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Question #43920
The Question
A horizontal cylinder of cross-sectional area contains an ideal monoatomic gas at an initial equilibrium pressure and volume . A movable piston of mass seals the gas and is connected to a fixed vertical wall by a linear spring of spring constant . The region on the other side of the piston is a vacuum. At equilibrium, the spring is in a stretched state. An external agent slowly displaces the piston outward by a small distance (where ). Let be the total work done by the external agent if the gas undergoes a reversible adiabatic process, and be the total work done if the gas undergoes a reversible isothermal process. Which of the following expressions correctly represents the difference to the leading order in ?
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Multi-Concept_Bridge
Master problems that require linking physics and math concepts to solve complex competitive questions.
Concepts Behind This Question
Effective Spring Constant|Adiabatic/Isothermal Processes via KTG|Integration
Why This Question Is Tricky
Common Mistakes
Students often misapply the work formula by only considering the first-order approximation , leading them to incorrectly conclude that the work done is identical for both processes (Option D).
Furthermore, they fail to recognize that the gas acts as a nonlinear spring whose 'effective spring constant' is derived from the second derivative of its potential energy function .
By equating the external work to the change in total potential energy, the first-order terms perfectly cancel due to the equilibrium condition ().
The difference in work strictly emerges from the second-order Taylor expansion terms, which depend on the polytropic index ( vs 1) governing the curvature of the potential well.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Oscillations - Class 11 Part 2 |
| Concept | Effective Spring Constant |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |