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Question #44216

Difficulty: 8/10 | 🟢 Free

The Question

A gaseous reaction \ce{A(g) -> 2B(g)} follows zero-order kinetics. When the reaction is carried out in a rigid $1.0 \text{ L}$ flask at $300 \text{ K}$ , the total pressure increases at a constant rate of $0.2463 \text{ atm min}^{-1}$ .

In a separate experiment, $1.0 \text{ mole}$ of pure \ce{A(g)} is placed in a cylinder fitted with a frictionless piston at $300 \text{ K}$ . The initial volume is $10.0 \text{ L}$ . The gas undergoes a reversible isothermal expansion at $300 \text{ K}$ , during which the volume of the cylinder increases at a constant rate of $0.5 \text{ L min}^{-1}$ .

Assuming ideal gas behavior, calculate the magnitude of the work done by the gas (in Joules) during the first $20 \text{ minutes}$ of the expansion.

(Given: $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ , $\ln 2 = 0.693$ )

Options

A.
1882 J
B.
1728 J
C.
2035 J
D.
18.6 J

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

Reversible Expansion Work|Units of Rate Constant (k)

Why This Question Is Tricky

Common Mistakes

This problem tests the synthesis of chemical kinetics and thermodynamics.

The core trap is a dimensional analysis mismatch.

Students must first use the gas constant $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$ to extract the molar rate of reaction from the pressure-time data.

Then, they must construct a time-dependent integral for reversible work that accounts for both the changing number of moles (due to the reaction) and the changing volume.

Finally, they must use $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ to calculate the work in Joules.

Common errors include ignoring the reaction during expansion (yielding 1728 J), miscalculating the stoichiometry of total moles (yielding 2035 J), or failing to convert L-atm to Joules by using the wrong $R$ value in the final step (yielding 18.6 J).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Reversible Expansion Work
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free