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. `2.67 kJ`
B. `1.33 kJ`
C. `3.45 kJ`
D. `4.25 kJ`

Answer

A

Explanation

The complete cyclic process can be visualized made up of tow cycles, ie., cycle `AEBA` (clockwise) and cycle `BDCB` (counter-clockwise).
Work done by the gas during the cycle `AEBA` should be positive
`W_(1)` = area of the loop in the `P-V` diagram.
`1/2 ("base")("altitude")=1/2(V_E-V_A)(P_B-P_(A))`
`1/2(20xx10^(-3)m^(3))(5-2)xx10^(5)Nm^(-2) = 3kJ`
Done by the gas the cycle `BDCB` should negative.
`W_(2) = - area of loop BDCB`
Now evidently traingale `ABE` and `BCD` are similar, the corresponding angles being equal (i.e.
`/_EBA = /_DBC=/_EAB = /_BCD`
`((V_E-V_A))/((P_B-P_A)) = ((V_C-V_D))/((P_C-P_B)) implies (20)/((5-2)) = ((V_C-V_D))/((6-5))`
`implies (V_C-V_D) = 20/ "litre"`
`W_(2)=-1/2xx20/3xx10^(-3)xx1xx10^(5)=-0.33kJ`
`:.` Total work done by the gas
`DeltaW = W_(1)+W_(2) = 3kJ - 0.33kJ=2.67kJ`.