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Question #42187

Difficulty: 8/10 | 🟢 Free

The Question

One mole of an ideal gas in a cylinder with a conducting piston of area AA undergoes a cyclic process 1→2→3→11 \to 2 \to 3 \to 1. Process 1→21 \to 2 is an isothermal expansion at temperature T0T_0 from volume V0V_0 to 2V02V_0. Process 2→32 \to 3 is isobaric compression at pressure P0P_0 back to volume V0V_0. Process 3→13 \to 1 is isochoric heating back to the initial state 1. The initial pressure is P1=2P0P_1 = 2P_0. During the expansion 1→21 \to 2, the piston moves through a magnetic field and experiences a retarding magnetic force F=αVF = \frac{\alpha}{V}, where α=P0V0A\alpha = P_0 V_0 A. The electrical energy from the resulting eddy currents is entirely dissipated as heat into the gas. The magnetic field is zero during 2→32 \to 3 and 3→13 \to 1. What is the net heat absorbed by the gas from the external thermostat during the complete cycle?

Options

  • A.

    P0V0(2ln⁡2−1)P_0 V_0 (2 \ln 2 - 1)

  • B.

    P0V0(ln⁡2−1)P_0 V_0 (\ln 2 - 1)

  • C.

    P0V0(3ln⁡2−1)P_0 V_0 (3 \ln 2 - 1)

  • D.

    P0V0ln⁡2P_0 V_0 \ln 2

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Isothermal Process Work|Cyclic Process Geometry|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of the First Law of Thermodynamics in a non-conservative hybrid system.

The classic trap is assuming that the net heat absorbed from the thermal reservoir (thermostat) in a cyclic process strictly equals the area enclosed by the P-V diagram (WgasW_{gas}).

Students often overlook that the total heat QnetQ_{net} entering the gas includes both the heat from the thermostat and the internal Joule heating generated by the non-conservative magnetic damping force (QeddyQ_{eddy}).

By failing to account for the mechanical work converted into heat via eddy currents, students incorrectly equate QthermQ_{therm} directly to ∮PdV\oint P dV, leading to the trap answer.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermodynamics - Class 11 Part 2
ConceptIsothermal Process Work
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree