The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

Options

A. the process during the path `ArarrB` is isothermal
B. heat flows out of the gas during the path `BrarrCrarrD`
C. work done during the path `ArarrBrarrC` is zero
D. positive work is done by the gas in the cycle ABCDA

Answer

B, D

Explanation

`DeltaQ=DeltaU+W`
For process `BrarrCrarrD`
`DeltaU` is negative and `W` is also negative, so `DeltaQ` is also negative, hance heat flows out during this process.
`A` to `B` , work is `+ve` . `B` to `C` work is `-ve` . But `+ve` work is more, so net work is not zero in process `ArarrBrarrC` .