Adiabatic Process
Within an insulated cylinder a diatomic gas at 300 K is compressed adiabatically until its volume becomes one-fourth of the original. Using γ=1.4, what is the final temperature of the gas?
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About This Question
- Subject
- physics
- Chapter
- thermodynamics
- Topic
- adiabatic process
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
522 K
An adiabatic process exchanges no heat, and for an ideal gas it obeys TVγ−1=constant, so Tf=Ti(VfVi)γ−1. Here the gas is compressed, so Vi/Vf=4 and γ−1=0.4. Then Tf=300×40.4. Evaluating 40.4=e0.4ln4=e0.4(1.386)=e0.554≈1.74, giving Tf≈300×1.74=522 K. The value 418 K uses an exponent near 0.25, mismatching γ. The value 600 K assumes temperature scales linearly with volume ratio, ignoring the exponent. The value 300 K wrongly treats the process as isothermal, but no heat leaves the insulated cylinder so the temperature must change. The relation TVγ−1=constant follows from combining the adiabatic law PVγ=constant with the ideal-gas equation PV=nRT to eliminate pressure. An equivalent route uses the First Law with Q=0, so the work done on the gas equals its rise in internal energy nCVΔT, which again forces the temperature upward. As a physical check, adiabatic compression always heats a gas because work done on it raises its internal energy, so the final temperature exceeding 300 K is exactly what we expect.
This medium difficulty physics question is from the chapter thermodynamics, covering the topic of adiabatic process. It appeared in the 2025 exam.
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