Before You Begin

Question #83547

Difficulty: 5/10 | 🟢 Free

The Question

A vertical diathermic cylinder contains $n$ moles of an ideal gas with adiabatic exponent $\gamma$ . A frictionless piston of mass $M$ and cross-sectional area $A$ rests on the gas, maintaining an equilibrium volume $V_0$ and pressure $P_0$ . The cylinder walls conduct heat to the surrounding environment (at constant temperature $T_0$ ) according to Newton's Law of Cooling: $\frac{dQ}{dt} = -K(T - T_0)$ , where $K$ is the thermal conductance. The piston is given a small displacement and released. Assuming $K$ is sufficiently small such that the oscillations are nearly adiabatic but experience weak damping, the thermal dissipation can be modeled as an effective viscous drag force $F_{\text{drag}} = -b v$ on the piston. Determine the effective damping coefficient $b$ .

Options

A.
$\frac{K M (\gamma - 1)^2}{n R \gamma}$
B.
$\frac{K M (\gamma - 1)}{n R \gamma}$
C.
$\frac{K (\gamma - 1)^2}{n R \gamma}$
D.
$\frac{K M^2 (\gamma - 1)}{n R \gamma}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Newton’s Law of Cooling Limits|Damped Oscillation Equation|Energy Partitioning|Phase Space Trajectory

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the conflation of steady-state heat transfer with dynamic thermal equilibrium.

Students often incorrectly assume that a small thermal conductance $K$ implies a perfectly adiabatic process (an undamped harmonic oscillator forming a closed ellipse in phase space).

However, Newton's Law of Cooling acts as a linear damping term, forcing the phase-space trajectory to spiral inward.

By requiring the mapping of a thermodynamic dissipation mechanism onto a mechanical viscous drag model, the problem traps students who attempt to use standard adiabatic or isothermal work formulas without integrating the first-order non-conservative thermal correction over a full cycle.

Additionally, it rewards dimensional analysis: since $K/nR$ has units of $s^{-1}$ , the damping coefficient $b$ (units $kg \cdot s^{-1}$ ) must be proportional to mass $M$ , immediately eliminating two options.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermal Properties of Matter - Class 11 Part 2
Concept	Newton’s Law of Cooling Limits
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free