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Question #42170

Difficulty: 8/10 | 🟢 Free

The Question

The internal energy of a certain non-ideal gas is given by the relation $U = 3P(V + V_0)$ , where $P$ is the pressure, $V$ is the volume, and $V_0$ is a positive constant. The gas is initially at a pressure $P_0$ and volume $7V_0$ . It undergoes a reversible adiabatic expansion until its volume becomes $26V_0$ . What is the work done BY the gas during this process?

Options

A.
$8 P_0 V_0$
B.
$-8 P_0 V_0$
C.
$12 P_0 V_0$
D.
$\frac{20}{3} P_0 V_0$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Sign Convention Trap|Adiabatic Process Constraint|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's fundamental understanding of the First Law of Thermodynamics ( $dU = dQ - dW$ ) and the sign convention for work.

A common trap is blindly applying the ideal gas adiabatic condition $PV^\gamma = \text{constant}$ instead of deriving the path equation from $dU = -P dV$ .

Furthermore, students often confuse work done BY the system with work done ON the system, leading to the incorrect relation $dU = +P dV$ .

This sign error yields the trap answer $12 P_0 V_0$ .

By using a non-ideal internal energy function $U = 3P(V+V_0)$ , the problem forces students to perform the exact differential integration and correctly apply the thermodynamic sign convention to find the final state and subsequent work done.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Sign Convention Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free