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Question #43946

Difficulty: 8/10 | 🟢 Free

The Question

Two moles of a non-ideal diatomic gas undergo an adiabatic expansion from an initial temperature T1=600 KT_1 = 600 \text{ K} to a final temperature T2=300 KT_2 = 300 \text{ K}. In this temperature range, the effective number of degrees of freedom of the gas, f(T)f(T), varies due to the partial activation of vibrational modes and is phenomenologically given by f(T)=5+2(TT0)2f(T) = 5 + 2 \left( \frac{T}{T_0} \right)^2, where T0=300 KT_0 = 300 \text{ K}. Calculate the work done by the gas during this expansion. (Assume the numerical coefficients in the options carry the appropriate units of mol⋅K\text{mol} \cdot \text{K}, and RR is the universal gas constant).

Options

  • A.

    2900R2900 R

  • B.

    3000R3000 R

  • C.

    2100R2100 R

  • D.

    3900R3900 R

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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

Adiabatic Processes via KTG|Energy Equipartition Theorem|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's fundamental understanding of the First Law of Thermodynamics and the definition of adiabatic work when the standard assumption of constant heat capacity fails.

A common trap is to blindly apply the standard adiabatic formula W=nR(T1−T2)γ−1W = \frac{nR(T_1 - T_2)}{\gamma - 1} or to assume an average value for the degrees of freedom.

Because the degrees of freedom f(T)f(T) vary with temperature, γ\gamma is not constant, rendering the phenomenological equation PVγ=constantPV^\gamma = \text{constant} invalid.

Students must revert to the fundamental relation dW=−dUdW = -dU and integrate the temperature-dependent heat capacity Cv(T)=f(T)R2C_v(T) = \frac{f(T)R}{2} over the given temperature range.

The incorrect options correspond to using the arithmetic mean of the degrees of freedom (3000R3000 R), the final degrees of freedom (2100R2100 R), or the initial degrees of freedom (3900R3900 R).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterKinetic Theory - Class 11 Part 2
ConceptAdiabatic/Isothermal Processes via KTG
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree