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Question #42499

Difficulty: 8/10 | 🟢 Free

The Question

An ideal Carnot refrigerator operates between a cold reservoir and a hot reservoir. Both reservoirs contain the same ideal gas of molar mass $M$ in thermal equilibrium. The most probable speed of the gas molecules in the cold reservoir is $v_0$ , while the root-mean-square (RMS) speed of the gas molecules in the hot reservoir is $2v_0$ . What is the coefficient of performance (COP) of this refrigerator?

Options

A.
$\frac{3}{5}$
B.
$\frac{1}{3}$
C.
$\frac{1}{5}$
D.
1

Check Your Answer

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

Refrigerator Coefficient of Performance|RMS vs Average vs Most Probable Speed|First Law Application

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to link the macroscopic thermodynamic definition of the Coefficient of Performance (COP) with the microscopic kinetic theory of gases.

The primary trap lies in the distinction between the different statistical measures of molecular speed: most probable speed ( $v_{mp} \propto \sqrt{2T}$ ) and root-mean-square speed ( $v_{rms} \propto \sqrt{3T}$ ).

A common mistake is to assume that temperature is simply proportional to the square of the given speed without accounting for the specific prefactors of the Maxwell-Boltzmann distribution moments.

If a student incorrectly treats both speeds as the same type (e.g., both RMS), they will find $T_H = 4T_C$ and erroneously calculate a COP of $1/3$ .

Another trap is directly using the speeds in the COP formula instead of temperatures, yielding a COP of $1$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Refrigerator Coefficient of Performance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free