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Question #42603

Difficulty: 8/10 | 🟢 Free

The Question

Two rigid vessels $A$ and $B$ , each of volume $V$ , contain an ideal monatomic gas. They are connected by a porous tube of length $L$ . The pore diameter of the tube is much smaller than the mean free path of the gas molecules. Initially, the entire system is at a uniform temperature $T_0$ and pressure $P_0$ . Vessel $A$ is maintained at $T_0$ , while vessel $B$ is heated and maintained at $4T_0$ . A steady-state non-linear temperature gradient $T(x)$ is established along the connecting tube. Neglecting the volume of the tube, what is the steady-state pressure in vessel $A$ ?

Options

A.
$\frac{4}{3} P_0$
B.
$\frac{8}{5} P_0$
C.
P_0
D.
$\frac{2}{3} P_0$

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

Effusion and Diffusion|Pressure-Kinetic Energy Relation|Steady State Temperature Gradient

Why This Question Is Tricky

Common Mistakes

This question tests the concept of thermal transpiration (Knudsen effect) in the kinetic theory of gases.

The primary trap is the assumption that mechanical equilibrium implies equal pressures ( $P_A = P_B$ ), which is only true for macroscopic fluid flow where the mean free path is much smaller than the connecting tube's diameter.

By specifying that the pore size is much smaller than the mean free path, the system operates in the effusion regime.

Students who incorrectly apply $P_A = P_B$ will obtain $P_A = 8P_0/5$ .

Furthermore, the mention of a 'non-linear temperature gradient' acts as a distractor to tempt students into unnecessary integration, whereas the steady-state zero net flux condition elegantly reduces to a boundary state relation $P_A/\sqrt{T_A} = P_B/\sqrt{T_B}$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Kinetic Theory - Class 11 Part 2
Concept	Effusion and Diffusion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free