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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2020
  • Hard
  • Numerical

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  \(.......\%\)

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Answer & Solution

Correct Answer

(A) \(19\)

Step-by-step Solution

Detailed explanation

\(W _{ ABCDA }=2 P _{0} V _{0}\)

\(Q_{\text {in }}=Q_{A B}+Q_{B C}\)

\(Q _{ AB }= nC \left( T _{ B }- T _{ A }\right)\)

\(=\frac{ n 3 R }{2}\left( T _{ B }- T _{ A }\right)\)

\(=\frac{3}{2}\left(P_{B} V_{B}-P_{A} V_{A}\right)\)

\(=\frac{3}{2}\left(3 P_{B} V_{0}=P_{0} V_{0}\right)=3 P_{0} V_{0}\)

\(Q _{ BC }= nC _{ P }\left( T _{ C }- T _{ B }\right)\)

\(=\frac{ n 5 R }{2}\left( T _{ C }- T _{ B }\right)\)

\(=\frac{5}{2}\left(P_{C} V_{C}-P_{B} V_{B}\right)\)

\(=\frac{5}{2}\left(6 P _{0} V _{0}-3 P _{0} V _{0}\right)=\frac{15}{2} P _{0} V _{0}\)

\(\eta=\frac{ W }{ Q _{ in }} \times 100=\frac{2 P _{0} V _{0}}{3 P _{0} V _{0}+\frac{15}{2} P _{0} V _{0}} \times 100\)

\(\eta=\frac{400}{21}=19.04 \approx 19\)

\(\eta=19\)

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