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Question #85168

Difficulty: 5/10 | 🟢 Free

The Question

A rigid, perfectly insulated, closed cylindrical container of length $L$ contains $n$ moles of an ideal gas of molar mass $M$ and adiabatic exponent $\gamma$ . Initially, the container is at rest and the gas is in thermal equilibrium at temperature $T_0$ . The container is then given a constant acceleration $a$ along its longitudinal axis. After a long time, internal dissipative forces and thermal conduction cause the gas to reach a new steady state (isothermal equilibrium) in the accelerating frame. Assuming the dimensionless parameter $\alpha = \frac{M a L}{R T_0} \ll 1$ , which of the following expressions correctly represents the change in temperature $\Delta T$ of the gas? (Note: $R$ is the universal gas constant. Neglect the heat capacity of the container.)

Options

A.
$\frac{(\gamma - 1) M^2 a^2 L^2}{12 R^2 T_0}$
B.
$0$
C.
$\frac{(\gamma - 1) M^2 a^2 L^2}{12 \gamma R^2 T_0}$
D.
$\frac{(\gamma - 1) M a L}{2 R}$

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

Sign Convention Trap|Frame of Reference Inertial vs Non-Inertial|Pseudo Force Directionality

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

The core trap tests the application of the First Law of Thermodynamics in a non-inertial reference frame.

Students often memorize $\Delta U = Q - W$ where $W = \int P dV$ , leading them to incorrectly conclude that in a rigid, adiabatic container ( $dV=0, Q=0$ ), the temperature change must be zero.

They fail to account for the macroscopic work done by the pseudo-force field (or equivalently, the change in pseudo-potential energy) as the gas redistributes its center of mass.

Furthermore, the options trap students who correctly identify the pseudo-work but incorrectly assume the gas completely collapses to the back wall, missing the nuanced integration of the exponential pressure gradient required for small $\alpha$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Sign Convention Trap
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free