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 Cv = 1.5 R, where R is gas constant)

Question

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 Cv = 1.5 R, where R is gas constant) (A) 0.24 (B) 0.15 (C) 0.32 (D) 0.08

Tardigrade Admin · Accepted Answer

w = P0V0 Heat given = QAB = QBC = nCVdTAB + nCPdTBC =(3/2)(nRTB-nRTA)+(5/2) (nRTC - nRTB) = (3/2)(2P0V0 - P0V0)+(5/2)(4P0V0 - 2P0V) = (13/2)P0V0 n = (w/Qgiven) = (2/13) = 0.15

Q. 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 Cv​=1.5R, where R is gas constant)

0.24

0.15

0.32

0.08

w=P0​V0​ Heat given = QAB​=QBC​ =nCV​dTAB​+nCP​dTBC​ =23​(nRTB​−nRTA​)+25​(nRTC​−nRTB​) =23​(2P0​V0​−P0​V0​)+25​(4P0​V0​−2P0​V) =213​P0​V0​ n=Qgivenw​=132​=0.15

Q. An engine operates by taking $n$ moles of an ideal gas through the cycle $A BC D A$ shown in figure. The thermal efficiency of the engine is :
(Take $C_{v} = 1.5 R,$ where $R$ is gas constant)