The diagram shows a P - V graph of a thermodynamic behaviour of an ideal gas. Find out from this graph.
Work done in process `A to B, B to C, C to D` and `D to A`.

Explanation

Process `A to B`.
The process is expansion, hence work is done by the gas.
Here, P = 12 bar `=12xx10^(5)(N)/(m^(2))` or (Pa)
`Delta V=7-2=5` litre `= 5xx10^(-3)m^(3)`
`therefore W=-P Delta V = -12xx10^(5)xx5xx10^(-3)=-6000 J`
negative sign shows that work is done by gas.
Process `B to C`.
Here, `Delta V=0` therefore W = 0.
Process `C to D`
The process is contraction, hence work is done on system.
Here, P = 2 bar `=2xx10^(5)Nm^(-2)`,
`Delta V=2-7=-5` litre `=-5xx10^(-3)m^(3)`.
`therefore W=-P Delta V=-2xx10^(5)xx-5xx10^(-3)=+1000 J`.
positive sign shows that work is done on gas.
Process `D to A`.
Here, `Delta V=0`, therefore, W = 0.