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Question #42548

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The Question

One mole of an ideal monatomic gas at initial temperature $T_0$ and volume $V_0$ is compressed adiabatically to a final volume $V_0/8$ . In Process I, the compression is performed quasi-statically. In Process II, the compression is performed suddenly against a constant external pressure equal to the final pressure of Process I. Let $\Delta S_I$ and $\Delta S_{II}$ be the change in entropy of the gas in Process I and Process II, respectively. Which of the following represents the final temperature $T_{II}$ in Process II and the correct relation between the entropies?

Options

A.
$T_{II} = \frac{59}{3}T_0$ , $\Delta S_{II} > \Delta S_I = 0$
B.
$T_{II} = 4T_0$ , $\Delta S_{II} = \Delta S_I = 0$
C.
$T_{II} = \frac{59}{3}T_0$ , $\Delta S_{II} = \Delta S_I = 0$
D.
$T_{II} = 4T_0$ , $\Delta S_{II} > \Delta S_I = 0$

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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|Entropy and Reversibility|Phase Space Trajectory

Why This Question Is Tricky

Common Mistakes

This question targets the lethal misconception that any adiabatic process ( $dQ=0$ ) is inherently isentropic ( $dS=0$ ).

Students often blindly apply the relation $PV^\gamma = \text{constant}$ to all adiabatic processes, failing to distinguish between quasi-static (reversible) and rapid (irreversible) compressions.

In a sudden compression against a constant external pressure, the work done on the gas is significantly greater than in the reversible case, leading to a higher final temperature.

Consequently, the irreversible nature of the rapid compression generates entropy, meaning $\Delta S > 0$ despite zero heat exchange with the surroundings.

Recognizing the limits of the isentropic assumption is a hallmark of JEE Advanced thermodynamics.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Adiabatic Process Constraint
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free