Class 11 Physics Chapter 12 Kinetic Theory – Extra Questions with Answers

Most questions in this set trace back to the ideal gas equation, PV=nRT, and the value of the universal gas constant R. Others probe degrees of freedom and the equipartition of energy across a molecule’s possible motions.

Last Updated: September 23, 2026

Very Short Answer Questions (1 mark)

Q1. Write the ideal gas equation.
Ans: PV=nRT.

Q2. What is the value of the universal gas constant R?
Ans: 8.314 J/(mol·K).

Q3. How many degrees of freedom does a monatomic gas molecule have?
Ans: 3 (all translational).

Q4. State the law of equipartition of energy.
Ans: Energy is distributed equally among all degrees of freedom, each contributing (1/2)k_BT per molecule.

Q5. Write the formula for pressure of an ideal gas from kinetic theory.
Ans: P=(1/3)ρv_rms².

Short Answer Questions (2–3 marks)

Q6. Find the number of moles in 44 g of CO₂ (molar mass 44 g/mol).
Ans: n=mass/molar mass=44/44=1 mole.

Q7. Two moles of an ideal gas occupy 0.05 m³ at 300 K. Find the pressure (R=8.314 J/mol·K).
Ans: P=nRT/V=2(8.314)(300)/0.05=99768 Pa ≈ 9.98×10⁴ Pa.

Q8. Calculate the average kinetic energy per molecule of a gas at 300 K (k_B=1.38×10⁻²³ J/K).
Ans: KE=(3/2)k_BT=(3/2)(1.38×10⁻²³)(300)=6.21×10⁻²¹ J.

Higher-Order Thinking / Application Questions

Q9. Explain, using the kinetic theory of gases, why the pressure of a gas increases when it is heated at constant volume, connecting this to molecular motion.
Ans: According to kinetic theory, temperature is a measure of the average kinetic energy of gas molecules: average KE=(3/2)k_BT. When a gas is heated, its temperature rises, so the average kinetic energy (and hence average speed) of the molecules increases. At constant volume, the molecules now collide with the container walls more frequently and with greater momentum change per collision (since they move faster). Since pressure is a result of the cumulative force from molecular collisions on the walls per unit area, this increase in collision frequency and force leads to a corresponding increase in pressure, consistent with P=(1/3)ρv_rms² (v_rms increasing with T).

Q10. Compare the number of degrees of freedom, and hence specific heat capacities, of a monatomic gas (like helium) and a diatomic gas (like oxygen) at moderate temperatures, explaining why diatomic gases have higher molar specific heats.
Ans: A monatomic gas has only 3 translational degrees of freedom, so by the law of equipartition of energy, its internal energy per mole is (3/2)RT, giving molar specific heat at constant volume C_v=(3/2)R. A diatomic gas, at moderate temperatures, has 5 degrees of freedom (3 translational + 2 rotational, since a diatomic molecule can rotate about two axes perpendicular to the bond axis), giving internal energy per mole = (5/2)RT and C_v=(5/2)R. Because diatomic gas molecules have more ways (degrees of freedom) to store thermal energy for the same rise in temperature, more heat is needed to raise their temperature by the same amount, giving them a higher molar specific heat than monatomic gases.

Quick visual: a worked diagram from the full Solutions page, for reference.

Frequently Asked Questions

How would you find the root mean square speed of gas molecules if the gas is heated to a higher temperature?
Since rms speed is proportional to the square root of absolute temperature, raising the temperature increases the rms speed by the square root of the temperature ratio.

If the pressure of an ideal gas doubles at constant volume, what happens to its temperature?
Since pressure is directly proportional to temperature at constant volume, the temperature also doubles.

Written by Satish

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