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Question #42503

Difficulty: 8/10 | 🟢 Free

The Question

A thermally insulated rigid container of volume $V$ initially contains $n_0$ moles of an ideal monatomic gas at temperature $T_0$ . A tiny pinhole is created, allowing the gas to slowly effuse into a surrounding vacuum. The process continues until the number of moles of gas remaining in the container becomes $n_0/e^2$ . What is the change in the molar entropy of the gas remaining inside the container? (Assume the hole is small enough that the Maxwell-Boltzmann velocity distribution is continuously maintained inside, and let $R$ be the universal gas constant).

Options

A.
$R$
B.
$2R$
C.
$0$
D.
$-R$

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

Entropy and Reversibility|Maxwell-Boltzmann Distribution|Effusion and Diffusion|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the understanding of the microscopic nature of effusion and its macroscopic thermodynamic consequences.

The lethal trap is assuming the process is either isentropic (due to the 'thermally insulated' keyword) or isothermal (assuming escaping particles carry the average internal energy of $1.5kT$ ).

In reality, effusion is a flux-dependent process where the probability of escape is proportional to velocity, skewing the energy distribution of escaping particles to $2kT$ .

This causes the remaining gas to cool down ( $T \propto N^{1/3}$ ).

Students who bypass the kinetic derivation and apply standard reversible adiabatic relations will incorrectly conclude $\Delta s = 0$ , while those ignoring the flux-weighting will get $2R$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Entropy and Reversibility
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free