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Question #42591

Difficulty: 8/10 | 🟢 Free

The Question

A planetary atmosphere consists of an ideal monatomic gas of molar mass $M$ . Due to a specific mixing mechanism, the atmosphere maintains a temperature profile such that the sum of the average microscopic kinetic energy per molecule and its macroscopic gravitational potential energy remains constant at all heights $h$ . Assuming a uniform gravitational acceleration $g$ , surface temperature $T_0$ , and surface pressure $P_0$ , what is the atmospheric pressure $P$ at a height $h = \frac{RT_0}{2Mg}$ ? (Here $R$ is the universal gas constant).

Options

A.
$P_0 \left( \frac{2}{3} \right)^{\frac{3}{2}}$
B.
$P_0 e^{-\frac{1}{2}}$
C.
$P_0 \left( \frac{1}{2} \right)^{\frac{5}{2}}$
D.
$P_0 \left( \frac{5}{6} \right)^{\frac{3}{2}}$

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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

Temperature and Kinetic Energy|Gravitational Potential Reference|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple microscopic kinetic theory with macroscopic hydrostatic equilibrium in a non-isothermal (polytropic) atmosphere.

The primary trap is the assumption of an isothermal environment, which leads to the standard exponential barometric formula $P = P_0 e^{-Mgh/RT_0}$ (Option B).

By explicitly defining a linear energy-height relationship, students must derive the temperature gradient and integrate the hydrostatic equation $\frac{dP}{P} = -\frac{Mg}{RT}dh$ with a variable $T$ , testing their fundamental understanding of energy conservation and fluid statics over rote memorization of the barometric law.

Quick Facts About This Question

Question Information

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