Before You Begin

Question #44692

Difficulty: 8/10 | 🟢 Free

The Question

A $40.4 \text{ g}$ sample of a noble gas mixture containing isotopes \ce{^{20}Ne} and \ce{^{22}Ne} in a $9:1$ molar ratio is confined in a thermally insulated container of volume $10 \text{ L}$ at $300 \text{ K}$ . The gas is allowed to undergo adiabatic free expansion into an evacuated chamber, increasing the total volume to $80 \text{ L}$ . Assuming ideal behavior, what is the change in internal energy ( $\Delta U$ ) and the change in entropy of the system ( $\Delta S_{sys}$ )?

Options

A.
$\Delta U = 0$ , $\Delta S_{sys} = 6 R \ln 2$
B.
$\Delta U = -675 R$ , $\Delta S_{sys} = 0$
C.
$\Delta U = 0$ , $\Delta S_{sys} = 6.06 R \ln 2$
D.
$\Delta U = -681.75 R$ , $\Delta S_{sys} = 0$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Isotopes-Isobars-Isotones-Isoelectronic|Adiabatic Free Expansion|Empirical vs Molecular Formula

Why This Question Is Tricky

Common Mistakes

This problem tests the conceptual understanding of adiabatic free expansion versus reversible adiabatic expansion, combined with isotopic average mass calculations.

The primary trap is the word 'adiabatic', which lures students into assuming $\Delta S_{sys} = 0$ (which is only true for reversible adiabatic processes) and calculating a temperature drop using $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$ to find $\Delta U = -675 R$ .

A secondary trap is ignoring the isotopic abundance and using a standard molar mass of $20 \text{ g/mol}$ , leading to an incorrect mole count ( $2.02 \text{ mol}$ ) and thus an incorrect entropy change of $6.06 R \ln 2$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Isotopes-Isobars-Isotones-Isoelectronic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free