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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Laws of Thermodynamics

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Laws of Thermodynamics

A gas is confined within a cylindrical vessel equipped with a...

A gas is confined within a cylindrical vessel equipped with a frictionless piston. The gas is heated slowly for some time. Throughout the process, 10 J10 \text{ J} of heat is added, and the piston moves outward by 10 cm10 \text{ cm}. Determine the increase in the internal energy of the gas. The cylinder's cross-sectional area is 4 cm24 \text{ cm}^2, and the atmospheric pressure is 100 kPa100 \text{ kPa}.

Options

  1. A6 J6 \text{ J}
  2. B14 J14 \text{ J}
  3. C4 J4 \text{ J}
  4. D10 J10 \text{ J}

Correct answer

A. 6 J6 \text{ J}

Step-by-step solution

The process is isobaric since the piston is frictionless and moves slowly, maintaining equilibrium with the atmospheric pressure.

The work done by the gas is given by W=PΔV=PAdW = P \Delta V = P A d.

Substituting the given values:

W=(100×103 Pa)×(4×10−4 m2)×(0.1 m)=4 JW = (100 \times 10^3 \text{ Pa}) \times (4 \times 10^{-4} \text{ m}^2) \times (0.1 \text{ m}) = 4 \text{ J}

According to the first law of thermodynamics:

Q=ΔU+WQ = \Delta U + W

Given that the heat added is Q=10 JQ = 10 \text{ J}, the increase in internal energy ΔU\Delta U is:

ΔU=Q−W=10 J−4 J=6 J\Delta U = Q - W = 10 \text{ J} - 4 \text{ J} = 6 \text{ J}
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