An ideal monoatomic gas is carried around the cycle ABCDA as shown in the figure. The efficiency of the gas cycle is

Options

A. ` 4/21`
B. `2/21`
C. `4/31`
D. `2/31`

Answer

A

Explanation

`W_(n et)=` area under the cycle
`=2p_0V_0`
`Q_(+ve)= Q_(ABC)=W_(ABC)+DeltaU_(ABC)`
= area under the graph `+nC_VDeltaT`
`=(3p_0V_0)+n(3/2R)(T_C-T_A)`
`=(3p_0V_0)+3/2(p_CV_C-p_AV_A)`
`10.5p_0V_0`
`eta=(W_(n et))/(Q_(+ve))=4/21`