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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

Occupying 100 cc , an ideal gas is initially at a pressure of 2.5 ×...

Occupying 100 cc100 \text{ cc}, an ideal gas is initially at a pressure of 2.5×105 Pa2.5 \times 10^5 \text{ Pa} and a temperature of 300 K300 \text{ K}. It is adiabatically compressed to half its original volume. Evaluate the work done by the gas during this process. Take γ=1.5\gamma = 1.5.

Options

  1. A21 J21 \text{ J}
  2. B−10 J-10 \text{ J}
  3. C−21 J-21 \text{ J}
  4. D−17 J-17 \text{ J}

Correct answer

C. −21 J-21 \text{ J}

Step-by-step solution

Given:
Initial volume V1=100 cc=10−4 m3V_1 = 100 \text{ cc} = 10^{-4} \text{ m}^3
Initial pressure P1=2.5×105 PaP_1 = 2.5 \times 10^5 \text{ Pa}
Final volume V2=V12=50 cc=0.5×10−4 m3V_2 = \dfrac{V_1}{2} = 50 \text{ cc} = 0.5 \times 10^{-4} \text{ m}^3
Ratio of specific heats γ=1.5\gamma = 1.5

For an adiabatic process, P1V1γ=P2V2γP_1 V_1^\gamma = P_2 V_2^\gamma

P2=P1(V1V2)γP_2 = P_1 \left( \dfrac{V_1}{V_2} \right)^\gamma

P2=2.5×105×(2)1.5=2.5×105×22=52×105 PaP_2 = 2.5 \times 10^5 \times (2)^{1.5} = 2.5 \times 10^5 \times 2\sqrt{2} = 5\sqrt{2} \times 10^5 \text{ Pa}

The work done by the gas in an adiabatic process is given by:

W=P1V1−P2V2γ−1W = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}

P1V1=(2.5×105)×(10−4)=25 JP_1 V_1 = (2.5 \times 10^5) \times (10^{-4}) = 25 \text{ J}

P2V2=(52×105)×(0.5×10−4)=2.52×10=252 JP_2 V_2 = (5\sqrt{2} \times 10^5) \times (0.5 \times 10^{-4}) = 2.5\sqrt{2} \times 10 = 25\sqrt{2} \text{ J}

Substituting the values:

W=25−2521.5−1=25(1−2)0.5W = \dfrac{25 - 25\sqrt{2}}{1.5 - 1} = \dfrac{25(1 - \sqrt{2})}{0.5}

W=50(1−1.414)=50(−0.414)=−20.7 JW = 50(1 - 1.414) = 50(-0.414) = -20.7 \text{ J}

Rounding to the nearest integer, we get W≈−21 JW \approx -21 \text{ J}.
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