Third Law of Thermodynamics - UNSOLVED PRACTICE SET

Q7. State the Third Law of Thermodynamics. What is meant by a 'perfect crystalline substance'?

Q8. Why does the Third Law of Thermodynamics allow us to calculate absolute entropy values, whereas the First and Second Laws only allow us to calculate entropy changes?

Q9. Explain what is meant by 'residual entropy.' Give an example of a substance that exhibits residual entropy.

Q10. Why is it impossible to reach absolute zero (0 K) in a finite number of steps? Explain with reference to the Third Law.

Q11. The entropy of CO(s) at very low temperatures is not zero, even though it is a solid. Explain this observation.

Q12. How does the Third Law of Thermodynamics help in calculating the standard entropy change (ΔS°) of a chemical reaction?

Q13. (a) State the Third Law of Thermodynamics clearly. Why is it considered the 'law of absolute entropy'?

(b) Explain why a perfect crystalline substance has zero entropy at absolute zero. Use the concept of molecular disorder in your explanation.

(c) A student argues that since entropy is a measure of disorder, the entropy of any substance at 0 K should be zero. Is the student always correct? Give examples to support your answer.

Q14. (a) Define absolute entropy and explain how it is calculated using the Third Law of Thermodynamics.

(b) Describe the process of calculating the absolute entropy of a substance from 0 K to temperature T using heat capacity data. Write the integral expression involved.

(c) Why is the Debye model used for calculating entropy at very low temperatures instead of direct calorimetric measurements?

Q15. (a) Explain the concept of residual entropy with reference to ice and carbon monoxide. Why do these substances have non-zero entropy even at temperatures approaching 0 K?

(b) The residual entropy of ice is approximately 3.4 J/K·mol. Explain the molecular origin of this value in terms of hydrogen bonding disorder.

(c) How does the presence of isotopes affect the entropy of a crystalline substance at very low temperatures? Explain.

Q16. The standard absolute entropies at 298 K are given below:

S° [H₂(g)] = 130.6 J/K·mol

S° [O₂(g)] = 205.0 J/K·mol

S° [H₂O(l)] = 69.9 J/K·mol

For the reaction: 2H₂(g) + O₂(g) → 2H₂O(l)

(a) Calculate the standard entropy change (ΔS°) for this reaction.

(b) Is the sign of ΔS° consistent with the decrease in the number of moles of gas? Explain.

(c) Given ΔH° = –571.6 kJ for this reaction, calculate ΔG° at 298 K.

(d) Is this reaction spontaneous under standard conditions? Explain.

Q17. The heat capacity of a solid substance is measured at various temperatures. The following data is obtained:

At 10 K: Cp = 0.12 J/K·mol

At 50 K: Cp = 8.5 J/K·mol

At 100 K: Cp = 18.2 J/K·mol

At 298 K: Cp = 24.5 J/K·mol

(a) Explain how you would use this data to calculate the absolute entropy of the substance at 298 K.

(b) Write the integral expression used for this calculation.

(c) Why must the Debye T³ law be used for temperatures below about 15 K instead of direct experimental data?

(d) If the substance is a perfect crystal, what would be its entropy at 0 K? State the law that supports your answer.

Q18. In cryogenic research facilities in India (such as those at ISRO and BARC), scientists work with substances at temperatures close to absolute zero.

(a) Explain why it is theoretically impossible to cool any substance to exactly 0 K, according to the Third Law of Thermodynamics.

(b) The lowest temperature achieved in a laboratory is about 100 picokelvin (100 × 10⁻¹² K). Explain why reaching even lower temperatures becomes increasingly difficult as temperature approaches 0 K.

(c) Liquid helium is used as a coolant in many cryogenic applications. At 4.2 K, liquid helium has a relatively high entropy compared to other substances at similar temperatures. Explain why this is so, and discuss its implications for cooling systems.

(d) Superconductors operate at very low temperatures. Explain how the Third Law of Thermodynamics helps scientists understand the entropy changes associated with the superconducting transition.