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Question #42587

Difficulty: 8/10 | 🟢 Free

The Question

One mole of an ideal monatomic gas undergoes a cyclic process $\mathcal{C}$ in the $P-V$ plane described by the equation $(P - 3P_0)^2 + \alpha^2 (V - 3V_0)^2 = P_0^2$ , where $\alpha = P_0/V_0$ , $P_0 = 10^5 \text{ Pa}$ , and $V_0 = 2 \times 10^{-3} \text{ m}^3$ . What is the net change in the total internal translational kinetic energy of the gas after completing one full cycle?

Options

A.
0 \text{ J}
B.
$200\pi \text{ J}$
C.
$300\pi \text{ J}$
D.
$-200\pi \text{ J}$

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

Pressure-Kinetic Energy Relation|Cyclic Process Geometry|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental distinction between state functions and path functions in thermodynamics.

The trap is set by providing a specific, integrable geometric path (an ellipse) on the P-V diagram, baiting students into calculating the area enclosed by the cycle (which equals the net work done, $200\pi \text{ J}$ ).

Students often erroneously equate this cyclic work or a multiple of it to the internal kinetic energy flux.

However, internal translational kinetic energy is a state function ( $K = \frac{3}{2}PV$ ).

Over any closed loop, the initial and final states are identical, meaning the net change in internal kinetic energy is strictly zero, rendering the complex path integration completely unnecessary.

Quick Facts About This Question

Question Information

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