A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is

Options

A. `P_0V_0`
B. `2P_0V_0`
C. `(P_0V_0)/(2)`
D. zero

Answer

D

Explanation

`DeltaW_(AODA)=(1)/(2)(2P_0-P_0)(2V_0-V_0)=(1)/(2)P_0V_0`
`AODA` is clockwise cycle on `P-V` diagram, `DeltaW=+ve`
`DeltaW_(COBC)=-(1)/(2)(3P_0-P_0)(2V_0-V_0)=(1)/(2)P_0V_0`
`COBC` is anticlockwise cycle on `P-V` diagram, `DeltaW=-ve`
`DeltaW_(ABCDA)=DeltaW_(AODA)+DeltaW_(COBC)=0`