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Question #42512

Difficulty: 8/10 | 🟢 Free

The Question

As part of a thermodynamic cycle, one mole of a gas undergoes a reversible adiabatic expansion. The gas obeys the ideal gas equation of state $PV = RT$ , but its internal energy is temperature-dependent and given by $U(T) = \frac{3}{2}RT + \frac{R}{2T_0}T^2$ , where $R$ is the universal gas constant and $T_0$ is a positive reference temperature. During the expansion, the root-mean-square (RMS) speed of the gas molecules drops to exactly half of its initial value. If the expansion begins at a temperature $T_1 = 2T_0$ , what is the expansion ratio $\frac{V_2}{V_1}$ ?

Options

A.
$8 e^{3/2}$
B.
$128$
C.
$8 e^{3/4}$
D.
$16$

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

RMS vs Average vs Most Probable Speed|Internal Energy Dependency|Cyclic Process Geometry

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental understanding of adiabatic processes when the degrees of freedom (and thus the heat capacity) are temperature-dependent.

The primary trap is blindly applying the standard Poisson equation $PV^\gamma = \text{constant}$ .

Students who calculate an initial $\gamma = \frac{9}{7}$ and assume it remains constant will incorrectly arrive at an expansion ratio of 128 (Option B).

Similarly, using the final $\gamma = \frac{3}{2}$ yields 16 (Option D).

Another subtle trap involves incorrectly differentiating $U = \frac{f(T)}{2}RT$ by treating $f(T)$ as a constant, missing the $T f'(T)$ term, which leads to Option C.

The correct approach requires abandoning $PV^\gamma = \text{constant}$ entirely and integrating the first law of thermodynamics $dU = -P dV$ from first principles.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Kinetic Theory - Class 11 Part 2
Concept	RMS vs Average vs Most Probable Speed
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free