Calculate the heat rejected by a system in going through the cyclic process shown in figure.

Explanation

`DeltaW_(cyclic)=`area of cycle
Since units on x and y axes are different, assume it as ellipse.
Area of ellipse `=piab`
`DeltaW_(cyclic)=-pi((300-100))/(2)xx10^3xx((300-100)/(2))xx10^-6`
`=-10pi=-31.4J`
`DeltaW` is negative because cycle clockwise on V-P diagram.
In cyclic process, `DeltaU=0`, hence,
`DeltaQ=DeltaW=-31.4J`
i.e., `31.4J` heat is released/rejected by the system (or gas)