Figure shows a cyclic process `ABCDBEA` performed on an ideal cycle. If `P_(A) = 2 atm, P_(B) = 5 atm and P_(6) = 6 atm. V_(E) - V_(A) = 20 "litre"`, find the work done by the gas in the complete, process (`1 atm. "Pressure" = 1 xx 10^(5) Pa`)

Options

A. `4.67kJ`
B. `3.67kJ`
C. `2.67kJ`
D. `1.67kJ`

Answer

C

Explanation

Evidently,the entire cycle can be visualised as made up of two cycles, I,c., cycle `ABEA` and cyles `BCDB`.
`W_(1)` = area of the loop in the `P-V` diagram

`W_(1)=(1)/(2)`(base) (alitude) `=(1)/(2)(V_(E)-V_(A)) (P_(B)-P_(A))` `W_(1)=(1)/(2)(20xx10^(-3)m^(3)) (5-2)xx10^(5) Nm^(-2)=3 kJ`
Work done in the cycle `BCDB` , by the gas
`W_(2)` =-(area of loop `BCDB`)
Now, evidently, triagles `ABE` and `BCD` are similar, the corresponding angles being equal (viz.,
`/_EBA=/_DBC,/_EAB=/_BCD,etc.)`
`:. ((P_(C)-P_(B))/(P_(B)-P_(A)))^(2)=((6atm-5atm)/(5atm-2atm))^(2)=(1)/(9)`
`:.` Area of `Delta BCD-(1)/(9) ("area of" Delta)=(1)/(9)xx3`
`kJ=0.33 kJ :.W_(2)=-0.33 kJ`
`DeltaW=W_(1)+W_(2)=3 kJ-0.33 kJ=2.67 kJ`