JEE Main 2017 · 8 Apr · Shift 1 · Physics Q10 · Dronacharya Exams

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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 $C_v=1.5$ R, where R is gas constant)

P-V diagram: a rectangle A(V₀, P₀) → B(V₀, 2P₀) → C(2V₀, 2P₀) → D(2V₀, P₀) → A, traversed with arrows up AB, right BC, down CD and left DA.

A

0.24

B

0.15

C

0.32

D

0.08

Answer and worked solution

Official answer: B

Work per cycle is the enclosed area, $P_0V_0$ . Heat enters on AB and BC:

AB (isochoric): $nC_v\Delta T=1.5(2P_0V_0-P_0V_0)=1.5P_0V_0$ ;
BC (isobaric): $nC_p\Delta T=2.5(4P_0V_0-2P_0V_0)=5P_0V_0$ .

$\eta=\frac{P_0V_0}{6.5P_0V_0}\approx0.15.$

Topics:Thermodynamic processes Work done by a gas Specific heats of gases

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