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Question #42597

Difficulty: 8/10 | 🟢 Free

The Question

An ionized ideal gas mixture is contained within a flexible, thermally insulated, non-conducting spherical shell of radius RR. The shell carries a uniform total charge QQ. Due to electrostrictive effects of the surrounding medium, the shell exerts a net inward pressure on the gas given by P=σ22ϵ0P = \frac{\sigma^2}{2\epsilon_0}, where σ\sigma is the surface charge density. The gas mixture consists of 11 mole of a monoatomic gas and 11 mole of a rigid diatomic gas.

Initially, the system is in mechanical and thermal equilibrium. If the total charge QQ undergoes a very small fractional increase such that ΔQQ=α\frac{\Delta Q}{Q} = \alpha, the system reaches a new equilibrium via a reversible adiabatic process. What is the corresponding fractional change in the gas pressure, ΔPP\frac{\Delta P}{P}?

Options

  • A.

    18α18\alpha

  • B.

    2α2\alpha

  • C.

    −6α-6\alpha

  • D.

    10α10\alpha

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Equation of State for Mixtures|Relative Error Propagation in Powers|Non-conducting Charged Sphere

Why This Question Is Tricky

Common Mistakes

This question tests the coupled thermodynamic and electrostatic equilibrium of a deformable system.

The primary trap is the assumption that the sphere's radius remains constant, leading students to incorrectly apply simple error propagation P∝Q2  ⟹  ΔP/P=2αP \propto Q^2 \implies \Delta P/P = 2\alpha.

However, because the pressure depends on the surface charge density σ\sigma, any change in QQ alters the pressure, which in turn changes the volume adiabatically, further modifying σ\sigma.

Students must establish a coupled proportionality by equating the electrostrictive boundary condition with the adiabatic equation of state for a gas mixture (γ=1.5\gamma = 1.5).

Failing to account for the volumetric expansion/contraction yields incorrect sensitivities like 2α2\alpha (rigid sphere assumption), −6α-6\alpha (isothermal assumption), or 10α10\alpha (incorrectly assuming a pure monoatomic gas).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterKinetic Theory - Class 11 Part 2
ConceptEquation of State for Mixtures
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree