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Question #44214

Difficulty: 8/10 | 🟢 Free

The Question

One mole of a real gas, obeying the van der Waals equation of state, undergoes an isothermal reversible expansion from an initial volume $V_1$ to a final volume $V_2$ at a constant temperature $T$ . Which of the following expressions correctly represents the heat absorbed ( $q$ ) by the gas during this process? (Given: $a$ and $b$ are van der Waals constants, and $R$ is the universal gas constant).

Options

A.
$RT \ln\left(\frac{V_2-b}{V_1-b}\right)$
B.
$RT \ln\left(\frac{V_2-b}{V_1-b}\right) + a\left(\frac{1}{V_2} - \frac{1}{V_1}\right)$
C.
$RT \ln\left(\frac{V_2}{V_1}\right)$
D.
$RT \ln\left(\frac{V_2-b}{V_1-b}\right) - a\left(\frac{1}{V_2} - \frac{1}{V_1}\right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Reversible Expansion Work|Internal Energy of Real Gases

Why This Question Is Tricky

Common Mistakes

This question tests a critical misconception in chemical thermodynamics: the assumption that $\Delta U = 0$ for ALL isothermal processes.

While true for ideal gases, real gases possess intermolecular attractive forces (represented by the van der Waals constant ' $a$ ').

As a real gas expands, work must be done against these internal cohesive forces, causing internal energy to increase even at constant temperature.

If a student blindly assumes $\Delta U = 0$ , they will incorrectly calculate $q = -w$ and choose the trap Option B.

The correct derivation requires either applying the First Law with the correct non-zero $\Delta U$ , or calculating $q_{rev}$ directly via the Maxwell relation for entropy ( $dS = (\partial P/\partial T)_V dV$ ).

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