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Question #42189

Difficulty: 8/10 | 🟢 Free

The Question

An ideal monatomic gas (γ=5/3\gamma = 5/3) is confined in a cylindrical container of radius RR by a movable piston. The flat ends of the cylinder (including the piston) are perfectly thermally insulated and non-radiating. The curved walls of the cylinder behave as a perfectly black body (ϵ=1\epsilon = 1) and have negligible heat capacity, instantaneously taking the temperature of the gas TT. Initially, the piston is at a distance x0x_0 from the closed end, and the gas temperature is T0T_0. The piston is then moved outward at a constant velocity vv. Assuming the expansion is reversible and adiabatic (energy lost via radiation is negligible compared to the work done), what is the total energy radiated by the cylinder as the piston moves from x=x0x = x_0 to x=8x0x = 8x_0? Let P0P_0 be the initial power radiated by the curved walls.

Options

  • A.

    9P0x08v\frac{9 P_0 x_0}{8 v}

  • B.

    93P0x0160v\frac{93 P_0 x_0}{160 v}

  • C.

    3P0x04v\frac{3 P_0 x_0}{4 v}

  • D.

    21P0x016v\frac{21 P_0 x_0}{16 v}

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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

Adiabatic Process Constraint|Stefan-Boltzmann Power Law|Differential Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of thermodynamic processes with heat transfer under time-varying boundary conditions.

The primary trap is neglecting the time-dependence of the radiating surface area (A∝xA \propto x) as the piston expands.

Students often blindly apply the Stefan-Boltzmann law P∝T4P \propto T^4 using only the adiabatic temperature drop (T∝x1−γT \propto x^{1-\gamma}), which leads to the incorrect integral of x−8/3x^{-8/3} (Option B).

To solve this correctly, one must formulate the instantaneous power as a product of both the changing area and the changing temperature P(x)∝x⋅x4(1−γ)P(x) \propto x \cdot x^{4(1-\gamma)}, and then integrate over time using the kinematic constraint dt=dx/vdt = dx/v.

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Question Information

SubjectPhysics
ChapterThermodynamics - Class 11 Part 2
ConceptAdiabatic Process Constraint
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree