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JEE Main 2017 (8 Apr, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2017 (8 Apr, Shift 1)

Q65JEE Main 2017 · 8 APR · SHIFT 1Difficult

Heat Engines, Refrigerators & Second Law

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is: (Take Cv = 1.5 R, where R is gas constant)

Figure

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Answer: (b)

  1. Net work = area of rectangle =(2P0−P0)(2V0−V0)=P0V0= (2P_0 - P_0)(2V_0 - V_0) = P_0V_0.
  2. Heat in A→B (isochoric): nCVΔT=1.5(2P0V0−P0V0)=1.5P0V0nC_V\Delta T = 1.5(2P_0V_0 - P_0V_0) = 1.5P_0V_0
  3. Heat in B→C (isobaric, CP=2.5RC_P = 2.5R): 2.5×2P0(2V0−V0)=5P0V02.5 \times 2P_0(2V_0 - V_0) = 5P_0V_0
  4. η=P0V06.5P0V0≈0.15\eta = \frac{P_0V_0}{6.5P_0V_0} \approx 0.15
  5. So the answer is (b) 0.15.
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