Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas)

Options

A. `15.4%`
B. `9.1%`
C. `10.5%`
D. `12.5%`

Answer

A

Explanation

`DeltaW_(cyclic)=`area of cycle ABCDA
`=(2P_0-P_0)(2V_0-V_B)=P_0V_0`
`DeltaQ_(AB)=nC_VDeltaT=n.(3R)/(2)(T_B-T_A)`
`=(3nR)/(2)((2P_0V_0)/(nR)-(P_0V_0)/(nR))=(3P_0V_0)/(2)`
`DeltaQ_(BC)=nC_PDeltaT=n.(5R)/(2)(T_C-T_B)`
`=(5nR)/(2)((2P_0.2V_0)/(nR)-(2P_0V_0)/(nR))`
`=5P_0V_0`
`DeltaQ_(CD)=-ve,DeltaQ_(DA)=-ve`
`eta=(DeltaW)/(DeltaQ_(AB)+DeltaQ_(BC))=(P_0V_0)/((3P_0V_0)/(2)+5P_0V_0)=(2)/(13)` i.e.,
`(200)/(13)=15.4%`