JEE Mains · Physics · STD 11 - 11. thermodynamics
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)

- A \(0.24\)
- B \(0.15\)
- C \(0.32\)
- D \(0.08\)
Answer & Solution
Correct Answer
(B) \(0.15\)
Step-by-step Solution
Detailed explanation
\(Work - done\left( W \right) = {P_0}{V_0}\) According to principle of calorimetry Heat given\( = {Q_{AB}} = {Q_{BC}}\) \( = n{C_V}d{T_{AB}} + n{C_p}d{T_{BC}}\) \( = \frac{3}{2}\left( {nR{T_B} - nR{T_A}} \right) + \frac{5}{2}\left( {nR{T_C} - nR{T_B}} \right)\)…
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