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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Thermodynamics

JIPMER 2018 · Physics · Thermodynamics

The efficiency of an ideal gas with adiabatic exponent γ for the...

The efficiency of an ideal gas with adiabatic exponent γ\gamma for the shown cyclic process would be The efficiency of an ideal gas with adiabatic exponent γ for the shown cyclic process would be

Options

  1. A(2ln⁡2−1)γ/(γ−1)\frac{(2 \ln 2-1)}{\gamma /(\gamma-1)}
  2. B(1−1ln⁡2)γ/(γ−1)\frac{(1-1 \ln 2)}{\gamma /(\gamma-1)}
  3. C(2ln⁡2+1)γ/(γ−1)\frac{(2 \ln 2+1)}{\gamma /(\gamma-1)}
  4. D(2ln⁡2−1)γ/(γ+1)\frac{(2 \ln 2-1)}{\gamma /(\gamma+1)}

Correct answer

A. (2ln⁡2−1)γ/(γ−1)\frac{(2 \ln 2-1)}{\gamma /(\gamma-1)}

Step-by-step solution

As,WBC=pΔ\mathrm{As}, \mathrm{W}_{\mathrm{BC}}=\mathrm{p} \Delta\mathrm{V}=\mathrm{nR}ΔT=−nR(2 T0−T0)=nRT0\Delta \mathrm{T}=-\mathrm{nR}\left(2 \mathrm{~T}_0-\mathrm{T}_0\right)=\mathrm{nRT}_0 and WCA=+2nRT0ln⁡2 Work done =WCA+WBC=2nRT0ln⁡2−nRT0=nRT⁡0(2ln⁡2−1)\begin{aligned} & \text{ and } \quad \mathrm{W}_{\mathrm{CA}}=+2 \mathrm{nRT}_0 \ln 2 \\ & \text{ Work done }=\mathrm{W}_{\mathrm{CA}}+\mathrm{W}_{\mathrm{BC}} \\ & =2 \mathrm{nRT}_0 \ln 2-\mathrm{nRT}_0 \\ & =\operatorname{nRT}_0(2 \ln 2-1) \\ & \end{aligned}$Also,inputheat,Also, input heat,\Delta \mathrm{Q}_{\mathrm{BC}}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}=\frac{\mathrm{nR} \gamma \mathrm{T}_0}{\gamma-1}EfficiencyEfficiency=\frac{\text{ Work }}{\text{ Input heat }}=\frac{(2 \ln 2-1)}{\gamma /(\gamma-1)}$
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