Explain why the entropy of an ideal gas increases when it undergoes a free expansion into a vacuum, even though no heat is exchanged with the surroundings and no work is done.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
An ideal gas is initially confined to one half of an insulated rigid container. A partition separating the gas from the evacuated half is suddenly removed, and the gas expands to fill the entire container.
Model answer (4 marks)
1. In a free expansion no work is done (W=0) and no heat is exchanged (Q=0), so the internal energy and temperature of an ideal gas remain unchanged.
2. The process is irreversible; the gas will not spontaneously return to the original half of the container, so the reverse process is never observed.
3. After expansion the gas molecules occupy a larger volume, giving many more possible microstates (ways to arrange positions and momenta of the molecules).
4. Entropy is S=k_B ln Ω; an increase in the number of microstates means S increases. The second law states that for an isolated system entropy cannot decrease, so the entropy of the gas rises.
2. The process is irreversible; the gas will not spontaneously return to the original half of the container, so the reverse process is never observed.
3. After expansion the gas molecules occupy a larger volume, giving many more possible microstates (ways to arrange positions and momenta of the molecules).
4. Entropy is S=k_B ln Ω; an increase in the number of microstates means S increases. The second law states that for an isolated system entropy cannot decrease, so the entropy of the gas rises.
Examiner tips
- Use the command word ‘Explain’ – give a clear cause and effect chain. Mention W=0, Q=0, irreversibility, increase in Ω, and the second law. Keep each point concise and use the exact terminology (entropy, microstates, isolated system).
Common mistakes
- Saying the temperature rises – it actually stays constant for an ideal gas. Forgetting to state that the process is irreversible. Using vague terms like ‘more disorder’ instead of ‘more microstates’.
Mark scheme (4 marks)
- In a free expansion, no work is done (W = 0) and no heat is exchanged (Q = 0), so the internal energy and temperature of an ideal gas remain unchanged.
- The process is irreversible because the gas will not spontaneously return to the original half of the container; the reverse process is never observed.
- After expansion the gas molecules occupy a greater volume, so there are significantly more possible microstates (ways to arrange positions and momenta of the molecules) available to the system.
- Since entropy is related to the number of microstates by S = k_B ln Ω, an increase in the number of microstates means entropy increases; the second law states that for an isolated system entropy cannot decrease, and here it increases.
Key terms in this question
Related
- All IB DP Physics Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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