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Law of Equipartition of Energy - UNSOLVED PRACTICE SET

Class 11

Chapter: Kinetic Theory of Gases | Topic: Law of Equipartition of Energy

Study Material.
Class 11

LAW OF EQUIPARTITION OF ENERGY - UNSOLVED PRACTICE SET

Topic: Law of Equipartition of Energy

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. According to the law of equipartition of energy, the energy associated with each degree of freedom is:

  1. kT
  2. (1/2)kT
  3. (3/2)kT
  4. 2kT

Q2. A monoatomic ideal gas has how many degrees of freedom per molecule?

  1. 1
  2. 2
  3. 3
  4. 5

Q3. The internal energy of one mole of a monoatomic ideal gas is:

  1. (1/2)RT
  2. RT
  3. (3/2)RT
  4. (5/2)RT

Q4. At high temperatures, a diatomic gas has 7 degrees of freedom because:

  1. Translational: 3, Rotational: 2, Vibrational: 2
  2. Translational: 3, Rotational: 3, Vibrational: 1
  3. Translational: 2, Rotational: 3, Vibrational: 2
  4. Translational: 3, Rotational: 2, Vibrational: 1

Q5. The ratio of specific heats γ for a monoatomic gas is:

  1. 5/3
  2. 7/5
  3. 4/3
  4. 9/7

Q6. When you heat a gas in a closed container, the molecules move faster. According to equipartition, the extra energy goes to:

  1. Only translational motion
  2. All available degrees of freedom equally
  3. Only rotational motion
  4. Only vibrational motion

Short Answer Questions

Q7. State the law of equipartition of energy. What is meant by a "degree of freedom"?

Q8. Explain why a monoatomic gas has 3 degrees of freedom while a diatomic gas has 5 degrees of freedom at moderate temperatures.

Q9. Using equipartition, show that the average kinetic energy per molecule of any ideal gas is (3/2)kT, regardless of its molecular structure.

Q10. What is the significance of the law of equipartition in explaining why C_v differs for monoatomic, diatomic, and polyatomic gases?

Q11. Explain why vibrational degrees of freedom are not active at room temperature but become active at very high temperatures.

Q12. A gas has γ = 7/5. Is it monoatomic, diatomic, or polyatomic? Justify your answer.

Long Answer Questions

Q13. State and explain the law of equipartition of energy. Using this law, derive expressions for the internal energy of:

(i) One mole of a monoatomic gas

(ii) One mole of a diatomic gas at moderate temperature (5 degrees of freedom)

(iii) One mole of a diatomic gas at high temperature (7 degrees of freedom)

(iv) One mole of a polyatomic gas

Also derive the corresponding values of C_v, C_p, and γ for each case.

Q14. Explain the concept of degrees of freedom in detail. For each of the following molecules, determine the number of degrees of freedom at room temperature and justify your answer:

(i) Helium (He)

(ii) Nitrogen (N₂)

(iii) Carbon dioxide (CO₂)

(iv) Methane (CH₄)

(v) Water vapor (H₂O)

Also explain why the equipartition theorem gives approximate rather than exact values for real gases.

Q15. The law of equipartition predicts C_v = (5/2)R for a diatomic gas, but experimental values for hydrogen at low temperatures are closer to (3/2)R, and at very high temperatures approach (7/2)R.

(i) Explain why C_v is (3/2)R at very low temperatures.

(ii) Explain why C_v becomes (5/2)R at moderate temperatures.

(iii) Explain why C_v approaches (7/2)R at very high temperatures.

(iv) What does this tell us about the validity and limitations of the classical equipartition theorem?

Numerical / Application-Based Problems

Q16. One mole of an ideal gas is contained in a vessel. Calculate the internal energy, C_v, C_p, and γ for:

(i) A monoatomic gas (e.g., argon)

(ii) A diatomic gas at moderate temperature (e.g., nitrogen at 300 K)

(iii) A diatomic gas at high temperature (e.g., nitrogen at 2000 K)

(iv) A polyatomic gas (e.g., CO₂)

(Given: R = 8.31 J mol⁻¹ K⁻¹)

Q17. A container holds a mixture of 2 moles of helium (monoatomic) and 3 moles of nitrogen (diatomic) at 400 K.

(i) Calculate the internal energy of helium.

(ii) Calculate the internal energy of nitrogen.

(iii) Calculate the total internal energy of the mixture.

(iv) Calculate the effective γ for the mixture.

(v) If the mixture is heated by 100 K at constant volume, calculate the total heat required.

(Given: R = 8.31 J mol⁻¹ K⁻¹)

Q18. The molar specific heat at constant volume for a certain gas is measured to be 3R.

(i) How many degrees of freedom does each molecule have?

(ii) Is the gas monoatomic, diatomic, or polyatomic? Explain.

(iii) Calculate C_p and γ for this gas.

(iv) If 2 moles of this gas are heated from 300 K to 400 K at constant pressure, calculate the heat supplied, work done, and change in internal energy.

(Given: R = 8.31 J mol⁻¹ K⁻¹)


Total: 30 Marks | Time: 40 mins

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