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Question #42586

Difficulty: 8/10 | 🟢 Free

The Question

One mole of an ideal diatomic gas is heated to a high temperature such that its vibrational modes become partially active. The internal energy of the gas is modeled empirically as U(T)=52RT+RT22T0U(T) = \frac{5}{2}RT + \frac{RT^2}{2T_0}, where RR is the universal gas constant and T0T_0 is a characteristic constant temperature. If the gas undergoes a reversible adiabatic expansion, which of the following relationships between temperature TT and volume VV is valid throughout the process?

Options

  • A.

    T5/2VeT/T0=constantT^{5/2} V e^{T/T_0} = \text{constant}

  • B.

    T5/2Ve−T/T0=constantT^{5/2} V e^{-T/T_0} = \text{constant}

  • C.

    T7/2VeT/T0=constantT^{7/2} V e^{T/T_0} = \text{constant}

  • D.

    T5/2V=constantT^{5/2} V = \text{constant}

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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

Pressure-Kinetic Energy Relation|Energy of Vibrational Modes|Adiabatic Process Constraint|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

Standard intuition assumes the adiabatic index γ\gamma is a constant, leading students to blindly apply the standard relation TVγ−1=constantTV^{\gamma-1} = \text{constant}.

This question introduces a temperature-dependent internal energy model, simulating the gradual activation of vibrational degrees of freedom at high temperatures.

To solve it, students must abandon the constant-γ\gamma shortcut and derive the T−VT-V relationship from first principles using the First Law of Thermodynamics (dU=−PdVdU = -P dV).

The primary trap is Option D, which assumes vibrational modes are negligible (constant f=5f=5), and Option B, which tests for sign convention errors during integration.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterKinetic Theory - Class 11 Part 2
ConceptPressure-Kinetic Energy Relation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree