Isothermal Process
One mole of an ideal gas is compressed isothermally at temperature 300 K, so what happens to its internal energy during this slow compression process?
Select the correct option:
Solution
Internal energy remains unchanged
Drawing on NCERT Class 11, Chapter 12 (Thermodynamics), for an ideal gas the internal energy depends only on temperature, since the molecules are assumed to have no intermolecular potential energy. In an isothermal process the temperature is held constant, here at 300 K, so the internal energy must stay constant: Δcup=0. This holds whether the gas is compressed or expanded, as long as temperature does not change. Applying the First Law, ΔQ=Δcup+ΔW=0+ΔW, so all heat exchanged equals the work done; during compression the surroundings do work on the gas and an equal amount of heat is released. The option that internal energy increases with pressure is wrong because internal energy of an ideal gas is not a function of pressure at fixed temperature. The option that it decreases with volume is wrong for the same reason. The option that it doubles when volume halves confuses internal energy with pressure behaviour under Boyle's law. A sanity check: with ΔT=0 and U∝T for an ideal gas, Δcup=0 is consistent and dimensionally trivial.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More isothermal process Practice Questions
About This Question
- Subject
- physics
- Chapter
- thermodynamics
- Topic
- isothermal process
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
Internal energy remains unchanged
Drawing on NCERT Class 11, Chapter 12 (Thermodynamics), for an ideal gas the internal energy depends only on temperature, since the molecules are assumed to have no intermolecular potential energy. In an isothermal process the temperature is held constant, here at 300 K, so the internal energy must stay constant: Δcup=0. This holds whether the gas is compressed or expanded, as long as temperature does not change. Applying the First Law, ΔQ=Δcup+ΔW=0+ΔW, so all heat exchanged equals the work done; during compression the surroundings do work on the gas and an equal amount of heat is released. The option that internal energy increases with pressure is wrong because internal energy of an ideal gas is not a function of pressure at fixed temperature. The option that it decreases with volume is wrong for the same reason. The option that it doubles when volume halves confuses internal energy with pressure behaviour under Boyle's law. A sanity check: with ΔT=0 and U∝T for an ideal gas, Δcup=0 is consistent and dimensionally trivial.
This easy difficulty physics question is from the chapter thermodynamics, covering the topic of isothermal process. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse thermodynamics questions on RankGuru.