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NEET
Exam
NEET
Subject
Physics
Chapter
Thermodynamics

NEET · Physics · Thermodynamics

An ideal gas is taken through the cycle A arrow B arrow C arrow A ,...

An ideal gas is taken through the cycle A→B→C→AA \rightarrow B \rightarrow C \rightarrow A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J5 \mathrm{~J}, the work done by the gas in the process C→AC \rightarrow A is, An ideal gas is taken through the cycle A arrow B arrow C arrow A , as shown in the figure. If the net heat supplied to

Options

  1. A−5 J-5 \mathrm{~J}
  2. B−10 J-10 \mathrm{~J}
  3. C−15 J-15 \mathrm{~J}
  4. D−20 J-20 \mathrm{~J}

Correct answer

A. −5 J-5 \mathrm{~J}

Step-by-step solution

Net work done by a gas, ΔWAB=p⋅ΔV=10×(2−1)=10JΔWBC=0 (as V= constant) \begin{array}{ll}\Delta W_{A B}=p \cdot \Delta V=10 \times(2-1)=10 \mathrm{ J}\\ \Delta W_{B C}=0 \text{ (as }V=\text{ constant) } \end{array}From first law of thermodynamics, we have, ΔQ=ΔW+ΔU△U=0 (process A\Delta Q=\Delta W+\Delta U \triangle U=0 \quad \text{ (process }AB C A  is  cyclic ) \text{\text{ is }\text{ cyclic }) }Therefore, ΔQ=ΔWAB+ΔWBC+ΔWCA\Delta Q=\Delta W_{A B}+\Delta W_{B C}+\Delta W_{C A}∴ΔWCA=(\DeltaQ−WAB−WBC)=(5−10−0)J=−5J\begin{aligned} \therefore \quad\Delta W_{C A} & =\left(\DeltaQ-W_{A B}-W_{B C}\right) \\ & =(5-10-0) \mathrm{J}=-5 \mathrm{ J} \end{aligned}$
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