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Question #42475

Difficulty: 8/10 | 🟢 Free

The Question

One mole of a substance, initially entirely in its liquid state, is contained in a cylinder fitted with a frictionless piston. It undergoes a quasi-static isothermal expansion at its boiling temperature $T_0$ until its final volume becomes $V_f = 3 \frac{RT_0}{P_0}$ , where $P_0$ is the saturation vapor pressure at $T_0$ . The molar latent heat of vaporization is $L$ . Assuming the vapor behaves as an ideal gas and the molar volume of the liquid is negligible, which of the following represents the total change in internal energy ( $\Delta U$ ) and the total work done ( $W$ ) by the substance?

Options

A.
$\Delta U = L - RT_0$ , $W = RT_0(1 + \ln 3)$
B.
$\Delta U = 0$ , $W = RT_0 \ln 3$
C.
$\Delta U = L$ , $W = RT_0 \ln 3$
D.
$\Delta U = L - 3RT_0$ , $W = 3RT_0$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Variable Force Integration|Isothermal Process Work|Latent Heat and Volume Change

Why This Question Is Tricky

Common Mistakes

Students often incorrectly assume that for any isothermal process, the change in internal energy is zero ( $\Delta U = 0$ ), forgetting that this only applies to ideal gases and not to phase changes.

Furthermore, they may fail to recognize the piecewise nature of the pressure-volume relationship: pressure remains constant at the saturation vapor pressure during vaporization, and only follows Boyle's Law ( $P \propto 1/V$ ) after the liquid has completely vaporized.

This leads to incorrect integration limits and missing the $RT_0$ work term from the phase transition.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Work, Energy, and Power - Class 11 Part 1
Concept	Variable Force Integration
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free