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Question #42186

Difficulty: 8/10 | 🟢 Free

The Question

A vertical cylinder of cross-sectional area $A$ contains a mixture of two non-reacting ideal gases with molar masses $M_1$ and $M_2$ , having $n_1$ and $n_2$ moles respectively. The cylinder is closed by a movable, frictionless piston initially at a height $H$ from the base. The system is maintained at a constant temperature $T$ in a uniform downward gravitational field $g$ . If the piston is slowly raised to an infinite height, what is the total work done by the gas mixture on the piston?

Options

A.
$-RT \left[ n_1 \ln\left(1 - e^{-\frac{M_1 g H}{RT}}\right) + n_2 \ln\left(1 - e^{-\frac{M_2 g H}{RT}}\right) \right]$
B.
$-(n_1 + n_2) RT \ln\left(1 - e^{-\frac{(n_1 M_1 + n_2 M_2) g H}{(n_1 + n_2) RT}}\right)$
C.
$RT \left[ n_1 \ln\left(1 + e^{-\frac{M_1 g H}{RT}}\right) + n_2 \ln\left(1 + e^{-\frac{M_2 g H}{RT}}\right) \right]$
D.
$(n_1 + n_2) RT \ln\left( \frac{(M_1 + M_2) g H}{2 RT} \right)$

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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

Isothermal Process Work|Dalton's Law of Partial Pressures|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Boltzmann distribution and Dalton's Law of partial pressures in a non-uniform density field.

A common trap for students is to treat the gas mixture as a single equivalent fluid with an average molar mass $M_{avg} = (n_1 M_1 + n_2 M_2)/(n_1 + n_2)$ , leading to an incorrect pressure gradient (Option B).

Because the gases have different molar masses, their density gradients decouple.

The total work must be calculated by integrating the superposition of their independent partial pressures over the spatial gradient, rather than using standard uniform-density boundary conditions.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Isothermal Process Work
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free