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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Specific Heat Capacities of Gases

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Specific Heat Capacities of Gases

When 50 J of heat is supplied to an ideal gas, it expands from 100...

When 50 J50 \text{ J} of heat is supplied to an ideal gas, it expands from 100 cm3100 \text{ cm}^3 to 200 cm3200 \text{ cm}^3 at a constant pressure of 2.0×105 Pa2.0 \times 10^5 \text{ Pa}. Calculate the change in internal energy of the gas.

Options

  1. A50 J50 \text{ J}
  2. B70 J70 \text{ J}
  3. C30 J30 \text{ J}
  4. D20 J20 \text{ J}

Correct answer

C. 30 J30 \text{ J}

Step-by-step solution

Given heat supplied to the gas, Q=50 JQ = 50 \text{ J}.

Initial volume, V1=100 cm3=100×10−6 m3V_1 = 100 \text{ cm}^3 = 100 \times 10^{-6} \text{ m}^3.

Final volume, V2=200 cm3=200×10−6 m3V_2 = 200 \text{ cm}^3 = 200 \times 10^{-6} \text{ m}^3.

Constant pressure, P=2.0×105 PaP = 2.0 \times 10^5 \text{ Pa}.

Work done by the gas is given by:

W=P(V2−V1)W = P(V_2 - V_1)

W=2.0×105×(200−100)×10−6 JW = 2.0 \times 10^5 \times (200 - 100) \times 10^{-6} \text{ J}

W=2.0×105×10−4 J=20 JW = 2.0 \times 10^5 \times 10^{-4} \text{ J} = 20 \text{ J}

According to the first law of thermodynamics:

Q=ΔU+WQ = \Delta U + W

ΔU=Q−W\Delta U = Q - W

ΔU=50 J−20 J=30 J\Delta U = 50 \text{ J} - 20 \text{ J} = 30 \text{ J}
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