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Question #43026

Difficulty: 8/10 | 🟢 Free

The Question

Consider a composite thermodynamic system consisting of nn moles of an ideal monoatomic gas and a macroscopic magnetic dipole of magnetic moment magnitude μ\mu. The system is confined in a perfectly insulated cylinder with a movable frictionless piston. The total internal energy of the system is U=Ugas+UmagU = U_{\text{gas}} + U_{\text{mag}}, where UgasU_{\text{gas}} is the thermal internal energy of the gas and Umag=−μ⃗⋅B⃗U_{\text{mag}} = -\vec{\mu} \cdot \vec{B} is the magnetic potential energy of the dipole in a uniform external magnetic field of magnitude BB. Due to an internal reversible mechanism, the angle θ\theta between μ⃗\vec{\mu} and B⃗\vec{B} is constrained by the volume VV of the gas such that cos⁡θ=(V0V)2\cos\theta = \left(\frac{V_0}{V}\right)^2, where V0V_0 is the initial volume. The system expands quasi-statically, doing external work PdVP dV. If the initial pressure of the gas is P0P_0 and the initial angle is θ=0\theta = 0, what is the gas pressure PP when the volume reaches 2V02V_0?

Options

  • A.

    P025/3−μBV0(125/3−18)\frac{P_0}{2^{5/3}} - \frac{\mu B}{V_0} \left( \frac{1}{2^{5/3}} - \frac{1}{8} \right)

  • B.

    P025/3\frac{P_0}{2^{5/3}}

  • C.

    P025/3+μBV0(125/3−18)\frac{P_0}{2^{5/3}} + \frac{\mu B}{V_0} \left( \frac{1}{2^{5/3}} - \frac{1}{8} \right)

  • D.

    P025/3−μBV0(125/3−14)\frac{P_0}{2^{5/3}} - \frac{\mu B}{V_0} \left( \frac{1}{2^{5/3}} - \frac{1}{4} \right)

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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

Magnetic Dipole Moment Orientation|Adiabatic Process Constraint|Integration

Why This Question Is Tricky

Common Mistakes

Students often blindly apply the standard adiabatic relation PVγ=constantPV^\gamma = \text{constant} whenever they see dQ=0dQ = 0.

However, in this composite system, the internal energy includes both the gas's thermal energy and the magnetic dipole's orientation energy.

The work done by the gas during expansion is partially used to increase the magnetic potential energy of the dipole against the external magnetic field.

This internal energy coupling modifies the effective adiabatic process equation.

A secondary trap is failing to find the correct integrating factor (V2/3V^{2/3}) when solving the modified First Law differential equation, leading to dimensionally inconsistent results or incorrect coefficients.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptMagnetic Dipole Moment Orientation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree