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 \(.......\%\)

Pick an option to check your answer.
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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