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)

Options

A. `0.24`
B. `0.15`
C. `0.32`
D. `0.08`

Answer

B

Explanation

`W=P_0V_0`
Heat given `=Q_(AB)=Q_(BC)`
`=nC_VdT_(AB)+nC_PdT_(BC)`
`=(3)/(2)(nRT_B-nRT_A)+(5)/(2)(nRT_(C)-nRT_(B))`
`=(3)/(2)(2P_0V_0-P_0V_0)+(5)/(2)(4P_0V_0-P_0V_0)`
`=(13)/(2)P_0V_0`
`n=(W)/(Q" given ")=(2)/(13)=0.15`.