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Question #42190

Difficulty: 8/10 | 🟢 Free

The Question

A thermally insulated rectangular box of length LL, width WW, and height HH contains nn moles of an ideal monoatomic gas of molar mass MM. The box is placed in a uniform downward gravitational field gg. Assume the gas is in thermal equilibrium at an initial temperature T0T_0, and the gravitational potential energy is small compared to the thermal energy (MgH≪RT0MgH \ll RT_0, where RR is the universal gas constant). The vertical wall at x=Lx=L (of area W×HW \times H) is flexible. An external agent quasi-statically pushes this wall inwards such that the inward displacement of the wall at a height yy from the bottom is given by δ(y)=δ0yH\delta(y) = \delta_0 \frac{y}{H}, where δ0≪L\delta_0 \ll L. Find the change in temperature ΔT\Delta T of the gas. (Keep terms up to first order in MgHRT0\frac{MgH}{RT_0}).

Options

  • A.

    T0δ03L(1−MgH6RT0)\frac{T_0 \delta_0}{3 L} \left( 1 - \frac{M g H}{6 R T_0} \right)

  • B.

    T0δ03L(1+MgH6RT0)\frac{T_0 \delta_0}{3 L} \left( 1 + \frac{M g H}{6 R T_0} \right)

  • C.

    T0δ03L(1−MgH3RT0)\frac{T_0 \delta_0}{3 L} \left( 1 - \frac{M g H}{3 R T_0} \right)

  • D.

    T0δ03L\frac{T_0 \delta_0}{3 L}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Adiabatic Process Constraint|Superposition Principle|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the superposition of infinitesimal work components in a system with a non-uniform pressure distribution.

Students typically memorize the global adiabatic constraint TVγ−1=constantTV^{\gamma-1} = \text{constant} and apply it using the average volume change, which implicitly assumes uniform pressure.

By introducing a gravitational pressure gradient and a spatially varying boundary displacement, the local work dW=P(y)dA⋅dx(y)dW = P(y) dA \cdot dx(y) must be integrated explicitly.

The trap lies in either ignoring the pressure gradient entirely (yielding the standard uniform-pressure result) or failing to normalize the base pressure P0P_0 against the total number of moles, which leads to an incorrect first-order correction.

Quick Facts About This Question

Question Information

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