JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Heat & Thermodynamics

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

Answer: 19

Solution

Over one cycle the net work is the area enclosed,

W = (3 P_{0} - P_{0}) (2 V_{0} - V_{0}) = 2 P_{0} V_{0}

. Heat enters along the isochoric step

A B

Q_{A B} = \frac{3}{2} (3 P_{0} V_{0} - P_{0} V_{0}) = 3 P_{0} V_{0}

, and along the isobaric step

B C

Q_{B C} = \frac{5}{2} (6 P_{0} V_{0} - 3 P_{0} V_{0}) = \frac{15}{2} P_{0} V_{0}

, giving a total intake

Q_{in} = \frac{21}{2} P_{0} V_{0}

. The efficiency is

η = \frac{W}{Q _{in}} = \frac{2 P _{0} V _{0}}{\frac{21}{2} P _{0} V _{0}} = \frac{4}{21} \approx 19%

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Integer answer

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