/2) + 1)R. This fundamental understanding is much faster and more reliable for solving problems than rote memorisation of values for different gases.
9. How do Vedantu's NCERT Solutions for Class 11 Physics Chapter 12 ensure complete coverage of the textbook exercises?
Our NCERT Solutions are structured to provide 100% coverage of all questions in the Class 11 Physics textbook for Chapter 12. Each solution is crafted by subject matter experts and is presented in a sequential, exercise-wise manner. This ensures that you find a detailed, step-by-step explanation for every numerical problem, short-answer question, and derivation listed at the end of the chapter, fully aligned with the 2025-26 CBSE syllabus.
10. Beyond just getting the correct answer, what is the main goal of using NCERT solutions for Kinetic Theory?
The primary goal is to build a strong conceptual bridge between the microscopic world of molecules and the macroscopic properties of gases you can measure. By following the step-by-step solutions, you learn how molecular motion creates pressure, how molecular speed is related to temperature, and how the internal structure of a molecule (monoatomic vs. diatomic) dictates its heat capacity. This reinforces the core principles of the kinetic model, which is essential for both board exams and competitive tests.

















