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JEE Main 2020 (6 Sep, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2020 (6 Sep, Shift 2)

Q55JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Heat Engines, Refrigerators & Second Law

An engine operates by taking a monoatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to __________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 19

  1. Monatomic gas: CV=32RC_V = \frac{3}{2}R, CP=52RC_P = \frac{5}{2}R. Net work =(3P0−P0)(2V0−V0)=2P0V0= (3P_0 - P_0)(2V_0 - V_0) = 2P_0V_0.
  2. Heat in A→B (isochoric): 32(3P0−P0)V0=3P0V0\frac{3}{2}(3P_0 - P_0)V_0 = 3P_0V_0
  3. Heat in B→C (isobaric): 52 (3P0)(V0)=7.5P0V0\frac{5}{2}\,(3P_0)(V_0) = 7.5P_0V_0
  4. η=2P0V010.5P0V0≈0.19\eta = \frac{2P_0V_0}{10.5P_0V_0} \approx 0.19
  5. So the efficiency is about 19%19\%.
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